Electronic Devices Circuits Pune University MCQs
Electronic Devices Circuits Pune University MCQs
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Basic Concepts”.
1. A solid copper sphere, 10 cm in diameter is deprived of 1020 electrons by a charging scheme. The charge on the sphere is _________
a) 160.2 C
b) -160.2 C
c) 16.02 C
d) -16.02 C
Answer: c
Explanation: n 10 20 , Q = ne = e 10 20 = 16.02 C.
Charge on sphere will be positive.
2. A lightning bolt carrying 15,000 A lasts for 100 s. If the lightning strikes an airplane flying at 2 km, the charge deposited on the plane is _________
a) 13.33 C
b) 75 C
c) 1500 C
d) 1.5 C
Answer: d
Explanation: dQ = i dt = 15000 x 100µ = 1.5 C.
3. If 120 C of charge passes through an electric conductor in 60 sec, the current in the conductor is _________
a) 0.5 A
b) 2 A
c) 3.33 mA
d) 0.3 mA
Answer: b
Explanation: i = dQ/dt = 120/60 = 2A.
4. The energy required to move 120 coulomb through 3 V is _________
a) 25 mJ
b) 360 J
c) 40 J
d) 2.78 mJ
Answer: b
Explanation: W = Qv = 360 J.
5. Consider the circuit graph shown in figure below. Each branch of circuit graph represent a circuit element. The value of voltage V1 is:
electronic-devices-circuits-questions-answers-basic-concepts-q5
a) 30 V
b) 25 V
c) 20 V
d) 15 V
Answer: d
Explanation: 100 = 65 + V2 => V2 = 35 V
V3 – 30 = V2 => V3 = 65 V
105 – V3 + V4 – 65 = 0 => V4 = 25 V
V4 + 15 – 55 + V1 = 0 => V1 = 15 V.
6. What will be the value of Req in the following Circuit?
electronic-devices-circuits-questions-answers-basic-concepts-q6
a) 11.86 ohm
b) 10 ohm
c) 25 ohm
d) 11.18 ohm
Answer: d
Explanation: electronic-devices-circuits-questions-answers-basic-concepts-q6a We infer from the given diagram that the same pattern is followed after every 10ohm resistor. The infinite pattern in parallel as a whole is considered as to be Rx.
Thus, Rx = R +
Solving for Rx, we get Rx = 1.62R, where R=10ohm.
So we have Rx = *, which gives Rx=16.2ohm
Therefore, Req = 5 +
=> 5 + [/] => Req = 5 + which gives Req=11.18 ohm.
7. In the circuit the dependent source __________
electronic-devices-circuits-questions-answers-basic-concepts-q7
a) supplies 16 W
b) absorbs 16 W
c) supplies 32 W
d) absorbs 32 W
Answer: d
Explanation: P = VIx = 2Ix Ix = 2 x 16 or 32 watt .
8. Twelve 6 ohm resistors are used to form an edge of a cube. The resistance between two diagonally opposite corner of the cube is ________
a) 5/6
b) 6/5
c) 5
d) 6
Answer: c
Explanation: electronic-devices-circuits-questions-answers-basic-concepts-q8a
The current I will be distributed to the cube branches symmetrically.
Connecting a voltage source Vab across the terminals and applying kvl in the outer loop, we have
=> -Vab + *R = 0
=> Vab = 5i
=> Req=Vab/i, which gives Req = 5ohm.
9. The energy required to charge a 10 µF capacitor to 100 V is ________
a) 0.1 J
b) 0.05 J
c) 5 x 10 J
d) 10 x 10 J
Answer: b
Explanation: Energy provided is equal to 0.5 CVxV.
10. A capacitor is charged by a constant current of 2 mA and results in a voltage increase of 12 V in a 10 sec interval. The value of capacitance is ________
a) 0.75 mF
b) 1.33 mF
c) 0.6 mF
d) 1.67 mF
Answer: d
Explanation: The capacitor current is given as i=C*, where dv/dt is the derivative of voltage, dt=t2-t1 given as 10 sec and dv is the change in voltage which is given as 12V.
So, we have C=i/
=> C = 2mA/ = 2mA/.
Hence C = 1.67mF.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Method of Analysis”.
1. What will be the value of V1 in the following circuit?
electronic-devices-circuits-questions-answers-method-analysis-q1
a) 0.4 Vs
b) 1.5 Vs
c) 0.67 Vs
d) 2.5 Vs
Answer: b
Explanation: Apply nodal analysis, V1 = Vs/ = 1.5 Vs.
2. What will be the value of Va in the following circuit?
electronic-devices-circuits-questions-answers-method-analysis-q2
a) -11 V
b) 11 V
c) 3 V
d) -3 V
Answer: b
Explanation: By current division, we get 4A and 1A in resistors 2ohm and 3ohm respectively, applying Kvl in the bottom loop, we have Va = 2 + 3 = 11V.
3. What will be the value of V1 in the following circuit?
electronic-devices-circuits-questions-answers-method-analysis-q3
a) 120 V
b) -120 V
c) 90 V
d) -90 V
Answer: d
Explanation: -V1/60 – V1/60 + 6 = 9 or V1 = -90 V.
4. What will be the value of Va in the following circuit?
electronic-devices-circuits-questions-answers-method-analysis-q4
a) 4.33 V
b) 4.09 V
c) 8.67 V
d) 8.18 V
Answer: c
Explanation: /4 + Va/2 = 4 or Va = 8.67 V.
5. What will be the value of V2 in the following circuit?
electronic-devices-circuits-questions-answers-method-analysis-q5
a) 0.5 V
b) 1.0 V
c) 1.5 V
d) 2.0 V
Answer: d
Explanation: V2/20 + V2+10/30 = 0.5 V or V2 = 2.0 V.
6. What will be the value of I1 in the following circuit?
electronic-devices-circuits-questions-answers-method-analysis-q6a
a) 0.6 A
b) 0.5 A
c) 0.4 A
d) 0.1 A
Answer: d
Explanation: By nodal analysis,
0.5=V1/30 + /20
i.e. V1=12V. So i1 = /20, which gives i1=0.1A.
7. What will be the value of i1 in the following circuit?
electronic-devices-circuits-questions-answers-method-analysis-q7
a) 0.6 A
b) 2.1 A
c) 1.3 A
d) 1.7 A
Answer: c
Explanation: By nodal analysis computing v1 and v2 where
/8 + V1/4 + /2=0 —
/2 + [V1-]*/ + V2/3=0 —
*converting current source to voltage source
Hence, i1 = /2 which gives i1 = 1.3A.
8. What will be the value of i1 in the following circuit?
electronic-devices-circuits-questions-answers-method-analysis-q8
a) 1 mA
b) 1.5 mA
c) 2 mA
d) 2.5 mA
Answer: b
Explanation: Applying kvl in the rightmost bottom loop, we have
=> 10k* – 75 + 90*i1 = 0
=> 10*i1-75 -75 + 90*i1 = 0
=> 100*i1=150
which gives i1=1.5 mA.
9. What will be the value of i1 in the following circuit?
electronic-devices-circuits-questions-answers-method-analysis-q9
a) 4 A
b) 3 A
c) 6 A
d) 5 A
Answer: b
Explanation: 3 – 5i1 – 12 => i1 = 3A.
10. What will be the value of i1 in the following circuit?
electronic-devices-circuits-questions-answers-method-analysis-q10
a) 20 mA
b) 15 mA
c) 10 mA
d) 5 mA
Answer: b
Explanation: 45 = 2ki1 + 500 => i1 = 15 mA.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Frequency Response – 1”.
A parallel resonant circuit has a resistance of 2k ohm and half power frequencies of 86 kHz and 90 kHz.
1. The value of capacitor is
a) 6 µF
b) 20 nF
c) 2 nF
d) 60 µF
Answer: b
Explanation: BW = 2pk = 1/RC or C = 19.89 nF.
2. The value of inductor is
a) 4.3 mH
b) 43 mH
c) 0.16 mH
d) 1.6 mH
Answer: c
Explanation: w = k/2 = 88 = or L = 0.16 mH.
3. The quality factor is
a) 22
b) 100
c) 48
d) 200
Answer: a
Explanation: Q = w/BW = 176pK/8pk = 22.
A parallel resonant circuit has a midband admittance of 25 X 10 S, quality factor of 80 and a resonant frequency of 200 krad s.
4. The value of R is
a) 40
b) 56.57
c) 80
d) 28.28
Answer: a
Explanation: At mid-band frequency Y = 1/R or R = 1000/25 or 40 ohm.
5. The value of C is
a) 2 µF
b) 28.1 µF
c) 10 µF
d) 14.14 µF
Answer: c
Explanation: Q = wRC or C = 80/ or 10 µF.
6. A parallel RLC circuit has R 1 k and C 1 F. The quality factor at resonance is 200. The value of inductor is
a) 35.4 H
b) 25 H
c) 17.7 H
d) 50 H
Answer: b
Explanation: Use Q = R 0.5 .
7. A parallel circuit has R = 1k ohm , C = 50 µF and L = 10mH. The quality factor at resonance is
a) 100
b) 90.86
c) 70.7
d) None of the above
Answer: c
Explanation: Use Q = R 0.5 .
8. A series resonant circuit has L = 1 mH and C = 10 F. The required R for the BW = 15 9 . Hz is
a) 0.1
b) 0.2
c) 0.0159
d) 500
Answer: a
Explanation: Use BW = R/L.
9. For the RLC parallel resonant circuit when R = 8k, L = 40 mH and C = 0.25 F, the quality factor
Q is
a) 40
b) 20
c) 30
d) 10
Answer: b
Explanation: use Q = R 0.5 .
10. A series resonant circuit has an inductor L = 10 mH. The resonant frequency w = 10^6 rad/s and bandwidth is BW = 10 3 rad/s. The value of R and C will be
a) 100 F, 10 ohm
b) 100 pF, 10 ohm
c) 100 pF, 10 Mega-ohm
d) 100 µF, 10 Meg-ohm
Answer: b
Explanation: Use wxw = 1/LC and BW = R/L.
This set of Basic Electronic Devices and Circuits Questions and Answers focuses on “Frequency Response – 2”.
1. The maximum voltage across capacitor would be
electronic-devices-circuits-basic-questions-answers-q1
a) 3200 V
b) 3 V
c) 3 V
d) 1600 V
Answer: a
Explanation: electronic-devices-circuits-basic-questions-answers-q1a
2. Find the resonnant frequency for the circuit.
electronic-devices-circuits-basic-questions-answers-q2
a) 346 kHz
b) 55 kHz
c) 196 kHz
d) 286 kHz
Answer: b
Explanation: electronic-devices-circuits-basic-questions-answers-q2a
3. Determine the resonant frequency of the circuit.
electronic-devices-circuits-basic-questions-answers-q3
a) 12 9 . kHz
b) 12.9 MHz
c) 2.05 MHz
d) 2.05 kHz
Answer: c
Explanation: electronic-devices-circuits-basic-questions-answers-q3a
4. The value of C and A for the given network function is
electronic-devices-circuits-basic-questions-answers-q4
a) 10 µF, 6
b) 5 µF, 10
c) 5 µF, 6
d) 10 µF, 10
Answer: c
Explanation: electronic-devices-circuits-basic-questions-answers-q4a
5. H = Vo/Vi = ?
electronic-devices-circuits-basic-questions-answers-q5
a) 0.6 / jw
b) 0.6 / jw
c) 3 / jw
d) 3 / jw
Answer: a
Explanation: electronic-devices-circuits-basic-questions-answers-q5a
6. H = Vo/Vi = ?
electronic-devices-circuits-basic-questions-answers-q6
a) 1 / jw
b) 1 / jw
c) 1 / jw
d) 5 / jw
Answer: a
Explanation: electronic-devices-circuits-basic-questions-answers-q6a
7. The value of input frequency is required to cause a gain equal to 1.5. The value is
electronic-devices-circuits-basic-questions-answers-q7
a) 20 rad/s
b) 20 Hz
c) 10 rad/s
d) No such value exists
Answer: d
Explanation: electronic-devices-circuits-basic-questions-answers-q7a
8. In the circuit shown phase shift equal to 45 degrees and is required at frequency w = 20 rad/s. The value of R is
electronic-devices-circuits-basic-questions-answers-q8
a) 200
b) 150
c) 100
d) 50
Answer: d
Explanation: electronic-devices-circuits-basic-questions-answers-q8a
9. For the circuit shown the input frequency is adjusted until the gain is equal to 0.6. The value of the frequency is
electronic-devices-circuits-basic-questions-answers-q9
a) 20 rad/s
b) 20 Hz
c) 40 rad/s
d) 40 Hz
Answer: a
Explanation: electronic-devices-circuits-basic-questions-answers-q9a
10. The bode diagram for the Vo/Vs for the given circuit is
electronic-devices-circuits-basic-questions-answers-q10
a) electronic-devices-circuits-basic-questions-answers-q10a
b) electronic-devices-circuits-basic-questions-answers-q10b
c) electronic-devices-circuits-basic-questions-answers-q10c
d) electronic-devices-circuits-basic-questions-answers-q10d
Answer: d
Explanation: electronic-devices-circuits-basic-questions-answers-q10e
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Two Port Network”.
1. [T] = ?
electronic-devices-circuits-questions-answers-two-port-network-q1
electronic-devices-circuits-questions-answers-two-port-network-q1a
Answer: d
Explanation: electronic-devices-circuits-questions-answers-two-port-network-q1b
2. [h] = ?
electronic-devices-circuits-questions-answers-two-port-network-q2
electronic-devices-circuits-questions-answers-two-port-network-q2a
Answer: a
Explanation: V2 = 2I2 + 4I1 and
I1 = 0.5I2 + 0.5 => V1 = 4I1 + 1.5V2.
3. [y] = ?
electronic-devices-circuits-questions-answers-two-port-network-q3
electronic-devices-circuits-questions-answers-two-port-network-q3a
Answer: a
Explanation: I1 = V1 – V2 and I2 = -V1 + V2.
4.The [y] parameter of a two port network is given by
electronic-devices-circuits-questions-answers-two-port-network-q4
electronic-devices-circuits-questions-answers-two-port-network-q4a
Answer: b
Explanation: The new parameter will be the sum of the previous network and the resistor parameter.
5. The [y] parameter for a 2-port network and the network itself are given below.
electronic-devices-circuits-questions-answers-two-port-network-q5
The value of Vo/vs is _______
a) 3/32
b) 1/16
c) 2/33
d) 1/17
Answer: a
Explanation: electronic-devices-circuits-questions-answers-two-port-network-q5a
6. [y] = ?
electronic-devices-circuits-questions-answers-two-port-network-q6
electronic-devices-circuits-questions-answers-two-port-network-q6a
Answer: b
Explanation: electronic-devices-circuits-questions-answers-two-port-network-q6b
7. A 2-port resistive network satisfy the condition A = D = 3/2B = 4/3C. The z11 of the network is
a) 4/3
b) 3/4
c) 2/3
d) 3/2
Answer: a
Explanation: z11 = A/C = 4/3.
8. A 2-port network is driven by a source Vs = 100 V in series with 5 ohm, and terminated in a 25 ohm resistor. The impedance parameters are
electronic-devices-circuits-questions-answers-two-port-network-q8
The Thevenin equivalent circuit presented to the 25 ohm resistor is
a) 80 V, 2.8 ohm
b) 160 V, 6.8 ohm
c) 100 V, 2.4 ohm
d) 120 V, 6.4 ohm
Answer: b
Explanation: 100 = 25I1 + 2I2, V2 = 40I1 + 10I2
Also V2 = 160I1 + 6.8I2
using above equations we get, Vth = 160 V and Rth = 6.8 ohm.
9. Find V1 and V2.
electronic-devices-circuits-questions-answers-two-port-network-q9
a) -68.6 V, 114.3 V
b) 68.6 V, -114.3 V
c) 114.3 V, -68.6 V
d) -114.3 V, 68.6 V
Answer: b
Explanation: 800 = 10V1 – V2 and 3V2 = 5V1 = 0.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Continous Time Signals”.
The number of cars arriving at ICICI bank drive-in window during 10-min period is Poisson random variable X with b=2.
1. The probability that more than 3 cars will arrive during any 10 min period is
a) 0.249
b) 0.143
c) 0.346
d) 0.543
Answer: b
Explanation: Evaluate 1 – P – P – P – P.
2. The probability that no car will arrive is
a) 0.516
b) 0.459
c) 0.246
d) 0.135
Answer: d
Explanation: Evaluate P.
Delhi averages three murder per week and their occurrences follow a Poisson distribution.
3. The probability that there will be five or more murder in a given week is
a) 0.1847
b) 0.2461
c) 0.3927
d) 0.4167
Answer: a
Explanation: P = 1 – P – P – P – P – P = 0.1847.
4. On the average, how many weeks a year can Delhi expect to have no murders ?
a) 1.4
b) 1.9
c) 2.6
d) 3.4
Answer: c
Explanation: P = 0.0498. Hence average number of weeks per year with no murder is 52 x P = 2.5889 week.
5. How many weeds per year can the Delhi expect the number of murders per week to equal or exceed the average number per week?
a) 15
b) 20
c) 25
d) 30
Answer: d
Explanation: P = 1 – P – P – P = 0.5768. Therefore average number of weeks per year = 52 x 0.5768 or 29.994 weeks.
The random variable X is defined by the density f = 0.5u e
6. The expect value of g = X 3 is
a) 48
b) 192
c) 36
d) 72
Answer: a
Explanation: Solve E[g] = E[X 3 ].
7. The mean of the random variable x is
a) 1/4
b) 1/6
c) 1/3
d) 1/5
Answer: a
Explanation: Solve integral dx) from negative infinity to x.
8. The variance of the random variable x is
a) 1/10
b) 3/80
c) 5/16
d) 3/16
Answer: b
Explanation: Variance is given by E[X 2 30] – 1/16.
A joint sample space for two random variable X and Y has four elements , , and . Probabilities of these elements are 0.1, 0.35, 0.05 and 0.5 respectively.
9. The probability of the event{X 2.5, Y 6} is
a) 0.45
b) 0.50
c) 0.55
d) 0.60
Answer: a
Explanation: The required answer is given by Fxy = 0.1 + 0.35 = 0.45.
10. The probability of the event that X is less than three is
a) 0.45
b) 0.50
c) 0.55
d) 0.60
Answer: b
Explanation: The required answer is given by Fx = Fxy = 0.1 + 0.35 + 0.05 = 0.50.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Random Process”.
1. For random process X = 6 and Rxx = 36 + 25 exp. Consider following statements:
X is first order stationary.
X has total average power of 36 W.
X is a wide sense stationary.
X has a periodic component.
Which of the following is true?
a) 1, 2, and 4
b) 2, 3, and 4
c) 2 and 3
d) only 3
Answer: c
Explanation: X Constant and Rxx is not a function of t, so X is a wide sense stationary. So is false & is true. Pxx = Rxx 36+25 = 61. Thus is false if X has a periodic component, then RXX will have a periodic component with the same period. Thus is false.
2. White noise with power density No/2 = 6 microW/Hz is applied to an ideal filter of gain 1 and bandwidth W rad/s. If the output’s average noise power is 15 watts, the bandwidth W is
a) 2.5 x 10
b) 2.5p x 10
c) 5 x 10
d) p5 x 10
Answer: b
Explanation: Pyy = 1/2p Integral |H|^2 dw ) from plus infinity to minus infinity. Hence solve for W.
The two-level semi-random binary process is defined by X A or -A where T < t < nt and the levels A and -A occur with equal probability. T is a positive constant and n = 0, ±1, ±2.
3. The mean value E[X] is
a) 1/2
b) 1/4
c) 1
d) 0
Answer: d
Explanation: E[X] = A P – P which is zero.
4. The auto correlation Rxx will be
a) 1
b) 0
c) A x A
d) 0.5
Answer: c
Explanation: Here Rxx is AxA if both t1 and t2 are different and zero if they are same. Hence the answer is AxA.
5. Air craft of Jet Airways at Ahmedabad airport arrive according to a Poisson process at a rate of 12 per hour. All aircraft are handled by one air traffic controller. If the controller takes a 2 – minute coffee break, what is the probability that he will miss one or more arriving aircraft?
a) 0.33
b) 0.44
c) 0.55
d) 0.66
Answer: a
Explanation: P = 1 – P = 1 – P.
6. A stationary random process X is applied to the input of a system for which h = u t 2 e . If E[X] = 2, the mean value of the system’s response Y is
a) 1/128
b) 1/64
c) 3/128
d) 1/32
Answer: c
Explanation: The mean value of Y is integral of hdt over negative infinity to positive infinity which gives the value equal to 3/128.
7. A random process is defined by X + A where A is continuous random variable uniformly distributed on
. The auto correlation function and mean of the process is
a) 1/2 & 1/3
b) 1/3 & 1/2
c) 1 & 1/2
d) 1/2 & 1
Answer: b
Explanation: E[XX] = 1/3 and E[X] = 1/2 respectively.
The auto correlation function of a stationary ergodic random process is shown below.
electronic-devices-circuits-questions-answers-random-process-q8
8. The mean value E[X] is
a) 50
b) sqrt
c) 20
d) sqrt
Answer: d
Explanation: Lim |t| tends to infinity, Rxx = 20 = X 2 . hence X is sqrt.
9. The E[X 2 ] is
a) 10
b) sqrt
c) 50
d) sqrt
Answer: c
Explanation: Rxx = X 2 = 50.
10. The variance is
a) 20
b) 50
c) 70
d) 30
Answer: d
Explanation: Here X = 0, y = 0, Rxx = 5, Ryy = 10. The only value that satisfies all the given conditions is 30.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Noise”.
1. In a receiver the input signal is 100 V, while the internal noise at the input is 10 V. With amplification the output signal is 2 V, while the output noise is 0.4 V. The noise figure of receiver is
a) 2
b) 0.5
c) 0.2
d) None of the mentioned
Answer: a
Explanation: NF = / or 2.
2. A receiver is operated at a temperature of 300 K. The transistor used in the receiver have an average output resistance of 1 k. The Johnson noise voltage for a receiver with a bandwidth of 200 kHz is
a) 1.8 µV
b) 8.4 µV
c) 4.3 µV
d) 12.6 µV
Answer: a
Explanation: v 2 = 4kBTR where symbols have their usual meanings, hence v = 1.8 µV.
3. A resistor R 1 k is maintained at 17C. The rms noise voltage generated in a bandwidth of 10 kHz is
a) 16 x 10 V
b) 0.4 µV
c) 4 µV
d) 16 x 10 V
Answer: b
Explanation: v 2 = 4kBTR where symbols have their usual meanings, hence v = 0.4 µV.
4. A mixer stage has a noise figure of 20 dB. This mixer stage is preceded by an amplifier which has a noise figure of 9 dB and an available power gain of 15 dB. The overall noise figure referred to the input is
a) 11.07
b) 18.23
c) 56.43
d) 97.38
Answer: a
Explanation: The required answer is 7.94 + /31.62 = 11.07.
5. A system has three stage cascaded amplifier each stage having a power gain of 10 dB and noise figure of 6 dB. the overall noise figure is
a) 1.38
b) 6.8
c) 4.33
d) 10.43
Answer: c
Explanation: The gain at each stage is 10db, hence the required answer is 4 + /10 + /100 or 4.33.
An amplifier when used with a source of average noise temperature 60 K, has an average operating noise figure of 5.
6. The Te is
a) 70 K
b) 110 K
c) 149 K
d) 240 K
Answer: d
Explanation: The required answer is 60 or 240K.
7. If the amplifier is sold to engineering public, the noise figure that would be quoted in a catalog is
a) 0.46
b) 0.94
c) 1.83
d) 2.93
Answer: c
Explanation: The required answer is 1 + .
8. What average operating noise figure results when the amplifier is used with an antenna of temperature 30 K?
a) 9.54 db
b) 10.96 db
c) 11.23 db
d) 12.96 db
Answer: a
Explanation: The required answer is 1 + 240/30 = 9 or 9.54 db.
9. What is the maximum average effective input noise temperature that an amplifier can have if its average standard noise figure is to not exceed 1.7?
a) 203 K
b) 215 K
c) 235 K
d) 255 K
Answer: a
Explanation: The required answer is 290 x or 203K.
10. If a matched attenuator with a loss of 3.2 dB is placed between the source and the amplifier’s input, what is the operating spot noise figure of the attenuator amplifier cascade if the attenuator’s physical temperature is 290 K?
a) 9 db
b) 11.3 db
c) 10.4 db
d) 13.3 db
Answer: d
Explanation: Here Te is 290[ + ] or 4544.4K. Hence the required answer is 1 + 4544.4/290 = 212 or 13.3 db.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Amplitde Modulation”.
An AM signal is represented by x = ) cos(2πt x 10 5 )V.
1. The modulation index is
a) 20
b) 4
c) 0.2
d) 10
Answer: c
Explanation: 20 + 4sin = 20), hence the modulation index is 0.2.
2. Total signal power is
a) 208 W
b) 204 W
c) 408 W
d) 416 W
Answer: b
Explantion: Pc = 20×20/2 or 200 W. Pt = Pc or 204 W.
3. Total sideband power is
a) 4 W
b) 8 W
c) 16 W
d) 2 W
Answer: a
Explanation: Pt – Pc = 204 – 200 = 4W.
4. An AM broadcast station operates at its maximum allowed total output of 50 kW with 80% modulation. The power in the intelligence part is
a) 12.12 kW
b) 31.12 kW
c) 6.42 kW
d) none of the mentioned
Answer: a
Explanation: Pi = Pt – Pc = 50 – 37.88 kW or 12.12 kW.
5. The aerial current of an AM transmitter is 18 A when unmodulated but increases to 20 A when modulated.The modulation index is
a) 0.68
b) 0.73
c) 0.89
d) 0.95
Answer: a
Explanation: 400/326 = 1 + (α 2 )/2, therefore α is 0.68.
6. A modulating signal is amplified by a 80% efficiency amplifier before being combined with a 20 kW carrier to generate an AM signal. The required DC input power to the amplifier, for the system to operate at 100% modulation, would be
a) 5 kW
b) 8.46 kW
c) 12.5 kW
d) 6.25 kW
Answer: c
Explanation: Pi = Pt -Pc = 30 – 20 = 10 kW. DC input = 10/0.8 or 12.5 kW.
7. A 2 MHz carrier is amplitude modulated by a 500 Hz modulating signal to a depth of 70%. If the unmodulated carrier power is 2 kW, the power of the modulated signal is
a) 2.23 kW
b) 2.36 Kw
c) 1.18 kW
d) 1.26 kW
Answer: a
Explanation:Pt = Pc .
8. A carrier is simultaneously modulated by two sine waves with modulation indices of 0.4 and 0.3. The resultant modulation index will be
a) 1.0
b) 0.7
c) 0.5
d) 0.35
Answer: c
Explanation: α 2 = 0.3 2 + 0.4 2 = 0.5 2 or α = 0.5.
9. In a DSB-SC system with 100% modulation, the power saving is
a) 100%
b) 55%
c) 75%
d) 100%
Answer: b
Explanation: This is so because the power is suppressed by two thirds of the total. hence the power saving is 66%.
10. A 10 kW carrier is sinusoidally modulated by two carriers corresponding to a modulation index of 30% and 40% respectively. The total radiated power is
a) 11.25 kW
b) 12.5 kW
c) 15 kW
d) 17 kW
Answer: a
Explanation: The required answer is 1 (1 + 0.4 2 + 0.3 2 ) or 11.25 kW.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Digital Transmission”.
1. Assuming that the signal is quantized to satisfy the condition of previous question and assuming the approximate bandwidth of the signal is W. The minimum required bandwidth for transmission of a binary PCM signal based on this quantization scheme will be
a) 5 W
b) 10 W
c) 20 W
d) None of the mentioned
Answer: b
Explanation: The minimum bandwidth requirement for transmission of a binary PCM signal is BW= vW. Since v 10, we have BW = 10 W.
2. In PCM system, if the quantization levels are increased form 2 to 8, the relative bandwidth requirement will
a) Remain same
b) Be doubled
c) Be tripled
d) Become four times
Answer: c
Explanation: If L = 2, then 2 = 2 n or n = 1 ND. If L = 8, then 8 = 2 n or n = 3. So relative bandwidth will be tripled.
3. A speech signal occupying the bandwidth of 300 Hz to 3 kHz is converted into PCM format for use in digital communication. If the sampling frequency is8 kHz and each sample is quantized into 256 levels, then the output bit the rate will be
a) 3 kb/s
b) 8 kb/s
c) 64 kb/s
d) 256 kb/s
Answer: c
Explanation: fs = 8 kHz, 2 n = 256 or n = 8. Bit Rate = 8 x 8k = 64 kb/s.
4. Analog data having highest harmonic at 30 kHz generated by a sensor has been digitized using 6 level PCM. What will be the rate of digital signal generated?
a) 120 kbps
b) 200 kbps
c) 240 kbps
d) 180 kbps
Answer: d
Explanation: Nyquist Rate = 2 x 30k = 60 kHz 2 n should be greater than or equal to 6. Thus n 3, Bit Rate = 60×3 = 18 kHz.
5. Four voice signals. each limited to 4 kHz and sampled at Nyquist rate are converted into binary PCM signal using 256 quantization levels. The bit transmission rate for the time-division multiplexed signal will be
a) 8 kbps
b) 64 kbps
c) 256 kbps
d) 512 kbps
Answer: c
Explanation: Nyquist Rate 2 x 4k = 8 kHz
Total sample 4 x 8 = 32 k sample/sec
256 = 2 8 , so that 8 bits are required
Bit Rate 32k x 8 = 256 kbps.
6. A TDM link has 20 signal channels and each channel is sampled 8000 times/sec. Each sample is represented by seven binary bits and contains an additional bit for synchronization. The total bit rate for the TDM link is
a) 1180 K bits/sec
b) 1280 K bits/sec
c) 1180 M bits/sec
d) 1280 M bits/sec
Answer: b
Explanation: Total sample 8000 x 20 = 160 k sample/sec
Bit for each sample 7 + 1 = 8
Bit Rate = 160k x 8 = 1280 kilobits/sec.
7. Four signals each band limited to 5 kHz are sampled at twice the Nyquist rate. The resulting PAM samples are transmitted over a single channel after time division multiplexing. The theoretical minimum transmissions bandwidth of the channel should be equal to
a) 5 kHz
b) 20 kHz
c) 40 kHz
d) 80 kHz
Answer: d
Explanation: fm = 5 kHz, Nyquist Rate 2 x 5 = 10 kHz Since signal are sampled at twice the Nyquist rate so sampling rate 2 x 10 = 20 kHz. Total transmission bandwidth 4 x 20 = 80 kHz.
8. A sinusoidal massage signal m is transmitted by binary PCM without compression. If the signal to-quantization-noise ratio is required to be at least 48 dB, the minimum number of bits per sample will be
a) 8
b) 10
c) 12
d) 14
Answer: a
Explanation: 3(L 2 )/2 = 48 db or L = 205.09. Since L is power of 2, so we select L = 256 Hence 256 = 2 8 , So 8 bits per sample is required.
9. A speech signal has a total duration of 20 sec. It is sampled at the rate of 8 kHz and then PCM encoded. The signal-to-quantization noise ratio is required to be 40 dB. The minimum storage capacity needed to accommodate this signal is
a) 1.12 KBytes
b) 140 KBytes
c) 168 KBytes
d) None of the mentioned
Answer: b
Explanation: q = 1.76 + 6.02 = 40 dB, n = 6.35
We take the n = 7.
Capacity = 20 x 8k x 7 = 1.12 Mbits = 140 Kbytes.
10. A linear delta modulator is designed to operate on speech signals limited to 3.4 kHz. The sampling rate is 10 time the Nyquist rate of the speech signal. The step size is 100 m V. The modulator is tested with a this test signal required to avoid slope overload is
a) 2.04 V
b) 1.08 V
c) 4.08 V
d) 2.16 V
Answer: b
Explanation: Amax = / or 1.08V.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Semiconductor Physics”.
In the problems assume the parameter given in following table. Use the temperature T= 300 K unless otherwise stated.
electronic-devices-circuits-questions-answers-semiconductor-physics
1. In germanium semiconductor material at T 400 K the intrinsic concentration is
a) 26.8
b) 18.4
c) 8.5
d) 3.6
Answer: c
Explanation: electronic-devices-circuits-questions-answers-semiconductor-physics-q1
2. The intrinsic carrier concentration in silicon is to be no greater than ni = 1 x 10^12 cc. The maximum temperature allowed for the silicon is
a) 300 K
b) 360 K
c) 382 K
d) 364 K
Answer: c
Explanation: electronic-devices-circuits-questions-answers-semiconductor-physics-q1
3. Two semiconductor material have exactly the same properties except that material A has a bandgap of 1.0
eV and material B has a bandgap energy of 1.2 eV. The ratio of intrinsic concentration of material A to that of
material B is
a) 2016
b) 47.5
c) 58.23
d) 1048
Answer: b
Explanation: electronic-devices-circuits-questions-answers-semiconductor-physics-q3
4. In silicon at T = 300 K the thermal-equilibrium concentration of electron is n0 = 5 x 10^4 cc. The hole concentration is
a) 4.5 x 10 15 cc
b) 4.5 x 10 15 m 3
c) 0.3 x 10 -6 cc
d) 0.3 x 10 -6 m 3
Answer: a
Explanation: ni x ni = no x po.
5. In silicon at T = 300 K if the Fermi energy is 0.22 eV above the valence band energy, the value of p0 is
a) 2 x 10 15 cm 3
b) 10 15 cm 3
c) 3 x 10 15 cm 3
d) 4 x 10 15 cm 3
Answer: a
Explanation: electronic-devices-circuits-questions-answers-semiconductor-physics-q5
6. The thermal-equilibrium concentration of hole p0 in silicon at T = 300 K is 1015 cm3. The value of n0 is
a) 3.8 x 10 8 cm 3
b) 4.4 x 10 4 cm 3
c) 2.6 x 10 4 cm 3
d) 4.3 x 10 8 cm 3
Answer: b
Explanation: electronic-devices-circuits-questions-answers-semiconductor-physics-q6
7. In germanium semiconductor at T 300 K, the acceptor concentrations is Na 1013 cm3 and donor concentration is Nd 0. The thermal equilibrium concentration p0 is
a) 2.97 x 10 9 cm 3
b) 2.68 x 10 12 cm 3
c) 2.95 x 10 13 cm 3
d) 2.4 cm 3
Answer: c
Explanation: electronic-devices-circuits-questions-answers-semiconductor-physics-q7
8. A silicon sample doped n type at 10^18 cm3 have a resistance of 10 ohm. The sample has an area of 10^ cm2 and a length of 10 µm . The doping efficiency of the sample is µ
a) 43.2%
b) 78.1%
c) 96.3%
d) 54.3%
Answer: b
Explanation: electronic-devices-circuits-questions-answers-semiconductor-physics-q8
9. Six volts is applied across a 2 cm long semiconductor bar. The average drift velocity is 104 cms. The electron mobility is
a) 4396 cm 2 /V -s 2
b) 3 x 10 4 cm 2 /V -s
c) 6 x 10 4 cm 2 V -s
d) 3333 cm 2 /V -s
Answer: d
Explanation: electronic-devices-circuits-questions-answers-semiconductor-physics-q9
10. A particular intrinsic semiconductor has a resistivity of 50 at T = 300 K and 5 at T = 330 K. If change in mobility with temperature is neglected, the bandgap energy of the semiconductor is
a) 1.9 eV
b) 1.3 eV
c) 2.6 eV
d) 0.64 eV
Answer: b
Explanation: electronic-devices-circuits-questions-answers-semiconductor-physics-q10
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Network Theroms”.
Circuit for Q.1-Q.2
electronic-devices-circuits-questions-answers-network-theorms-q1
1. Vth = ?
a) 1 V
b) 2 V
c) 3 V
d) 4 V
Answer: d
Explanation: Vth = / or 4 V.
2. Rth = ?
a) 2
b) 3
c) 4
d) 5
Answer: c
Explanation: Rth = + 2 or 4 ohm.
3. A battery has a short-circuit current of 30 A and an open circuit voltage of 24 V. If the battery is connected to an electric bulb of resistance 2 ohm -1 , the power dissipated by the bulb is
a) 80 W
b) 1800 W
c) 112.5 W
d) 228 W
Answer: c
Explanation: r = Voc/Isc = 1.2 ohm
Power used by bulb = x 2/ x or 11.5 Watt.
Circuit for Q.4-Q-5
electronic-devices-circuits-questions-answers-network-theorms-q4
4. In = ?
a) 1.5 A
b) 3 A
c) 6 A
d) 10 A
Answer: b
Explanation: V1 = /
In = Isc = V1/2 = 3 A.
5. Rn = ?
a) 10/3
b) 4
c) 6
d) 10
Answer: a
Explanation: Rn = + 2.
Circuit for
electronic-devices-circuits-questions-answers-network-theorms-q6
6. Vth = ?
a) -2 V
b) -1 V
c) 1 v
d) 2 V
Answer: c
Explanation: Vth = /3+3 = 1V.
7. Rth = ?
a) 5/6
b) 6/5
c) 5/3
d) 3/5
Answer: a
Explanation: Rth = 1||5 or 5/6 ohm.
8. The equivalent to the given circuit is
electronic-devices-circuits-questions-answers-network-theorms-q8
electronic-devices-circuits-questions-answers-network-theorms-q8a
electronic-devices-circuits-questions-answers-network-theorms-q8b
Answer: b
Explanation: After killing all source equivalent resistance is R. Open circuit voltage is v1.
9. V1 = ?
electronic-devices-circuits-questions-answers-network-theorms-q9
a) 6 V
b) 7 V
c) 8 V
d) 10 V
Answer: a
Explanation: electronic-devices-circuits-questions-answers-network-theorms-q9a
10. i1 = ?
electronic-devices-circuits-questions-answers-network-theorms-q10
a) 3 A
b) 0.75 mA
c) 2 mA
d) 1.75 mA
Answer: b
Explanation: electronic-devices-circuits-questions-answers-network-theorms-q10a
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Circuit analysis in S domain”.
1. The network function (s 2 + 4s)/ represents
a) RC impedance
b) RL impedance
c) LC impedance
d) None of the mentioned
Answer: d
Explanation: Poles and zero does not interlace on negative real axis so it is not a immittance function.
2. The network function (3s 2 + 8s)/ represents
a) RC impedance
b) RL impedance
c) LC impedance
d) None of the mentioned
Answer: c
Explanation: The singularity nearest to origin is a zero. So it may be RL impedance or RC admittance function. Because of option it is required to check that it is a valid RC admittance function. The poles and zeros interlace along the negative real axis. The residue of Yrc/s is positive.
3. The network function /s represents
a) RC impedance
b) RL impedance
c) LC impedance
d) All of the mentioned
Answer: b
Explanation: The singularity nearest to origin is a pole. So it may be RC impedance or RL admittance function.
4. The network function s^2 + 10s + 24/s 2 + 8s + 15 represents
a) RC impedance
b) RL impedance
c) LC impedance
d) None of the mentioned
Answer: d
Explanation: The singularity is near to origin is pole. So it may be RC impedance or RL admittance function.
5. A valid immittance function is
a) /
b) s/
c) s/
d) s/
Answer: d
Explanation:
a) pole lie on positive real axis
b) poles and zero does not interlace on axis.
c) poles and zero does not interlace on axis.
d) is a valid immittance function.
6. The network function (s 2 + 8s +15)/(s 2 + 6s + 8) is
a) RL admittance
b) RC admittance
c) LC admittance
d) All of the mentioned
Answer: a
Explanation: The singularity nearest to origin is a pole. So it may be a RL admittance or RC impedance function.
7. The voltage response of a network to a unit step input is Vo = 10/s(s 2 + 8s + 16). The response is
a) under damped
b) over damped
c) critically damped
d) can’t be determined
Answer: c
Explanation: The characteristic equation has real and repeated roots . Hence it is critically damped.
8. The current response of a network to a unit step input is Io = 10/s(s 2 + 11s + 30). The response is
a) Under damped
b) Over damped
c) Critically damped
d) None of the mentioned
Answer: b
Explanation: The roots are real and unequal for the characteristic equation. Hence it is over damped.
Circuit for q.9-Q.10
electronic-devices-circuits-questions-answers-circuit-analysis-s-domain-q9
9. The ratio of the transfer function Io/Is is
a) s/(s 2 + 3s + 4)
b) s/
c) (s 2 + 3s + 4)/s
d) /s
Answer: b
Explanation: Io/Is = / = s/.
10. The response is
a) Over damped
b) Under damped
c) Critically damped
d) can’t be determined
Answer: a
Explanation: The characteristic roots are real and unequal , therefore it is over damped.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Magneticaly coupled circuits”.
For the circuit given below i1 = 4 sin and i2 = 0
electronic-devices-circuits-questions-answers-magnetically-coupled-circuits-q1
1. v1 = ?
a) -16 cos 2t V
b) 16 cos 2t V
c) 4 cos 2t V
d) -4 cos 2t V
Answer: b
Explanation: v1 = 2 d/dt + d/dt.
2. v2 = ?
a) 2 cos 2t V
b) -2 cos 2t V
c) 8 cos 2t V
d) -8 cos 2t V
Answer: c
Explanation: v2 = d/dt + d/dt.
Circuit for Q.3-Q.4
electronic-devices-circuits-questions-answers-magnetically-coupled-circuits-q3
3. If i1 = 0 and i2 = 2 sin, the voltage v1 is
a) -24 cos V
b) 24 cos V
c) 1.5 cos V
d) -1.5 cos V
Answer: b
Explanation: v1 = 3 d/dt – 3 d/dt.
4. If i1 = e and i2 = 0, the voltage v1 is
a) -6e V
b) 6e V
c) 1.5e V
d) -1.5e V
Answer: c
Explanation: v1 = 4 d/dt – 3 d/dt.
Circuit for Q.5-Q.8
electronic-devices-circuits-questions-answers-magnetically-coupled-circuits-q5
5. If i1 = 3 cos and and i2 = 0. Find v1.
a) -24 sin V
b) 24 sin V
c) 1.5 sin V
d) -1.5 sin V
Answer: a
Explanation: v1 = 2 d/dt – 2 d/dt.
6. If i1 = 3 cos and and i2 = 0. Find v2.
a) -24 sin V
b) -36 sin V
c) sin V
d) -sin V
Answer: a
Explanation: v2 = -3 d/dt + 2 d/dt.
7. If i1 = 4 cos and and i1 = 0. Find v2.
a) 12 cos V
b) -12 cos V
c) -24 cos V
d) 24 cos V
Answer: c
Explanation: v1 = 2 d/dt – 2 d/dt.
8. If i2 = 4 cos and and i1 = 0. Find v2.
a) -12 cos V
b) -24 cos V
c) -36 cos V
d) -48 cos V
Answer: c
Explanation: v2 = 3 d/dt + 2 d/dt.
9. Leq = ?
electronic-devices-circuits-questions-answers-magnetically-coupled-circuits-q9
a) 4 H
b) 6 H
c) 7 H
d) 0 H
Answer: c
Explanation: Leq = L1 + L2 – 2M.
10. Leq = ?
electronic-devices-circuits-questions-answers-magnetically-coupled-circuits-q10
a) 2 H
b) 4 H
c) 6 H
d) 8 H
Answer: a
Explanation: Leq = L1 + L2 – 2M.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Electrons and Holes in Semiconductor”.
1. Which of the following expressions represents the correct distribution of the electrons in the conduction band? =density of quantum states, fF=Fermi dirac probability
a) n=gc*fF
b) n=gc*fF
c) n=gc*fF
d) n= gc*fF
Answer: a
Explanation: The distribution of the electrons in the conduction band is given by the product of the density into Fermi-dirac distribution.
2. What is the value of the effective density of states function in the conduction band at 300k?
a) 3*10 19 cm -3
b) 0.4*10 -19 cm -3
c) 2.5*10 19 cm -3
d) 2.5*10 -19 cm -3
Answer: c
Explanation: electronic-devices-circuits-questions-answers-electrons-holes-semiconductor-q2
Substituting the values of m n =m0 ,h=6.626*10 -34 J/s ,k=1.38*10 -23 and T=300K, we get
Nc=2.5*10 19 cm -3 .
3. In a semiconductor which of the following carries can contribute to the current?
a) Electrons
b) Holes
c) Both
d) None
Answer: c
Explanation: In a semiconductor, two types of charges are there by which the flow of the current takes place. So, both the holes and electrons take part in the flow of the current.
4. Which of the following expressions represent the Fermi probability function?
a) fF=exp
b) fF=exp
c) fF=exp
d) fF=exp
Answer: b
Explanation: It is the correct formula for the Fermi probability function.
5. Electrons from valence band rises to conduction band when the temperature is greater than 0 k. Is it True or False?
a) True
b) False
Answer: a
Explanation: As the temperature rises above 0 k, the electrons gain energy and rises to the conduction band from the valence band.
6. What is the intrinsic electrons concentration at T=300K in Silicon?
a) 1.5*10 10 cm -3
b) 1.5*10 -10 cm -3
c) 2.5*10 19 cm -3
d) 2.5*10 -19 cm -3
Answer: a
Explanation: Using the formula,
electronic-devices-circuits-questions-answers-electrons-holes-semiconductor-q6
We get, ni=1.5*10 10 cm -3 .
7. The intrinsic Fermi level of a semiconductor depends on which of the following things?
a) E midgap
b) m p *
c) m n *
d) All of the mentioned
Answer: d
Explanation: electronic-devices-circuits-questions-answers-electrons-holes-semiconductor-q7
From the above formula, E fi depends on all of the options given.
8. What is the difference between the practical value and theoretical value of ni?
a) Factor of 1
b) Factor of 2
c) Factor of 3
d) Factor of 4
Answer: b
Explanation: This is practically proved.
9. The thermal equilibrium concentration of the electrons in the conduction band and the holes in the valence band depends upon?
a) Effective density of states
b) Fermi energy level
c) Both A and B
d) Neither A nor B
Answer: c
Explanation: The electrons and holes depends upon the effective density of the states and the Fermi energy level given by the formula,
electronic-devices-circuits-questions-answers-electrons-holes-semiconductor-q6 .
10. In which of the following semiconductor, the concentration of the holes and electrons is equal?
a) Intrinsic
b) Extrinsic
c) Compound
d) Elemental
Answer: a
Explanation: In the intrinsic semiconductor, ni=pi that is the number of the electrons is equal to the number of the holes. Whereas in the extrinsic conductor ni is not equal to pi.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Conductivity of a Semiconductor”.
1. What is the SI unit of conductivity?
a) Ωm
b) -1
c) Ω
d) m
Answer: b
Explanation: The formula of the conductivity is the σ=1/ρ.
So, the unit of resistivity is Ωm.
Now, the unit of conductivity becomes the inverse of resistivity.
2. Which of the following expressions doesn’t represent the correct formula for Drift current density?
a) J=σE
b) J=qnµE
c) J=µE
d) None
Answer : c
Explanation: The following formulae represent the correct expression for drift current density,
J=σE
And J=qnµE.
3. Does a semiconductor satisfy the ohm’s law?
a) True
b) False
Answer: a
Explanation: V=IR
J=σE
I/A=σ
V=*I=*I/A=IR
Thus, above equation satisfies Ohm’s law.
4. In which range of temperature, freeze out point begins to occur?
a) Higher range
b) Lower range
c) Middle range
d) None
Answer: b
Explanation: At lower range of temperature, the concentration and conductivity decreases with lowering of the temperature.
5. Which of the following expression represents the correct formula for the conductivity in an intrinsic material?
a) ρ=e(μ n+μ p )ni
b) σ=e(μ n+μ p )ni
c) σ=1/(e(μ n+μ p )ni)
d) ρ=1/(e(μ n+μ p )ni)
Answer: b
Explanation: Option b is the correct formula.
6. What is the voltage difference if the current is 1mA and length and area is 2cm and 4cm2 respectively?
a) 0.025V
b) 25V
c) 0.25V
d) None
Answer: d
Explanation: V=IR
R=ρl/=2*2/4=100Ω
V=1mA*100
=0.1V.
7. Is resistivity is a function of temperature?
a)True
b)False
Answer: a
Explanation: Resistance depends on the temperature and the resistivity depends on the resistance, so now the resistivity depends on the temperature.
8. What is the electric field when the voltage applied is 5V and the length is 100cm?
a) 0.5V/m
b) 5V/m
c) 50V/m
d) None
Answer: c
Explanation: E=V/L=5/100cm=5V/m.
9. Calculate the average random thermal energy at T=300K?
a) 0.038eV
b) 3.8eV
c) 38eV
d) 0.38eV
Answer: a
Explanation: Average random thermal energy=3/2*k*T=0.038eV.
10.
electronic-devices-circuits-questions-answers-conductivity-semiconductor-q10
In the above figure, a semiconductor having an area ‘A’ and length ‘L’ and carrying current ‘I’ applied a voltage of ‘V’ volts across it. Calculate the relation between V and A?
a) V= /A)*I
b) V= /L)*I
c) V= /)
d) V=
Answer: a
Explanation: Option A, satisfies the Ohm’s law which is V=IR where R=/A.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Donor and Acceptor impurities”.
1. At what temperature the donor states are completely ionized?
a) 0 K
b) ROOM
c) 300K
d) 900K
Answer: b
Explanation: At room temperature, the donors have donated their electrons to the conduction band.
2. The opposite of ionization takes place at which temperature?
a) 0 K
b) ROOM
c) 300 K
d) 900K
Answer: a
Explanation: AT 0 K, all the donors and acceptors are in their lowest energy levels.
3. What do you mean by the tem ‘FREEZE-OUT’?
a) All the electrons are frozen at room temperature
b) None of the electrons are thermally elevated to the conduction band
c) All the electrons are in the conduction band
d) All the holes are in the valence band
Answer: b
Explanation: Freeze out means none of the electrons are transmitted to the conduction band.
4. Which of the following expressions represent the correct formula for the density of electrons occupying the donor level?
a) n d =N d -N d +
b) n d =N d -N d –
c) n d =N d +N d +
d) n d =N d +N d –
Answer: a
Explanation: The density of the electrons is equal to the electrons present in the substrate minus the number of donors present.
5. Which of the following band is just above the intrinsic Fermi level for n-type semiconductor?
a) Donor band
b) Valence band
c) Acceptor band
d) Conduction band
Answer: a
Explanation: For n-type semiconductors, the donor band is just above the intrinsic Fermi level.
6. At absolute zero temperature, which level is above the Fermi energy level in the case of donors?
a) Donor energy level
b) Acceptor energy level
c) Conduction Band
d) Valence Band
Answer: c
Explanation: At T=0 K, the tem exp=0 in the expression of
electronic-devices-circuits-questions-answers-donor- acceptor-impurities-q6
Thus, EF>ED
So, only conduction band lies above the Fermi energy level.
7. At T=0 K, the location of Fermi level with respect to the Ec and Ed for the n type material is?
a) Above than conduction band
b) Midway
c) Lower than Ed
d) Greater than Ed
Answer: b
Explanation: At T=0 K, the Fermi level of n band lies between the midway of Ec and Ed as intrinsic Fermi level always lies between the Ec and Ev.
8. At absolute zero temperature, which level is below the Fermi energy level in the case of acceptors?
a) Donor energy level
b) Valence Band
c) Conduction band
d) Acceptor energy level
Answer: b
Explanation: At T=0 K, the tem exp=0 in the expression of
electronic-devices-circuits-questions-answers-donor- acceptor-impurities-q6
So, only valence band lies below the Fermi energy level of the acceptors.
9.
electronic-devices-circuits-questions-answers-donor- acceptor-impurities-q9
For the above n-type semiconductor, what is B knows as?
a) Valence Band
b) Conduction Band
c) Donor Energy level
d) Acceptor energy level
Answer: c
Explanation: For n-type semiconductors, the donor energy level is always greater than the Fermi level energy.
10.
electronic-devices-circuits-questions-answers-donor- acceptor-impurities-q10
For the above given figure, identify the correct option for satisfying the above semiconductor figure?
a) P type, A-Conduction band, B-donor energy band, C- Valence band
b) P type, A-Conduction band, B-acceptor energy band, C- Valence band
c) n type, A-Conduction band, B-donor energy band, C- Valence band
d) n type, A-Conduction band, B-acceptor energy band, C- Valence band
Answer: b
Explanation: The given figure has B band below the intrinsic Fermi level, so that would be acceptor energy band and will be a p-type semiconductor.
This set of Electronic Devices and Circuits Interview Questions and Answers focuses on “Charge Densities in a Semiconductor impurities”.
1. Is n/p=ni 2 is a correct formula?
a) True
b) False
Answer: b
Explanation: The correct formula is n*p=ni 2 .
2. Calculate the number of electrons is the number of holes are 15*10 10 ?
a) 15*10 10
b) 1.5*10 8
c) 1.5*10 9
d) 1.5*10 10
Answer: c
Explanation: n*p=(1.5*10 10 ) 2
n*15*10 10 =1.5*1.5*10 10 *10 10
n=1.5*10 9 electrons.
3. For which type of material, the number of free electron concentration is equal to the number of donor atoms?
a) P type semiconductor
b) Metal
c) N-type semiconductor
d) Insulator
Answer: c
Explanation: The n-type semiconductor has equal concentration of free electron and donor atoms.
4. Identify the correct condition for a semiconductor to be electrically neutral.
a) N d +p=N a +n
b) N d -p=N a +n
c) N d +p=N a -n
d) N d -p=N a -n
Answer: a
Explanation: The sum of the number of donors and the holes is equal to the sum of the number of the acceptors and the electrons.
5. Do the Fermi energy level changes in a semiconductor?
a) True
b) False
Answer: a
Explanation: The Fermi energy level changes as the electron and hole concentrations change because of the formula which defines the position of the Fermi level depending on the concentration of holes and electrons.
6. Consider a silicon wafer having Nc=2.8*10 19 cm -3 and the Fermi energy is .25eV below the conduction band. Calculate the equilibrium concentrations of electrons at T=300K?
a) 18*10 16 cm -3
b) 1.8*10 16 cm -3
c) 1.8*10 14 cm -3
d) 180*10 16 cm -3
Answer: a
Explanation: n0=Nc*exp=2.8*10 19 *exp
=18*10 16 cm -3 .
7. If Ef>Efi, then what is the type of the semiconductor?
a) n-type
b) P-type
c) Elemental
d) Compound
Answer: a
Explanation: For n-type, the Fermi energy level is greater than the intrinsic Fermi energy level because in an energy band, Fermi level of donors is always greater than that of the acceptors.
8. The 1-f F increases in which of the following band for n type semiconductor?
a) Conduction band
b) Donor band
c) Acceptor band
d) Valence band
Answer: d
Explanation: For an n-type semiconductor, the probability of f F decreases in the valence band. The probability of finding the electron in the conduction band is more.
9. The f F decreases in which of the following band for p-type semiconductor?
a) Conduction band
b) Donor band
c) Acceptor band
d) Valence band
Answer: a
Explanation: The probability of finding the electron in the conduction band decreases for a p-type semiconductor because in a p-type semiconductor, the holes will be in conduction band rather than the electrons.
10. Do the intrinsic Fermi energy level changes with the addition of dopants and acceptors?
a) True
b) False
Answer: b
Explanation: The intrinsic Fermi energy level always remains constant because it is an imaginary level taken to distinguish between the Fermi level of the types of semiconductor.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Fermi Level in a Semiconductor having Impurities”.
1. Which states get filled in the conduction band when the donor-type impurity is added to a crystal?
a) Na
b) Nd
c) N
d) P
Answer: b
Explanation: When the donor-type impurity is added to a crystal, first Nd states get filled because it is of the highest energy.
2. Which of the following expression represent the correct formulae for calculating the exact position of the Fermi level for p-type material?
a) E F = E V + kTln(N D / N A )
b) E F = -E V + kTln(N D / N A )
c) E F = E V – kTln(N D / N A )
d) E F = -E V – kTln(N D / N A )
Answer: a
Explanation: The correct position of the Fermi level is found with the formula in the ‘a’ option.
3. Where will be the position of the Fermi level of the n-type material when N D =N A ?
a) Ec
b) Ev
c) Ef
d) Efi
Answer: a
Explanation: When N D =N A , kTln(N D /N A )=0
So,
Ef=Ec.
4. When the temperature of either n-type or p-type increases, determine the movement of the position of the Fermi energy level?
a) Towards up of energy gap
b) Towards down of energy gap
c) Towards centre of energy gap
d) Towards out of page
Answer: c
Explanation: whenever the temperature increases, the Fermi energy level tends to move at the centre of the energy gap.
5. Is it true, when the temperature rises, the electrons in the conduction band becomes greater than the donor atoms?
a) True
b) False
Answer: a
Explanation: When the temperature increases, there is an increase in the electron-hole pairs and all the donor atoms get ionized, so now the thermally generated electrons will be greater than the donor atoms.
6. If the excess carriers are created in the semiconductor, then identify the correct energy level diagram.
a) electronic-devices-circuits-questions-answers-fermi-level-semiconductor-impurities-q6
b) electronic-devices-circuits-questions-answers-fermi-level-semiconductor-impurities-q6a
c) electronic-devices-circuits-questions-answers-fermi-level-semiconductor-impurities-q6b
d) electronic-devices-circuits-questions-answers-fermi-level-semiconductor-impurities-q6c
Answer: a
Explanation: The diagram A refers the most suitable energy level diagrams because Efp>Ef>Efi>Efp>Ev.
7. If excess charge carriers are created in the semiconductor then the new Fermi level is known as Quasi-Fermi level. Is it true?
a) True
b) False
Answer: a
Explanation: Quasi-fermi level is defined as the change in the level of the Fermi level when the excess chare carriers are added to the semiconductor.
8. Ef lies in the middle of the energy level indicates the unequal concentration of the holes and the electrons?
a) True
b) False
Answer: b
Explanation: When the Ef is in the middle of the energy level, it indicates the equal concentration of the holes and electrons.
9. Consider a bar of silicon having carrier concentration n0=10 15 cm -3 and ni=10 10 cm -3 . Assume the excess carrier concentrations to be n=10 13 cm -3 , calculate the quasi-fermi energy level at T=300K?
a) 0.2982 eV
b) 0.2984 eV
c) 0.5971 eV
d) 1Ev
Answer: b
Explanation: electronic-devices-circuits-questions-answers-fermi-level-semiconductor-impurities-q9
=1.38*10 -23 *300*ln(10 13 +10 15 /10 13 )
=0.2984 eV.
10. From the above equation, assuming the same values for the for ni, n= p and T. Given that p0=10 5 cm -3 . Calculate the quasi-fermi energy level in eV?
a) 0.1985
b) 0.15
c) 0.1792
d) 0.1
Answer: c
Explanation: Using the same equation,
electronic-devices-circuits-questions-answers-fermi-level-semiconductor-impurities-q10
Substituting the respective values,
E Fi – E Fp =0.1792 eV.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Diffusion”.
1. What is the SI unit of electron diffusion constant?
a) cm 2 /s
b) m 2 /s
c) m/s
d) none
Answer: b
Explanation: J=eDdn/dx
So D=q*m/ )
=m 2 /s.
2. Calculate the diffusion current density when the concentration of electron varies from the 1*10 18 to 7*10 17 cm -3 over a distance of 0.10 cm.D=225cm 2 /s
a) 100 A/cm 2
b) 108 A/cm 2
c) 0.01A/cm 2
d) None
Answer: b
Explanation: J=eDdn/dx
J=1.6*10 -19 *225*(10 18 -(7*10 17 ))/0.1
=108A/cm 2 .
3. Which is the correct formula for the Jp?
a) Jp=qDdp/dx
b) Jp=pDdn/dx
c) Jp=-qDdp/dx
d) None
Answer: c
Explanation: Jp is negative for the p type of semiconductors.
4. What is the direction of the electron diffusion current density relative to the electron flux?
a) Same direction
b) Opposite to each other
c) Perpendicular to each other
d) At 270 degrees to each other
Answer: b
Explanation: From the graph between electron concentration and the distance, we can see that the direction of the electron diffusion current density is opposite to the electron flux.
5. In diffusion, the particles flow from a region of _______ to region of ___________
a) High, low
b) Low , high
c) High , medium
d) Low, medium
Answer: a
Explanation: Diffusion is the process of flow of particles form the region of the high concentration to a region of low concentration.
6. Which of the following parameter describes the best movement of the electrons inside a semiconductor?
a) Velocity gradient
b) Diffusion
c) Mobility
d) Density gradient
Answer: c
Explanation: Mobility is defined as the movement of the electrons inside a semiconductor. On the other hand, velocity gradient is the ratio of velocity to distance.
7. Which of the following term isn’t a part of the total current density in a semiconductor?
a) Temperature
b) µ
c) e
d) E
Answer: a
Explanation: J=enµE+epµE+eDdn/dx-eDdp/dx
So, temperature isn’t a part of the equation.
8. What does dn/dx represent?
a) Velocity gradient
b) Volume gradient
c) Density gradient
d) None
Answer: c
Explanation: dn/dx represent velocity gradient.
9. Calculate the diffusion constant for the holes when the mobility of the holes is 400cm 2 /V-s and temperature is 300K?
a) 1.035m m 2 /s
b) 0.035m m 2 /s
c) 1.5m m 2 /s
d) 1.9m m 2 /s
Answer: a
Explanation: D p =V T *μn
= (1.38*10 -23 *300*400*10 -2 )/ (1.6*10 -19 )
= 1.035m m 2 /s.
10. Calculate the diffusion constant for the electrons when the mobility of the electrons is 325cm 2 /V-s and temperature is 300K?
a) 0.85 m 2 /s
b) 0.084 m 2 /s
c) 0.58 m 2 /s
d) 0.95 m 2 /s
Answer: c
Explanation: D n =V T *μn
= (1.38*10 -23 *300*325*10 -2 )/ (1.6*10 -19 )
= 0.084 m 2 /s.
This set of Electronic Devices and Circuits Questions and Answers for Freshers focuses on “Carrier Life Time”.
1. What is the range of the carrier lifetime?
a) Nanoseconds to microseconds
b) Nanoseconds to hundreds of microseconds
c) Nanoseconds to tens of microseconds
d) Nanoseconds to milliseconds
Answer: b
Explanation: Carrier lifetime is defined as the existence of any carrier for τ seconds. Carrier lifetime ranges from nanoseconds to hundreds of microseconds.
2. What is the process number of Schokley-Read-Hall Theory processes?
Process-‘ The capture of an electron from the conduction band by an initially neutral empty trap’
a) Process1
b) Process2
c) Process3
d) Process4
Answer: a
Explanation: This is the first process of Schokley-Read-Hall theory of Recombination.
3. Calculate the recombination rate if the excess carrier concentration is 10 14 cm -3 and the carrier lifetime is 1µseconds.
a) 10 8
b) 10 10
c) 10 20
d) 10 14
Answer: c
Explanation: The recombination rate, R=δn/τ.
So, R=10 14 / 10 -6
R=10 20 .
4. Calculate the capture rate where Cn=10, Nt=10 10 cm -3 , n=10 20 and f F =0.4.
a) 6*10 30
b) 5*10 30
c) 36*10 30
d) 1.66*10 29
Answer: a
Explanation: R cn =C n *N_t*(1-f F )*n
Substituting the values,
R cn =6*10 30 .
5. Calculate the emission rate where En=2.5, Nt=10 10 cm -3 and f F =0.6 .
a) 15*10 10
b) 1.5*10 10
c) 15*10 11
d) 1.5*10 11
Answer: b
Explanation: R en =E n *N t *(f F )
Substituting the values,
R en =1.5*10 10 .
6. At what condition, the rate of electron capture from the conduction band and the rate of the electron emission back into the conduction band must be equal?
a) Thermal equilibrium
b) At room temperature
c) T=250K
d) At boiling temperature
Answer: a
Explanation: At thermal equilibrium, the electron capture rate and the emission rate will be same in the conduction band.
7. Calculate the carrier lifetime when Cp=5 and Nt=10 10 cm -3 .
a) 2*10 11
b) 2*10 -11
c) 20*10 -11
d) 20*10 11
Answer: b
Explanation: τ p =1/(C p *N t )
=1/(5*10 10 )
=2*10 -11 .
8. The number of majority carriers that are available for recombining with excess minority carriers decreases as the excess semiconductor becomes intrinsic. Is it true?
a) True
b) False
Answer: a
Explanation: With the increase in the number of the majority carriers, the carriers for the recombination will be decreasing with the excess minority carriers and will finally become intrinsic as the concentrations will be same.
9. Which of the following is used as the recombination agent by semiconductor device manufactures?
a) Silver
b) Gold
c) Platinum
d) Aluminium
Answer: b
Explanation: Gold is used as a recombination agent because of its chemical properties as it was used in the Bohr’s experiment. Thus the device designer can obtain the desired carrier lifetimes by introducing gold into silicon under controlled conditions.
10. The rate of change of the excess density is proportional to the density. Is it true of false?
a) True
b) False
Answer: a
Explanation: The rate of change of the excess density depends on the density of the semiconductor and the rate with respect to time is also dependent on it.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Continuity Equation”.
1. What does p/τ represent?
a) holes
b) time
c) holes per second lost
d) p per unit time
Answer: c
Explanation: Option represents the holes per second lost by recombination per unit volume.
2. Which of the following is the Taylor’s expression?
a) electronic-devices-circuits-questions-answers-continuity-equation-q2
b) electronic-devices-circuits-questions-answers-continuity-equation-q2a
c) electronic-devices-circuits-questions-answers-continuity-equation-q2b
d) electronic-devices-circuits-questions-answers-continuity-equation-q2c
Answer: a
Explanation: Option a represents the correct formula for the Taylor’s expression.
3. Calculate the number of coulombs per second if the area is 4cm 2 , recombination rate of hole is 1000 cm -3 /s and the differential length is 2mm.
a) 1.28*10 -23
b) 1.28*10 -22
c) 1.28*10 -21
d) 1.28*10 -20
Answer: b
Explanation: number of coulombs per second= eAdxp/τ
=1.6*10 -19 *4*10 -4 *2*10 -3 *1000
=1.28*10 -22 .
4. The current entering the volume at x is I and leaving is I+Δi , the number of coulombs per second will be equal to δI. Is it true or false?
a) True
b) False
Answer: a
Explanation: Coulombs per second is known as the current. The differential current will be the current through the semiconductor.
5. The change in the carrier density is due to
a) Flow of incoming flux
b) Flow of outgoing flux
c) Difference of flow between incoming and outgoing flux
d) Difference of flow between incoming and outgoing flux plus generation and minus recombination
Answer: d
Explanation: The change in the carrier density describes the continuity equation which is equal to the difference between the incoming and outgoing flux plus generation and minus recombination.
6. What of the following conditions satisfies when the number of holes which are thermally generated is equal to the holes lost by recombination?
a) I≠0
b) dp/dt≠0
c) g=p/τ
d) g≠p/τ
Answer: c
Explanation: Under the equilibrium conditions, I will be zero and then the dp/dt will aso be equal to zero in the continuity equation. Then, g= p/τ is left which is option c.
7. What is the diffusion length for holes when Dp=25cm 2 /s and τ p =25s?
a) 25cm
b) 1cm
c) 0.04cm
d) 50cm
Answer: a
Explanation: L p =√(D p *τ p )
=√
=25cm.
8. Which of the following represents the continuity equation?
a) dp/dt=-/τp+Dp(d 2p /dx 2 )-µpd/dx
b) dp/dt=-/τp-Dp(d 2p /dx 2 )-µpd/dx
c) dp/dt=-/τp+Dp(d 2p /dx 2 )+µpd/dx
d) dp/dt=/τp-Dp(d 2p /dx 2 )-µpd/dx
Answer: a
Explanation: Option a represents the correct equation of the continuity equation for holes.
9. What is the diffusion length for electrons when Dn=10cm 2 /s and τ n =40s?
a) 50cm
b) 25cm
c) 20cm
d) 15cm
Answer: c
Explanation: L n =√(D n *τ n )
=√
=20cm.
10. Which of the following represents the best definition for the diffusion length for holes?
a) Average distance which an electron is injected travels before recombining with an electron
b) Average distance which a hole is injected travels before recombining with an electron
c) Average distance which a hole is injected travels before recombining with a hole
d) Average distance which an electron is injected before recombining with a hole
Answer: b
Explanation: Diffusion length for holes is represented as the average distance which a hole is injected travels before recombining with an electron. It is the distance into the semiconductor at which the injected concentration falls to 1/ϵ of its value at x=0.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Hall Effect”.
1. In the Hall Effect, the directions of electric field and magnetic field are parallel to each other.
The above statement is
a) True
b) False
Answer: b
Explanation: To make Lorentz force into the effect, the electric field and magnetic field should be perpendicular to each other.
2. Which of the following parameters can’t be found with Hall Effect?
a) Polarity
b) Conductivity
c) Carrier concentration
d) Area of the device
Answer: d
Explanation: The Hall Effect is used for finding the whether the semiconductor is of n-type or p-type, mobility, conductivity and the carrier concentration.
3. In the Hall Effect, the electric field is in x direction and the velocity is in y direction. What is the direction of the magnetic field?
a) X
b) Y
c) Z
d) XY plane
Answer: c
Explanation: The Hall Effect satisfies the Lorentz’s Force which is
E=vxB
So, the direction of the velocity, electric field and magnetic field should be perpendicular to each other.
4. What is the velocity when the electric field is 5V/m and the magnetic field is 5A/m?
a) 1m/s
b) 25m/s
c) 0.2m/s
d) 0.125m/s
Answer: a
Explanation: E=vxB
v=E/B
=5/5
=1m/s.
5. Calculate the hall voltage when the Electric Field is 5V/m and height of the semiconductor is 2cm.
a) 10V
b) 1V
c) 0.1V
d) 0.01V
Answer: c
Explanation: Vh=E*d
=5*2/100
=0.1V.
6. Which of the following formulae doesn’t account for correct expression for J?
a) ρv
b) I/wd
c) σE
d) µH
Answer: d
Explanation: B=µH
So, option d is correct option.
7. Calculate the Hall voltage when B=5A/m, I=2A, w=5cm and n=10 20 .
a) 3.125V
b) 0.3125V
c) 0.02V
d) 0.002V
Answer: d
Explanation: Vh=BI/wρ
=5=2/ (5*10 -2 *10 5 *1.6*10 -19 )
=0.002V.
8. Calculate the Hall Effect coefficient when number of electrons in a semiconductor is 10 20 .
a) 0.625
b) 0.0625
c) 6.25
d) 62.5
Answer: b
Explanation: R=1/ρ
=1/(1.6*10 -19 *10 20 )
=0.0625.
9. What is the conductivity when the Hall Effect coefficient is 5 and mobility is 5cm 2 /s.
a) 100 S/m
b) 10 S/m
c) 0.0001S/m
d) 0.01 S/m
Answer: c
Explanation: µ=σR
σ =µ/R
=5*10 -4 /5
=0.0001 S/m.
10. In Hall Effect, the electric field applied is perpendicular to both current and magnetic field?
a) True
b) False
Answer: a
Explanation: In Hall Effect, the electric field is perpendicular to both current and magnetic field so that the force due to magnetic field can be balanced by the electric field or vice versa.
This set of Electronic Devices and Circuits Interview Questions and Answers for freshers focuses on “Qualitative Theory of the p-n junction”.
1. The donor ions is represented by a positive plus sign. Is it True or False?
a) True
b) False
Answer: a
Explanation: The donor atom donates the extra ion to the semiconductor. Therefore, it is represented by the positive plus sign.
2. Initially, the p-type carriers are located to the____________of the semiconductor.
a) Right
b) Left
c) Middle
d) Top
Answer: b
Explanation: The p-type carriers are nominally located to the left of the junction and n-type carriers are to the right.
3. The displacement of the charges results in
a) Magnetic field
b) Electric field
c) Rust
d) Hall effect
Answer: The flow of carriers in a semiconductor results in the electric field across the junction. The electric field thus makes the current flow in the device.
4. What is the value of 1 micron?
a) 10 -6 cm
b) 10 -5 cm
c) 10 -4 cm
d) 10 -3 cm
Answer: c
Explanation: 1 micron=10 -4 cm=10 -6 cm.
5. Which of the following results when the equilibrium established in a semiconductor?
a) Restrain the process of diffusion
b) Electric field becomes very high
c) Both a and b
d) None of these
Answer: c
Explanation: As the electric field is very high, the flow of the carries will be restricted and the equilibrium will be obtained.
6. Which of the following options doesn’t defined for the necessity for the existence of the potential barrier?
a) Contact
b) Potential
c) Diffusion
d) Fermi dirac
Answer: d
Explanation: The potential barrier is formed at the junction of the semiconductor. It’s necessity of the potential barrier is known as the contact, potential or diffusion.
7. Under the open-circuited conditions the net hole current must be zero. Is this statement is True or false?
a) True
b) False
Answer: a
Explanation: The net hole current is zero because if this wasn’t true, the hole density at one end of the semiconductor would continue to increase indefinitely with time, a situation which is obviously physically impossible.
8. The un-neutralised ions in the neighbourhood of the junction are known as
a) Depletion charges
b) Uncovered charges
c) Mobile ions
d) Counter ions
Answer: b
Explanation: The un-neutralised ions in the neighbourhood of the junction are known as uncovered ions because they are not mobile.
9. Which of the following doesn’t defines for the junction which is depleted of mobile charges?
a) Depletion region
b) Uncovered region
c) Space charge region
d) Transition region
Answer: b
Explanation: The junction which is depleted of mobile charges is known as depletion region or space charge region and transition region.
10. Convert 10 micron to meters.
a) 10 -5 m
b) 10 7 m
c) 10 -6 m
d) 10 -4 m
Answer: a
Explanation: Since, 1 micron=10 -6 m
10 micron =10 -5 m.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The P-N Junction as a Diode”.
1. How many junction/s do a diode consist?
a) 0
b) 1
c) 2
d) 3
Answer: b
Explanation: Diode is a one junction semiconductor device which has one cathode and anode. The junction is of p-n type.
2. If the positive terminal of the battery is connected to the anode of the diode, then it is known as
a) Forward biased
b) Reverse biased
c) Equilibrium
d) Schottky barrier
Answer: a
Explanation: When a positive terminal is connected to the anode, the diode is forward biased which lets the flow of the current in the circuit.
3. During reverse bias, a small current develops known as
a) Forward current
b) Reverse current
c) Reverse saturation current
d) Active current
Answer: c
Explanation: When the diode is reverse biased, a small current flows between the p-n junction which is of the order of the Pico ampere. This current is known as reverse saturation current.
4. If the voltage of the potential barrier is V 0 . A voltage V is applied to the input, at what moment will the barrier disappear?
a) V< V 0
b) V= V 0
c) V> V 0
d) V<< V 0
Answer: b
Explanation: When the voltage will be same that of the potential barrier, the potential barrier disappears resulting in flow of current.
5. During the reverse biased of the diode, the back resistance decrease with the increase of the temperature. Is it true or false?
a) True
b) False
Answer: a
Explanation: Due to the increase in the reverse saturation current due to the increase in the temperature, the back resistance decrease with the increasing temperature.
6. What is the maximum electric field when V bi =2V , V R =5V and width of the semiconductor is 7cm?
a) -100V/m
b) -200V/m
c) 100V/m
d) 200V/m
Answer: b
Explanation: Emax=-2(V bi +V R )/W
=-2/ (7*10 -2 )
=-200V/m.
7. When the diode is reverse biased with a voltage of 6V and V bi =0.63V. Calculate the total potential.
a) 6V
b) 6.63V
c) 5.27V
d) 0.63V
Answer: b
Explanation: Vt=V bi +V R
=0.63+6
=6V.
8. It is possible to measure the voltage across the potential barrier through a voltmeter?
a) True
b) False
Answer: b
Explanation: The contacts of the voltmeter have some resistance which will not accurately measure the voltage across the potential barrier. Thus, it is not possible to measure the voltage across the potential barrier.
9. What will be the output of the following circuit?
electronic-devices-circuits-questions-answers-pn-junction-diode-q9
a) 12V
b) 12.7V
c) 11.3V
d) 0V
Answer: c
Explanation: V=12-0.7
=11.3V.
10. Which of the following formula represents the correct formula for width of the depletion region?
a) electronic-devices-circuits-questions-answers-pn-junction-diode-q10
b) electronic-devices-circuits-questions-answers-pn-junction-diode-q10a
c) electronic-devices-circuits-questions-answers-pn-junction-diode-q10b
d) electronic-devices-circuits-questions-answers-pn-junction-diode-q10c
Answer: a
Explanation: Option a is the correct formula.
This set of Electronic Devices and Circuits Questions and Answers for Experienced people focuses on “Band Structure of an Open-Circuited p-n junction”.
1. The conduction band edge in the p material is not at the same level to that of conduction band edge in the n material. Is it true or false?
a) True
b) False
Answer: a
Explanation: In a p-n junction diode, the energy levels of the p material and n material will not be at same level. They will be different. So, the conduction band edge as well as the valence band edge of the p material will not be same to that of the n material.
2. Which of the following equations represent the correct expression for the shift in the energy levels for the p-n junction?
a) E o = E cn – E cp
b) E o = E cp – E cn
c) E o = E cp + E cn
d) E o = -E cp – E cn
Answer: b
Explanation: The shift in the energy of the energy level will be the difference of the conduction band edge of the p material and conduction band edge of n material. In the energy level diagram, the conduction band edge of p material is higher than that of the n material.
3. Calculate the E o given that N d =1.5*10 10 cm -3 , N a =1.5*10 10 cm -3 at temperature 300K?
a) 1.5*10 10 eV
b) 0.256eV
c) 0eV
d) 4.14*10 -21 eV
Answer: c
Explanation: E o =kTln((N d *N a )/(n i ) 2 )
Substituting k=1.38*10 -23 /K, T=300k and the values ofN d ,N a and ni,
We get
E o =0eV.
4. In a p-n junction, the valence band edge of the p material is greater than which of the following band?
a) Conduction band edge of n material
b) Valence band edge of n material
c) Conduction band edge of p material
d) Fermi level of p material
Answer: b
Explanation: When the p-n junction is formed, the energy levels of the p- material go higher than the n material. That’s why the valence band of the p material will be greater than that of the n material.
5. Which of the following equations represent the correct expression for the band diagram of the p-n junction?
a) E cn – E f = *E G – E 1
b) E cn – E f = *E G – E 2
c) E f – E cp = *E G – E 1
d) E cn – E f = *E G + E 1
Answer: a
Explanation: From the energy band diagram of the p-n junction, the option ‘a’ satisfies that band diagram.
6. Calculate the value of E o when p no =10 4 cm -3 and p po =10 16 cm -3 at T=300K.
a) 1meV
b) 0.7meV
c) 0.7eV
d) 0.1meV
Answer: c
Explanation: E o =kTln(p po /p no )
Substituting the values, we get
E o =0.7eV.
7. Calculate the value of Dp when µp=400cm/s and V T =25mV.
a) 1
b) 0.01
c) 0.1
d) 10
Answer: c
Explanation: Dp= µp*V T
=400*10 -2 *25*10 -3
=0.1.
8. What is the value of kT at room temperature?
a) 0.0256eV
b) 0.25eV
c) 25eV
d) 0.0025eV
Answer: a
Explanation: kT=1.38*10 -23 *300K
=4.14*10 -21 / (1.6*10 -19 )
=0.0256eV.
9. Is Vo depends only on the equilibrium concentrations. Is it true or false?
a) True
b) False
Answer: a
Explanation: Vo is the contact potential of the junction when the junction is in equilibrium. If, the junction is not in the equilibrium, Vo can’t be calculated.
10. Calculate Vo when p po =10 16 cm -3 , p no =10 4 cm -3 and Vt=25mV.
a) 69V
b) 6.9V
c) 0.69V
d) 0.069V
Answer: c
Explanation: V o =V T ln(p po /p no )
=25*10 -3 *ln(10 16 /10 4 )
=0.69V.
This set of Electronic Devices and Circuits Interview Questions and Answers for Experienced people focuses on “The Current Components in a P-N junction diode”.
1. When a forward biased is applied to a diode, the electrons enter to which region of the diode?
a) P-region
b) N-region
c) P-n junction
d) Metal side
Answer: a
Explanation: When the forward biased is applied, the electrons enter to the p-region and the holes enter to the n-region so that holes can flow from p-region to n-region. Whereas, the electrons can travel from n-region to p-region.
2. The number of injected minority carriers falls off linearly with the increase in the distance from the junction. Is it true or false?
a) True
b) False
Answer: b
Explanation: The number of minority carriers fall off exponentially rather than linearly with the increase in the distance from the junction.
3. What is the total current in a diode when x=0?
a) I = I pn – I np
b) I = I pn + I np
c) I = -I pn – I np
d) I = -I pn + I np
Answer: b
Explanation: At junction, the total current is equal to the minority hole current plus the minority electron current.
4. The current in the diode is
1. Unipolar
2. Bipolar
a) I only
b) II only
c) I and II both
d) Neither I nor II
Answer: b
Explanation: The current in the diode consists of both the electrons and holes. So, it is bipolar.
5. The current is constant throughout the device. Is it true or false?
a) True
b) False
Answer: a
Explanation: The current in the device is constant but the proportion due to the electrons and holes varies with distance.
6. Which of the following statements is correct under forward biased p-n diode?
a) current enters n side as hole current and leaves p side as electron current
b) current enters n side as electron current and leaves p side as hole current
c) current enters p side as hole current and leaves n side as electron current
d) current enters p side as hole current and leaves p side as electron current
Answer: c
Explanation: When the current flows in a p-n diode, the current enters p side as hole current and leaves n side as electron current.
7. Calculate the total current when I pn =1mA and I np =2mA.
a) 1mA
b) -1mA
c) 0
d) 3mA
Answer: d
Explanation: I=I pn +I np
=1mA+2mA
=3mA.
8. What is the hole current in the p region of the diode?
a) I pp =I-I np
b) I pp =I+I np
c) I pp =-I-I np
d) I pp =-I+I np
Answer: a
Explanation: The hole current in the p region is equal to the total current minus the minority electrons in the p region.
9. What does I np represent?
a) Hole current in n region
b) Hole current in p region
c) Electron current in n region
d) Electron current in p region
Answer: d
Explanation: I np constitutes of the electron current in the p region. It is the minority electron carrier in the p region.
10. Deep into the p side the current is a drift current I pp of holes sustained by the small electric field in the semiconductor. Is the statement true or false?
a) True
b) False
Answer: a
Explanation: In the p region, the drift current is sustained into the p region by the small electric field which is formed at the junction in the semiconductor. So, the above statement is true.
This set of Electronic Devices and Circuits online test focuses on “Quantitative Theory of the P-N diode Currents”.
1. What is the thickness of ‘space charge region’ or ‘transition region’ in P-N junction diode?
a) 1 micron
b) 5 micron
c) 10 micron
d) 2.876 micron
Answer: a
Explanation: The region of the junction is depleted by mobile charges, hence it is called space charge region or depletion region or transition region which is 10 -4 cm = 10 -6 m= 1 micron.
2. If what of the following is doped into a semiconductor say germanium a P-N junction is formed.
a) Electrons and Protons
b) Protons and Neutrons
c) Neutrons and Electrons
d) Gallium and Phosphorus
Answer: d
Explanation: A P-N junction is formed only when a donor impurities and acceptor impurities are added to either side of a semiconductor like silicon and germanium.
3. Which of the factors doesn’t change the diode current.
a) Temperature
b) External voltage applied to the diode
c) Boltzmann‘s constant
d) Resistance
Answer: d
Explanation: I = Io [e -1], as shown in this equation the diode current is dependent on temperature , voltage applied on the diode , Boltzmann’s constant but diode current is not dependent on resistance as it is independent of resistance.
4. The product of mobility of the charge carriers and applied Electric field intensity is known as
a) Drain velocity
b) Drift velocity
c) Push velocity
d) Pull velocity
Answer: b
Explanation: When the semiconductors like silicon and germanium is implied by an electric field the charge carriers when get drifted by certain velocity known as drift velocity. Drift velocity is product of mobility of charge carriers and field intensity.
5. The tendency of charge carriers to move from a region of heavily concentrated charges to region of less concentrated charge is known as.
a) Depletion current
b) Drain current
c) Diffusion current
d) Saturation current
Answer: c
Explanation: In a semiconductor the charge will always have a tendency to move from higher concentrated area to less concentrated area to maintain equilibrium this movement of charges will result in diffusion current.
6. If the drift current is 100mA and diffusion current is 1A what is the total current in the semiconductor diode.
a) 1.01 A
b) 1.1 A
c) 900m A
d) 10 A
Answer: b
Explanation: We know that the total current in a semiconductor is equal to sum of both drift current and diffusion current. Total current = 1A + 100mA =1.1A.
7. Which of the following is reverse biased?
electronic-devices-circuits-questions-answers-online-test-q7
a) A)
b) B)
c) C)
d) D)
Answer: c
Explanation: The P-N junction diode is forward bias when the voltage applied to p type is greater than the n type and vice versa, since the voltage applied to p type is less in C) it is the answer.
8. The drift velocity is 5V and the applied electric field intensity 20v/m what will be the mobility of charge carriers.
a) 100 m 2 /
b) 4 m 2 /
c) 15 m 2 /
d) 0.25 m 2 /
Answer: d
Explanation: We know that mobility of charge carriers is drift velocity divide by applied electric field intensity. Mobility = drift velocity / field intensity.
9. When there is an open circuit what will be the net hole current.
a) 5A
b) 0.05A
c) 0.5A
d) 0A
Answer: d
Explanation: If there is any current present under open circuit there will be an indefinite growth of holes at one end of the semiconductor which is practically not possible hence zero amperes.
10. Rate of change of concentration per unit length in a semiconductor is called as.
a) Concentration change
b) Concentration mixture
c) Concentration gradient
d) Concentration variant
Answer: c
Explanation: In a semiconductor the holes as well as electrons which are the charge carriers is not equally concentrated on all regions of the semiconductor the change in their rate is referred as concentration gradient.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Volt Ampere Characteristics”.
1. The voltage equivalent of temperature in a P-N junctions is given by.
a) T/1000 volts
b) T/300 volts
c) T/1600 volts
d) T/11600 volts
Answer: d
Explanation: We know that the P-N junction is temperature dependent, it varies with the change in temperature the measure of change that is the voltage equivalent of temperature is given by Vt = T/11600 volts.
2. At room temperature what will be voltage equivalent of temperature.
a) 10 mV
b) 4.576 mV
c) 26 mV
d) 98 V
Answer: c
Explanation: Room temperature is 27 o C = 300 k .We know that V t = T/11600 volts by substituting the value of T we get 300/11600 = 26mV.
3. In a P-N junction the positive voltage at which the diode starts to conduct consequently is called.
a) Cut off voltage
b) Saturation voltage
c) Knee voltage
d) Breakdown voltage
Answer: c
Explanation: At a certain critical voltage, a large reverse current flows and the diode is said to be in breakdown region, at this region the diode will be forward biased and starts to conduct consequently.
4. In volt ampere characteristics the current increases with voltage _________
a) Exponentially
b) Equally
c) Sinusoidal
d) Unequally
Answer: a
Explanation: The current in the volt ampere characteristics increases exponentially with respect to voltage I = eV.
5. The cut off voltage for diode of silicon semiconductor and germanium semiconductor is ____ volts.
a) 0.5 and 0.1
b) 0.7 and 0.3
c) 1 and 0.5
d) 0.5 and 1
Answer: b
Explanation: The cut off voltage is the voltage only after which the semiconductors conduct, the cut off voltage for silicon is 0.7V in the sense the silicon diode will conduct only when voltage is more than 0.7V and 0.3 for germanium.
6. What would be the current and voltage when there is no external voltage applied on the diode?
a) 0
b) 0.7
c) 0.3
d) 1
Answer: a
Explanation: When there is no external voltage applied on the circuit it acts as an open circuit and there will be no flow of charges hence the current and voltage will be zero.
7. In P-N junction V-I characteristics during forward biased, at what region the current increase is very low.
a) Saturation
b) Depletion
c) Cut off
d) Breakdown
Answer: b
Explanation: In the V-I characteristics the change in the current with respect to voltage is very less in depletion region due to the large resistance in the circuit as the resistance deceases by a certain value the current increases exponentially with voltage.
8. In a P-N junction diode during forward bias if the current increases more than the value that is rated will destroy the diode.
a) True
b) False
Answer: a
Explanation: If the current in the P-N junction diode during forward bias increases beyond the value rated on it will destroy the diode because voltage is directly proportional to current so extreme voltage will burn the diode down.
9. When the P-N junction diode is forward bias the current in circuit is controlled by.
a) External voltage
b) Capacitance
c) Resistance
d) Internal voltage
Answer: c
Explanation: When the P-N junction is in forward bias that is the p side connected to the positive terminal of voltage source the current in the circuit can be varied by varying the resistance, the current flow decreases as the resistance increases and vice versa.
10. The P-N junction diode conducts in which direction.
a) Reverse direction
b) Forward direction
c) Both Forward and Reverse direction
d) Neither Forward nor Reverse direction
Answer: b
Explanation: The P-N junction diode conducts only in forward direction, it will not conduct in reverse direction so only Zener Diode was introduced as it conducts in both forward and reverse direction.
This set of Electronic Devices and Circuits test focuses on “The Temperature Dependence of P-N Characteristics”.
1. The magnitude of the electric charge is given by ____________
a) -1.6*10 -19 C
b) 1.6*10- 19 C
c) 9.11*10- 31 C
d) 1.637*10 -37 C
Answer: a
Explanation: The charge of the electron is the magnitude of electric force that an electron exerts on other particles which is equal to -1.6*10 -19 C, the negative sign indicates the direction of force.
2. What is the forbidden gap voltage for silicon material?
a) 1.46 V
b) 1.56 V
c) 10 V
d) 1.21 V
Answer: d
Explanation: The forbidden gap voltage of a material is numerically equal to forbidden gap energy of the material which is 1.21 joules for silicon so forbidden gap voltage will be 1.21 V.
3. Which of the following parameters of P-N junction diode increases with temperature.
a) Cut in voltage
b) Reverse saturation current.
c) Ideality factor
d) Resistance
Answer: b
Explanation: Reverse saturation current at temperature T 2 is 2 [/10] times greater than reverse saturation current at temperature T 1 where T 2 is greater than T 1 .
4. Which of the following diodes do not exhibits a constant reverse saturation current with the change in reverse saturation voltage.
a) 1N909
b) 1N405
c) 1N207
d) 1N676
Answer: c
Explanation: 1N207 is the germanium diode for which the reverse saturation current is not constant which the change in voltage due to the leakage in the surface of the diode and due to the generation of new current carriers.
5. Which of these P-N junction characteristics are not dependent on temperature.
a) Junction resistance
b) Reverse saturation current
c) Bias current
d) Barrier voltage
Answer: a
Explanation: As the temperature of the P-N junction increases the current increases and the voltage decreases so the barrier voltage, reverse saturation current, bias current changes with temperature but junction resistance is independent of temperature.
6. As the temperature to the P-N junction increases the current increases due to?
a) Leakage in bias region
b) Electron-hole pair
c) Leakage in P region
d) Leakage in N region
Answer: b
Explanation: As the temperature to the P-N junction increases the mobility of charges increases thus increases the electron-hole pair which proportionally increases the current in the P-N junction diode.
7. By what percentage the reverse saturation current increases with 10 C rise in the temperature.
a) 25%
b) 12.5%
c) 50%
d) 7%
Answer: d
Explanation: As the temperature to the P-N junction diode increases the mobility of charges increases thus increasing the current, the reverse saturation current increases by 7% with 10C rise in temperature and doubles with every 100C rise in temperature.
8. What will be the decrease of barrier voltage with the rise in 10C in temperature?
a) 10V
b) 1mV
c) 10mV
d) 2mV
Answer: d
Explanation: As the temperature to the P-N junction diode increases the voltage across the junction decreases and the current increases with every degree rise in temperature the barrier voltage increases by 2mV.
9. What will be the reverse saturation current in the junction when the voltage across the junction is 0?
a) 0.3A
b) 0.7A
c) 0A
d) 1.24A
Answer: c
Explanation: When the voltage across the junction is zero in the sense there will be potential difference between the junctions hence there will be no movement of electrons and holes, hence the current will be 0.
10. The breakdown voltage of the P-N junction diode decreases due to the increase in.
a) Reverse saturation current
b) Reverse leakage current
c) Bias voltage
d) Barrier voltage
Answer: b
Explanation: Breakdown voltage of the diode is inversely proportional to the reverse leakage current so it decreases with the increase in reverse leakage current.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Diode Resistance”.
1. The static resistance R of the diode is given by __________
a) V/I
b) V*I
c) V+I
d) V-I
Answer: a
Explanation: According to Ohms law the electric current in the circuit is directly proportion to voltage and inversely proportional to resistance so, R=V/I.
2. In the volt ampere characteristics of the diode, the slope of the line joining the operating point to the origin at any point is equal to reciprocal of the _________
a) resistance
b) conductance
c) voltage
d) current
Answer: a
Explanation: In the diode’s volt ampere characteristics, the line joining the operating point and the origin, at any point of the line is equal to the conductance so, it is reciprocal of the resistance.
3. At room temperature (V T = 26) what will be the approximate value of r when n=1 and I=100mA?
a) 26 ohms
b) 2.6 ohms
c) 260 ohms
d) 2600 ohms
Answer: c
Explanation: We know that R= (n*V T ) /I, by substituting the value of n, V T , I we get R= 260 ohms, /100*10 -3 = 260 ohms.
4. In the diode volt ampere characteristics what will be the resistance if a slope is drawn between the voltages 50 to 100 and corresponding current 5 to 10?
a) 5 ohms
b) 10 ohms
c) 50 ohms
d) 100 ohms
Answer: b
Explanation: We know that, in volt ampere characteristics the resistance is equal to the reciprocal of the line joining the origin and operating point, R = dV/dI, by substituting the value of dV and dI we get R= 10ohms.
5. In piecewise linear characteristics what will be the R F value if the slope is 0.5?
a) 25 m ohms
b) 50 m ohms
c) 2 ohms
d) 10 ohms
Answer: c
Explanation: In piecewise linear characteristics the forward resistance will be equal to reciprocal of the slope so, R F = 1/slope, R F = 1/0.5 which is equal to 2 ohms.
6. A diode will behave as an open circuit if the voltage in the circuit is less than __________
a) cut off voltage
b) saturation voltage
c) leakage voltage
d) threshold voltage
Answer: d
Explanation: The diode made up of semiconductor has a certain threshold voltage only after which it behave as closed circuit in the sense it performs some operation if the threshold voltage is greater than the voltage in circuit.
7. What will be the approximate value of thermal voltage of diode?
a) 25mV at 300K
b) 30mV at 180K
c) 25mV at 180K
d) 30mV at 300K
Answer: a
Explanation: We know that the thermal voltage of diode is approximately equal to room temperature which is 300K then for all practical purpose the thermal voltage of diode is taken as 25mV so it will be 25mV at 300K.
8. What will be the thermal voltage of the diode if the temperature is 300K?
a) 25.8 mV
b) 50 mV
c) 50V
d) 19.627 mV
Answer: a
Explanation: The thermal voltage of the diode is given by, V T = KT/q, by substituting the values of T, K which is Boltzmann constant and q which is the charge of the electron we get V T = (300*1.38*10 -23 )/ (1.602*10 -19 ), V T = 25.8mV.
9. What will be the diode resistance if the current in the circuit is zero?
a) 0 ohms
b) 0.7 ohms
c) 0.3 ohms
d) 1 ohms
Answer: a
Explanation: When the current in the circuit is zero there will be no flow of charges to resist hence the diode resistance will be zero.
10. Which of these following is not a characteristic of an ideal diode?
a) Perfect conductor when forward bias
b) Zero voltage across it when forward bias
c) Perfect insulator when reverse bias
d) Zero current through it when forward bias
Answer: d
Explanation: The diode acts as an ideal diode when it is a perfect conductor and has zero voltage across it during forward bias, a perfect insulator and zero current through it during reverse bias.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Diode Capacitances”.
1. Compared to a PN junction with N A =10 14 /CM 3 , which one of the following is true for N A =N D = 10 20 /CM 3 ?
a) depletion capacitance decreases
b) depletion capacitance increases
c) depletion capacitance remains same
d) depletion capacitance can’t be predicted
Answer: b
Explanation: We know, C T =Aε/W and
W ∝ (1/N A +1/N D ) 1/2 . So, C T ∝ (1/N A +1/N D ) -1/2
So when N A and N D increases, depletion capacitance C T increases.
2. If C T is the transition capacitance, which of the following are true?
1) in forward bias, C T dominates
2) in reverse bias, C T dominates
3) in forward bias, diffusion capacitance dominates
4) in reverse bias, diffusion capacitance dominates
a) 1 only
b) 2only
c) 2 and 3
d) 3 only
Answer: c
Explanation: In reverse bias condition, depletion region increases and acts as an insulator or dielectric medium. So, the transition capacitance increases. In forward bias condition, due to stored charge of minority carriers, diffusion capacitance increases.
3. For an abrupt PN junction diode, small signal capacitance is 1nF/cm 2 at zero bias condition.If the built in voltage, V bi is 1V, the capacitance at reverse bias of 99V is?
a) 0.1nF/cm 2
b) 1nF/cm 2
c) 1.5nF/cm 2
d) 2nF/cm 2
Answer: a
Explanation: C jo is the capacitance at zero bias, that is V R =0V, C jo =C j for V R =0V. We know, C j = C jo /(1+(V R /V bi ))m , m=1/2 for abrupt. So, putting C j =0.1nF/cm 2 where, V R =99V and V bi =1V we get, C jo = 0.1 1/2 = 0.1nF/cm 2 .
4. The built in capacitance V 0 for a step graded PN junction is 0.75V. Junction capacitance C j at reverse bias when V R =1.25V is 5pF. The value of C j when V R =7.25V is?
a) 0.1pF
b) 1.7pF
c) 1pF
d) 2.5Pf
Answer: d
Explanation: We know, C j1 / C j2 =[(V 0 +V R2 )/(V 0 +V R2 )] 1/2
So, C j2 =C j1 / {/} 1/2 we get C j2 =C j1 /2 =5/2=2.5Pf.
5. Consider an abrupt PN junction. Let V0 be the built in potential of this junction and VR be the reverse bias voltage applied. If the junction capacitance C j is 1pF for V 0 +V R =1V, then for V 0 +V R =4V what will be the value of C j ?
a) 0.1pF
b) 1.7pF
c) 1pF
d) 0.5Pf
Answer: d
Explanation: We know, C j1 / C j2 =[(V 0 +V R2 )/(V 0 +V R1 )] 1/2
C j2 =C j1 1/2 =1/2 .
We get C j2 =1/2=0.5pF.
6. A silicon PN junction diode under revers bias has depletion width of 10µm, relative permittivity is 11.7 and permittivity, ε0 =8.85×10 -12 F/m. Then depletion capacitance /m 2 =?
a) 0.1µF/m 2
b) 1.7µF/m 2
c) 10µF/m 2
d) 0.5µF/m 2
Answer: c
Explanation: We know, C T =Aε 0 ε r /W
C T /A= (8.85×10 -12 )/10
=10
By putting the values we get 10µF/m 2 .
7. The transition capacitance, C T of a PN junction having uniform doping in both sides, varies with junction voltage as ________
a) 1/2
b) -1/2
c) 1/4
d) -1/4
Answer: b
Explanation: C T = K/(V 0 +V B ) 1/2
As it’s having uniform doping on both sides, the voltage V 0 will be zero. So, C T =K/ 1/2 . The variation of transition capacitance with built in capacitance is (V B ) -1/2 .
8. The C T for an abrupt PN junction diode is ________
a) C T = K/(V 0 +V B ) 1/2
b) C T = K/(V 0 +V B ) -1/2
c) C T = K/(V 0 +V B ) 1/3
d) C T = K/(V 0 +V B ) -1/3
Answer: a
Explanation: For an abrupt PN junction diode, C T = K/(V 0 +V B ) n . Here, n=1/2 for abrupt PN junction diode and 1/3 for linear PN junction diode. When the doping concentration of a diode varies within a small scale of area, then the diode is called as an abrupt diode.
9. The diffusion capacitance of a PN junction _______
a) decreases with increasing current and increasing temperature
b) decreases with decreasing current and increasing temperature
c) increasing with increasing current and increasing temperature
d) doesnot depend on current and temperature
Answer: b
Explanation: C D =τ I /n 0 V T
Where, I is the current and VT is temperature factor. The diffusion capacitance is directly proportional to current and indirectly proportional to the temperature.
10. Transition capacitance is also called as _______
a) diffusion capacitance
b) depletion capacitance
c) conductance capacitance
d) resistive capacitance
Answer: b
Explanation: Transition capacitance occurs in reverse bias. We obtain a depletion layer in that case. Hence it’s also called as depletion capacitance. The diffusion capacitance occurs in forward bias.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “PN Diode Switching Times”.
1. Diode acts as a short circuit when switched from forward to reverse bias for some time due to______
a) Accumulation of minority charge carriers when it’s in forward bias
b) Accumulation of majority charge carriers when it’s in forward bias
c) Accumulation of minority charge carriers when it’s in reverse bias
d) Accumulation of majority charge carrier when it’s in reverse bias
Answer: a
Explanation: When a diode is switched suddenly, it persists the conducting property for a short time in its reverse bias also. This leads to excess minority charge carrier settlement at potential barrier. Hence acts as a short circuit.
2. Reverse recovery time for a diode is?
a) Time taken to eliminate excess minority charge carriers
b) Sum of storage time (T S ) and transition time (T T )
c) Time taken to eliminate excess majority charge carriers
d) Time elapsed to return to non conduction state
Answer: a
Explanation: The time period for which diode remains in conduction state even in reverse direction is called storage time. The time elapsed to return the non conduction state is called transition time. Their sum is called reverse recovery time.
3. Switching speed of P+ junction depends on.
a) Mobility of minority carriers in P junction
b) Life time of minority carriers in P junction
c) Mobility of majority carriers in N junction
d) Life time of minority carriers in N junction
Answer: d
Explanation: Switching leads to move holes in P region to N region as minority carriers. Removal of this accumulation determines switching speed. P+ regards to a diode in which the p type is doped excessively.
4. Time taken for a diode to reach 90% of its final value when switched from steady state is______
a) 2.3*time constant
b) 2.2*time constant
c) 1.5*time constant
d) equals the time constant
Answer: b
Explanation: Time constant = RC. To reach 90% of the final value, time taken is 2.2 of RC. Time constant is the time required to discharge the capacitor, through the resistor, by 36.8%.
5. Which of the following are true?
1) In reverse bias, the diode undergoes stages of storage and transition times
2) Minority charge carriers accumulation makes the diode as a short circuit
3) Storage time is the sum of recovery and transition times
a) 1 only
b) 2 and 3
c) 3 only
d) 1 and 2 only
Answer: d
Explanation: When a diode is switched from forward to reverse bias, storage and transition times takes place. The accumulation time or the life time of minority carriers makes it a short circuit. The conduction property is holds for a short period of time in reverse bias also.
6. In a circuit below, the switch is at position 1 at t<0 and at position 2 when t=0. Assume diode has zero voltage drop and storage time. For 0<t<t s , the V R at 1k ohm resistor is given by_____
electronic-devices-circuits-questions-answers-pn-diode-switching-times-q6
a) 5V
b) -5V
c) 0v
d) 10V
Answer: b
Explanation: At position ‘1’ when connected to +5V, the diode is forward biased and acts as a short circuit. So, V R is 5V. For 0<t<t s V R is -5V as the diode is in reverse bias. But it holds the conductive property within the storage time period. So, V is -5V.
7. The switch is at position shown in the figure initially and steady state is from t=0 to t=to. The switch suddenly is thrown to the other position. The current flowing through the 10K resistor from t=0 is?
electronic-devices-circuits-questions-answers-pn-diode-switching-times-q7
a) 1mA
b) 2mA
c) -2mA
d) -1mA
Answer: c
Explanation: Initially, the diode is in forward bias. When suddenly switched to reverse bias, upto a storage time limit, it conducts during storage time period.
We know that, current I=V/R=-20/10K=-2mA.
8. A PN junction diode with 100Ω resistor is forward biased such that 100A current flows. If voltage across this combination is instantaneously reversed to 10V at t=0, the reverse current that flows through diode at t=0 is?
electronic-devices-circuits-questions-answers-pn-diode-switching-times-q8
a) 10mA
b) 100mA
c) -100mA
d) -10mA
Answer: b
Explanation: At t=0, V=-10V. During storage time, current still flows.
We know that,
current I=V/R=10/100Ω=100mA from N to P region.
9. The delay in switching between the ON and OFF states is due to _________
a) The time required to change amount of excess minority carriers stored in quasi-neutral regions
b) The time required to change amount of excess majority carriers stored in quasi-neutral regions
c) The conduction between storage time and recovery time
d) The exponential increase in carriers in N region
Answer: a
Explanation: When switched instantaneously it stays in a quasi state i.e.., temporary state which stores charges. The delay is produced due to this charge settlement. The diode needs to discharge these excess carriers in order to return the non conduction stage.
10. The delay time can be reduced by?
a) decreasing lifetime and increasing ratio of reverse to forward current
b) increasing lifetime and decreasing ratio of reverse to forward current
c) increasing lifetime and increasing ratio of reverse to forward current
d) decreasing lifetime and decreasing ratio of reverse to forward current
Answer: a
Explanation: When the current increases the depletion layer decreases and the storage and transition time decreases. A decreased depletion layer can easily discharge the excess carrier and thereby lessens the delay time.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Breakdown Diodes”.
1. A zener diode works on the principle of_________
a) tunneling of charge carriers across the junction
b) thermionic emission
c) diffusion of charge carriers across the junction
d) hopping of charge carriers across the junction
Answer: a
Explanation: Due to zener effect in reverse bias under high electric field strength, electron quantum tunneling occurs. It’s a mechanical effect in which a tunneling current occurs through a barrier. They usually cannot move through that barrier.
2. Which of the following are true about a zener diode?
1) it allows current flow in reverse direction also
2) it’s used as a shunt regulator
3) it operates in forward bias condition
a) 3 only
b) 1 and 2
c) 2 and 3
d) 2 only
Answer: b
Explanation: The operation of a zener diode is made in reverse bias when breakdown occurs. So, it allows currnt in reverse direction. The most important application of a zener diode is voltage or shunt regulator.
3. When the voltage across the zener diode increases_________
a) temperature remains constant and crystal ions vibrate with large amplitudes
b) temperature increases and crystal ions vibrate with large amplitudes
c) temperature remains constant and crystal ions vibrate with smaller amplitudes
d) temperature decreases and crystal ions vibrate with large amplitudes
Answer: b
Explanation: When voltage is increased, the tunnelling at reverse bias increases. The voltage rises temperature. The crystal ions with greater thermal energy tend to vibrate with larger amplitudes.
4. For the zener diode shown in the figure, the zener voltage at knee is 7V, the knee current is negligible and the zener dynamic resistance is 10Ω. If the input voltage (V i ) ranges from 10 to 16 volts, the output voltage (V o ) ranges from?
electronic-devices-circuits-questions-answers-breakdown-diodes-q4
a) 7 to 7.29V
b) 6 to 7V
c) 7.14 to 7.43V
d) 7.2 to 8V
Answer: c
Explanation: If i is the current flowing, then V 0 =10i+7
i=(V I -7)/210. By substituting, if V I =10V then i=1/70 and V 0 =+7=7.14V
if V I =16V then i=3/70 and V 0 =+7=7.43V.
5. In the circuit below, the knee current of ideal zener diode is 10mA. To maintain 5V across the R L , the minimum value of R L is?
electronic-devices-circuits-questions-answers-breakdown-diodes-q5
a) 120
b) 125
c) 250
d) 100
Answer: b
Explanation: Here, I KNEE =10mA, V Z =5V. I=I L +I Z . I= /100=50mA
Now, 50=10+ILMAX .
I LMAX =40mA. R LMIN =5/40mA=125 Ω.
6. The zener diode in the circuit has a zener voltage of 5.8V and knee current of 0.5mA. The maximum load current drawn with proper function over input voltage range between 20 and 30V is?
electronic-devices-circuits-questions-answers-breakdown-diodes-q6
a) 23.7mA
b) 20mA
c) 26mA
d) 48.3mA
Answer: a
Explanation: Here, I 1MAX =I ZMIN +I LMAX .
I ZMIN =0.5mA, I 1MAX =(V 1MAX -V Z )/R S . Putting the values we get , I 1MAX =24.2mA.
So, 24.2-0.5=23.7mA.
7. In the given limiter circuit, an input voltage Vi=10sin100πt is applied. Assume that the diode drop is 0.7V when it’s forward biased. The zener breakdown voltage is 6.8V.The maximum and minimum values of outputs voltage are _______
electronic-devices-circuits-questions-answers-breakdown-diodes-q7
a) 6.1V,-0.7V
b) 0.7V,-7.5V
c) 7.5V,-0.7V
d) 7.5V,-7.5V
Answer: c
Explanation: With V I = 10V when maximum, D 1 is forward biased, D 2 is reverse biased. Zener is in breakdown region. V OMAX =sum of breakdown voltage and diode drop=6.8+0.7=7.5V. V OMIN =negative of voltage drop=-0.7V. There will be no breakdown voltage here.
8. The 6V Zener diode shown has zener resistance and a knee current of 5mA. The minimum value of R so that the voltage does not drop below 6V is?
electronic-devices-circuits-questions-answers-breakdown-diodes-q8
a) 1.2Ω
b) 80 Ω
c) 50 Ω
d) 70 Ω
Answer: b
Explanation: Here, V z =6V, I ZMIN =5mA.I S =I ZMIN +I LMAX .
80=5+I LMAX . I LMAX =75Ma.R LMIN =V I /I LMAX =6/75mA
=80 Ω.
9. Avalanche breakdown in zener diode is ______
a) electric current multiplication takes place
b) phenomenon of voltage multiplication takes place
c) electrons are decelerated for a period of time
d) sudden rise in voltage takes place.
Answer: a
Explanation: The carriers in transition region are accelerated by electric field to energies. That energies are sufficient to create electron current multiplication. A single carrier that is energized will collide with another by gaining energy. Thus an avalanche multiplication takes place.
10. The zener diode is heavily doped because______
a) to have low breakdown voltage
b) to have high breakdown voltage
c) to have high current variations
d) to maintain perfect quiescent point
Answer: a
Explanation: The value of reverse breakdown voltage at which zener breakdown occurs is controlled by amount of doping. If the amount of doping is high, the value of voltage at which breakdown occurs will decrease. Better doping gives a sooner breakdown voltage.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Tunnel Diodes and its Characteristics”.
1. If ‘X’ corresponds to a tunnel diode and ‘Y’ to an avalanche diode, then__________
a) X operates in reverse bias and Y operates in forward bias
b) X operates in reverse bias and Y operates in reverse bias
c) X operates in forward bias and Y operates in forward bias
d) X operates in forward bias and Y operates in reverse bias
Answer: d
Explanation: In forward bias, negative resistance helps for tunnel diode to operate. Here, the current decreases with increase in voltage. If they are used in reverse bias, they are called as back diodes. Avalanche diode operates in reverse bias at breakdown region.
2. The range of tunnel diode voltage V D , for which slope of its V-I characteristics is negative would be? (The V P is the peak voltage and V V is the valley voltage).
a) V D > 0
b) 0 D < V P
c) V V > V D > V P
d) V D > V V
Answer: c
Explanation: In tunnel diode characteristics, the slope is negative in the region between V V and V P . Here, it offers negative resistance. The characteristics are depicted below: electronic-devices-circuits-questions-answers-tunnel-diodes-characteristics-q2
3. Tunnel diode has a very fast operation in__________
a) gamma frequency region
b) ultraviolet frequency region
c) microwave frequency region
d) radio frequency region
Answer: c
Explanation: Tunnel diode is a type of semiconductor which works on tunneling effect of electrons in microwave region. So, tunnel diode has a very fast operation in microwave region.
4. Which of the following are true about a tunnel diode?
1) it uses negative conductance property
2) it operates at high frequency
3) fermilevel of p side becomes higher than the n side in forward bias
a) 1 only
b) 1 and 2
c) 3 only
d) 2 and 3
Answer: b
Explanation: The negative resistance property helps in the operation of tunnel diode. As the tunnel diode works at high frequency, its applications are mostly in that range. High frequency oscillators are based on the resonant tunneling diode.
5. The depletion layer of tunnel diode is very small beacause______
a) its abrupt and has high dopants
b) uses positive conductance property
c) its used for high frequency ranges
d) tunneling effect
Answer: a
Explanation: When the P and N regions are very highly doped, the depletion layer comes closer. The tunnel diode is also highly doped. Its doping concentration varies within a small scale. So it’s an abrupt diode. For these reasons, the depletion region is small.
6. With interments of reverse bias, the tunnel current also increases because________
a) electrons move from balance band of pside to conduction band of nside
b) fermi level of pside becomes higher than that of nside
c) junction currrent decreases
d) unequality of n and p bandedge
Answer: a
Explanation: When the forward bias is increased, the tunnel current is also increased upto a certain limit. This happens when the electron movement takes place from P to N side.
7. The tunnneling involves_______
a) acceleration of electrons in p side
b) movement of electrons from n side conduction band to p side valance band
c) charge distribution managementin both the bands
d) positive slope characteristics of diode
Answer: b
Explanation: Tunneling means a direct flow of electrons across small depletion region from N side conduction band to P side valance band. The electrons begin to accelerate in the N side of the semiconductor.
8. Tunnel diodes are made up of________
a) Germanium and silicon materials
b) AlGaAs
c) AlGaInP
d) ZnTe
Answer: a
Explanation: Germanium and silicon materials have low band gaps and flexibility. That matches tunnel diode requirements. The remaining materials emits the energy in terms of light or heat.
9. For a tunnel diode, when ‘p’ is probability that carrier crosses the barrier, ’e’ is energy,’w’ is width.
a) p ∝ e
b) p ∝ 1/ e
c) p ∝ e
d) p ∝ 1/e
Answer: a
Explanation: The carrier jump occurs without any loss of energy due to small depletion layer. The probability of the carrier to jump across a barrier depends on the energy and width of the band. This variess exponentially for a given carrier.
10. In the construction of tunnnel diode,why is the pellet soldered to anode contact and a tindot to cathode contact via a mesh screen?
a) for better conduction and reduce inductance respectively
b) for heat dissipation and increase conduction respectively
c) for heat dissipation and reduce induction respectively
d) for better conduction and reduce inductance respectively
Answer: c
Explanation: Anode goes through better heat dissipation. So the pellet is used for the purpose. The tindot via mesh screen resists inductive effects caused at the cathode. Conduction is an independent factor which can’t be controlled.
11. What happens to a tunnel diode when the reverse bias effect goes beyond the valley point?
a) it behaves as a normal diode
b) it attains increased negative slope effects
c) reverse saturation current increases
d) beacomes independent of temperature
Answer: a
Explanation: After the valley point is crossed, the tunnel diode obtains positive slope resistance. That is similar to the characteristics of a normal diode. So it behaves like a normal diode after beyond valley point.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “p-i-n Diode and its Characteristics”.
1. PIN diode is a photosensitive diode because of _______
a) large current flow in p and n region
b) depletion layer increases giving a larger surface area
c) stronger covalent bonds
d) low carrier storage
Answer: b
Explanation: An intrinsic layer that is sandwiched between p and n layers. This gives a larger surface area making it compatible for photosensitivity. Reverse bias causes an increased depleted region in a PIN diode.
2. During forward bias, the PIN diode acts as _______
a) a variable resistor
b) a variable capacitor
c) a switch
d) an LED
Answer: a
Explanation: In forward bias, the forward resistance decreases and acts as a variable resistor. The low frequency model of a PIN diode neglects the input capacitive values.
3. During reverse bias, the PIN diode acts as _______
a) Variable resistor
b) Switch
c) Variable capacitor
d) LED
Answer: c
Explanation: In reverse bias, the intrinsic layer is completely covered by depletion layer. The stored charges vanishes acting like a variable capacitor. The high frequency model of a PIN diode neglects the input resistances.
4. When the p and n regions are used for high resistivity, the depletion region at the respective places is called _________
a) Q and ϒ regions
b) ϒ and π regions
c) Q and π regions
d) π and ϒ regions
Answer: d
Explanation: When p region is used for high resistance, the depletion layer is high at p side.When n side is used the depletion layer is high at n side. They are called as π and ϒ regions respectively.
5. The applications for PIN diode are __________
a) Microwave switch
b) LED
c) Voltage regulator
d) Amplifier
Answer: a
Explanation: Being employed at 300Hz, the swept voltage is attained at π region.Then it’s used as a microwave switch. Swept voltage is nothing but, the voltage at which the complete intrinsic layer is swept out as a depleted one.
6. In high frequency model, the values of resistance ‘R’ and capacitance ‘C’ are _______
a) 0.1 to 10KΩ and 0.02 to 2pF respectively
b) 1 to 10KΩ and 0.02 to 2pF respectively
c) 10 to 100KΩ and 0.02 to 2pF respectively
d) 0.1 to 10KΩ and 2 to 20pF respectively
Answer: a
Explanation: At high frequency, the applied values for resistance and capacitance is 0.1 to 10KΩ and 0.02 to 2pF respectively. At high frequencies, it almost acts as a perfect resistor.
7. What happens in PIN diode for low frequency model?
a) reactance decreases
b) conductance increases
c) resistance increases
d) reactance increases
Answer: d
Explanation: In a low frequency model, the resistance decreases and reactance increases.Here the variable resistance is neglected. At low frequencies, the charge can be removed and the diode can be turned off.
8. Which of the following is true about a PIN diode?
a) it’s photosensitive in reverse bias
b) it offers low resistance and low capacitance
c) it has a decreased reversed breakdown voltage
d) carrier storage is low
Answer: a
Explanation: Due to increased depletion region, the covalent bonds break and increase the surface area for photosensitivity. This property is used in fields of light sensors, image scanners, artificial retina systems.
9. In the application of frequency models, the value of forward current is _____
a) IF = A(µ P P + µ N N)q
b) IF = A(µ P N + µ N P)q
c) IF = A(µ P P – µ N N)q
d) IF = A(µ P N – µ N P)q
Answer: a
Explanation: The forward current depends on mobility and carrier concentration. In frequency models, the value of forward current is IF = A*(µ P P + µ N N)q. Where, µ P and µ N are the mobility of p and n type charge carriers respectively.
10. The forward resistance for a PIN diode is given by ________
a) R F = W/σP
b) R F = W/σN
c) R F = WσP
d) R F = WσN
Answer: b
Explanation: Forward resistance for a PIN diode depends on the width, current density and positive carrier concentration of the diode. No diode is perfectly ideal. In practise, a diode offers a small resistance in forward bias which is called as forward resistance.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Point Contact Diode”.
1. The materials that are used in the construction of point contact diode are _________
a) Silicon
b) SnTe or Bi 2 Te 3
c) GaS or CdS
d) HgI
Answer: b
Explanation: The diode base of SnTe or Bi 2 Te 3 is highly detection sensitive. They are mechanically stable over long periods of use either as harmonic generators or mixers. They are emphasized in the 2-200THz region.
2. Which of the following are true?
1) point contact diode has a metal whisker to make pressure contact during its operation
2) it has high voltage rating
3) its V-I characteristics are constant
4) it has low breakdown voltage
a) 1 and 2
b) 3 only
c) 1 and 4
d) 2 only
Answer: c
Explanation: The diode contains a die of a semiconductor material on which an epitaxial layer is deposited. It uses the metal whisker to make pressure contact against that layer. It has a low breakdown voltage.
3. In the forward bias condition, the resistance of point contact diode is_________
a) less than that of a general PN diode
b) greater than that of a general PN diode
c) equal to that of a general PN diode
d) varies exponentially than that of a general PN diode
Answer: a
Explanation: The current flow of the point contact diode is not independent of voltage applied to the crystal unlikely to a general PN diode. This characteristic of contact diode makes its capacitance high at high frequency. A small capacitive current flows in the circuit.
4. The barrier layer capacitance of a point contact diode is_________
a) 0.1pF to 1pF
b) 5pF to 50pF
c) 0.2pF to 2pF
d) 0.008µF to 20µF
Answer: a
Explanation: The barrier capacitance at the point is very low about 0.1pF to 1pF. The capacitance between the cat whisker and crystal is less compared to junction diode capacitance between both sides of the diode. For a general PN diode is 0.008µF to 20µF.
5. The cat whisker wire present in the contact diode is used for_________
a) for heat dissipation
b) for charge transfer between sections
c) maintaining the pressure between sections
d) preventing current flow
Answer: c
Explanation: The operation of a contact diode depends on the pressure of contact between semiconductor crystal and point. The cat whisker wire presses against the crystal to form a section and the section allows the current flow. This is similar to the behaviour of PN diode.
6. The semiconductor junctions those are present in a contact diode_________
a) beryllium-copper and bronze-phosphor
b) beryllium-phosphor and bronze-copper
c) mercury-iodine
d) tin-tungsten
Answer: a
Explanation: The diode contains two sections having a small rectangular crystal of N type silicon and a fine beryllium-copper and bronze-phosphor. It has a tungsten wire which is called as a cat whisker wire. This helps in pressing one section to other.
7. The application of a contact diode is_________
a) Clampers and clippers
b) Voltage multipliers
c) Rectifiers
d) AM detectors
Answer: d
Explanation: The point contact diodes are the oldest microwave semiconductor devices. They were developed during world war 2. They have excessive applications in microwave fields and used as receivers and detectors.
8. The operating frequencies of the point contact diode is_________
a) 30KHz or above
b) 10GHz or above
c) 30GHz or above
d) 10KHz or above
Answer: b
Explanation: It’s used in high frequency conversions and circuits in the order of 10KHz or above. The reactance due to capacitance is high and at high frequency a very small capacitive current flows.
9. During the manufacture of point contact diode, why is a relatively large current passed from cat whisker to silicon crystal?
a) to control the amount of current flow
b) to form small region of p type material
c) to allow mechanical support for the sections
d) to form anode and cathode regions
Answer: b
Explanation: The behaviour of a contact diode is similar to that of a PN diode. It has a tungsten wire which is called as a cat whisker wire. This helps in pressing one section to other.
10. What is the capacitive reactance across the point contact diode when compared to normal PN junction diode
a) lower
b) higher
c) equal
d) cannot be determined
Answer: a
Explanation: The current flow of the point contact diode is not independent of voltage applied to the crystal unlikely to a general PN diode. A small capacitive current flows in the circuit.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Ideal Diode”.
1. For an ideal diode which of the following is true?
a) It allows the passage of current in the forward bias with zero potential drop across the diode
b) It does not allow the flow of current in reverse bias
c) All of the mentioned
d) None of the mentioned
Answer: c
Explanation: Both of the facts mentioned hold true for an ideal operational amplifier.
2. Consider an AC sine wave voltage signal being used to connect a diode and a resistor as shown in the figure. The variation of the voltage across the diode (V d ) with respect to time is given by
electronic-devices-circuits-questions-answers-ideal-diode-q2
a) electronic-devices-circuits-questions-answers-ideal-diode-q2a
b) electronic-devices-circuits-questions-answers-ideal-diode-q2b
c) electronic-devices-circuits-questions-answers-ideal-diode-q2c
d) None of the mentioned
Answer: c
Explanation: V d = V i – V o .
3. The figure below shows a circuit for charging a 12-V battery. If V s is a sinusoid with 24-V peak amplitude, the fraction of each cycle during which the diode conducts is
electronic-devices-circuits-questions-answers-ideal-diode-q3
a) One quarter of a cycle
b) One-third of a cycle
c) One half of the cycle
d) Three quarters of a cycle
Answer: b
Explanation: For one half of the cycle the diode conducts when 24 cos = 12 or Θ is 60 0 . Hence for a complete cycle the diode does not conducts for 2Θ or 120 0 . Hence the diode does not conducts for a third of a cycle.
4. Diodes can be used in the making of
a) Rectifiers
b) LED lamps
c) Logic gates
d) All of the mentioned
Answer: d
Explanation: Diodes are used to make rectifiers , LED lamps uses diodes and logic gates can also be made using diodes using the fact that diodes are conducting only in the forward biased configuration.
5. For the connections shown below, the equivalent logic gate is
electronic-devices-circuits-questions-answers-ideal-diode-q5
a) OR gate
b) AND gate
c) XOR gate
d) NAND gate
Answer: a
Explanation: The following circuit behaves as a OR logic gate.
6. For the connections shown below, the equivalent logic gate is
electronic-devices-circuits-questions-answers-ideal-diode-q6
a) OR gate
b) AND gate
c) XOR gate
d) NAND gate
Answer: b
Explanation: The circuit shown behaves as an AND logic gate.
7. The figure below shows a circuit for an AC voltmeter. It utilizes a moving-coil meter that gives a full-scale reading when the average current flowing through it is 1 mA. The moving-coil meter has a 50-Ω resistance. The value of R that results in the meter indicating a full-scale reading when the input sine-wave voltage V I is 20 V peak-to-peak is
electronic-devices-circuits-questions-answers-ideal-diode-q7
a) 3.183 kΩ
b) 3.133 kΩ
c) 3.183 Ω
d) 56.183 Ω
Answer: b
Explanation: Average value of input voltage= 20/π or 6.366 V.
Since there is diode half of the time there is no voltage across the voltmeter hence the average value of the input voltage is 10/π or 3.183V.
The total resistance required to generate 1mA current is 3.183 kΩ out of which 50Ω is provided by the moving coil. Therefore, a resistance of 3.133 kΩ is required.
8. The value of I and V for the circuit shown below are
electronic-devices-circuits-questions-answers-ideal-diode-q8
a) 2A and 5V respectively
b) -2A and 5V respectively
c) -2A and -5V respectively
d) 2A and 0V respectively
Answer: d
Explanation: I = V/R or 5/2.5 or 2A
For an ideal diode when it is conducting the potential drop across its terminal is zero.
9. The units frequently used to measure the forward bias and reverse bias current of a diode are
a) µA and µA respectively
b) µA and mA respectively
c) mA and µA respectively
d) mA and mA respectively
Answer: c
Explanation: The currents in forward bias is generally in MA and in the reverse bias it is very small and is generally measured in µA.
10. A diode
a) Is the simplest of the semiconductor devices
b) Has a characteristic that closely follows that of a switch
c) Is two terminal device
d) All of the mentioned
Answer: d
Explanation: Both the statements are true for a diode.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Modelling the Diode Forward Characteristic”.
1. Which of the following is method to model a diode’s forward characteristics?
a) Iteration method
b) Graphical method
c) Constant-voltage drop model
d) All of the mentioned
Answer: d
Explanation: All of the mentioned are different methods used to model a diode’s forward characteristics.
2. A voltage regulator needs to provide a constant voltage in spite of the fact that there may be
a) Change in the load current drawn from the terminals of the regulator
b) Change of the DC power supply that feeds the regulator circuit
c) None of the mentioned
d) All of the mentioned
Answer: d
Explanation: Voltage regulators are required in both of the situations mentioned.
For the circuit shown below calculate
electronic-devices-circuits-questions-answers-modelling-diode-forward-characteristic-q3
3. Calculate the %age change in the regulated voltage caused by a change of ±10% in the input voltage. (R L is not connected to the circuit)
a) ± 0.5%
b) ± 1%
c) ± 5%
d) ± 10%
Answer: a
Explanation: Current I in the circuit = / 1 or 7.9 mA
Incremental resistance of each diode is r d = 25mV/7.9mA at room temperature or 3.2Ω
Total resistance of the diode connection = r = 9.6Ω
The series combination of R and r acts as a voltage divider, hence
▲V 0 = ±r/r+R or 9.5 mA or ± 0.5%.
4. Calculate the change in the voltage when R L is connected as shown
a) -10 mA
b) -15 mA
c) -20 mA
d) -25 mA
Answer: c
Explanation: When R L is connected it draws approx. 2.1v/1000 A current or 2.1 mA. Hence the current in the diode decreases by approximately 2.1 mA. The voltage change across the diode is thus -2.1mA * 9.6 Ω or -20mA.
5. The value of the diode small-signal resistance r d at bias currents of 0.1 mA is
a) 250 Ω
b) 25 Ω
c) 2.5 Ω
d) 0.25 Ω
Answer: a
Explanation: Thermal voltage at room temperature is 25mV and the current is 0.1mA. r d = 25mV/0.1mA or 250 Ω.
6. In many applications, a conducting diode is modeled as having a constant voltage drop, usually approximately
a) 1 V
b) 0.1 V
c) 0.7 V
d) 7V
Answer: c
Explanation: Normally a silicon diode has a voltage drop of 0.7V.
7. The graphical method of modeling a diode characteristics is based on
a) Iteration method
b) Constant voltage drop method
c) Small signal approximation
d) Exponential method
Answer: d
Explanation: Graphical method uses equations obtained in the exponential method and then the Answer of these equations is obtained by plotting them on a graph.
8. For small-signal operation around the dc bias point, the diode is modeled by a resistance equal to the
a) Slope of the i-v characteristics of the diode at the bias point
b) Square root of the slope of the i-v characteristics of the diode at the bias point
c) Inverse of the slope of the i-v characteristics of the diode at the bias point
d) Square root of the inverse of the slope of the i-v characteristics of the diode at the bias point
Answer: d
Explanation: It is a required characteristic for small-ground operations.
9. The other name for bias point is
a) Quiescent point
b) Node point
c) Terminal point
d) Static point
Answer: a
Explanation: Quiescent point is also called base point. It is due to this name that is also called Q–point.
10. In iteration method for modelling a diode the answer obtained by each subsequent iteration is
a) Is close to the true value
b) Is close to the value obtained by exponential method
c) All of the mentioned
d) None of the mentioned
Answer: c
Explanation: True value is obtained by the exponential method and therefore both of the statements are true.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Zener Diode”.
1. Zener diodes are also known as
a) Voltage regulators
b) Forward bias diode
c) Breakdown diode
d) None of the mentioned
Answer: c
Explanation: Zener diodes are used as voltage regulators but they aren’t called voltage regulators. They are called breakdown diodes since they operate in breakdown region.
2. Which of the following is true about the resistance of a Zener diode?
a) It has an incremental resistance
b) It has dynamic resistance
c) The value of the resistance is the inverse of the slope of the i-v characteristics of the Zener diode
d) All of the mentioned
Answer: d
Explanation: All of the statements are true for the resistance of the zener diode.
3. Which of the following is true about the temperature coefficient or TC of the Zener diode?
a) For Zener voltage less than 5V, TC is negative
b) For Zener voltage around 5V, TC can be made zero
c) For higher values of Zener voltage, TC is positive
d) All of the mentioned
Answer: d
Explanation: All of the mentioned are true for the TC of a zener diode.
4. Which of the following can be used in series with a Zener diode so that combination has almost zero temperature coefficient?
a) Diode
b) Resistor
c) Transistor
d) MOSFET
Answer: a
Explanation: If a Zener diode of TC of about -2mV is connected with a forward diode in series, the combination can be used to obtain a very low TC.
5. In Zener diode, for currents greater than the knee current, the v-i curve is almost
a) Almost a straight line parallel to y-axis
b) Almost a straight line parallel to x-axis
c) Equally inclined to both the axes with a positive slope
d) Equally inclined to both the axes with a negative slope
Answer: b
Explanation: Note that the curve is v-I curve and not an i-v curve.
6. Zener diodes can be effectively used in voltage regulator. However, they are these days being replaced by more efficient
a) Operational Amplifier
b) MOSFET
c) Integrated Circuits
d) None of the mentioned
Answer: c
Explanation: ICs have been widely adapted by the industries over conventional zener diodes as their better replacements for a voltage regulators.
7. A 9.1-V zener diode exhibits its nominal voltage at a test current of 28 mA. At this current the incremental resistance is specified as 5 Ω. Find VZ0 of the Zener model.
a) 8.96V
b) 9.03V
c) 9.17V
d) 9.24V
Answer: b
Explanation: V Z = V Zo + M Z I ZT
9.1 = V Zo + 5 * 28 * 10 -3
V Zo = 8.96v
V Z = V Zo + 5I Z = 8.96 * 5I Z .
8. A shunt regulator utilizing a zener diode with an incremental resistance of 5 Ω is fed through an 82-Ω resistor. If the raw supply changes by 1.0 V, what is the corresponding change in the regulated output voltage?
a) 72.7 mV
b) 73.7 mV
c) 74.7 mV
d) 75.7 mV
Answer: c
Explanation: electronic-devices-circuits-questions-answers-zener-dioide-q8
9. A designer requires a shunt regulator of approximately 20 V. Two kinds of Zener diodes are available: 6.8-V devices with r z of 10 Ω and 5.1-V devices with r z of 30 Ω. For the two major choices possible, find the load regulation. In this calculation neglect the effect of the regulator resistance R.
a) -30mV/mA and 120mV/mA respectively
b) 30mV/mA and 60mV/mA respectively
c) -60mV/mA and +60mV/mA respectively
d) -30mV/mA and -120mV/mA respectively
Answer: d
Explanation: Three 6.8v zeners provide 3*6.8 = 20.4v with 3 * 10 =30Ω Resistance, neglecting R, we have
load Regulation = -30mV/mA.
For 5.1 Zeners we need 4 diodes to provide 20.4v with 4 * 30 =120Ω Resistance.
load Regulation = -120mV/mA .
10. Partial specifications of a Zener diode is provided. V Z = 10.0 V, V ZK = 9.6 V, and I ZT = 50 mA. Assuming that the power rating of a breakdown diode is established at about twice the specified Zener current (I ZT ), what is the power rating of each of the diodes described above?
a) 1.04 W
b) 0.104 W
c) 10.4 mW
d) 1.04 mW
Answer: a
Explanation: electronic-devices-circuits-questions-answers-zener-dioide-q10
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Rectifiers”.
1. Which of the following isn’t a type of rectifier?
a) Precision Half-wave Rectifier
b) Bridge Rectifier
c) Peak Rectifier
d) None of the mentioned
Answer: d
Explanation: All of the mentioned are different types of a rectifier.
2. For a half wave or full wave rectifier the Peak Inverse Voltage of the rectifier is always
a) Greater than the input voltage
b) Smaller than the input voltage
c) Equal to the input voltage
d) Greater than the input voltage for full wave rectifier and smaller for the half wave rectifier
Answer: b
Explanation: The peak input voltage is smaller than the input voltage due to the presence of diode. A single diode reduces the output voltage by approximately 0.7V.
3. For a half-wave rectifier having diode voltage V D and supply input of V I , the diode conducts for π – 2Θ, where Θ is given by
a) tan -1 V D /V I
b) tan -1 V D /V I – V I
c) sin -1 V D /V I
d) sin -1 V D /V I – V I
Answer: c
Explanation: The diode doesn’t conducts when V D ≥V I . Hence Θ = sin -1 ( D /V I ).
4. Bridge rectifier is an alternative for
a) Full wave rectifier
b) Peak rectifier
c) Half wave rectifier
d) None of the mentioned
Answer: a
Explanation: Bridge rectifier is a better alternative for a full wave rectifier.
5. Which of the following is true for a bridge rectifier?
a) The peak inverse voltage or PIV for the bridge rectifier is lower when compared to an identical center tapped rectifier
b) The output voltage for the center tapped rectifier is lower than the identical bridge rectifier
c) A transistor of higher number of coil is required for center tapped rectifier than the identical bridge rectifier
d) All of the mentioned
Answer: d
Explanation: All of the given statements are true for a bridge rectifier.
6. The diode rectifier works well enough if the supply voltage is much than greater than 0.7V. For smaller voltage input which of the following can be used?
a) Superdiode
b) Peak rectifier
c) Precision rectifier
d) None of the mentioned
Answer: a
Explanation: For the supply voltages less than 0.7V super diodes are used.
7. A simple diode rectifier has ‘ripples’ in the output wave which makes it unsuitable as a DC source. To overcome this one can use
a) A capacitor in series with a the load resistance
b) A capacitor in parallel to the load resistance
c) Both of the mentioned situations will work
d) None of the mentioned situations will work
Answer: b
Explanation: A capacitor is parallel with a resistor can only makes ripples go away. Series connection will become equal to an open circuit once the capacitor is fully charged.
8. Consider a peak rectifier fed by a 60-Hz sinusoid having a peak value Vp = 100 V. Let the load resistance R = 10 kΩ. Calculate the fraction of the cycle during which the diode is conducting
a) 1.06 %
b) 2.12 %
c) 3.18%
d) 4.24%
Answer: c
Explanation: w Δt ~ √
Θ = √
Θ = 0.2 rad or 3.18% of the cycle
The op amp in the precision rectifier circuit is ideal with output saturation levels of ±12 V. Assume that when conducting the diode exhibits a constant voltage drop of 0.7 V.
electronic-devices-circuits-questions-answers-rectifiers-q9
9. Find V – when V I is -1V.
a) 0V
b) 0.7V
c) 1V
d) 1.7V
Answer: a
Explanation: V I = -1v
V o = 1v
V A = 1.7v
V – = 0v
Virtual gnd as negative feedback is closed through R.
V I > 0 D 1 conducts D 2 cutoff.
V I < 0 D 2 conducts D 1 cutoff.
V 0 ⁄ V I = -1.
electronic-devices-circuits-questions-answers-rectifiers-q9a
10. Find V 0 when V I is 2V.
a) 0V
b) 0.7V
c) 1V
d) 1.7V
Answer: a
Explanation: V I = 2v
V o = 0v
V A = -0.7v
V – = 0v
Virtual gnd as negative feedback is closed through R.
V I > 0 D 1 conducts D 2 cutoff.
V I < 0 D 2 conducts D 1 cutoff.
V 0 ⁄ V I = -1.
electronic-devices-circuits-questions-answers-rectifiers-q9a
This set of Electronic Devices and Circuits Question Bank focuses on “Limiting and Clamping Circuits and Some Special Types of diodes”.
1. Which of the following circuits can be used as limiter or clamper or both?
a) electronic-devices-circuits-questions-bank-q1
b) electronic-devices-circuits-questions-bank-q1a
c) electronic-devices-circuits-questions-bank-q1b
d) All of the mentioned
Answer: d
Explanation: Each of the circuit can be either as a clamper or limiter or both.
2. The V 0 vs V I curve for the below circuit is
electronic-devices-circuits-questions-bank-q2
a) electronic-devices-circuits-questions-bank-q2a
b) electronic-devices-circuits-questions-bank-q2b
c) electronic-devices-circuits-questions-bank-q2c
d) electronic-devices-circuits-questions-bank-q2d
Answer: d
Explanation: Only this curve represents the characteristics correctly.
3. For the circuit and the input signal shown which of the following is true for the output voltage?
electronic-devices-circuits-questions-bank-q3
a) The output waveform is a square wave with lowest peak clamped to 0V
b) The output wave is a square wave with lowest peak clamped to -6V
c) The output waveform is a square wave with highest peak clamped to 4V
d) The output waveform is a straight line with the value of output voltage equal to 10V
Answer: a
Explanation: It is a clamped capacitor circuit.
4. Which of the following is not true for the duty cycle of a waveform?
a) Duty cycles can be used to describe the percent time of an active signal in an electrical device
b) Duty cycle can be used to determine the percentage of the time a signal is active
c) 60% duty cycle means that the waveform is active for 40% of the total time
d) 50% duty cycle means that the waveform is non-active for 50% of the total time
Answer: c
Explanation: 60% duty time means that a signal is active for 60% of the total time.
5. The maximum and the minimum voltage across the diode D 1 respectively is
electronic-devices-circuits-questions-bank-q5
a) 2V p and 0V respectively
b) 0V and -2V p respectively
c) V p and -V p respectively
d) 2V p and V p respectively
Answer: b
Explanation: It is a voltage doubler circuit.
6. Limiting and clamping circuits are employed in
a) FM transmitters
b) Television receivers and transmitter
c) Production of various signal waveforms such as trapezoidal, square or rectangular waves
d) All of the mentioned
Answer: d
Explanation: All of the given are uses of limiting and clamping circuits.
7. Which of the following is not true for a Schottky-Barrier Diode ?
a) It is formed by bringing metal into contact with a moderately doped n-type semiconductor material.
b) In the SBD, current is conducted by mainly by minority carriers.
c) The forward voltage drop of a conducting SBD is lower than that of a pn-junction diode.
d) SBD are used in the design of a special form of bipolar-transistor logic circuits
Answer: d
Explanation: Current is conducted largely by electrons which are majority current carriers.
8. Photodiodes are part of
a) Optoelectronics
b) High Intensity Discharge
c) Low pressure Discharge
d) None of the mentioned
Answer: a
Explanation: Optoelectronics are also called photodiodes.
9. Which of the following is true for LEDs
a) The light emitted by a LED is inversely proportional to the current flowing through the diode
b) LED operate in a manner opposite to the working of the optoelectronics or photodiodes
c) LED cannot be used to generate coherent source of light
d) None of the mentioned
Answer: b
Explanation: The working of the optoelectronics is opposite to that of the LEDs.
10. Optoisolator is
a) A combination of LED and varactor in the same package
b) A combination of a varactor and photodiode in the same package
c) A combination of LED and a photodiode in the same package
d) A combination of photodiode and Schottky-Barrier Diode in the same package
Answer: c
Explanation: By definition, an optoisolator is a package containing both LED as well as a photodiode.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Diode Circuits”.
For the given circuits and input waveform determine the output waveform
1. electronic-devices-circuits-questions-answers-diode-circuits-q1
electronic-devices-circuits-questions-answers-diode-circuits-q1a
Answer: d
Explanation: Diode is off when Vi < 5 and for Vi > 5, Vo = 5V.
2. electronic-devices-circuits-questions-answers-diode-circuits-q2
electronic-devices-circuits-questions-answers-diode-circuits-q2a
Answer: c
Explanation: Diode is off when Vi + 2 > 0.
3. electronic-devices-circuits-questions-answers-diode-circuits-q3
electronic-devices-circuits-questions-answers-diode-circuits-q3a
Answer: d
Explanation: For V1 < 4 the diode is on otherwise its off.
4. electronic-devices-circuits-questions-answers-diode-circuits-q4
electronic-devices-circuits-questions-answers-diode-circuits-q4a
electronic-devices-circuits-questions-answers-diode-circuits-q4b
Answer: c
Explanation: During +ve half cycle when Vi < 8 both the diodes are off. For V1 > 8, D1 is on. Similarly for the negative half cycle.
5. The maximum load current that can be drawn is
electronic-devices-circuits-questions-answers-diode-circuits-q5
a) 1.4 mA
b) 2.3 mA
c) 1.8 mA
d) 2.5 mA
Answer: a
Explanation: At regulated poer supply, Is = 30-9/15k or 1.4 mA. The load current will remain less than this current.
6. For the circuit shown diode cutting voltage is Vin = 0. The ripple voltage is to be no more than vrip = 4 V. The minimum load resistance, that can be connected to the output is
electronic-devices-circuits-questions-answers-diode-circuits-q6
a) 6.25
b) 12.5
c) 25
d) 30
Answer: a
Explanation: electronic-devices-circuits-questions-answers-diode-circuits-q6a
7. The circuit inside the box in fig. P3.1.31. contains only resistor and diodes. The terminal voltage vo is
connected to some point in the circuit inside the box. The largest and smallest possible value of vo most nearly to is respectively
electronic-devices-circuits-questions-answers-diode-circuits-q7
a) 15 V, 6 V
b) 24 V, 0 V
c) 24 V, 6 V
d) 15 V, 9 V
Answer: d
Explanation: The output voltage cannot exceed the positive power supply voltage and cannot be lower than the negative power supply voltage.
8. The Q-point of the zener diode in the circuit shown below is
electronic-devices-circuits-questions-answers-diode-circuits-q8
a)
b)
c)
d)
Answer: a
Explanation: electronic-devices-circuits-questions-answers-diode-circuits-q8a
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Half-Wave Rectifier”.
1. The diode in a half wave rectifier has a forward resistance RF. The voltage is V m sinωt and the load resistance is RL. The DC current is given by _________
a) V m /√2R L
b) V m /(R F +R L )π
c) 2V m /√π
d) V m /R L
Answer: b
Explanation: For a half wave rectifier, the I DC =I AVG =I m /π
I= V m sinωt/(R F +R L )=I m sinωt
I m =V m / R F +R L So, I DC =I m /π=V m /(R F +R L ).
2. The below figure arrives to a conclusion that _________
electronic-devices-circuits-questions-answers-half-wave-rectifier-q2
a) for V i > 0, V 0 =-(R 2 /R 1 )V i
b) for V i > 0, V 0 =0
c) V i < 0, V 0 =-(R 2 /R 1 )V i
d) V i < 0, V 0 =0
Answer: b
Explanation: The given op-amp is in inverting mode and this makes the output voltage to have a phase shift of 180°. The output voltage is now negative. So, the diode 1 is reverse biased and diode 2 is forward biased. Then output is clearly zero.
3. What is the output as a function of the input voltage for the given figure. Assume it’s an ideal op-amp with zero forward drop (D i =0)
electronic-devices-circuits-questions-answers-half-wave-rectifier-q3
a) 0
b) -V i
c) V i
d) 2V i
Answer: c
Explanation: When the input of the inverted mode op-amp is positive, the output is negative.
The diode is reverse biased. The input appears at the output.
4. In a half wave rectifier, the sine wave input is 50sin50t. If the load resistance is of 1K, then average DC power output will be?
a) 3.99V
b) 2.5V
c) 5.97V
d) 6.77V
Answer: b
Explanation: The standard form of a sine wave is V m sinωt. BY comparing the given information with this equation, V m =50.
Power=V m 2 /R L =50*50/1000=2.5V.
5. In a half wave rectifier, the sine wave input is 200sin300t. The average value of output voltage is?
a) 57.876V
b) 67.453V
c) 63.694V
d) 76.987V
Answer: c
Explanation: Comparing with the standard equation, V m =200V.
Average value is given by, V avg =V m /π.
So, 200/π=63.694.
6. Efficiency of a half wave rectifier is
a) 50%
b) 60%
c) 40.6%
d) 46%
Answer: c
Explanation: Efficiency of a rectifier is the effectiveness to convert AC to DC. For half wave it’s 40.6%. It’s given by, V out /V in *100.
7. If peak voltage for a half wave rectifier circuit is 5V and diode cut in voltage is 0.7, then peak inverse voltage on diode will be?
a) 5V
b) 4.9V
c) 4.3V
d) 6.7V
Answer: c
Explanation: PIV is the maximum reverse bias voltage that can be appeared across a diode in the given circuit, If the PIV rating is less than this value of breakdown of diode will occur. For a rectifier, PIV=V m -V d =5-0.7=4.3V.
8. Transformer utilisation factor of a half wave rectifier is _________
a) 0.234
b) 0.279
c) 0.287
d) 0.453
Answer: c
Explanation: Transformer utilisation factor is the ratio of AC power delivered to load to the DC power rating. This factor indicates effectiveness of transformer usage by rectifier. For a half wave rectifier, it’s low and equal to 0.287.
9. If the input frequency of a half wave rectifier is 100Hz, then the ripple frequency will be_________
a) 150Hz
b) 200Hz
c) 100Hz
d) 300Hz
Answer: c
Explanation: The ripple frequency of the output and input is same. This is because, one half cycle of input is passed and other half cycle is seized. So, effectively the frequency is the same.
10. Ripple factor of a half wave rectifier is_________(I m is the peak current and RL is load resistance)
a) 1.414
b) 1.21
c) 1.4
d) 0.48
Answer: b
Explanation: The ripple factor of a rectifier is the measure of disturbances produced in the output. It’s the effectiveness of a power supply filter to reduce the ripple voltage. The ratio of ripple voltage to DC output voltage is ripple factor which is 1.21.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Full-wave Rectifier”.
1. Efficiency of a centre tapped full wave rectifier is _________
a) 50%
b) 46%
c) 70%
d) 81.2%
Answer: d
Explanation: Efficiency of a rectifier is the effectiveness to convert AC to DC. It’s obtained by taking ratio of DC power output to maximum AC power delivered to load. It’s usually expressed in percentage. For centre tapped full wave rectifier, it’s 81.2%.
2. A full wave rectifier supplies a load of 1KΩ. The AC voltage applied to diodes is 220V . If diode resistance is neglected, what is the ripple voltage?
a) 0.562V
b) 0.785V
c) 0.954V
d) 0.344V
Answer: c
Explanation: The ripple voltage is (V ϒ ) RMS =ϒV DC /100.
V DC =0.636*V RMS * √2=0.636*220* √2=198V and ripple factor ϒ for full wave rectifier is 0.482.
Hence, (V ϒ ) RMS =0.482*198 /100=0.954V.
3. A full wave rectifier delivers 50W to a load of 200Ω. If the ripple factor is 2%, calculate the AC ripple across the load.
a) 2V
b) 5V
c) 4V
d) 1V
Answer: a
Explanation: We know that, P DC =V DC 2 /R L . So, V DC =(P DC *RL) 1/2 =10000 1/2 =100V.
Here, ϒ=0.02
ϒ=V AC /V DC =V AC /100.So, V AC =0.02*100=2V.
4. A full wave rectifier uses load resistor of 1500Ω. Assume the diodes have R f =10Ω, R r =∞. The voltage applied to diode is 30V with a frequency of 50Hz. Calculate the AC power input.
a) 368.98mW
b) 275.2mW
c) 145.76mW
d) 456.78mW
Answer: b
Explanation: The AC power input P IN =I RMS 2 (R F +R r ).
I RMS =I m /√2=V m /(R f +R L )√2=30/*1.414=13.5mA
So, P IN =2*=275.2mW.
5. In a centre tapped full wave rectifier, R L =1KΩ and for diode Rf=10Ω. The primary voltage is 800sinωt with transformer turns ratio=2. The ripple factor will be _________
electronic-devices-circuits-questions-answers-full-wave-rectifier-q5
a) 54%
b) 48%
c) 26%
d) 81%
Answer: b
Explanation: The ripple factor ϒ= [(I RMS /I AVG ) 2 – 1] 1/2 . I RMS =I m /√2=V m /(R f +R L )√2=200/1.01=198.
(Secondary line to line voltage is 800/2=400. Due to centre tap V m =400/2=200)
I RMS =198/√2=140mA, I AVG =2*198/π=126mA. ϒ=[ 2 -1] 1/2 =0.48. So, ϒ=48%.
6. If input frequency is 50Hz for a full wave rectifier, the ripple frequency of it would be _________
a) 100Hz
b) 50Hz
c) 25Hz
d) 500Hz
Answer: a
Explanation: In the output of the centre tapped rectifier, one of the half cycle is repeated. The frequency will be twice as that of input frequency. So, it’s 100Hz.
7. Transformer utilization factor of a centre tapped full wave rectifier is_________
a) 0.623
b) 0.678
c) 0.693
d) 0.625
Answer: c
Explanation: Transformer utilisation factor is the ratio of AC power delivered to load to the DC power rating. This factor indicates effectiveness of transformer usage by rectifier. For a half wave rectifier, it’s low and equal to 0.693.
8. In the circuits given below, the correct full wave rectifier is _________
a) electronic-devices-circuits-questions-answers-full-wave-rectifier-q8
b) electronic-devices-circuits-questions-answers-full-wave-rectifier-q8a
c) electronic-devices-circuits-questions-answers-full-wave-rectifier-q8b
d) electronic-devices-circuits-questions-answers-full-wave-rectifier-q8c
Answer: c
Explanation: When the input is applied, a full wave rectifier should have a current flow. The flow should be in the same direction for both positive and negative half cycles. Only the third circuit satisfies the above condition.
9. If the peak voltage on a centre tapped full wave rectifier circuit is 5V and diode cut in voltage is 0.7. The peak inverse voltage on diode is_________
a) 4.3V
b) 9.3V
c) 5.7V
d) 10.7V
Answer: b
Explanation: PIV is the maximum reverse bias voltage that can be appeared across a diode in the given circuit, if PIV rating is less than this value of breakdown of diode will occur. For a rectifier, PIV=2V m -V d = 10-0.7 = 9.3V.
10. In a centre tapped full wave rectifier, the input sine wave is 250sin100t. The output ripple frequency will be _________
a) 50Hz
b) 100Hz
c) 25Hz
d) 200Hz
Answer: b
Explanation: The equation of sine wave is in the form V m sinωt. So, by comparing we get ω=100. Frequency, f =ω/2=50Hz. The output of centre tapped full wave rectifier has double the frequency of inpu. Hence, f out = 100Hz.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Bridge Rectifier”.
1. DC average current of a bridge full wave rectifier (where I m is the maximum peak current of input).
a) 2I m
b) I m
c) I m /2
d) 1.414I m
Answer: b
Explanation: Average DC current of half wave rectifier is I m . Since output of half wave rectifier contains only one half of the input. The average value is the half of the area of one half cycle of sine wave with peak I m . This is equal to I m .
2. DC power output of bridge full wave rectifier is equal to (I m is the peak current and RL is the load resistance).
a) 2 I m 2 R L
b) 4 I m 2 R L
c) I m 2 R L
d) I m 2 R L /2
Answer: b
Explanation: DC output power is the power output of the rectifier. We know V DC for a bridge rectifier is 2V m and I DC for a bridge rectifier is 2I m . We also know V DC =I DC /R L . Hence output power is 4I m 2 R L .
3. Ripple factor of bridge full wave rectifier is?
a) 1.414
b) 1.212
c) 0.482
d) 1.321
Answer: c
Explanation: Ripple factor of a rectifier measures the ripples or AC content in the output. It’s obtained by dividing AC rms output with DC output. For full wave bridge rectifier it is 0.482.
4. If input frequency is 50Hz then ripple frequency of bridge full wave rectifier will be equal to_________
a) 200Hz
b) 50Hz
c) 45Hz
d) 100Hz
Answer: d
Explanation: Since in the output of bridge rectifier one half cycle is repeated, the frequency will be twice as that of input frequency. So, f=100Hz.
5. Transformer utilization factor of bridge full wave rectifier _________
a) 0.623
b) 0.812
c) 0.693
d) 0.825
Answer: b
Explanation: Transformer utilization factor is the ratio of AC power delivered to load to the DC power rating. This factor indicates effectiveness of transformer usage by rectifier. For bridge full wave rectifier it’s equal to 0.693.
6. If peak voltage on a bridge full wave rectifier circuit is 5V and diode cut in voltage os 0.7, then the peak inverse voltage on diode will be_________
a) 4.3V
b) 9.3V
c) 8.6V
d) 3.6V
Answer: d
Explanation: PIV is the maximum reverse bias voltage that can be appeared across a diode in the circuit. If PIV rating of diode is less than this value breakdown of diode may occur.. Therefore, PIV rating of diode should be greater than PIV in the circuit, For bridge rectifier PIV is V m -V D = 5-1.4=3.6.
7. Efficiency of bridge full wave rectifier is_________
a) 81.2%
b) 50%
c) 40.6%
d) 45.33%
Answer: a
Explanation: It’s obtained by taking ratio of DC power output to maximum AC power delivered to load. Efficiency of a rectifier is the effectiveness of rectifier to convert AC to DC. It’s usually expressed inn percentage. For bridge full wave rectifier, it’s 81.2%.
8. In a bridge full wave rectifier, the input sine wave is 40sin100t. The average output voltage is_________
a) 22.73V
b) 16.93V
c) 25.47V
d) 33.23V
Answer: c
Explanation: The equation of sine wave is in the form E m sinωt.
Therefore, E m =40. Hence output voltage is 2E m =80V.
9. Number of diodes used in a full wave bridge rectifier is_________
a) 1
b) 2
c) 3
d) 4
Answer: d
Explanation: The model of a bridge rectifier is same as Wein Bridge. It needs 4 resistors. Bridge rectifier needs 4 diodes while centre tap configuration requires only one.
10. In a bridge full wave rectifier, the input sine wave is 250sin100t. The output ripple frequency will be_________
a) 50Hz
b) 200Hz
c) 100Hz
d) 25Hz
Answer: c
Explanation: The equation of sine wave is in the form of E m sinωt. So, ω=100 and frequency =ω/2=50Hz. Since output of bridge rectifier have double the frequency of input, f=100Hz.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Rectifier Voltmeter”.
1. _________ diodes are preferred in AC rectifier voltmeter arrangements.
a) Silicon
b) Germanium
c) Gallium
d) Arsenide
Answer: a
Explanation: Silicon diodes are preferred because of their low reverse current and high forward current ratings. A PMMC movement is also used by the rectifier type instruments along with a rectifier arrangement.
2. AC voltmeter consists of_________
a) half wave rectifier
b) full wave rectifier
c) center tap rectifier
d) bridge wave rectifier
Answer: d
Explanation: The bridge rectifier provides a full wave pulsating dc. Due to the inertia of the movable coil, the meter indicates a steady deflection proportional to the average value of the current. The meter scale is usually calibrated to give the RMS value an alternating sine wave input.
3. In reverse bias, rectifier behaves as a___________
a) Resistor
b) Capacitor
c) Inductor
d) Amplifier
Answer: b
Explanation: The rectifier exhibits capacitance properties when reverse biased, and tends to bypass higher frequencies. The meter reading may be in error by as much as 0.5% decrease for every 1 kHz rise in frequency.
4. In a general rectifier voltmeter, the meter has low sensitivity because of ___________ of the diode.
a) low forward resistance
b) high forward resistance
c) low reverse impedance
d) high reverse impedance
Answer: b
Explanation: The diode offers high resistance when forward biased. Thus, offering more resistance to the current flow in the circuit. Thus making the meter less sensitive.
5. Which of the following instruments cannot be applied for ac measurements?
a) Hot wire
b) PMMC
c) Electrostatic
d) Induction type
Answer: b
Explanation: The moving coil instrument can only be used on D.C supply as the reversal of current produces a reversal of torque on the coil. It’s very delicate and sometimes uses AC circuit with a rectifier. It’s costly as compared to moving coil iron instruments.
6. The resistance of an ideal voltmeter is __________
a) low
b) infinite
c) zero
d) high
Answer: b
Explanation: The internal resistance of an ideal voltmeter is infinity and the internal resistance of an ideal ammeter is zero. Ammeter is connected in series and voltmeter is connected in parallel with the electric appliance. The resistance of an idea voltmeter is infinite so that it draws no current from the circuit under test.
7. Electronic voltmeters can be designed to measure ____________
a) Only very small voltages
b) Only very high voltages
c) Both very small and very high voltages
d) Neither high nor small voltages
Answer: c
Explanation: A voltmeter is an instrument that measures the difference in electrical potential between two points in an electric circuit. An analog voltmeter moves a pointer across a scale in proportion to the circuit’s voltage; a digital voltmeter provides a numerical display. Thus it is suitable for measuring both small and high voltages.
8. In electronic voltmeter, the range of input voltages can be extended by using _______
a) Functional switch
b) Input attenuator
c) Rectifier
d) Balanced bridge dc amplifier
Answer: b
Explanation: The function of the attenuator is that it helps to select a particular range of voltage values. The rectifier is essential in a voltmeter for the conversion of AC voltage into DC voltage.
9. The scale of a voltmeter is uniform. Its type is _________
a) Moving Iron
b) Induction
c) Moving coil permanent magnet
d) Moving coil dynamometer
Answer: c
Explanation: Moving coil permanent magnet instruments have permanent magnets. It is suited for DC measurement because here deflection is proportional to the voltage because resistance is constant for a material of the meter and hence if voltage polarity is reversed, deflection of the pointer will also be reversed so it is used only for DC measurement.
10. Which of the instruments is most accurate?
a) PMMC
b) Moving iron
c) Thermo couple
d) Induction type
Answer: a
Explanation: This torque in PMMC ensures the pointer comes to an equilibrium position i.e. at rest in the scale without oscillating to give an accurate reading. In PMMC as the coil moves in the magnetic field, eddy current sets up in a metal former or core on which the coil is wound or in the circuit of the coil itself which opposes the motion of the coil resulting in the slow swing of a pointer and then come to rest quickly with very little oscillation. It has great accuracy.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Inductor Filters”.
1. What is the effect of an inductor filter on a multi frequency signal?
a) Dampens the AC signal
b) Dampens the DC signal
c) To reduce ripples
d) To change the current
Answer: a
Explanation: Presence of inductor usually dampens the AC signal. Due to self induction induces opposing EMF or changes in the current.
2. The ripple factor ϒ of inductor filter is_________
a) ϒ = R Z 3/√2ωL
b) ϒ = R Z /3√2ωL
c) ϒ = R Z 3√2/ωL
d) ϒ = R Z 3/√2ωL
Answer: b
Explanation: Ripple factor will decrease when L is increased and R L . Inductor has a higher dc resistance. It depends on property of opposing the change of direction of current.
3. The inductor filter gives a smooth output because_________
a) It offers infinite resistance to ac components
b) It offers infinite resistance to dc components
c) Pulsating dc signal is allowed
d) The ac signal is amplified
Answer: a
Explanation: The inductor does not allow the ac components to pass through the filter. The main purpose of using an inductor filter is to avoid the ripples. By using this property, the inductor offers an infinite resistance to ac components and gives a smooth output.
4. A full wave rectifier with a load resistance of 5KΩ uses an inductor filter of 15henry. The peak value of applied voltage is 250V and the frequency is 50 cycles per second. Calculate the dc load current.
a) 0.7mA
b) 17mA
c) 10.6mA
d) 20mA
Answer: c
Explanation: For a rectifier with an inductor filter,
V DC =2V m /π, I dc =V DC /R L =2V m /R L π
I DC =2*250/=10.6mA.
5. The output of a rectifier is pulsating because_________
a) It has a pulse variations
b) It gives a dc output
c) It contains both dc and ac components
d) It gives only ac components
Answer: c
Explanation: For any electronic devices, a steady dc output is required. The filter is used for this purpose. The ac components are removed by using a filter.
6. A dc voltage of 380V with a peak ripple voltage not exceeding 7V is required to supply a 500Ω load. Find out the inductance required.
a) 10.8henry
b) 30.7henry
c) 28.8henry
d) 15.4henry
Answer: c
Explanation: Given the ripple voltage is 7V. So, 7=1.414V RMS
ϒ=V RMS /V DC =4.95/380=0.0130. ϒ=1/3√2(R L /L ω )
So, L=28.8henry.
7. The inductor filter should be used when RL is consistently small because_________
a) Effective filtering takes place when load current is high
b) Effective filtering takes place when load current is low
c) Current lags behind voltage
d) Current leads voltage
Answer: a
Explanation: When R L is infinite, the ripple factor is 0.471. This value is close to that of a rectifier. So, the resistance should be small.
8. The output voltage VDC for a rectifier with inductor filter is given by_________
a) (2V m /π)-I DC R
b) (2V m /π)+I DC R
c) (2V m π)-I DC R
d) (2V m π)+I DC R
Answer: a
Explanation: The inductor with high resistance can cause poor voltage regulation. The choke resistance, the resistance of half of transformer secondary is not negligible.
9. What causes to decrease the sudden rise in the current for a rectifier?
a) the electrical energy
b) The ripple factor
c) The magnetic energy
d) Infinite resistance
Answer: c
Explanation: When the output current of a rectifier increases above a certain value, magnetic energy is stored in the inductor. This energy tends to decrease the sudden rise in the current. This also helps to prevent the current to fall down too much.
10. A full wave rectifier with a load resistance of 5KΩ uses an inductor filter of 15henry. The peak value of applied voltage is 250V and the frequency is 50 cycles per second. Calculate the ripple factor ϒ.
a) 0.1
b) 0.6
c) 0.5
d) 0.4
Answer: d
Explanation: ϒ=I AC /I DC , I AC =2√2V m /3π(R L 2 +4ω 2 L 2 ) 1/2
By putting the values, I AC =4.24Ma. V DC =2V m /π, I DC =V DC /R L =2V m /R L π
I DC =2*250/(3.14*15*10 3 )=10.6mA. ϒ=4.24/10.6=0.4.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Capacitor Filters”.
1. In a shunt capacitor filter, the mechanism that helps the removal of ripples is_________
a) The current passing through the capacitor
b) The property of capacitor to store electrical energy
c) The voltage variations produced by shunting the capacitor
d) Uniform charge flow through the rectifier
Answer: b
Explanation: Filtering is frequently done by shunting the load with capacitor. It depends on the fact that a capacitor stores energy when conducting and delivers energy during non conduction. Throughout this process, the ripples are eliminated.
2. The cut-in point of a capacitor filter is_________
a) The instant at which the conduction starts
b) The instant at which the conduction stops
c) The time after which the output is not filtered
d) The time during which the output is perfectly filtered
Answer: a
Explanation: The capacitor charges when the diode is in ON state and discharges during the OFF state of the diode. The instant at which the conduction starts is called cut-in point. The instant at which the conduction stops is called cut-out point.
3. The rectifier current is a short duration pulses which cause the diode to act as a_________
a) Voltage regulator
b) Mixer
c) Switch
d) Oscillator
Answer: c
Explanation: The diode permits charge to flow in capacitor when the transformer voltage exceeds the capacitor voltage. It disconnects the power source when the transformer voltage falls below that of a capacitor.
4. A half wave rectifier, operated from a 50Hz supply uses a 1000µF capacitance connected in parallel to the load of rectifier. What will be the minimum value of load resistance that can be connected across the capacitor if the ripple% not exceeds 5?
a) 114.87Ω
b) 167.98Ω
c) 115.47Ω
d) 451.35Ω
Answer: c
Explanation: For a half wave filter,
ϒ=1/2√3 fCR L =1/2√3*50*10 -3 *R L
R L =10 3 /5√3=115.47Ω.
5. A 100µF capacitor when used as a filter has 15V ac across it with a load resistor of 2.5KΩ. If the filter is the full wave and supply frequency is 50Hz, what is the percentage of ripple frequency in the output?
a) 2.456%
b) 1.154%
c) 3.785%
d) 3.675%
Answer: b
Explanation: For a full wave rectifier, ϒ=1/4√3 fCR L
=1/4√3*50*10 -3 *2.5
=0.01154. So, ripple is 1.154%.
6. A full wave rectifier uses a capacitor filter with 500µF capacitor and provides a load current of 200mA at 8% ripple. Calculate the dc voltage.
a) 15.56V
b) 20.43V
c) 11.98V
d) 14.43V
Answer: d
Explanation: The ripple factor ϒ=I L / 4√3 fCV DC
V DC =200*10 -3 / 4√3 *50*500*8
=14.43.
7. The charge lost by the capacitor during the discharge time for shunt capacitor filter.
a) I DC *T
b) I DC /T
c) I DC *2T
d) I DC /2T
Answer: a
Explanation: The ‘T’ is the total non conducting time of capacitor. The charge per unit time will give the current flow.
8. Which of the following are true about capacitor filter?
a) It is also called as capacitor output filter
b) It is electrolytic
c) It is connected in parallel to load
d) It helps in storing the magnetic energy
Answer: b
Explanation: The rectifier may be full wave or half wave. The capacitors are usually electrolytic even though they are large in size.
9. The rms ripple voltage (V rms ) of a shunt filter is_________
a) I DC /2√3
b) I DC 2√3
c) I DC /√3
d) I DC √3
Answer: a
Explanation: The ripple waveform will be triangular in nature. The rms value of this wave is independent of slopes or lengths of straight lines. It depends only on the peak value.
10. A shunt capacitor of value 500µF fed rectifier circuit. The dc voltage is 14.43V. The dc current flowing is 200mA. It is operating at a frequency of 50Hz. What will be the value of peak rectified voltage?
a) 18.67V
b) 16.43V
c) 15.98V
d) 11.43V
Answer: b
Explanation: We know, V m =V dc +I dc /4fC
=14.43+ {200/ } 10 3
=14.43+2=16.43V.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “L-Section Filter”.
1. An L section filter with L=2henry and C= 49µF is used in the output of a full wave single phase rectifier that is fed from a 40-0-40 V peak transformer. The load current is 0.2A. Calculate the output dc voltage.
a) 20.76V
b) 24.46V
c) 34.78V
d) 12.67V
Answer: b
Explanation: Given, V L = 40V.
V DC =2/π*V L =2/π*40=25.46V.
2. Calculate LC for a full wave rectifier which provides 10V dc at 100mA with a maximum ripple of 2%. Input ac frequency is 50Hz.
a) 40*10 -6
b) 10*10 -6
c) 30*10 -6
d) 90*10 -6
Answer: a
Explanation: LC=1/6√2ω 2 ϒ
ω=2πf=314
By putting the values, LC=1/6√2 2 0.02=40*10 -6 .
3. The value of inductance at which the current in a choke filter does not fall to zero is_________
a) peak inductance
b) cut-in inductance
c) critical inductance
d) damping inductance
Answer: c
Explanation: When the value of inductance is increased, a value is reached where the diodes supplies current continuously. This value of inductance is called critical inductance.
4. The condition for the regulation curve in a choke filter is_________
a) LC≥R L /3ω
b) LC≤R L /3ω
c) L≥R L /3ω
d) LC≥R L 3ω
Answer: a
Explanation: IDC should not exceed the negative peak of ac component. So, the regulation curve in direct output voltage against load current for a filter is given the relation LC≥R L /3ω.
5. The ripple factor for an l section filter is_______
a) ϒ= 1/6√2ω 2 LC
b) ϒ= 6√2ω 2 LC
c) ϒ= 6√3ω 2 LC
d) ϒ= 1/6√3ω 2 LC
Answer: a
Explanation: The ripple factor is the ratio of root mean square value of ripple voltage to absolute value of dc component. It can also be expressed as the peak to peak value.
6. The output dc voltage of an LC filter is_______
a) V DC =2V m /π + I DC R
b) V DC =V m /π – I DC R
c) V DC =2V m /π – 2I DC R
d) V DC =2V m /π – I DC R
Answer: d
Explanation: The value for V DC is same as that of inductor filter. If inductor has no dc resistance, then V DC =2V m /π. If R is the series resistance of transformer, then V DC =2V m /π – I DC R.
7. The rms value of ripple current for an L section filter is_______
a) I RMS =√2/3*X L *V DC
b) I RMS =√2/3*X L *V DC
c) I RMS =√2/3*X L *V DC
d) I RMS =√2/3*X L *V DC
Answer: a
Explanation: The ac current through L is determined primarily by X L =2ωL. It is directly proportional to voltage produced and indirectly proportional to the reactance.
8. What makes the load in a choke filter to bypass harmonic components?
a) capacitor
b) inductor
c) resistor
d) diodes
Answer: a
Explanation: When the capacitor is shunted across the load, it bypasses the harmonic components. This is because it offers low reactance to ac ripple component. In another hand the inductor offers high impedance to harmonic terms.
9. The ripple to heavy loads by a capacitor is_______
a) high
b) depends on temperature
c) low
d) no ripple at all
Answer: c
Explanation: Ripple factor is inversely proportional to load resistance for a capacitor filter.
So, the ripples that are produced are low when the load is high.
10. In a choke l section filter_______
a) the inductor and capacitor are connected across the load
b) the inductor is connected in series and capacitor is connected across the load
c) the inductor is connected across and capacitor is connected in series to the load
d) the inductor and capacitor are connected in series
Answer: b
Explanation: The choke filter is sometimes also called as L section filter because the inductor and capacitor are connected in the shape ’L’ or inverted manner. But several sections of this choke filter are employed to smooth the output curve and make it free from ripples.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “CLC Filter”.
1. What is the number of capacitors and inductors used in a CLC filter?
a) 1, 2 respectively
b) 2, 1 respectively
c) 1, 1 respectively
d) 2, 2 respectively
Answer: b
Explanation: A very smooth output can be obtained by a filter consisting of one inductor and two capacitors connected across each other. They are arranged in the form of letter ‘pi’. So, these are also called as pi filters.
2. Major part of the filtering is done by the first capacitor in a CLC filter because _________
a) The capacitor offers a very low reactance to the ripple frequency
b) The capacitor offers a very high reactance to the ripple frequency
c) The inductor offers a very low reactance to the ripple frequency
d) The inductor offers a very high reactance to the ripple frequency
Answer: a
Explanation: The CLC filters are used when high voltage and low ripple frequency is needed than L section filters. The capacitor in a CLC filter offers very low reactance to the ripple frequency. So, maximum of the filtering is done by the first capacitor across the L section part.
3. At f=50Hz, the ripple factor of CLC filter is_________
a) ϒ=5700RL / (LC 1 C 2 )
b) ϒ=5700/ (LC 1 C 2 R L )
c) ϒ=5700LC 1 / (C 2 R L )
d) ϒ=5700C 1 C 2 / (LR L )
Answer: b
Explanation: The ripple factor of a rectifier is the measure of disturbances produced in the output. It’s the effectiveness of a power supply filter to reduce the ripple voltage. The ratio of ripple voltage to DC output voltage is ripple factor which is 5700 / (LC 1 C 2 R L ) at 50Hz.
4. A single phase full wave rectifier makes use of pi section filter with 10µF capacitors and a choke of 10henry. The secondary voltage is 280V and the load current is 100mA. Determine the dc output voltage when f=50Hz.
a) 345V
b) 521V
c) 243V
d) 346V
Answer: d
Explanation: Given, V RMS =280V
So, V¬m = 1.414*280=396V.
From theory of capacitor filter, V DC = V m –I DC /4fC=396-0.1/ (4*50*10*10 -6 )=346V.
5. For a given CLC filter, the operating frequency is 50Hz and 10µF capacitors used. The load resistance is 3460Ω with an inductance of 10henry. Calculate the ripple factor.
a) 0.165%
b) 0.142%
c) 0.178%
d) 0.321%
Answer: a
Explanation: We have, ϒ=5700 / (LC 1 C 2 R L )
=5700 / (10*100*10 -12 *3460)
=0.165%.
6. The inductor is placed in the L section filter because_________
a) It offers zero resistance to DC component
b) It offers infinite resistance to DC component
c) It bypasses the DC component
d) It bypasses the AC component
Answer: a
Explanation: The inductor offers high reactance to ac component and zero resistance to dc component. So, it blocks the ac component which cannot be bypassed by the capacitors.
7. The voltage in case of a full wave rectifier in a CLC filter is_________
a) V ϒ = I DC /2fC
b) V ϒ = I DC fC
c) V ϒ = I DC /fC
d) V ϒ = 2I DC fC
Answer: a
Explanation: T he filter circuit is a combination of capacitors and inductors. The RMS value depends on the peak value of charging and discharging magnitude, V PEAK .
8. The advantages of a pi-filter is_________
a) low output voltage
b) low PIV
c) low ripple factor
d) high voltage regulation
Answer: c
Explanation: Due to the use of two capacitors with an inductor, an improved filtering action is provided. This leads to decrement in ripple factor. A low ripple factor signifies regulated and ripple free DC voltage.
9. What is the relation between time constant and load resistance?
a) They don’t depend on each other
b) They are directly proportional
c) They are inversely proportional
d) Cannot be predicted
Answer: c
Explanation: If the load resistance value is large, the discharge time constant will be of a high value. Thus the capacitors time to discharge will get over soon. This lowers the amount of ripples in the output and increases the output voltage. If the load resistance is small, the discharge time constant will be more with decrease in output voltage.
10. The output waveform of CLC filter is superimposed by a waveform referred to as_________
a) Square wave
b) Triangular wave
c) Saw tooth wave
d) Sine wave
Answer: c
Explanation: Since the rectifier conducts current only in the forward direction, any energy discharged by the capacitor will flow into the load. This result in a DC voltage upon which is superimposed a waveform referred to as a saw tooth wave.
This set of Electronic Devices and Circuits Questions and Answers for Entrance exams focuses on “Voltage Regulation Using Zener Diode”.
1. The percentage voltage regulation (V L ) is given by_________
a) (V NL -V L )/V NL *100
b) (V NL +V L )/V NL *100
c) (V NL -V L )/V L *100
d) (V NL +V L )/V L *100
Answer: a
Explanation: The change in the output voltage from no load to full load condition is called as voltage regulation, where V NL is the voltage at no load condition. It is used to maintain a nearly constant output voltage. If the regulation is high, the output voltage is stable.
2. The limiting value of the current resistor used in a Zener diode
a) min =[(V in ) max + V Z /R
b) min =[(V in ) max -V Z ]/R
c) min =[(V in ) max -V Z ]R
d) min =[(V in ) max + V Z ]R
Answer: b
Explanation: When the input voltage is maximum, the load current is minimum, the Zener current should not increase the maximum rated value. Therefore there should be a minimum value of resistor.
3. When the regulation by a Zener diode is with a varying input voltage, what happens to the voltage drop across the resistance?
a) Decreases
b) Has no effect on voltage
c) Increases
d) The variations depend on temperature
Answer: c
Explanation: When the input voltage varies, the input current also varies. This makes more current to flow in the diode. This increase in the current should balance a change in the load current. Hence the voltage drop increases across the resistor.
4. In the given limiter circuit, an input voltage V i =10sin100πt is applied. Assume that the diode drop is 0.7V when it’s forward biased. The zener breakdown voltage is 6.8V.The maximum and minimum values of outputs voltage are _______
electronic-devices-circuits-questions-answers-entrance-exams-q4
a) 6.1V,-0.7V
b) 0.7V,-7.5V
c) 7.5V,-0.7V
d) 7.5V,-7.5V
Answer: c
Explanation: With V I = 10V when maximum, D 1 is forward biased, D 2 is reverse biased. Zener is in breakdown region. V OMAX =sum of breakdown voltage and diode drop=6.8+0.7=7.5V. V OMIN =negative of voltage drop=-0.7V. There will be no breakdown voltage here.
5. Determine the maximum and minimum values of load current for which the Zener diode shunt regulator will maintain regulation when V IN =24V and R=500Ω. The Zener diode has a V Z =12V and (I Z )MAX=90mA.
a) 40mA, 0mA respectively
b) 36mA, 5mA respectively
c) 10mA, 6mA respectively
d) 21mA, 0mA respectively
Answer: d
Explanation: The current through the resistance R is given by, I=(V IN -V Z )/R= /500=24mA. (I L ) MAX =I-(I Z ) MIN =24-3=21mA .This current is less than (I Z ) MAX . So, we assume that all the input current flows through the Zener diode. Under this condition, (I L ) MIN is 0mA.
6. Determine the minimum value of load resistance that can be used in the circuit with (I Z )Min=3mA. The input voltage is 10V and the resistance R is 500Ω. The Zener diode has a V Z =6V 0and (I Z )MAX=90mA.
a) 1KΩ
b) 2.4KΩ
c) 1.2KΩ
d) 3.6KΩ
Answer: c
Explanation: The I=(V IN -V Z )/R=/500=8mA. (I L ) MAX =I-(I Z ) MIN =8-3=5mA. (R L ) MIN =V Z /(I L ) MAX =6/5m=1.2KΩ.
7. A Zener regulator has to handle supply voltage variation from 19.5V to 22.5V. Find the magnitude of regulating resistance, if the load resistance is 6KΩ. The Zener diode has the following specifications: breakdown voltage =18V, (I Z ) Min =2µA, maximum power dissipation=60mW and Zener resistance =20Ω.
a) 0 < R < 500Ω
b) 77.8 < R < 500Ω
c) 77.8 < R < 100Ω
d) 18 < R < 500Ω
Answer: b
Explanation: (P Z ) MAX /r Z =(I Z ) MAX 2 . So, (I Z ) MAX =60m/20=54.8µA. I L =V O /R L =18/6000=3mA.
R MAX =(V Min -V Z )/[( I Z ) Min +( I L ) MAX ]=/µ=500Ω.
R Min =(V MAX -V Z )/[( I Z ) MAX +( I L ) Min ]=/=77.8Ω.
8. A transistor series regulator has the following specifications: V IN =15V, V Z =8.3V, β=100, R=1.8KΩ, R L =2KΩ. What will be the Zener current in the regulator circuit?
a) 4.56mA
b) 3.26mA
c) 4.56mA
d) 3.68mA
Answer: d
Explanation: We know, V O =V Z -V BE =8.3-0.7=7.6V. V CE =V IN -V 0 =15-7.6=7.4V. So, I R =(V IN -V Z )/R=/1.8m=3.72mA. I L =V O /R L =7.6/2000=3.8mA. I B =I L / β=3.8mA/100=0.038mA. Finally, I Z =I R -I B =3.72-0.038=3.682mA.
9. When is a regulator used?
a) when there are small variations in load current and input voltage
b) when there are large variations in load current and input voltage
c) when there are no variations in load current and input voltage
d) when there are small variations in load current and large variations in input voltage
Answer: a
Explanation: The regulator has following limitations: 1.It has low efficiency for heavy load currents 2. The output voltage changes slightly due to Zener impedance. Hence, it is used when there are small variations in load current and input voltage.
10. A transistor in a series voltage regulator acts like a variable resistor. The value of its resistance is determined by _______
a) emitter current
b) base current
c) collector current
d) it is not controlled by the transistor terminals
Answer: b
Explanation: The principle is based on the fact that a large fraction of the increase in input voltage appears across the transistor so that the output voltage remains to be constant. When input voltage is increased, the output voltage also increases which biases the transistor towards less current.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Junction Transistor”.
1. The advantages over the vacuum triode for a junction transistor is_________
a) high power consumption
b) high efficiency
c) large size
d) less doping
Answer: b
Explanation: A junction transistor is an analogous to a vacuum triode. The main difference between them is that a transistor is a current device while a vacuum triode is a voltage device. The advantages of a transistor over a vacuum triode are long life, high efficiency, light weight, smaller in size, less power consumption.
2. What is the left hand section of a junction transistor called?
a) base
b) collector
c) emitter
d) depletion region
Answer: c
Explanation: The main function of this section is to supply majority charge carriers to the base. Hence it is more heavily doped in comparison to other regions. This forms the left hand section of the transistor.
3. In an NPN transistor, the arrow is pointed towards_________
a) the collector
b) the base
c) depends on the configuration
d) the emitter
Answer: d
Explanation: As regards to the symbols, the arrow head is always at the emitter. The direction indicates the conventional direction of current flow. In case of PNP transistor, it is from base to emitter.
4. Which of the following is true in construction of a transistor?
a) the collector dissipates lesser power
b) the emitter supplies minority carriers
c) the collector is made physically larger than the emitter region
d) the collector collects minority charge carriers
Answer: c
Explanation: In most of the transistors, the collector is made larger than emitter region. This is due to the fact that collector has to dissipate much greater power. The collector and emitter cannot be interchanged.
5. In the operation of an NPN transistor, the electrons cross which region?
a) emitter region
b) the region where there is high depletion
c) the region where there is low depletion
d) P type base region
Answer: d
Explanation: The electrons in the emitter region are repelled by the negative terminal of the battery towards the emitter junction. The potential barrier at the junction is reduced due to forward bias and base region is very thin and lightly doped, electrons cross the P type base region.
6. Which of the following are true for a PNP transistor?
a) the emitter current is less than the collector current
b) the collector current is less than the emitter current
c) the electrons are majority charge carriers
d) the holes are the minority charge carriers
Answer: b
Explanation: The 2 – 5% of holes is lost in recombination with electrons in the base region. The majority charge carriers are holes for a PNP transistor. Thus the collector current is slightly less than the emitter current.
7. In the saturated region, the transistor acts like a_________
a) poor transistor
b) amplifier
c) open switch
d) closed switch
Answer: d
Explanation: In saturated mode, both emitter and collector are forward biased. The negative of the battery is connected to emitter and similarly the positive terminals of batteries are connected to the base. The transistor now acts like a closed switch.
8. When does the transistor act like an open switch?
a) cut off region
b) inverted region
c) saturated region
d) active region
Answer: a
Explanation: In cut off region, both the junctions are reverse biased. The transistor has practically zero current because the emitter does not emit charge carriers to the base. So, the transistor acts as open switch.
9. If the emitter-base junction is forward biased and the collector-base junction is reverse biased, what will be the region of operation for a transistor?
a) cut off region
b) saturated region
c) inverted region
d) active region
Answer: d
Explanation: When the emitter-base junction is forward biased and the collector-base junction is reverse biased, the transistor is used for amplification. A battery is connected to collector base circuit. The positive terminal is connected to the collector while the negative is connected to the base.
10. The transfer of a signal in a transistor is_________
a) low to high resistance
b) high to low resistance
c) collector to base junction
d) emitter to base junction
Answer: a
Explanation: A forward biased emitter base junction has a low resistance path. A reversed biased junction has a high resistance path. The weak signal is introduced in a low resistance circuit and the output is taken from the high resistance circuit.
This set of Electronic Devices and Circuits Questions and Answers for Aptitude test focuses on “The Transistor as an Amplifier”.
1. The emitter current consist of_________
a) electrons passing from collector to emitter
b) holes crossing from base to collector
c) electron current Ine constituted by electrons
d) immobile charge carriers
Answer: c
Explanation: The emitter current consists of two parts. It consists of hole current IpE constituted by holes. The other part is that it consists the electron current InE constituted by electrons.
2. The total emitter current (I E ) is given by_________
a) I E = I pE * I nE
b) I E = I pE – I nE
c) I E = I pE / I nE
d) I E = I pE + I nE
Answer: d
Explanation: The total emitter current is the sum of I nE and I pE . In commercial transistors, the doping of emitter region is made much heavier than base. Hence current by majority charge carriers I nE is negligible when compared to current by minority charge carriers I pE .
3. A common base transistor amplifier has an input resistance of 20Ω and output resistance of 100kΩ. If a signal of 400mV is applied between emitter and base, find the voltage amplification. Assume αac to be one.
a) 20
b) 50
c) 30
d) 25
Answer: b
Explanation: I E = V/R=400M/20=20mA
I C =αI E = 1*20mA=20mA. V O =I C *R L =20m*1k=20V
Amplification, A= V O /signal voltage=20V/400m=50.
4. The amplification factor for a transistor is given by_________
a) A=αR L /r e
b) A=αR L r e
c) A=r e / αR L
d) A=R L /r e α
Answer: a
Explanation: One of the most important application of a transistor is an amplifier. A small change in signal voltage produces an appreciable change in emitter current because the input circuit has low resistance (α=∆I C /I E ).
5. Why is the silicon mostly chosen when compared to germanium?
a) low power consumption
b) high efficiency
c) greater working temperature
d) large I CBO
Answer: c
Explanation: The normal working temperature of germanium is approximately 70°C .The normal working temperature of silicon is approximately 150°C. The other advantages of using a silicon material are, it has a smaller I CBO and its variations are smaller with temperature.
6. The change in output voltage across the load resistor for a transistor during amplification is_________
a) R L *α*∆I E
b) R L *∆I E /α
c) R L *α 2 *∆I E
d) R L *α 1/2 *∆I E
Answer: a
Explanation: A small change of voltage ∆V i between emitter and base causes a relatively large emitter current change ∆I E . We define by the symbol α that fraction of this current change which is collected and passes through R L .
7. A transistor has an I C of 100mA and I B of 0.5mA. What is the value of α dc ?
a) 0.565
b) 0.754
c) 1.24
d) 0.995
Answer: d
Explanation: Emitter current I E =I C +I B =100+0.5=100.5mA.
αdc=I C /I E =100/100.5=0.995.
8. A germanium transistor used as an amplifier has a collector cut off current I CBO =10µA at a temperature 27°C and β=50. What is the collector current when the base current is 0.25mA?
a) 10.76mA
b) 13.01mA
c) 15.67mA
d) 11.88Ma
Answer: b
Explanation: I C =βI B +I CBO
I C =50*0.25/1000+51/100000=13.01mA.
9. In a PNP germanium transistor, the cut in voltage is about_________
a) -0.1V
b) -0.01V
c) -0.05V
d) -0.07V
Answer: a
Explanation: The cut in voltage of germanium is lower than that of silicon. If both germanium and silicon are in parallel, Ge starts conducting earlier and stops silicon from conducting.
10. In a PNP transistor operating in active region, the main stream of current is_________
a) drift of holes
b) drift of electrons
c) diffusion of holes
d) diffusion of electrons
Answer: c
Explanation: The emitter-base junction is forward biased while collector-base junction is reversed biased. The transistor now operates in active region. Here, it can be used for amplification purpose.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Transistor Construction”.
1. Which gas is used to fill the chamber in the grown junction type transistor construction?
a) helium
b) boron
c) nitrogen
d) oxygen
Answer: c
Explanation: In the process of transistor construction, a crucible is placed in the chamber. This chamber consists of hydrogen or nitrogen. These gases help in the prevention of oxidation. It also contains purified Ge or Si at a temperature few degrees above its melting point.
2. In a grown junction type construction, the method used form a junction transistor is_________
a) alloy type diffusion
b) mesa type
c) speed variation method
d) fused junction type
Answer: c
Explanation: The grown junction may be formed by suddenly varying the rate of pulling the seed crystal from the melt. This method is based on the fact that proportion in which N and P type impurities crystallise i.e.., enter the grown crystal depends on the rate of pulling.
3. Which of the following methods take impurity variation method for transistor construction?
a) alloy type diffusion
b) grown junction type
c) epitaxial type
d) mesa type
Answer: b
Explanation: In impurity variation method, the impurity content of the semiconductor is altered in its type as well as the quantity. For example, in making NPN germanium grown junction transistor, a small type of N type impurity is added to molten germanium and the crystal growth is started.
4. Which of the following is true about grown junction type construction?
a) N type impurity is added to P type impurity
b) Boron helps in the prevention of oxidation
c) The seed is pulled to a large distance for a correct growth
d) Slow pulling leads to the formation of P type crystal
Answer: d
Explanation: This method is based on the fact that proportion in which N and P type impurities crystallise i.e.., enter the grown crystal depends on the rate of pulling. If the pulling rate is small, a P type crystal is grown. If the pulling rate is fast, an N type crystal is grown.
5. What is the melting point of indium in alloy type transistors?
a) 300°C
b) 200°C
c) 155°C
d) 100°C
Answer: c
Explanation: This is similar to soldering and PNP transistor is generally is made by this process. In this method, first of all N type germanium is obtained. The N type wafer and indium dots are placed in a furnace and heated to about 500°C.
6. The non rectifying base contact is made from_________
a) welding a strip
b) germanium
c) indium
d) graphite
Answer: a
Explanation: Leads for emitter and collector are soldered to the dots making non rectifying contacts. Further, non rectifying base contact is usually made from a welding a strip or loop of gold plated wire to the base plate.
7. What is the thickness of wafer in the alloy type transistors?
a) 1-2m inch
b) 3-5m inch
c) 5-6m inch
d) 4-7m inch
Answer: b
Explanation: The wafer of crystal has a 3-5m inch thickness and 80m inch square. This is placed in a graphite jig with a dot of prepared indium. One dot of an indium is 3 times larger than the other.
8. The larger dot of the indium is used as_________
a) base
b) emitter
c) control pin
d) collector
Answer: d
Explanation: The wafer is placed in a graphite jig with a dot of prepared indium. One dot of an indium is 3 times larger than the other. Finally the larger dot is used as collector. The smaller dot is used as emitter.
9. The electrical properties of a transistor in alloy type construction is determined by_________
a) space between the junctions in the wafer
b) proportions of N and P type impurities
c) the pulling rate of crystal
d) uniformity of the crystal lattice
Answer: a
Explanation: Large area collector junction helps in collecting most of the holes emitted from the emitter ensuring that the collector current almost equals the emitter current. The spacing between two junctions inside germanium wafer is very small and determines the electrical properties.
10. The grown junction type transistors is generally used for_________
a) PNP transistors
b) NPN transistors
c) Both transistors
d) Depends on the material used
Answer: b
Explanation: Grown junction type transistors are manufactured through growing single large crystal which is slowly pulled from the melt in crystal growing furnace. This is generally used for NPN transistors.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Common Base Configuration”.
1. The AC current gain in a common base configuration is_________
a) -∆I C /∆I E
b) ∆I C /∆I E
c) ∆I E /∆I C
d) -∆I E /∆I C
Answer: a
Explanation: The AC current gain is denoted by α ac . The ratio of change in collector current to the change in emitter current at constant collector base voltage is defined as current amplification factor.
2. The value of αac for all practical purposes, for commercial transistors range from_________
a) 0.5-0.6
b) 0.7-0.77
c) 0.8-0.88
d) 0.9-0.99
Answer: d
Explanation: For all practical purposes, αac=αdc=α and practical values in commercial transistors range from 0.9-0.99. It is the measure of the quality of a transistor. Higher is the value of α, better is the transistor in the sense that collector current approaches the emitter current.
3. A transistor has an I C of 100mA and I B of 0.5mA. What is the value of αdc?
a) 0.787
b) 0.995
c) 0.543
d) 0.659
Answer: b
Explanation: Emitter current I E =I C +I B =100+0.5=100.5mA
α dc =I C /I E =100/100.5=0.995.
4. In CB configuration, the value of α=0.98A. A voltage drop of 4.9V is obtained across the resistor of 5KΩ when connected in collector circuit. Find the base current.
a) 0.01mA
b) 0.07mA
c) 0.02mA
d) 0.05mA
Answer: c
Explanation: Here, I C =4.9/5K=0.98mA
α = I C /I E .So,
I E =I C /α=0.98/0.98=1mA.
I B =I E -I C =1-0.98=0.02Ma.
5. The emitter current I E in a transistor is 3mA. If the leakage current I C BO is 5µA and α=0.98, calculate the collector and base current.
a) 3.64mA and 35µA
b) 2.945mA and 55µA
c) 3.64mA and 33µA
d) 5.89mA and 65µA
Answer: b
Explanation: I C =αI E + I C BO
=0.98*3+0.005=2.945mA.
I E =I C +I B . So, I B =3-2.495=0.055mA=55µA.
6. Determine the value of emitter current and collector current of a transistor having α=0.98 and collector to base leakage current I C BO=4µA. The base current is 50µA.
a) 1.5mA
b) 3.7mA
c) 2.7mA
d) 4.5mA
Answer: c
Explanation: Given, I B =50µA=0.05mA
I CBO =4µA=0.004Ma
I C =α/I B +1/I CBO =2.45+0.2=2.65Ma
I E =I C +I B =2.65+0.05=2.7mA.
7. The negative sign in the formula of amplification factor indicates_________
a) that I E flows into transistor while I C flows out it
b) that I C flows into transistor while I E flows out it
c) that I B flows into transistor while I C flows out it
d) that I C flows into transistor while I B flows out it
Answer: a
Explanation: When no signal is applied, the ratio of collector current to emitter current is called dc alpha, α dc of a transistor. α dc =-I C /I E . It is the measure of the quality of a transistor. Higher is the value of α, better is the transistor in the sense that collector current approaches the emitter current.
8. The relation between α and β is _________
a) β=α/
b) α= β/
c) β=α/
d) α= β/
Answer: b
Explanation: β is an ac base amplification factor. α is called as current amplification factor. The relation of I C and I B change as I C = βI B + I CBO .
9. A transistor has an I E of 0.9mA and amplification factor of 0.98. What will be the I C ?
a) 0.745mA
b) 0.564mA
c) 0.236mA
d) 0.882mA
Answer: d
Explanation: Given, I E = 0.9mA, α=0.98
We know, α= I C /I E
So, I C =0.98*0.9=0.882mA.
10. The collector current is 2.945A and α=0.98. The leakage current is 2µA. What is the emitter current and base current?
a) 3mA and 55µA
b) 2.945mA and 55µA
c) 3.64mA and 33µA
d) 5.89mA and 65µA
Answer: a
Explanation: (I C – I CBO )/α=I E
= /0.98=3mA.
I E =I C +I B . So, I B =3-2.495=0.055mA=55µA.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Common Emitter Configuration”.
1. The base current amplification factor β is given by_________
a) I C /I B
b) I B /I C
c) I E /I B
d) I B /I E
Answer: a
Explanation: The current amplification factor is given by I C //I B . When no signal is applied, then the ratio of collector current to the base current is called current amplification factor of a transistor.
2. In an NPN silicon transistor, α=0.995, I E =10mA and leakage current I CBO =0.5µA. Determine I CEO .
a) 10µA
b) 100µA
c) 90µA
d) 500µA
Answer: b
Explanation: I C =α I E +I CBO =0.995*10mA+0.5µA=9.9505mA.
I B =I E -I C =10-9.9505=0.0495mA. β=α/=0.995/=199
I CEO =9.9505-199*0.0495=0.1mA==100µA.
3. A germanium transistor with α=0.98 gives a reverse saturation current I CBO =10µA in a CB configuration. When it is used in CE configuration with a base current of 0.22µA, calculate the collector current.
a) 0.9867mA
b) 0.7654mA
c) 0.51078mA
d) 0.23456mA
Answer: c
Explanation: Given, I CBO =10µA, α=0.98 and I B =0.22µA. I C =α/ I B + 1/ I CBO
0.01078+0.5=0.51078mA.
4. In CE configuration, if the voltage drop across 5kΩ resistor connected in the collector circuit is 5V. Find the value of I B when β=50.
a) 0.01mA
b) 0.25mA
c) 0.03mA
d) 0.02mA
Answer: d
Explanation: I C =V across R L /R L =5V/5KΩ=1mA.
I B =I C /β=1/50=0.02mA.
5. A transistor is connected in CE configuration. Collector supply voltage V cc =10V, R L =800Ω, voltage drop across R L =0.8V, α=0.96. What is base current?
a) 41.97µA
b) 56.78µA
c) 67.67µA
d) 78.54µA
Answer: a
Explanation: Here, I C =0.8/800=1mA
β= α/ =0.96/1-0.96=24.
Now, I B =I C / β=1/24=41.67µA.
6. The collector supply voltage for a CE configured transistor is 10V. The resistance R L =800Ω. The voltage drop across R L is 0.8V. Find the value of collector emitter voltage.
a) 3.7V
b) 9.2V
c) 6.5V
d) 9.8V
Answer: b
Explanation: Here, I C =0.8/800=1mA.
We know, V CE =V CC -I C R L
=10-0.8=9.2V.
7. The relation between α and β is_________
a) β = α/
b) α = β/
c) β = α/
d) α = β/
Answer: b
Explanation: β is an ac base amplification factor. α is called as current amplification factor. The relation of I C and I B change as I C = βI B + I CBO .
8. In I CEO , wt does the subscript ‘CEO’ mean?
a) collector to base emitter open
b) emitter to base collector open
c) collector to emitter base open
d) emitter to collector base open
Answer: c
Explanation: The subscript ‘CEO’ means that it is collector to emitter base open. It is called as the leakage current. It occurs in a reverse bias in PNP transistor. The total current can be calculated by I C =βI B +I C .
9. When the signal is applied, the ratio of change of collector current to the ratio of change of base current is called_________
a) dc current gain
b) base current amplification factor
c) emitter current amplification factor
d) ac current gain
Answer: d
Explanation: The ac current gain is given by β=∆I C /∆I B . When the signal is applied, the ratio of change of collector current to the ratio of change of base current is called ac current gain.
10. The range of β is _________
a) 20 to 500
b) 50 to 300
c) 30 to 400
d) 10 to 20
Answer: a
Explanation: Almost in all the transistors, the base current is less than 5% of the emitter current. Due to this fact, it is generally greater than 20. Usually it ranges from 20 to 500. Hence this configuration is frequently used when appreciable current gain as well as voltage gain is required.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Common Collector Configuration”.
1. The current amplification factor ϒdc is given by_________
a) I E /I B
b) I B /I E
c) I C /I E
d) I E /I C
Answer: a
Explanation: When no signal is applied, then the ratio of emitter current to base current is called as ϒ dc of the transistor. As the collector is common to both input and output circuits, hence the name common collector configuration.
2. The relation between α and β is given by _________
a) 1/=1- β
b) 1/=1+ β
c) 1/=1+ β
d) 1/=1- β
Answer: c
Explanation: The current amplification factor is given by I C //I B . When no signal is applied, then the ratio of collector current to the base current is called current amplification factor of a transistor. β is an ac base amplification factor. α is called as current amplification factor. The relation of I C and I B change as I C = βI B + I CBO .
3. The CC configuration has an input resistance_________
a) 500kΩ
b) 750kΩ
c) 600kΩ
d) 400kΩ
Answer: b
Explanation: It has a high input resistance and very low output resistance so the voltage gain is always less one. It is used for driving a low impedance load from a high impedance source.
4. The application of a CC configured transistor is_________
a) voltage multiplier
b) level shifter
c) rectification
d) impedance matching
Answer: d
Explanation: The most important use of CC transistor is an impedance matching device. It is seldom used for amplification purposes. The current gain is same as that of CE configured transistor.
5. What is the output resistance of CC transistor?
a) 25 Ω
b) 50 Ω
c) 100 Ω
d) 150 Ω
Answer: a
Explanation: The CC transistor has a very low value of output resistance of 25 Ω. The voltage gain is always less one. It is used for driving a low impedance load from a high impedance source.
6. Increase in collector emitter voltage from 5V to 8V causes increase in collector current from 5mA to 5.3mA. Determine the dynamic output resistance.
a) 20kΩ
b) 10kΩ
c) 50kΩ
d) 60kΩ
Answer: b
Explanation: r o =∆V CE /∆I C
=3/0.3m=10kΩ.
7. A change in 300mV in base emitter voltage causes a change of 100µA in the base current. Determine the dynamic input resistance.
a) 20kΩ
b) 10kΩ
c) 30kΩ
d) 60kΩ
Answer: c
Explanation: r o =∆V BE /∆I B
=300m/100µ=30kΩ.
8. The point on the DC load line which is represented by ‘Q’ is called _________
a) cut off point
b) cut in point
c) breakdown point
d) operating point
Answer: d
Explanation: The point which represents the values of I C and V CE that exist in a transistor circuit when no signal is applied is called as operating point. This is also called as working point or quiescent point.
9. When is the transistor said to be saturated?
a) when V CE is very low
b) when V CE is very high
c) when V BE is very low
d) when V BE is very high
Answer: a
Explanation: When V CE is very low, the transistor said to be saturated and it operates in saturated region of characteristic. The change in base current I B does not produce a corresponding change in the collector voltage I C .
10. The input resistance is given by _________
a) ∆V CE /∆I B
b) ∆V BE /∆I B
c) ∆V BE /∆I C
d) ∆V BE /∆I E
Answer: b
Explanation: The ratio of change in base emitter voltage (∆V BE ) to resulting change in base current (∆I B ) at constant collector emitter voltage (V CE ) is defined as input resistance. This is denoted by r i .
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The CE Characteristics”.
1. The input characteristics of a CE transistor is_________
a) electronic-devices-circuits-questions-answers-ce-characteristics-q1
b) electronic-devices-circuits-questions-answers-ce-characteristics-q1a
c) electronic-devices-circuits-questions-answers-ce-characteristics-q1b
d) electronic-devices-circuits-questions-answers-ce-characteristics-q1c
Answer: b
Explanation: A graph of I B against V BE is drawn. The curve so obtained is known as input characteristics. The collector emitter voltage (V CE ) is kept constant.
2. The input resistance is given by _________
a) ∆V CE /∆I B
b) ∆V BE /∆I B
c) ∆V BE /∆I C
d) ∆V BE /∆I E
Answer: b
Explanation: The ratio of change in base emitter voltage (∆V BE ) to resulting change in base current (∆I B ) at constant collector emitter voltage (V CE ) is defined as input resistance. This is denoted by r i .
3. Which of the following depicts the output characteristics of a CE transistor?
a) electronic-devices-circuits-questions-answers-ce-characteristics-q3
b) electronic-devices-circuits-questions-answers-ce-characteristics-q3a
c) electronic-devices-circuits-questions-answers-ce-characteristics-q3b
d) electronic-devices-circuits-questions-answers-ce-characteristics-q3c
Answer: d
Explanation: A graph of I C against V CE is drawn. The curve so obtained is known as output characteristics. The base current (I B ) is kept constant.
4. The output resistance is given by _________
a) ∆V CE /∆I B
b) ∆V BE /∆I B
c) ∆V BE /∆I C
d) ∆V CE /∆I C
Answer: d
Explanation: The ratio of change in collector emitter voltage (∆V CE ) to resulting change in collector current (∆I C ) at constant base current (I B ) is defined as output resistance. This is denoted by r o .
5. Which of the following cases damage the transistor?
a) when V CE is increased too far
b) when V CE is decreased too far
c) when V BE is increased too far
d) when V BE is decreased too far
Answer: a
Explanation: When V CE is increased too far, collector base junction completely breaks down and due to this avalanche breakdown, collector current increases rapidly. This is not shown in the characteristic. In this case, the transistor is damaged.
6. When the collector junction is reverse biased and emitter junction is forward biased, the operating region of the transistor is called_________
a) inverted region
b) active region
c) cut off region
d) cut in region
Answer: b
Explanation: In the active region, for small values of base current, the effect of collector voltage over collector current is small while for large base currents this effect increases. The shape of characteristic here is same as that of CB transistors.
7. The small amount of current which flows even when base current I B =0 is called_________
a) I BEO
b) I CBO
c) I CEO
d) I C
Answer: c
Explanation: In the cut off region, a small amount of collector current flows even when base current I B is zero. This is called I CEO . Since the main current is also zero, the transistor is said to be cut off.
8. A change in 700mV in base emitter voltage causes a change of 200µA in the base current. Determine the dynamic input resistance.
a) 2kΩ
b) 10kΩ
c) 3kΩ
d) 3.5kΩ
Answer: c
Explanation: r o =∆V BE /∆I B
=700m/200µ=3.5kΩ.
9. The change in collector emitter voltage from 6V to 9V causes increase in collector current from 6mA to 6.3mA. Determine the dynamic output resistance.
a) 20kΩ
b) 10kΩ
c) 50kΩ
d) 60kΩ
Answer: b
Explanation: r o =∆V CE /∆I C
=3/0.3m=10kΩ.
10. Which of the following points locates the quiescent point?
a) (I C , V CB )
b) (I E , V CE )
c) (I E , V CB )
d) (I C , V CE )
Answer: a
Explanation: The quiescent point is best located between the cut off and saturation point. I E = V EE /R E , V CB =V CC -I C R L . It is denoted by ‘Q’.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The CB Characteristics”.
1. The input resistance in a CB transistor is given by _________
a) ∆V CE /∆I B
b) ∆V BE /∆I B
c) ∆V BE /∆I C
d) ∆V EB /∆I E
Answer: d
Explanation: The ratio of change in emitter base voltage (∆V EB ) to resulting change in emitter current (∆I E ) at constant collector base voltage (V CB ) is defined as input resistance. This is denoted by r i .
2. The output resistance of CB transistor is given by _________
a) ∆V CB /∆I C
b) ∆V BE /∆I B
c) ∆V BE /∆I C
d) ∆V EB /∆I E
Answer: a
Explanation: The ratio of change in collector base voltage (∆V CB ) to resulting change in collector current (∆I C ) at constant emitter current (I E )¬ is defined as output resistance. This is denoted by r o .
3. Which one of the following depicts the output characteristics for a CB transistor?
a) electronic-devices-circuits-questions-answers-cb-characteritics-q3
b) electronic-devices-circuits-questions-answers-cb-characteritics-q3a
c) electronic-devices-circuits-questions-answers-cb-characteritics-q3b
d) electronic-devices-circuits-questions-answers-cb-characteritics-q3c
Answer: b
Explanation: A graph of I C against V CB is drawn. The curve so obtained is known as output characteristics. The emitter current (I E ) is kept constant.
4. The input characteristics of a CE transistor is_________
a) electronic-devices-circuits-questions-answers-cb-characteritics-q4
b) electronic-devices-circuits-questions-answers-cb-characteritics-q4a
c) electronic-devices-circuits-questions-answers-cb-characteritics-q4b
d) electronic-devices-circuits-questions-answers-cb-characteritics-q4c
Answer: c
Explanation: A graph of I E against V EB is drawn. The curve so obtained is known as input characteristics. The collector base voltage (V BC ) is kept constant.
5. A transistor is connected in CB configuration. The emitter voltage is changed by 200mV, the emitter by 5mA. During this transition the collector base voltage is kept constant. What is the input dynamic resistance?
a) 30Ω
b) 60Ω
c) 40Ω
d) 50Ω
Answer: c
Explanation: The ratio of change in emitter base voltage (∆V EB ) to resulting change in emitter current (∆I E ) at constant collector base voltage (V CB ) is defined as input resistance. This is denoted by r i .
We know, ∆V EB /∆I E =r i
=200/5=40Ω.
6. When the collector junction is reverse biased and emitter junction is forward biased, the operating region of the transistor is called_________
a) inverted region
b) active region
c) cut off region
d) cut in region
Answer: b
Explanation: In the active region, for small values of base current, the effect of collector voltage over collector current is small while for large base currents this effect increases. The shape of characteristic here is same as that of CB transistors.
7. Which of the following corresponds to the output circuit of a CB transistor?
a) V BE
b) I B
c) V CB
d) V CE
Answer: c
Explanation: Here, the quantity collector to base voltage corresponds to the output circuit of a CB transistor. The complete electrical behaviour of a transistor can be described by stating the relation between these quantities.
8. The input of a CB transistor is given between_________
a) collector and emitter terminals
b) base ad collector terminals
c) ground and emitter terminals
d) emitter and base terminals
Answer: d
Explanation: The name of the CB transistor says that it’s a common based one. The input is given between the emitter and base terminals and the output is taken between collector and base terminals.
9. The current gain of the CB transistor is_________
a) less than or equal to unity
b) equal to unity
c) greater than unity
d) remains same
Answer: a
Explanation: The input current flowing into the emitter terminal must be higher than the base current and collector current to operate the transistor. Therefore the output collector current is less than the input emitter current.
10. The input characteristics of a CB transistor resembles_________
a) Forward biased diode
b) Illuminated photo diode
c) LED
d) Zener diode
Answer: b
Explanation: The input characteristics resemble the illuminated photo diode and the output characteristics resemble the forward biased diode. This transistor has low input impedance and high output impedance.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “DC Load Lines”.
1. Which of the following depicts the DC load line?
a) electronic-devices-circuits-questions-answers-dc-load-lines-q1
b) electronic-devices-circuits-questions-answers-dc-load-lines-q1a
c) electronic-devices-circuits-questions-answers-dc-load-lines-q1b
d) electronic-devices-circuits-questions-answers-dc-load-lines-q1c
Answer: a
Explanation: In transistor circuit analysis, sometimes it is required to know the collector currents for various collector emitter voltages. The one way is to draw its load line. We require the cut off and saturation points.
2. For the circuit shown, find the quiescent point.
electronic-devices-circuits-questions-answers-dc-load-lines-q2
a)
b)
c)
d)
Answer: c
Explanation: We know, I E =V EE /R E =30/10kΩ=3mA
I C =α I E =I E =3mA
V CB =V CC -I C R L =25-15=10V. So, quiescent point is .
3. Which of the following depicts the load line for the circuit shown below?
electronic-devices-circuits-questions-answers-dc-load-lines-q3
a) electronic-devices-circuits-questions-answers-dc-load-lines-q3a
b) electronic-devices-circuits-questions-answers-dc-load-lines-q3b
c) electronic-devices-circuits-questions-answers-dc-load-lines-q3c
d) electronic-devices-circuits-questions-answers-dc-load-lines-q3d
Answer: d
Explanation: We know, I E =V EE /R E =15/5kΩ=3mA
I C =α I E =I E =3mA
V CB =V CC -I C R L =20-15=5V. So, quiescent point is .
4. For the circuit shown, find the quiescent point.
electronic-devices-circuits-questions-answers-dc-load-lines-q4
a)
b)
c)
d)
Answer: c
Explanation: We know, V CE =12V
(I C ) SAT =V CC /R L =12/6K=2mA. I B =10V/0.5M=20µA. I C = βI B =1mA. I
V CE =V CC -I C R L =12-1*6=6V. So, quiescent point is .
5. Which of the following depicts the load line for the given circuit?
electronic-devices-circuits-questions-answers-dc-load-lines-q5
a) electronic-devices-circuits-questions-answers-dc-load-lines-q5a
b) electronic-devices-circuits-questions-answers-dc-load-lines-q5b
c) electronic-devices-circuits-questions-answers-dc-load-lines-q5c
d) electronic-devices-circuits-questions-answers-dc-load-lines-q5d
Answer: d
Explanation: We know, V CE =6V
(I C ) SAT =V CC /R L =10/2K=5mA. I B =10V/0.5M=20µA. I C = βI B =1mA. I
V CE =V CC -I C R L =10-1*2=8V. So, quiescent point is .
6. The DC equivalent circuit for an NPN common base circuit is.
electronic-devices-circuits-questions-answers-dc-load-lines-q6
electronic-devices-circuits-questions-answers-dc-load-lines-q6a
electronic-devices-circuits-questions-answers-dc-load-lines-q6b
Answer: a
Explanation: In the common base circuit, the emitter diode acts like a forward biased ideal diode, while collector diode acts as a current source due to transistor action. Thus an ideal transistor may be regarded as a rectifier diode in the emitter and a current source at collector.
7. The DC equivalent circuit for an NPN common emitter circuit is.
electronic-devices-circuits-questions-answers-dc-load-lines-q7
electronic-devices-circuits-questions-answers-dc-load-lines-q7a
electronic-devices-circuits-questions-answers-dc-load-lines-q7b
Answer: b
Explanation: In the common emitter circuit, the ideal transistor may be regarded as a rectifier diode in the base circuit and a current source in the collector circuit. In the current source, the direction of arrow points in direction of conventional current.
8. What is the other representation of the given PNP transistor connected in common emitter configuration?
electronic-devices-circuits-questions-answers-dc-load-lines-q8
electronic-devices-circuits-questions-answers-dc-load-lines-q8a
electronic-devices-circuits-questions-answers-dc-load-lines-q8b
Answer: d
Explanation: The emitter junction is forward biased with the help of battery V EE by which, negative of the battery is connected to the emitter while positive is connected to base. R E is the emitter resistance. The collector junction is reversed biased.
9. What is the DC characteristic used to prove that the transistor is indeed biased in saturation mode?
a) I C = βI B
b) I C > βI B
c) I C >> βI B
d) I C < βI B
Answer: d
Explanation: When in a transistor is driven into saturation, we use V CE as another linear parameter. In, addition when a transistor is biased in saturation mode, we have I C < βI B . This characteristic used to prove that the transistor is indeed biased in saturation mode.
10. For the circuit shown, find the quiescent point.
electronic-devices-circuits-questions-answers-dc-load-lines-q10
a)
b)
c)
d)
Answer: c
Explanation: We know, I E =V EE /R E =10/5kΩ=2mA
I C =α I E =I E =2mA
V CB =V CC -I C R L =20-10=10V. So, quiescent point is .
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Transistor as a Switch”.
1. In which region a transistor acts as an open switch?
a) cut off region
b) inverted region
c) active region
d) saturated region
Answer: a
Explanation: In this mode, both the junctions are reverse biased. The transistor has practically zero current because the emitter does not emit charge carriers to the base. There is negligibility current due to minority carriers. In this mode the transistor acts as an open switch.
2. In which region a transistor acts as a closed switch?
a) cut off region
b) inverted region
c) active region
d) saturated region
Answer: d
Explanation: In this mode, both the junctions are forward biased. The negative terminal of the battery is connected to the emitter. The collector current becomes independent of base current. In this mode the transistor acts as a closed switch.
3. Which of the following circuits act as a switch?
a) electronic-devices-circuits-questions-answers-transistor-switch-q3
b) electronic-devices-circuits-questions-answers-transistor-switch-q3a
c) electronic-devices-circuits-questions-answers-transistor-switch-q3b
d) electronic-devices-circuits-questions-answers-transistor-switch-q3c
Answer: b
Explanation: This is an inverter, in which the transistor in the circuit is switched between cut off and saturation. The load, for example, can be a motor or a light emitting diode or any other electrical device.
4. The current which is helpful for LED to turn on is_________
a) emitter current
b) base current
c) collector current
d) depends on bias
Answer: c
Explanation: Depending on the type of load, a collector current is induced that would turn on the motor or LED. The transistor in the circuit is switched between cut off and saturation. The load, for example, can be a motor or a light emitting diode or any other electrical device.
5. Which of the following statements is true?
a) Solid state switches are applications for an AC output
b) LED’s can be driven by transistor logics
c) Only NPN transistor can be used as a switch
d) Transistor operates as a switch only in active region
Answer: b
Explanation: Output devices like LED’s only require a few milliamps at logic level DC voltages and can therefore be driven directly by the output of a logic gate. However, high power devices such as motors or lamps require more power than that supplied by an ordinary logic gate so transistor switches are used.
6. The base emitter voltage in a cut off region is_________
a) greater than 0.7V
b) equal to 0.7V
c) less than 0.7V
d) cannot be predicted
Answer: c
Explanation: From the cut off characteristics, the base emitter voltage (V BE ) in a cut off region is less than 0.7V. The cut off region can be considered as ‘off mode’. Here, V BE > 0.7 and I C =0. For a PNP transistor, the emitter potential must be negative with respect to the base.
7. In saturation region, the depletion layer_________
a) increases linearly with carrier concentration
b) decreases linearly with carrier concentration
c) increases by increasing the emitter current
d) decreases by decreasing the emitter voltage drop
Answer: d
Explanation: Here, the transistor will be biased so that maximum amount of base current is applied, resulting in maximum collector current resulting in minimum emitter voltage drop which results in depletion layer as small as possible and maximum current flows through the transistor.
8. The base emitter voltage in a saturation region is_________
a) greater than 0.7V
b) equal to 0.7V
c) less than 0.7V
d) cannot be predicted
Answer: d
Explanation: From the saturation mode characteristics, the transistor acts as a single pole single throw solid state switch. A zero collector current flows. With a positive signal applied to the base of transistor it turns on like a closed switch.
9. The switching of power with a PNP transistor is called_________
a) sourcing current
b) sinking current
c) forward sourcing
d) reverse sinking
Answer: a
Explanation: Sometimes DC current gain of a bipolar transistor is too low to directly switch the load current or voltage, so multiple switching transistors is used. The load is connected to ground and the transistor switches the power to it.
10. The switching of power with a NPN transistor is called_________
a) sourcing current
b) sinking current
c) forward sourcing
d) reverse sinking
Answer: b
Explanation: Sometimes DC current gain of a bipolar transistor is too low to directly switch the load current or voltage, so multiple switching transistors is used. The load is connected to supply and the transistor switches the power to it.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Transistor Switching Times”.
1. The collector current will not reach the steady state value instantaneously because of_________
a) stray capacitances
b) resistances
c) input blocking capacitances
d) coupling capacitance
Answer: a
Explanation: When a pulse is given, the collector current will not reach the steady state value instantaneously because of stray capacitances. The charging and discharging of capacitance makes the current to reach a steady state value after a given time constant.
2. For the BJT, β=∞, V BEon =0.7V V CEsat =0.7V. The switch is initially closed. At t=0, it is opened. At which time the BJT leaves the active region?
electronic-devices-circuits-questions-answers-transistor-switching-times-q2
a) 20ms
b) 50ms
c) 60ms
d) 70ms
Answer: b
Explanation: At t < 0, the BJT is OFF in cut off region. I B =0 as β=∞, so I C =I E . When t > 0, switch opens and BJT is ON. The voltage across capacitor increases. From the input loop, -5-V BE -I+10=0 and gives I=1mA. I C1 =1-0.5=0.5mA. V C1 =0.7+4.3+10=-5V. I C1 =C 1 dV C1 /dt. From this equation, we get t=50ms.
3. The technique used to quickly switch off a transistor is by_________
a) reverse biasing its emitter to collector junction
b) reverse biasing its base to collector junction
c) reverse biasing its base to emitter junction
d) reverse biasing any junction
Answer: c
Explanation: The technique used to quickly switch off a transistor is by reverse biasing its base to collector junction. It is demonstrated in a high voltage switching circuit. The advantage of this circuit is that it is not necessary to have high voltage control signal.
4. The disadvantage of using the method of reverse biasing base emitter junction is_________
a) high voltage control signal
b) low voltage control signal
c) output swing
d) incomplete switching of output
Answer: d
Explanation: This method is used to quickly switch off a transistor is by reverse biasing its base to collector junction. It is demonstrated in a high voltage switching circuit. The disadvantage of using the method of reverse biasing base emitter junction is that the output does not switch completely to GND due to forward voltage drop of the diode.
5. Which of the following circuits helps in the applications of switching times?
a) electronic-devices-circuits-questions-answers-transistor-switching-times-q5a
b) electronic-devices-circuits-questions-answers-transistor-switching-times-q5b
c) electronic-devices-circuits-questions-answers-transistor-switching-times-q5c
d) electronic-devices-circuits-questions-answers-transistor-switching-times-q5d
Answer: b
Explanation: This is an inverter, in which the transistor in the circuit is switched between cut off and saturation. The load, for example, can be a motor or a light emitting diode or any other electrical device.
6. Which of the following helps in reducing the switching time of a transistor?
a) a resistor connected from base to ground
b) a resistor connected from emitter to ground
c) a capacitor connected from base to ground
d) a capacitor connected from emitter to ground
Answer: a
Explanation: Connecting a resistor connected from base of a transistor to ground/negative voltage helps in reducing the switching the switching time of the transistor. When transistor saturate, there is stored charge in the base that must be removed before it turns off.
7. The time taken for a transistor to turn from saturation to cut off is _________
a) inversely proportional to charge carriers
b) directly proportional to charge carriers
c) charging time of the capacitor
d) discharging time of the capacitor
Answer: b
Explanation: When sufficient charge carriers exist, the transistor goes into saturation. When the switch is turned off, in order to go into cut off, the charge carriers in the base region need to leave. The longer it takes to leave, the longer it takes for a transistor to turn from saturation to cut off.
8. The switching of power with a PNP transistor is called _________
a) sourcing current
b) sinking current
c) forward sourcing
d) reverse sinking
Answer: a
Explanation: Sometimes DC current gain of a bipolar transistor is too low to directly switch the load current or voltage, so multiple switching transistors is used. The load is connected to ground and the transistor switches the power to it.
9. The base emitter voltage in a cut off region is _________
a) greater than 0.7V
b) equal to 0.7V
c) less than 0.7V
d) cannot be predicted
Answer: c
Explanation: From the cut off characteristics, the base emitter voltage (V BE ) in a cut off region is less than 0.7V. The cut off region can be considered as ‘off mode’. Here, V BE < 0.7 and I C =0. For a PNP transistor, the emitter potential must be negative with respect to the base.
10. Switching speed of P+ junction depends on _________
a) Mobility of minority carriers in P junction
b) Life time of minority carriers in P junction
c) Mobility of majority carriers in N junction
d) Life time of minority carriers in N junction
Answer: d
Explanation: Switching leads to move holes in P region to N region as minority carriers. Removal of this accumulation determines switching speed. P+ regards to a diode in which the p type is doped excessively.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Operating Point”.
1. Where should be the bias point set in order to make transistor work as an amplifier?
a) Cut off
b) Active
c) Saturation
d) Cut off and Saturation
Answer: b
Explanation: To operate transistor as an amplifier, it requires more current amplification factor and in cut off and saturation, the current amplification is less, therefore active region is better to fix the Q point.
2. Q point can be set to work on active region requires particular conditions. What are they?
a) BE reverse biased and BC forward biased
b) BE reverse biased and BC reverse biased
c) BE forward biased and BC reverse biased
d) BE forward biased and BC forward biased
Answer: c
Explanation: BJT requires the forward voltage nearly equal to 0.7v and the p-junction should be more positive in BE junction and n region should be more positive in BC junction. This will make the current to flow through emitter which is the sum of current through base and emitter.
3. The Q-point of a transistor is made to shift between Active and cut off Region, then how does the transistor behave?
a) Switch
b) Amplifier
c) Inverter
d) Bulb
Answer: a
Explanation: When the Q point lies in cut off, No current flows and hence it acts as a closed switch. When the Q point is shifted to saturation, Current flows through the circuit creating a closed switch. Thus the current flow makes the turn on and off of switch.
4. For a Fixed bias circuit having R C =2.2KΩ, R B =240Ω, V CC =12v and current amplification factor is 100 and the current flowing through the base is 20µA, the value if Collector current in saturation is_____________
a) 5.4mA
b) 3mA
c) 1mA
d) 0A
Answer: a
Explanation: V ce = V CC – I C R C
For saturation, V ce = 0 I Csat = V CC /R C = 5.4mA.
5. The bias point of a transistor occurs when the supply voltage exceeds the breakdown voltage of a transistor.
a) True
b) False
Answer: b
Explanation: Bias point can be set on the basis of DC load line that is Q point can be found out only without applying any input. The DC load line is defined as the line drawn in response of collector current and base emitter voltage when no input is applied.
6. For a fixed bias circuit having R C =4.7KΩ and R B =1KΩ, V CC =10V, and base current at Bias point was found to be 0.2µA, Find β?
a) 100
b) 106
c) 125
d) 0
Answer: b
Explanation: I CQ = V cc /R c = 2.12mA
β=I CQ /I BQ =106.
7. For a Voltage divider bias circuit, having R1=R2=10KΩ, R C =4.7 KΩ, R E =1 KΩ, What is the value of collector current at saturation if V CC =10V?
a) 1A
b) 10mA
c) 0.87mA
d) 1ma
Answer: b
Explanation: By using Thevenin’s law Vth=VccR2/=10V
I Csat =V CC /(R C +R E )=0.87mA.
8. For a Voltage divider circuit having R C =R1=R2=R E =1KΩ, if V CC =20V, find I C when V ce = V CC ?
a) 1mA
b) 2mA
c) 20mA
d) 0
Answer: 0
Explanation: when V ce = V CC ,
V CC = V ce -I C (R C +R E ) => V CC -V ce = 0 = I C .
9. Changes in the temperature will not affect the bias point.
a) True
b) False
Answer: b
Explanation: The temperature changes the β value of the Transistor which wills in turn shifts the Q-point of the Transistor. Once the temperature changes, it will increase the mobility of electrons resulting in a change of system current, hence temperature does affect the transistor parameter.
10. For the voltage divider circuit, if a diode is connected in reverse direction across the base Find the value of β? (V d =1.2v and input is 1V)
a) 0
b) 50
c) 100
d) cannot be determined
Answer: d
Explanation: Since the input voltage is less than the diode voltage, transistor never turns on during the positive half cycle, hence it is difficult to measure the β value. Also during a negative half cycle, the diode will be turned off and hence no current flows, so the value if beta cannot be determined, it can be 0 because the manufacturer will certainly provide the value.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Bias Stability”.
1. What is Stability factor?
a) Ratio of change in collector current to change in a current amplification factor
b) Ratio of change in collector current to change in base current
c) Current amplification factor
d) Ratio of base current to collector current
Answer: a
Explanation: Stability factor is defined as the rate at which collector current changes when Base to emitter voltage changes, keeping base current constant. It can also be defined as the ratio of change in collector current to change in base current when temperature changes occur.
2. The base current for a BJT remains constant at 5mA, the collector current changes from 0.2mA to 0.3 mA and beta was changed from 100 to 110, then calculate the value of S.
a) 0.01m
b) 1m
c) 100m
d) 25m
Answer: a
Explanation: Since the current in the above case, remains constant, therefore stability factor is 0.01 as it is defined as the ratio of change in collector current to change in beta.
S=change in collector current/change in beta=0.1mA/10=0.01m.
3. For a n-p-n transistor, the collector current changed from 0.2mA to 0.22mA resulting a change of base emitter voltage from 0.8v to 0.8005V. What is the value of Stability factor?
a) 0
b) 0.25
c) 0.04
d) 0.333
Answer: c
Explanation: Change in V be = 0.0005V
Change in collector current = 0.02mA
S = 0.02m/0.0005 = 0.04.
4. There are two transistors A and B having ‘S’ as 25 and 250 respectively, on comparing the value of S, we can say B is more stable than A.
a) True
b) False
Answer: b
Explanation: More the value of S, lesser the stability, since A has lesser S value the change in beta does not affect much on the collector current. When S is high, even if I B changes by a small value, the I C current will drastically vary. Hence stability factor must possess lesser value for the proper working of a transistor.
5. What is the value of Stability factor for an ideal transistor?
a) 100
b) 1000
c) infinite
d) 0
Answer: 0
Explanation: For a transistor, the ideal value of S is 0 which interprets that for a change in beta, there should not be changing. In Ideal transistor, the collector current will vary only if either base or emitter current varies or hence for an ideal transistor the value of S is zero.
6. For a fixed bias circuit having I c = 0.3mA and In=0.0003mA, S is______________
a) 100
b) 0
c) 11
d) 111
Answer: c
Explanation: For fixed bias S=1+beta
Beta=I C /I B =10
S=1+10=11.
7. For a fixed bias circuit having R C =2Kohm and V CC =60V, I B =0.25mA and S=101, find V ce .
a) 12V
b) 10V
c) 5V
d) 2.5V
Answer: b
Explanation: S = 1 + beta,
=> 100 = I C /I B => I c = 25mA
V ce = V CC – I c R C
V ce = 10V.
8. For an ideal transistor having a fixed bias configuration, what will be the value of Beta?
a) 0
b) 2
c) -1
d) 1
Answer: c
Explanation: S = 1 + Beta
S = 0
Beta = -1.
9. The temperature changes do not affect the Stability.
a) True
b) False
Answer: b
Explanation: The temperature changes the value of beta which in turn changes the stability of the transition. The temperature changes affect the mobility of the charge carries which results in a change of the current parameters affecting stability.
10. Comparing fixed and collector to base bias which of the following statement is true?
a) Fixed bias is more stable
b) Collector to base bias is more stable
c) Both are the same in terms of stability
d) Depends on the design
Answer: b
Explanation: For fixed bias circuit, S = 1+beta, more the beta, lesser the stability
For collector to base bias S = /(1+beta(R C /R C +R B ))
Hence collector to base bias is more stable.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Collector-to-Base Bias”.
1. The collector current (I C ) that is obtained in a collector to base biased transistor is_________
a) (V CC -V BE )/R B
b) (V CC +V BE )/R B
c) (V CE -V BE )/R B
d) (V CE +V BE )/R B
Answer: a
Explanation: The collector current is analysed by the DC analysis of a transistor. It involves the DC equivalent circuit of a transistor. The base current is first found and the collector current is obtained from the relation, I C =I B β.
2. The collector to emitter voltage (V CE ) is obtained by_________
a) V CC – RC(I C -I B )
b) V CC – RC(I C +I B )
c) V CC + RC(I C +I B )
d) V CC + RC(I C -I B )
Answer: b
Explanation: The collector to emitter voltage is obtained in order to find the operating point of a transistor. It is taken when there is no signal applied to the transistor. The point thus obtained lies in the cut off region when the transistor is used as a switch.
3. What is the DC characteristic used to prove that the transistor is indeed biased in saturation mode?
a) I C = βI B
b) I C > βI B
c) I C >> βI B
d) I C < βI B
Answer: d
Explanation: When in a transistor is driven into saturation, we use V CE as another linear parameter. In, addition when a transistor is biased in saturation mode, we have I C < βI B . This characteristic used to prove that the transistor is indeed biased in saturation mode.
4. The thermal runway is avoided in a collector to base bias because_________
a) of its independence of β
b) of the positive feedback produced by the base resistor
c) of the negative feedback produced by the base resistor
d) of its dependence of β
Answer: c
Explanation: The self destruction of a transistor due to increase temperature is called thermal run away. It is avoided by the negative feedback produced by the base resistor in a collector to base bias. The I C which is responsible for the damage is reduced by decreased output signal.
5. When the temperature is increased, what happens to the collector current after a feedback is given?
a) it remains same
b) it increases
c) it cannot be predicted
d) it decreases
Answer: d
Explanation: Before the feedback is applied, when the temperature is increased, the reverse saturation increases. The collector current also increases. When the feedback is applied, the base current increases with decreasing collector current and the thermal runway too.
6. The demerit of a collector to base bias is_________
a) its need of high resistance values
b) its dependence on β
c) its independence on β
d) the positive feedback produced by the base resistor
Answer: a
Explanation: When the stability factor S=1, the collector resistor value should be very large when compared to the base resistor. So, when RC is large we need to provide large power supply which increases the cost. At the same time, as the base resistor is small we need to provide small power supply.
7. The negative feedback does good for DC signal by_________
a) decreasing the gain
b) increasing the gain
c) stabilising the operating point
d) increasing the stability factor
Answer: c
Explanation: The resistor R B can provide negative feedback for both AC and DC signals. The negative feedback for DC signal is done good as it can provide stable operating point. On the other side, the negative feedback is badly done for AC signal by decreasing the voltage gain.
8. In the circuit, transistor has β =60, V BE =0.7V. Find the collector to emitter voltage drop V CE .
electronic-devices-circuits-questions-answers-collector-base-bias-q8
a) 5V
b) 3V
c) 8V
d) 6V
Answer: d
Explanation: We know, I C =(V CC -V BE )/R B
By putting the values, we have I C =5.9mA. I E =I C /α. So, I E =5.99mA.
V CE = V CC -RC(I C +I B ). We have V CE =6V.
9. In the circuit shown below, β =100 and V BE =0.7V. The Zener diode has a breakdown voltage of 6V. Find the operating point.
electronic-devices-circuits-questions-answers-collector-base-bias-q9
a)
b)
c)
d)
Answer; a
Explanation: We know, by KVL -12+(I C +I B )1K+6+V BE =0
We have I E =5.3. I C = αIE=5.24mA. From another loop, -12+I E I K +V BE =0
We have, V CE =12-5.3m*1000=6.7V. Hence the Q point is .
10. When the β value is large for a given transistor, the I C and V CE values are given by_________
a) (V CC -V BE )/R B , V CC -RCI C
b) (V CC +V BE )/R B , V CC -RC(I C +I B )
c) (V CC +V BE )/R B , V CC +RC(I C +I B )
d) (V CC +V BE )/R B , V CC +RC(I C -I B )
Answer: a
Explanation: The base current I B is zero when β value is large. So, the V CE changes to V CC -R C I C . The collector current I C is changed to (V CC -V BE )/R B from β(V CC -V BE )/R E + R B .
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Emitter Feedback Bias”.
1. For emitter feedback bias, to make I C independent of DC current gain, which of the following condition is required?
a) R C >> R B /dc current gain
b) R E >> R B /dc current gain
c) R B >> R C /dc current gain
d) R e >> R C /dc current gain
Answer: a
Explanation: In order to make the I C stable, the dc current gain has to be maintained in a proper constant value, and the Re value must be very much greater than (R B / current gain). Since the value of R C becomes very high, this results in I C independent of beta.
2. For an emitter feedback bias circuit, the value of V CC = 10V, R B = 10Kohm, R E =1Kohm and R C =2Kohm, if current flowing through collector is 0.5mA, what is the voltage difference between collector and emitter?
a) 8.5V
b) 9V
c) 10V
d) 10.5V
Answer: a
Explanation: V CC = V ce + I C R C + I E R E
V ce = V CC – I C R C – I E R E
V ce = 10-0.5-1
V ce = 8.5V.
3. For an emitter feedback bias circuit, R C is 10Kohm, Re= 5Kohm, R B = 1Kohm, If I C = 1mA, current gain is
_________________
a) 4
b) 2
c) 5
d) 6
Answer: b
Explanation: For emitter feedback bias circuit, current gain=R C /R E
There will be a change in phase of 180 degrees, current gain=10Kohm/5Kohm=2
Hence while designing an amplifier circuit depending on the gain value, the resistors R C and Re can be selected.
4. The feedback helps to maintain a constant gain value.
a) True
b) False
Answer: a
Explanation: From the equation I E = I C + I B , once the I E increases, the base current will reduce because of emitter feedback, resulting in a constant gain. In emitter feedback circuits, the resistance R E is used to provide negative feedback which is given to the base to maintain a constant gain.
5. What will be the temperature changes effects on the emitter feedback circuit?
a) Increases voltage gain
b) Increases current gain
c) Does not affect the gain
d) Decreases both current and voltage gain
Answer: c
Explanation: Since the temperature changes I E value, these results in the increase of feedback through R_E, which in turn reduces Ib, resulting in constant gain, even though temperature effects beta and other transistor parameters, due to the feedback these effects are neutralized.
6. For an emitter feedback bias Circuit having, R E =1Kohm, R C = 4.7Kohm, I B =0.005mA, I E = 1mA, V CC = 12V and V ce = 5V. Find the value of beta.
a) 254
b) 100
c) 1000
d) 500
Answer: a
Explanation: V CC =V ce +I C R C + I E R E
I C =1.27mA
Beta=1.27/0.005=254.
7. Which of the following statement is the main disadvantage of emitter feedback bias?
a) Reduces the gain
b) Positive feedback
c) Design is difficult
d) High output impedance
Answer: a
Explanation: Due to the negative feedback, the voltage gain will reduce drastically compared to other biasing techniques. Hence there will be a huge amount of power loss in the form of heat dissipated across emitter and collector. Setting Q- Point is also difficult for emitter feedback circuit.
8. Among the following statements which one is true according to feedback bias?
a) Fixed bias is more stable than emitter feedback bias
b) Fixed bias produces more feedback than emitter feedback bias
c) Emitter feedback bias is more stable than the fixed bias
d) Fixed and emitter bias have same stability and feedback
Answer: c
Explanation: Since S is having a linear relationship with beta in fixed bias, it is less stable, but in emitter feedback bias, S is not exactly linear with a beta. In fixed bias, even a small change in I Co changes the I C value very much, but due to negative feedback in emitter bias, I Co does not change the I C value maintaining its stability.
9. In order to make an amplifier which of the following biasing technique is used more?
a) Fixed bias
b) Self bias
c) Collector to base bias
d) Emitter feedback bias
Answer: d
Explanation: Since emitter feedback bias circuit, the output swing is very much stable and the design shows more stability to changes in temperature. Voltage divider circuit is the most used among all of the biasing technique because of its gain stability and impedance parameters.
10. What will happen if a capacitor is connected in parallel with RE in the amplifier design which uses emitter feedback bias circuit?
a) No changes
b) Gain value increases
c) Feedback increases
d) Gain value remains the same but feedback doubles
Answer: b
Explanation: Since the capacitor acts as a short circuit during high frequency, there will be no feedback and hence gain increases. All of the current will flow to ground through a capacitor which acts as a short circuit.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Collector-Emitter Feedback Bias”.
1. Why do we need collector emitter feedback bias?
a) To provide a non – linear output
b) To maintain transistor in active region
c) To maintain transistor in saturation region
d) To maintain transistor in cut – off region
Answer: b
Explanation: We use a collector emitter feedback to collect negative feedback and to maintain the transistor in an active region. To maintain stability in active region, the DC base biased voltage is resultant from the collector voltage V C .
2. What is the function of R E in the collector emitter feedback circuit?
a) To improve stability and decrease positive feedback
b) To improve stability and increase positive feedback
c) To improve stability and decrease negative feedback
d) To improve stability and increase negative feedback
Answer: d
Explanation: In a collector feedback circuit, the emitter resistor provides additional stability along with increasing the negative feedback sent to the collector. The addition of the emitter resistance enables the transistor’s emitter to no longer be grounded to zero – volt potential.
3. In the circuit given below, assume V CC = 5V, V BE = 0.7V, R E = 10kΩ, R B = 20kΩ and β = 50. How much is the current I E ?
a) 0.561mA
b) 0.335mA
c) 0.413mA
d) 0.513mA
Answer: c
Explanation: Given values are V CC = 5V, V BE = 0.7V, R E = 10kΩ, R B = 20kΩ and β = 50.
To find the value of I E , we substitute the values in the below equation.
I E = (V CC – V BE ) / (R E + (R B / β) )
I E = / )
I E = 4.3 / = 0.413mA.
4. What happens if collector current increases in a collector emitter feedback circuit?
a) Emitter voltage increases therefore base voltage increases
b) Emitter voltage decreases therefore base voltage decreases
c) Emitter voltage increases therefore base voltage decreases
d) Emitter voltage decreases therefore base voltage increases
Answer: a
Explanation: If the collector current V C is increased, the corresponding emitter current also increases. Which in turn causes the voltage across R E to increase. This in turn causes a proportional rise in the base voltage since V B = V E + V BE .
5. Why is collector emitter feedback better for linear circuits?
a) Independent of β
b) Dependent on β
c) Highly predictable
d) Not stable
Answer: b
Explanation: Collector emitter feedback is better for linear circuits as compared to self – bias circuits as it is dependent on β. Voltage divider bias circuits are highly predictable whereas self – bias circuits are independent of β. Therefore, for a collector emitter feedback bias linear circuits are preferred.
6. How does emitter resistor R E provides stability?
a) Consumes less power
b) Has an easier circuit design
c) Automatically biases the circuit
d) It does not provide stability
Answer: c
Explanation: In a collector emitter feedback, the emitter resistor provides stability by automatically biasing the circuit using negative feedback. The negative feedback negates any change due to the collector current with an opposing change provided by the base bias voltage and thus helps maintain circuit stability.
7. In the circuit given below, assume V CC = 12V, V BE = 0.7V, R B = 330kΩ, R C = 3.3kΩ, R E = 2.7kΩ and β = 50. What is the base current I B ?
a) 16.432µA
b) 17.856µA
c) 20.542µA
d) 17.936µA
Answer: d
Explanation: Given values are V CC = 12V, V BE = 0.7V, R B = 330kΩ, R C = 3.3kΩ, R E = 2.7kΩ and β = 50.
To find the value of I B , we substitute the values in the below equation.
I B = (V CC – V BE ) / (R B + β × (R C + R E ))
I B = ) = 11.3 / = 11.3 / 630 = 17.936µA.
8. How do you calculate the value of V CE ?
a) V CE = V CC + V C (R C + R E )
b) V CE = V CC – V C (R C + R E )
c) V CE = V CC – V C (R B + R E )
d) V CE = V CC + V C (R B + R E )
Answer: b
Explanation: The value of V CE can be calculated using this equation: V CE = V CC – V C (R C + R E ) . It is the voltage between the collector and emitter terminal of the transistor and is measured as the output of the transistor.
9. How does the negative feedback help a collector emitter feedback circuit?
a) Helps make it more predictable
b) Provides opposing change in base voltage
c) Helps make it more predictable, provides opposing change in base voltage
d) It doesn’t affect
Answer: c
Explanation: The negative feedback in a collector emitter feedback circuit provides a negative feedback which in turn helps make the circuit more predictable as it provides opposing change in the base voltage which cancels out any change in the collector current.
10. What are the disadvantages of collector emitter feedback bias circuits?
a) Requires few resistors
b) Provides a lot of stability
c) Provides negative feedback
d) Provides positive feedback
Answer: c
Explanation: A collector emitter bias circuit provides negative feedback as well as requires multiples resistors for a small change. The negative feedback limits the frequency range it will work in. Higher frequencies will provide poor performance. It also requires a greater number of resistors just to provide a stability against a small parameter.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Self-Bias”.
1. The collector current (I C ) that is obtained in a self biased transistor is_________
a) (V TH – V BE )/R E
b) (V TH + V BE )/R E
c) (V TH – V BE )/R E
d) (V TH + V BE )/R E
Answer: a
Explanation: The collector current is analysed by the DC analysis of a transistor. It involves the DC equivalent circuit of a transistor. The base current is first found and the collector current is obtained from the relation, I C =I B β.
2. The collector to emitter voltage (V CE ) is obtained by_________
a) V CC – R C I C +RBI B
b) V CC – R C I C -R E I E
c) V CC + R C I C
d) V CC + R C I B
Answer: b
Explanation: The collector to emitter voltage is obtained in order to find the operating point of a transistor. It is taken when there is no signal applied to the transistor. The point thus obtained lies in the cut off region when the transistor is used as a switch.
3. The thermal runway is avoided in a self bias because_________
a) of its independence of β
b) of the positive feedback produced by the emitter resistor
c) of the negative feedback produced by the emitter resistor
d) of its dependence of β
Answer: c
Explanation: The self destruction of a transistor due to increase temperature is called thermal run away. It is avoided by the negative feedback produced by the emitter resistor in a self bias. The I C which is responsible for the damage is reduced by decreased output signal.
4. When the temperature is increased, what happens to the collector current after a feedback is given?
a) it remains same
b) it increases
c) it cannot be predicted
d) it decreases
Answer: d
Explanation: Before the feedback is applied, when the temperature is increased, the reverse saturation increases. The collector current also increases. When the feedback is applied, the drop across the emitter resistor increases with decreasing collector current and the thermal runway too.
5. What is the Thevenin’s voltage (V TH ) in a self bias shown below?
electronic-devices-circuits-questions-answers-self-bias-q5
a) V CC R 2 /R 1 +R 2
b) V CC R 1 /R 1 +R 2
c) V CC R 2 /R 1 -R 2
d) V CC R 2 /R 1 -R 2
Answer: a
Explanation: The base current cannot be obtained directly from the KVL or KCL applications. The V CC and V BE cannot come under a single equation. So, the circuit is changed with a Thevenin’s voltage (V TH ) and Thevenin’s resistance.
6. What is the Thevenin’s resistance (R TH ) in a self bias shown below?
electronic-devices-circuits-questions-answers-self-bias-q5
a) R 1 R 2 /R 1 +R 2
b) R 2 /R 1 +R 2
c) R 1 R 2 /R 1 -R 2
d) R 1 /R 1 -R 2
Answer: a
Explanation: The base current cannot be obtained directly from the KVL or KCL applications. A potential divider network is formed by R 1 and R 2 .The V CC and V BE cannot come under a single equation. So, the circuit is changed with a Thevenin’s resistance.
7. The stability factor for a self biased transistor is_________
a) 1 – R TH /R E
b) 1 + R TH /R E
c) 1 + R E /R TH
d) 1 – R E /R TH
Answer: b
Explanation: The stability of the circuit is inversely proportional to the stability factor. The emitter resistor is very large when compared to the Thevenin’s resistance. When β is not that large, then S=( R TH + R E )/ R E + R TH .
8. In the circuit, the transistor has a large β value (V BE =0.7V). Find the current through RC.
electronic-devices-circuits-questions-answers-self-bias-q8
a) 0.5mA
b) 2mA
c) 1mA
d) 1.6mA
Answer: c
Explanation: We know, I C =V TH -V BE /R E
=9*3/9=3V. I C =3-0.7/2.3=1mA.
9. A silicon NPN transistor is used and it has a large value of β. Find the required value of R 2 when I C =1mA.
electronic-devices-circuits-questions-answers-self-bias-q9
a) 10kΩ
b) 20kΩ
c) 30kΩ
d) 40kΩ
Answer: d
Explanation: For silicon, V BE =0.8V, V CE =0.2V. I C =V TH -V BE /R E . By pitting the values, we have V TH =1.3V. R 2 can be found from, V CC R 2 /R 1 +R 2 . We get R 2 =40KΩ.
10. The value of αac for all practical purposes, for commercial transistors range from_________
a) 0.5 to 0.6
b) 0.7 to 0.77
c) 0.8 to 0.88
d) 0.9 to 0.99
Answer: d
Explanation: For all practical purposes, αac=αdc=α and practical values in commercial transistors range from 0.9-0.99. It is the measure of the quality of a transistor. Higher is the value of α, better is the transistor in the sense that collector current approaches the emitter current.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Stabilization against Variations in VBE and Beta for Self Bias Circuit”.
1. What is the function of a bias circuit?
a) To simplify the circuit
b) To provide a non – linear output
c) To optimize the power
d) To provide steady current or voltage
Answer: d
Explanation: The Q – point of a device is the direct current or voltage of a device when no input is applied. The bias circuit is a part of the device with provides the steady current or voltage. It is designed by determining the necessary voltage and current levels across each resistor.
2. How many main types of bias circuits are there in bipolar transistors?
a) 3
b) 4
c) 5
d) 6
Answer: c
Explanation: There are five main type of biasing circuits. They are Fixed Bias, Collector – to – Base Bias, Voltage Divider Bias, Fixed Bias with Emitter Resistor and Emitter Bias.
3. Why do we require R E for a good stable bias circuit?
a) To obtain a current I E sensitive to β and V BE
b) To obtain a current I B sensitive to β and V BE
c) To obtain a current I E insensitive to β and V BE
d) To obtain a current I B insensitive to β and V BE
Answer: c
Explanation: The aim of having a bias circuit is to maintain a stable current or voltage, regardless of other parameters changing. Hence, we need R E to get a stable I E current, without R E the value of I E changes drastically with a small change in β.
4. In the circuit given below, assume V CC = 4V, V BB = 5V, V BE = 0.7V, R E = 3.3kΩ, R C = 10kΩ, R B = 40kΩ and β = 100. What is the current I E ?
a) 1.388mA
b) 0.012mA
c) 10.47mA
d) 1.165mA
Answer: a
Explanation: Given the values V CC = 4V, V BB = 5V, V BE = 0.7V, R E = 3.3kΩ, R C = 10kΩ, R B = 40kΩ and β = 100. To find the value of I E , we substitute the values in this equation:
I E = (V BB – V BE ) / (R E + (R B / β + 1))
I E = / )
I E = 4.3 / = 1.388mA.
5. Why is it essential to stabilize the operating point of a circuit?
a) Thermal runaway, power efficiency, individual variations
b) Temperature dependency of I B , individual variations, power efficiency
c) Thermal runaway, Temperature dependency of I C , individual variations
d) Temperature dependency of I C , thermal runaway, cost efficiency
Answer: c
Explanation: It is essential to stabilize the operating point of a circuit to prevent thermal runaway, temperature dependency of I C and individual variations. Thermal runaway occurs because of the self – destruction of an unstable transistor. Temperature dependency on I C causes major variations in I E . Individual variations of parameters such as β and V BE cause huge fluctuations in I E .
6. In an ideal stable self – bias circuit, what should be the current I B with respect to I C ?
a) I B should be 20% of I C
b) I B should be 5% of I C
c) I B should be 10% of I C
d) I B should be 50% of I C
Answer: c
Explanation: Ideally, the resistor values are designed in a way that the voltage drops across R E is approximately 10% of V cc and I B should be 10% of I c . Hence, stable self – bias circuits work best at low power supply voltages.
7. What values of V BE and β would provide maximum stability?
a) V BE = 5V, β = 50
b) V BE = 4V, β = 100
c) V BE = 4.5V, β = 50
d) It is irrelevant
Answer: d
Explanation: In a stable bias circuit, the variations in the value of V BE and β do not affect the stability of the system. The current through the emitter remains unchanged.
8. What is the stability factor if R E = 8kΩ and R TH = 11kΩ?
a) 1.335
b) 2.375
c) 1.727
d) 0.272
Answer: b
Explanation: Stability factor is calculated by the following equation. On plugging in the given values R E = 8kΩ and R TH = 11kΩ, we get:
S = 1 + R TH / R E
S = 1 + 11 / 8 = 2.375.
9. Collector base feedback and emitter feedback combined together provide stability for self – bias circuits.
a) True
b) False
Answer: a
Explanation: Collector base feedback and emitter feedback combined provide stability for self – bias circuits. This is because the emitter base junction is forward biased due to the voltage drop across R E .
10. Where does degeneration take place in a self – bias circuit?
a) Across R B
b) Across R E
c) No degeneration occurs
d) Across R C
Answer: b
Explanation: Degeneration or negative feedback occurs across R E which in turn stabilizes the fluctuations of current I E due to temperature changes and variations in V BE and β.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Bias Compensation”.
1. The compensation techniques are used to_________
a) increase stability
b) increase the voltage gain
c) improve negative feedback
d) decrease voltage gain
Answer: b
Explanation: Usually, the negative feedback is used to produce a stable operating point. But it reduces the voltage gain of the circuit. This sometimes is intolerable and should be avoided in some applications. So, the biasing techniques are used.
2. Compensation techniques refer to the use of_________
a) diodes
b) capacitors
c) resistors
d) transformers
Answer: a
Explanation: Compensation techniques refer to the use of temperature sensitive devices such as thermistors, diodes, transistors, sensistors etc to compensate variation in currents. Sometimes for excellent bias and thermal stabilization, both stabilization and compensation techniques are used.
3. In a silicon transistor, which of the following change significantly to the change in I C ?
a) V CE
b) I B
c) V BE
b) I E
Answer: c
Explanation: For germanium transistor, changes in I CO with temperature contribute more serious problem than for silicon transistor. On the other hand, in a silicon transistor, the changes of V BE with temperature possesses significantly to the changes in I C .
4. What is the compensation element used for variation in V BE and I CO ?
a) diodes
b) capacitors
c) resistors
d) transformers
Answer: a
Explanation: A diode is used as the compensation element used variation in V BE and I CO . The diode used is of the same material and type as that of transistor. Hence, the voltage across the diode has same temperature coefficient as V BE of the transistor.
5. The expression for I C in the compensation for instability due to I CO variation_________
a) βI+βI O +βI CO
b) βI+βI O
c) βI O +βI CO
d) βI+βI CO
Answer: a
Explanation: In this method, diode is used for the compensation in variation of I CO . The diode used is of the same material and type as that of transistor. Hence, the reverse saturation current I O of the diode will increase with temperature at the same rate as the transistor collector saturation current I CO .
6. Which of the following has a negative temperature coefficient of resistance?
a) sensistor
b) diode
c) thermistor
d) capacitor
Answer: c
Explanation: The thermistor has a negative temperature coefficient of resistance. It means, its resistance decreases exponentially with increasing T. The thermistor RT is used to minimize the increase in collector current.
7. Which of the following has a negative temperature coefficient of resistance?
a) capacitor
b) diode
c) thermistor
d) sensistor
Answer: d
Explanation: The sensistor has a positive temperature coefficient of resistance. It is a temperature sensitive resistor. It is a heavily doped semiconductor. When voltage is decreased, the net forward emitter voltage decreases. As a result the collector current decreases.
8. Increase in collector emitter voltage from 5V to 8V causes increase in collector current from 5mA to 5.3mA. Determine the dynamic output resistance.
a) 20kΩ
b) 10kΩ
c) 50kΩ
d) 60kΩ
Answer: b
Explanation: r o =∆V CE /∆I C
=3/0.3m=10kΩ.
9. The output resistance of CB transistor is given by _________
a) ∆V CB /∆I C
b) ∆V BE /∆I B
c) ∆V BE /∆I C
d) ∆V EB /∆I E
Answer: a
Explanation: The ratio of change in collector base voltage (∆V CB ) to resulting change in collector current (∆I C ) at constant emitter current (I E ) is defined as output resistance. This is denoted by r o .
10. The negative sign in the formula of amplification factor indicates_________
a) that I E flows into transistor while I C flows out it
b) that I C flows into transistor while I E flows out it
c) that I B flows into transistor while I C flows out it
d) that I C flows into transistor while I B flows out it
Answer: a
Explanation: When no signal is applied, the ratio of collector current to emitter current is called dc alpha, αdc of a transistor. α dc =-I C /I E . It is the measure of the quality of a transistor. Higher is the value of α, better is the transistor in the sense that collector current approaches the emitter current.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Thermistor and Sensistor Compensation”.
1. Which type of temperature dependent resistor exhibits a positive temperature coefficient of resistivity?
a) Thermistor
b) Sensistor
c) Varistor
d) Photoresistor
Answer: b
Explanation: Sensistor is a temperature dependent device whose resistance is dependent upon temperature. It has a direct variation with respect to change in temperature & hence usually appropriate for bias compensation.
Its functionality is accurately opposite to that of thermistor which exhibits negative temperature coefficient of resistivity.
As the resistance of sensistor increases due to increase in temperature, it is also supposed to have the property of positive temperature coefficient of resistivity.
2. The compensation technique uses __________
a) transformers
b) inductors
c) diodes
d) capacitors
Answer: c
Explanation: Compensation techniques require the use of temperature sensitive devices such as thermistors, diodes, transistors, sensistors etc to compensate variation in currents. However, for excellent bias and thermal stabilization, both stabilization and compensation techniques are used.
3. Which of the following has a negative temperature coefficient of resistance?
a) Thtermistor
b) capacitor
c) sensistor
d) diode
Answer: a
Explanation: The Thermistor decreases exponentially with respect to T. The Thermistor RT is used to minimize the increase in collector current. Thus it has a negative temperature coefficient.
4. Which process plays a crucial role in devising the independency of operating point over the variations in temperature or transistor parameters?
a) Bias stabilization
b) Bias compensation
c) Bias stabilization & compensation
d) NO process
Answer: a
Explanation: Bias stabilization is a process which makes the operating point independent of change in temperature or any change in transistor parameters.
As Q-point exhibits its independent nature over the numerous variations, it plays a significant role in providing the stability to greater level especially for self-bias, fixed-bias and collector to base bias circuits.
Since the stability factor of circuit gets affected due to several variations, bias stabilization has a provision of maintaining the Q-point condition irrespective of the changes in temperature or transistor parameters and thereby ensuring the confined level of bias stability to circuits.
5. On which factor/s do/does the values of thermal resistance depend?
a) Size of transistor
b) Type of cooling system
c) Transistor size & cooling system type
d) Number of holes
Answer: c
Explanation: The value of thermal resistance usually depends on the size of transistor, type of heat transfer mechanism and the type of cooling systems . It also depends on the thermal connectivity of device to a heat sink or metal chassis. Thermal resistance of a power transistor is always less than that of logic level transistor. The value of thermal resistance should be essentially small so as to allow the smooth flow of heat from the junction of the power transistor to the surrounding.
6. Which among the below mentioned parameters of transistors is/are likely to get affected or exhibit/s variations due to an increase in temperature?
a) Base-to-Emitter voltage (V BE )
b) Current Gain (β dc )
c) Base-to-Emitter voltage & Current Gain
d) Forward resistance
Answer: c
Explanation: In a transistor, the junction temperature mainly depends on the quantity of current passing through it. As the temperature increases, various parameters of transistors exhibit variations. These parameters include; base-to-emitter voltage (V BE ), current gain (β dc ) & reverse saturation current (I ICBO ).
Due to an increase in temperature, base-to-emitter voltage (V BE ) decreases and eventually tends to change the Q-point. Since the current gain is a function of collector current (I c ), variation in current gain also ultimately varies collector current.
7. Generally, the resistance of the thermistor decreases _______
a) Exponentially with an increase in temperature
b) Linearly with an increase in temperature
c) Linearly with the decrease in temperature
d) Exponentially with the decrease in temperature
Answer: a
Explanation: Thermistor is widely applicable for several compensation techniques. It exhibits variation in its resistance with respect to change in temperature. As the resistance of the thermistor decreases with an increase in temperature, this property of thermistor is also regarded as negative temperature coefficient of resistivity.
8. What is /are the purpose/s of adopting stabilization and compensation techniques?
a) To provide maximum bias
b) To provide thermal stabilization
c) To provide maximum bias & thermal stabilization
d) To provide minimum bias
Answer: c
Explanation: In some of the negative feedback circuits, the amplification level of AC signals gets ablated rapidly. Since it becomes complicated for the circuits to abide the loss of signals, it is essential to condense the drift in an operating point by means of compensation and stabilization. Compensation techniques comprise diode compensation, bias compensation using thermistor, sensistor & so on. They play a major role in given that maximum bias in addition to the thermal stabilization to the circuits.
9. Thermistors provide precise temperature measurements.
a) True
b) False
Answer: a
Explanation: Thermistors can be used for precise temperature measurements. It controls the temperature and the compensation due to a large variation in resistance with temperature.
10. A Thermistor has low resistance.
a) True
b) False
Answer: b
Explanation: Thermistors generally have an extremely high value of resistance. Cables with a shield are required for usage to minimize the interference.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Thermal Runaway”.
1. Thermal runaway is_________
a) an uncontrolled positive feedback
b) a controlled positive feedback
c) an uncontrolled negative feedback
d) a controlled negative feedback
Answer: a
Explanation: Thermal runaway is a self destruction process in which an increase in temperature creates such a condition which in turn increases the temperature again. This uncontrolled rise in temperature causes the component to get damaged.
2. The thermal runway is avoided in a self bias because_________
a) of its independence on β
b) of the positive feedback produced by the emitter resistor
c) of the negative feedback produced by the emitter resistor
d) of its dependence on β
Answer: c
Explanation: The self destruction of a transistor due to increase temperature is called thermal run away. It is avoided by the negative feedback produced by the emitter resistor in a self bias. The IC which is responsible for the damage is reduced by decreased output signal.
3. When the temperature is increased, what happens to the collector current after a feedback is given?
a) it remains same
b) it increases
c) it cannot be predicted
d) it decreases
Answer: d
Explanation: Before the feedback is applied, when the temperature is increased, the reverse saturation increases. The collector current also increases. When the feedback is applied, the drop across the emitter resistor increases with decreasing collector current and the thermal runway too.
4. The thermal runway is avoided in a collector to base bias because_________
a) of its independence on β
b) of the positive feedback produced by the base resistor
c) of the negative feedback produced by the base resistor
d) of its dependence on β
Answer: c
Explanation: The self destruction of a transistor due to increase temperature is called thermal run away. It is avoided by the negative feedback produced by the base resistor in a collector to base bias. The I C which is responsible for the damage is reduced by decreased output signal.
5. When the temperature is increased, what happens to the collector current after a feedback is given?
a) it remains same
b) it increases
c) it cannot be predicted
d) it decreases
Answer: d
Explanation: Before the feedback is applied, when the temperature is increased, the reverse saturation increases. The collector current also increases. When the feedback is applied, the base current increases with decreasing collector current and the thermal runway too.
6. Discrete transistors T1 and T2 having maximum collector current rating of 0.75A are connected in parallel as shown in the figure. This combination is treated as a single transistor to carry a single current of 1A, when biased with a self bias circuit. When the circuit is switched ON, T1 had draws 0.55A and T2 draws 0.45A. If the supply is kept ON continuously, it is very likely that_________
electronic-devices-circuits-questions-answers-thermal-runaway-q6
a) both T1 and T2 get damaged
b) both T1 and T2 will be safe
c) only T1 gets damaged
d) only T2 gets damaged
Answer: c
Explanation: The T1 transistor is having more power dissipation as it is drawing 0.55A. When power dissipation increases, the temperature increases and this leads to the ultimate further increase in the current drawn by T1. The current drawn by T2 will be reduced as the sum of currents drawn by T1 and T2 should be constant.
7. When the collector current is increased in a transistor_________
a) the reverse current is increased
b) the temperature is increased
c) collisions of electrons decrease
d) the emitter does not emit electrons
Answer: b
Explanation: As the collector current is increased, the emitter releases more number of electrons. This causes more collisions of electrons at collector. This happens in a cycle and produces such a condition in which temperature is further more increased.
8. Which of the following are true?
a) T J – T A = θP d
b) T J – T A = θ/P d
c) T J – T A = θ+P d
d) T J – T A = θ-P d
Answer: a
Explanation: The T J is called as junction temperature which varies and T A is called as the ambient temperature which is fixed. The difference between these temperatures is directly proportional to the power dissipation. Here, θ is called as thermal resistance which is proportionality constant.
9. When the power dissipation increases in a transistor, the thermal resistance_________
a) increases
b) cannot be predicted
c) decreases
d) remains same
Answer: c
Explanation: The power dissipation is directly proportional to thermal resistance. We have, T J – T A = θP d in which we can observe θ ∝ 1/P d . So, a device with low power dissipation has high thermal resistance.
10. Which of the following biasing techniques are prone to thermal runaway?
a) self bias
b) collector to base bias
c) fixed bias
d) the biasing technique is identified by temperature effect
Answer: c
Explanation: The collector current of a fixed bias transistor is I C = β(V CC -V BE )/R B . When the temperature is increased, the reverse saturation increases. The collector current also increases. This in turn increases the current again which leads to damage of transistor.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Thermal Stability”.
1. For a given transistor, the thermal resistance is 8°C/W and for the ambient temperature T A is 27°C. If the transistor dissipates 3W of power, calculate the junction temperature (T J ).
a) 51°C
b) 27°C
c) 67°C
d) 77°C
Answer: a
Explanation: We know, T J -T A =HP D
T J =T A +HP D =27+8*3=51°C.
2. Which of the following are true?
a) T J -T A =θp d
b) T J -T A =θ/p d
c) T J -T A =θ+p d
d) T J -T A =θ-p d
Answer: a
Explanation: The T J ¬ is called as junction temperature which varies and T A is called as the ambient temperature which is fixed. The difference between these temperatures is directly proportional to the power dissipation. Here, θ is called as thermal resistance which is proportionality constant.
3. A silicon power transistor is operated with a heat sink H S -A=1.5°C/W. The transistor rated at 150W has H J -C=0.5°C/W and the mounting insulation has HC-S=0.6°C/W. What maximum power can be dissipated if the ambient temperature is 40°C and (T J )MAX=200°C?
a) 70.6W
b) 61.5W
c) 37.8W
d) 56.9W
Answer: b
Explanation: P D =(T J -T A )/ H J-C +H C-S +H S-A
=200-40/0.5+0.6+1.5=61.5W.
4. The total thermal resistance of a power transistor and heat sink is 20°C/W. The ambient temperature is 25°C and (T J )MAX=200°C. If V CE =4V, find the maximum collector current that the transistor can carry without destruction.
a) 3.67A
b) 7.56A
c) 2.19A
d) 4.16A
Answer: c
Explanation: P D =(T J -T A )/ H
=200-25/20=8.75W.
Now, V CE I C = 8.75/4=2.19A.
5. The total thermal resistance of a power transistor and heat sink is 20°C/W. The ambient temperature is 25°C and (T J )MAX=200°C. If V CE =4V, find the maximum collector current that the transistor can carry without destruction. What will be the allowed value of collector current if ambient temperature rises to 75°C?
a) 3.67A
b) 7.56A
c) 2.19A
d) 1.56A
Answer: d
Explanation: P D =(T J -T A )/ H
=200-75/20=6.25W.
Now, I C = 6.25/4=1.56A.
6. Which of the following is true?
a) H C-A = H J-C – H J-A
b) H C-A = H J-C + H J-A
c) H J-A = H J-C – H C-A
d) H J-A = H J-C + H C-A
Answer: d
Explanation: H J-C is thermal resistance between junction and case and H C-A is thermal resistance between case and ambient. The circuit designer has no control over H J-C . So, a proper approach to dissipate heat from case to ambient is through heat sink.
7. The condition to be satisfied to prevent thermal runaway?
a) ∂P C /∂T J > 1/Q
b) ∂P C /∂T J < 1/Q
c) ∂P C /∂T J > 1/Q
d) ∂P C /∂T J < 1/Q
Answer: b
Explanation: P C is the power dissipated at the collector junction. T J is junction temperature which varies. The difference between these temperatures is directly proportional to the power dissipation. Here, Q is called as thermal resistance which is proportionality constant.
8. Thermal stability can be obtained by_________
a) shifting operating point
b) increasing power supply
c) heat sink
d) decreasing current at collector
Answer: c
Explanation: As power transistors handle large currents, they always heat up during operation. Generally, power transistors are mounted in large metal case to provide a large area from which the heat generated by the device radiates.
9. Thermal stability is dependent on thermal runaway which is_________
a) an uncontrolled positive feedback
b) a controlled positive feedback
c) an uncontrolled negative feedback
d) a controlled negative feedback
Answer: a
Explanation: Thermal runaway is a self destruction process in which an increase in temperature creates such a condition which in turn increases the temperature again. This uncontrolled rise in temperature causes the component to get damaged.
10. Which of the following biasing techniques are affected by thermal runaway?
a) self bias
b) collector to base bias
c) fixed bias
d) the biasing technique is identified by temperature effect
Answer: c
Explanation: The collector current of a fixed bias transistor is IC= β/RB. When the temperature is increased, the reverse saturation increases. The collector current also increases. This in turn increases the current again which leads to damage of transistor.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Signals”.
1. A cosine wave voltage signal has a 10V RMS value and 60Hz frequency. Also at time, t=0, the value of the voltage signal is equal to its RMS value. Which of the following is the correct mathematical representation of the voltage signal?
a) 10 cos
b) 10 cos
c) 14.14 cos
d) 14.14 cos
Answer: d
Explanation: Only equation 14.14 cos satisfies all the given parameters.
2. Which of the following is a characteristic of digital signal?
a) It takes quantized value
b) Its waveform is a continuous function
c) The maximum number of signals that can be produced by N bits is 2 N-1
d) There is no loss of value after converting an analog signal to digital signal
Answer: a
Explanation: Digital signal is non continuous and has discrete sets of possible value which it can take.
3. Consider an N-bits ADC whose analog input varies from 0 to Vmax, then which of the following is not true?
a) The least significant bit correspond to a change of V max /2 N -1 in the analog signal
b) The resolution of the ADC is V max /2 N -1
c) The maximum error in the conversion is V max /2(2 N -1)
d) None of the mentioned
Answer: d
Explanation: None of the statements are true.
4. In compact disc audio technology, the signal is sampled at 44kHz. Each sample is represented by 16bits. What is the speed of the system in bits/second?
a) 1.34 bits/second
b) 2,750 bits/second
c) 704,000 bits/second
d) 1,441,792,000 bits/second
Answer: c
Explanation: 44000 X 16 = 704000 bits/s
5. An electrical signal can be represented in either Thevenin form or Norton From. However, Thevenin representation is preferred when
a) The load resistance is very large as compared to the source resistance
b) The load resistance is very low compared to the source resistance
c) There is no preferred case.
d) Both of the cases mentioned are the preferred case
Answer: a
Explanation: a is preferred in Thevenin’s case and b in Norton’s case
6. An electrical signal can be represented in either Thevenin form or Norton From. However, Norton representation is preferred when
a) The load resistance is very large as compared to the source resistance
b) The load resistance is very low as compared to the source resistance
c) There is no preferred case
d) Both of the cases mentioned are the preferred case
Answer: b
Explanation: a is preferred in Thevenin’s case and b in Norton’s case
electronic-devices-circuits-questions-answers-signals-q7
Consider the DAC shown below. Where b 1 , b 2 ….b n can be either 0 or 1.
7. What will be the value of the output current io, if the digital signal is in the form of b1b2b3…bn
a) Vref/2R (b1/2+ b2/4 + b3/8….+bn/(2 N ))
b) 2Vref/R (b1/2+ b2/4 + b3/8….+bn/(2 N ))
c) Vref/R (b1/2+ b2/4 + b3/8….+bn/(2 N ))
d) none of the above
Answer: c
Explanation: It is the total current that will flow individual resistances.
8. Which is the Most Significant Bit and Least Significant Bit in this case?
a) MSB: b N , LSB: b N
b) MSB: b N , LSB: b 1
c) MSB: b 1 , LSB: b N
d) MSB: b 1 , LSB: b 1
Answer: b
Explanation: LSB will have maximum resistance and MSB will have least resistance .
9. For v ref = 10V, n = 6, and R= 5000 ohm, which of the following is true?
a) The maximum value of the output current is 1.9375 mA
b) The change in the output current if the LSB is changed from 0 to 1 is 0.03125 mA
c) The change in the output current is the MSB is change from 0 to 1 is 0.5 mA
d) None of the mentioned
Answer: b
Explanation: According to the figure, the LSB is given by 10 / 5000 6 mA.
10. What is the binary representation of 57?
a) 0011 1101
b) 0101 1001
c) 0111 1001
d) 0011 1001
Answer: d
Explanation: The binary equivalent of 57 is 00111001.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Amplifiers”.
1. An amplifier operating from ±3V provide a 2.2V peak sine wave across a 100 ohm load when provided with a 0.2V peak sine wave as an input from which 1.0mA current is drawn. The average current in each supply is measured to be 20mA. What is the amplifier efficiency?
a) 20.2%
b) 25.2%
c) 30.2%
d) 35.2%
Answer: a
Explanation:
electronic-devices-circuits-questions-answers-amplifiers-q1
2. In order to prevent distortion in the output signal after amplification, the input signal must be
a) Higher than the positive saturation level of the amplifier
b) Lower than the negative saturation level of the amplifier
c) Must lie with the negative and the positive saturation level of the amplifier
d) Both higher than the positive saturation level of the amplifier and lower than the negative saturation level of the amplifier
Answer: c
Explanation: Higher than the positive saturation and lower than the negative saturation level of the amplifier are the desired characteristics in order to prevent distortion.
3. The voltage gain of the amplifier is 8 and the current gain is 7. The power gain of the amplifier is
a) 56 db
b) 17.481 db
c) 34.963 db
d) 1 db
Answer: b
Explanation: The power gain is given by 10 log db.
4. Statement 1: Voltage gain of -5 means that the output voltage has been attenuated.
Statement 2: Voltage gain of -5db means that the output voltage has been attenuated.
a) Statement 1 and Statement 2 are true
b) Statement 1 and Statement 2 are false
c) Only Statement 1 is true
d) Only Statement 2 is true
Answer: d
Explanation: A negative voltage gain means that a phase difference of 1800 has been introduced in the output waveform when compared to the input waveform. A voltage gain of -5db means that the signal has been attenuated.
5. Which of the following isn’t true?
a) Both transformer and amplifier can provide voltage gain
b) Both transformer and amplifier can provide current gain
c) Both transformer and amplifier can provide power gain
d) None of the mentioned
Answer: c
Explanation: For an ideal transformer the power input is always equal to the power output. In real conditions there is slight loss of power when transferring the power from an input source to an output source. Amplifiers only provide power gain.
6. Symmetrically saturated amplifiers operating in clipping mode can be used to convert a sine wave to a
a) Square wave
b) Pseudo Square wave
c) Sawtooth wave
d) Triangular wave
Answer: b
Explanation: Clipping circuits with low peak values of the output signals are used to generate pseudo square waves if the input signal is very large as compared to the output signal.
7. What is meant by stability of the an amplified signal?
a) The amplified signal must have a finite amplitude
b) The amplified signal should not have self oscillation
c) The input and the output signal must be proportional
d) The ratio of the input and the output signal must be finite
Answer: b
Explanation: The ability of the amplifier to prevent self oscillation is a measure of its stability.
8. If A v , A i and A p represents the voltage gain, current gain and power gain ratio of an amplifier which of the below is not the correct expression for the corresponding values in decibel?
a) Current gain: 20 log A i db
b) Voltage gain: 20 log A v db
c) Power gain: 20 log A p db
d) Power gain: 10 log A p
Answer: c
Explanation: Power gain is given by 10 log A p db.
9. An amplifier has a voltage gain of 100 V/V and a current gain of 1000A/A. the value of the power gain decibel is
a) 30 db
b) 40 db
c) 50 db
d) 60 db
Answer: c
Explanation: Power gain in db is given by 10 log db.
10. The units of voltage gain is
a) It has no units, it is a ratio
b) Decibels
c) All of the mentioned
d) None of the mentioned
Answer: a
Explanation: Voltage gain (V o ) = output voltage/input voltage (V i ). It is also expresses as 20 log (V o /V i ) db.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Circuit Models for Amplifier”.
1. Buffer amplifier needs to have
a) Low input resistance and low output resistance
b) High Input resistance and high output resistance
c) Low input resistance and high output resistance
d) High input resistance and low output resistance
Answer: d
Explanation: Buffer amplifiers are used to connect high input resistance source to a low output resistance load.
2. The ideal values for the input resistance and the output resistance of a transconductance amplifier are
a) R i = 0 and R o = 0
b) R i = ∞ and R 0 = ∞
c) R i = 0 and R 0 = ∞
d) R i = ∞ and R i = 0
Answer: b
Explanation: It is a desired characteristics of transconductance amplifier ideally.
3. An amplifier has a voltage gain of 40db. The value of A VO is
a) 10
b) 100
c) 20
d) 200
Answer: b
Explanation: The expression is given by 10 log A VO = 40. Solving for A vo gives 100 as the answer.
4. The output voltage of a voltage amplifier has been found to decrease by 20% when a load resistance of 1 kΩ is connected. What is the value of the amplifier output resistance?
a) 50Ω
b) 200Ω
c) 250Ω
d) 350Ω
Answer: c
Explanation: 250 / X 100% = 20%. Hence the output resistance is 250 ohm.
5. Which of the following is a transresistance amplifier?
a) electronic-devices-circuits-questions-answers-circuit-models-amplifier-q5
b) electronic-devices-circuits-questions-answers-circuit-models-amplifier-5a
c) electronic-devices-circuits-questions-answers-circuit-models-amplifier-q5b
d) electronic-devices-circuits-questions-answers-circuit-models-amplifier-q5c
Answer: d
Explanation: Figure d is the correct representation rest are voltage, current, transconductance amplifiers.
6. The signal to be amplified is current signal and the output desired is a voltage signal. Which of the following amplifier can perform this task?
a) Voltage amplifier
b) Current amplifier
c) Transconductance amplifier
d) Transresistance amplifier
Answer: d
Explanation: It is a characteristic of transconductance amplifier.
7. You are given two amplifiers, A and B, to connect in cascade between a 10-mV, 100-kΩ source and a 100-Ω load . The amplifiers have voltage gain, input resistance, and output resistance as follows: for A, 100 V/V, 10 kΩ, 10 kΩ, respectively; for B, 1 V/V, 100 kΩ, 100 Ω, respectively. Your problem is to decide how the amplifiers should be connected so that the voltage gain is maximum.
a) SABL
b) SBAL
c) Both have the same voltage gain
d) None as neither combination is able to amplify the input signal
Answer: a
Explanation: None
8. A transconductance amplifier with Ri = 2 kΩ, Gm = 40 mA/V, and Ro = 20 kΩ is fed with a voltage source having a source resistance of 2 kΩ and is loaded with a 1-kΩ resistance. Find the voltage gain realized.
a) 18.05 V/V
b) 19.05 V/V
c) 20.05 V/V
d) 21.05V/V
Answer: b
Explanation:
electronic-devices-circuits-questions-answers-circuit-models-amplifier-q8
9. The ratio of the short circuit current gain of a current amplifier (A i ) to the open circuit voltage gain of a voltage amplifier (A V ), given that both amplifiers have the same value of the input resistance (R i ) and output resistance (R 0 ), is
a) R i
b) R o
c) R i / R 0
d) R o / R i
Answer: c
Explanation: It is a standard mathematical relation.
10. The ratio of the open circuit voltage of a voltage amplifier (A V ) to the short circuit transconductance of a (G m ) of a transconductance amplifier, given that both have the same value of the internal resistance (R i ) and the output resistance (R 0 ), is
a) R i
b) R 0
c) 1/R i
d) 1/R 0
Answer: b
Explanation: It is a standard mathematical relation.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Frequency Response of Amplifier”.
1. Consider a voltage amplifier having a frequency response of the low-pass STC type with a dc gain of 60 dB and a 3-dB frequency of 1000 Hz. Then the gain db at
a) f = 10 Hz is 55 db
b) f = 10 kHz is 45 db
c) f = 100 kHz is 25 db
d) f = 1Mhz is 0 db
Answer: d
Explanation: Use standard formulas for frequency response and voltage gain.
2. STC networks can be classified into two categories: low-pass and high-pass . Then which of the following is true?
a) HP network passes dc and low frequencies and attenuate high frequency and opposite for LP network
b) LP network passes dc and low frequencies and attenuate high frequency and opposite for HP network
c) HP network passes dc and high frequencies and attenuate low frequency and opposite for LP network
d) LP network passes low frequencies only and attenuate high frequency and opposite for HP network
Answer: b
Explanation: By definition a LP network allows dc current and an LP network does the opposite, that is, allows high frequency ac current.
3. Single-time-constant networks are those networks that are composed of, or can be reduced to
a) One reactive component and a resistance
b) Only capacitive component and resistance
c) Only inductive component and resistance
d) Reactive components and resistance
Answer: a
Explanation: STC has only one reactive component and one resistive component.
4. The signal whose waveform is not effected by a linear circuit is
a) Triangular Waveform signal
b) Rectangular waveform signal
c) Sine/Cosine wave signal
d) Sawtooth waveform signal
Answer: c
Explanation: Only sine/cosine wave are not affected by a linear circuit while all other waveforms are affected by a linear circuit.
5. Which of the following is not a classification of amplifiers on the basis of their frequency response?
a) Capacitively coupled amplifier
b) Direct coupled amplifier
c) Bandpass amplifier
d) None of the mentioned
Answer: d
Explanation: None of the options provided are correct.
6. General representation of the frequency response curve is called
a) Bode Plot
b) Miller Plot
c) Thevenin Plot
d) Bandwidth Plot
Answer: a
Explanation: General representation of frequency response curves are called Bode plot. Bode plots are also called semi logarithmic plots since they have logarithmic values values on one of the axes.
7. Under what condition can the circuit shown be called a compensated attenuator.
electronic-devices-circuits-questions-answers-frequency-response-amplifier-q7
a) C1R1 = C2R2
b) C1R2 = C2R1
c) C1C2 = R1R2
d) R1 = 0
Answer: a
Explanation: Standard condition of a compensated attenuator. Here is the derivation for the same.
electronic-devices-circuits-questions-answers-frequency-response-amplifier-q7a
electronic-devices-circuits-questions-answers-frequency-response-amplifier-q7b
8. When a circuit is called compensated attenuator?
a) Transfer function is directly proportional to the frequency
b) Transfer function is inversely proportional to the frequency
c) Transfer function is independent of the frequency
d) Natural log of the transfer function is proportional to the frequency
Answer: c
Explanation: Transfer function does not has frequency in its mathematical formula.
9. Which of the following is true?
a) Coupling capacitors causes the gain to fall off at high frequencies
b) Internal capacitor of a device causes the gain to fall off at low frequencies
c) All of the mentioned
d) None of the mentioned
Answer: d
Explanation: Both the statements are false.
10. Which of the following is true?
a) Monolithic IC amplifiers are directly coupled or dc amplifiers
b) Televisions and radios use tuned amplifiers
c) Audio amplifiers have coupling capacitor amplifier
d) All of the mentioned
Answer: d
Explanation: These all are practical applications of different types of amplifiers.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Sinusoidal Steady State Analysis”.
1. i = ?
electronic-devices-circuits-questions-answers-sinusoidal-steady-state-analysis-q1
a) 20 cos A
b) 20 cos A
c) 2.48 cos A
d) 2.48 cos A
Answer: d
Explanation: electronic-devices-circuits-questions-answers-sinusoidal-steady-state-analysis-q1a
2. Vc = ?
electronic-devices-circuits-questions-answers-sinusoidal-steady-state-analysis-q2
a) 0.89 cos V
b) 0.89 cos V
c) 0.45 cos V
d) 0.45 cos V
Answer: a
Explanation: electronic-devices-circuits-questions-answers-sinusoidal-steady-state-analysis-q2a
3. Vc = ?
electronic-devices-circuits-questions-answers-sinusoidal-steady-state-analysis-q3
a) 2.25 cos V
b) 2.25 cos V
c) 2.25 cos V
d) 2.25 cos V
Answer: d
Explanation: electronic-devices-circuits-questions-answers-sinusoidal-steady-state-analysis-q3a
4. i = ?
electronic-devices-circuits-questions-answers-sinusoidal-steady-state-analysis-q4
a) 2 sin A
b) cos A
c) 2 sin A
d) cos A
Answer: b
Explanation: electronic-devices-circuits-questions-answers-sinusoidal-steady-state-analysis-q4a
5. In the bridge shown, Z1 = 300 ohm, Z2 = 400 – j300 ohm, Z3 = 200 + j100 ohm. The Z4 at balance is
electronic-devices-circuits-questions-answers-sinusoidal-steady-state-analysis-q5
a) 400 + j300 ohm
b) 400 – j300 ohm
c) j100 ohm
d) -j900 ohm
Answer: b
Explanation: Use Z1 x Z4 = Z2 x Z3.
Circuit for Q.6 and Q.7
electronic-devices-circuits-questions-answers-sinusoidal-steady-state-analysis-q6
6. i1 = ?
a) 2.36 cos A
b) 2.36 cos A
c) 1.37 cos A
d) 2.36 cos A
Answer: c
Explanation: electronic-devices-circuits-questions-answers-sinusoidal-steady-state-analysis-q6a
7. i2 = ?
a) 2.04 sin A
b) 2.04 sin A
c) 2.04 cos A
d) 2.04 cos A
Answer: b
Explanation: electronic-devices-circuits-questions-answers-sinusoidal-steady-state-analysis-q7
8. In a two element series circuit, the applied voltage and the resulting current are v = 60 + 66 sin V, i = 2.3sin 3 A. The nature of the elements would be
a) R C
b) L C
c) R L
d) R R
Answer: a
Explanation: RC circuit causes a positive shift in the circuit.
9. P = 269 W, Q = 150 VAR . The power in the complex form is
a) 150 – j269 VA
b) 150 + j269 VA
c) 269 – j150 VA
d) 269 + j150 VA
Answer: c
Explanation: S = P – jQ.
10. Q = 2000 VAR, pf = 0.9 . The power in complex form is
a) 4129.8 j2000 VA
b) 2000 j4129.8 VA
c) 2000 j41.29.8 VA
d) 4129.8 j2000 VA
Answer: d
Explanation: Use cos T = 0.9 or T = 25.84 degrees.
Q = S sin T or S = 4588.6 VA
p = S cos T or P = 0.9 X 4588.6 4129.8 VA.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Ideal Operational Amplifiers”.
1. What is the minimum number of terminals required in an IC package containing four operational amplifiers ?
a) 12
b) 13
c) 14
d) 15
Answer: c
Explanation: The minimum no of pins required by dual-op-amp is 8. Each op-amp has 2 input terminals and one output terminal. Another 2 pins are required for power.
Similarly, The minimum no of pins required by dual-op-amp is 14: 4*2 + 4*1 + 2 = 14.
2. Which of the following is not a property of an ideal operational amplifier?
a) Zero input impedance
b) Infinite bandwidth
c) Infinite open loop gain
d) Zero common-mode gain or conversely infinite common mode-rejection.
Answer: a
Explanation: An ideal operational amplifier does not has a zero input impedance.
3. In an ideal op amp the open-loop gain is 10 3 . The op amp is used in a feedback circuit, and the voltages appearing at two of its three signal terminals are measured as v 2 = 0V and v 3 = 2V where it is assumed that v 1 and v 2 are input terminals and v 3 is the output terminal. The value of the differential (v d ) and common-mode (v cm )signal is
a) V d = 2 mV and v cm = 1 mv
b) V d = 2 mV and v cm = -1 mV
c) V d = 2 mV and v cm = 2mV
d) V d = 2 mV and v cm = -2mV
Answer: b
Explanation: V c = 0.5(V 1 + V 2 ) and
V d = V 2 – V 1 .
4. Consider the figure given below. Known that v o = 4V and v i = 2V, determine the gain for the op amp assuming that it is ideal except for the fact that it has finite gain
electronic-devices-circuits-questions-answers-ideal-operational-amplifiers-q4
a) 1001
b) 2002
c) 3003
d) 4004
Answer: b
Explanation: The Voltage at the positive input has to be -3.000v, v i = -3.020v
A = v o / v i – v r = -2 / -3.020 - = 100.
5. Which of the following is not a terminal for the operational amplifier?
a) Inverting terminal
b) Non-inverting terminal
c) Output terminal
d) None of the mentioned
Answer: d
Explanation: There are three terminals for the operational amplifier.
6. Operational amplifiers are
a) Differential input and single-ended output type amplifier
b) Single-ended input and single-ended output type amplifier
c) Single-ended input and differential output type amplifier
d) Differential input and differential output type amplifier
Answer: a
Explanation: It is another way to refer to op amps based on its terminal characteristics.
7. Express the input voltages v 1 and v 2 in terms of differential input (v d ) and common-mode input(v c ). Given v 2 > v 2 .
a) V d = V 1 – V 2 , V c = 0.5(V 1 + V 2 )
b) V d = V 2 – V 1 , V c = V 1 + V 2
c) V d = V 1 – V 2 , V c = V 1 + V 2
d) V d = V 2 – V 1 , V c = 0.5(V 1 + V 2 )
Answer: d
Explanation: This is the correct mathematical representation.
8. What is the minimum number of pins for a dual operational amplifier IC package?
a) 4
b) 6
c) 8
d) 10
Answer: c
Explanation: The minimum no of pins required by dual-op-amp is 8. Each op-amp has 2 input terminals and one output terminal. Another 2 pins are required for power.
9. For an ideal operational amplifier one set of the value for the input voltages (v 2 is the positive terminal v 1 is the negative terminal) and output voltage (v 0 ) as determined experimentally is v 1 = 2.01V, v 2 =2.00V and v 0 = -0.99V. Experiment was carried with different values of input and output voltages. Which of the following is not possible considering experimental error?
a) v 1 = 1.99V, v 2 = 2.00V, v 0 = 1.00V
b) v 1 = 1.00V, v 2 = 1.00V, v 0 = 0V
c) v 1 = 1.00V, v 2 = 1.10V, v 0 = 10.1V
d) v 1 = 0.99V, v 2 = 2.00V, v 0 = 1.00V
Answer: d
Explanation: Only option d does not satisfies the mathematical relation between the given quantities.
10. What are the units of slew rate?
a) Second/Volt
b) Volt/second
c) It is a ratio, no units
d) Ohm/second
Answer: b
Explanation: These units are obtained from the definition of the term slew rate.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Inverting Configuration”.
1. When does a resistance provide a negative feedback to an amplifier?
a) Resistance is connected between the positive input terminal and the output terminal
b) Resistance is connected between the negative input terminal and the output terminal
c) Resistance is connected between the input terminals
d) Resistance is connected between the negative input terminal and ground
Answer: b
Explanation: An op amp is said to have a negative feedback when a resistance is connected between the input and output terminals respectively.
2. The effect of the inverting configuration is
a) The output signal and the input signal are out of phase by 180 o
b) The output signal and the input signal are in phase
c) The output phase is leading the input phase by 90 o
d) The output phase is lagging behind the input phase by 90 o
Answer: a
Explanation: Inverting introduces a phase shift of 180 o or it ‘inverts’ a peak.
3. For an ideal negative feedback configuration which of the following is true?
a) There is a virtual open circuit between the input terminals
b) The closed loop gain for a negative feedback does not depend only on the external parameters
c) There is a virtual short circuit between the input terminals
d) There is a virtual ground at the negative input terminal
Answer: c
Explanation: There is always a virtual short circuit in this type of case. There will be a virtual ground if and only if one of the terminals is grounded.
4. The negative feedback causes
a) The voltage between the two input terminals to the very small, ideally zero
b) The voltage between the two input resistance very high, ideally infinite
c) Current flow through the positive input terminal and no current flows through the negative input terminal
d) Both a and c
Answer: a
Explanation: Ideally the input terminals are at the same potential but in real practice there is a very small potential between the two terminals.
5. The non-inverting closed loop configuration features a high resistance. Therefore in many cases unity gain follower called buffer amplifier is often used to
a) Connect a high resistance source to high resistance load
b) Connect low resistance source to low resistance load
c) Connect low resistance source to a high resistance source
d) Connect high resistance source to a low resistance load
Answer: d
Explanation: Buffer amplifiers are required to connect a high resistance load to a low input resistance output.
6. The advantage of a weighted summer operational amplifier is
a) It is capable of summing various input voltages together
b) Each input signal may be independently adjusted by adjusting the corresponding input resistance
c) If one needs both sign of a voltage signal then two operational amplifiers are needed
d) All of the mentioned
Answer: d
Explanation: All of the mentioned are characteristics of a weighted summer operational amplifier over the traditional amplifier.
7. The following is a circuit of weighted summer capable of summing coefficients of both sign. The expressions for the output voltage v 0 is
electronic-devices-circuits-questions-answers-inverting-configuration-q7
a) v 0 = v 1 + v 1 – v 1 – v 1
b) v 0 = – v 1 – v 1 + v 1 + v 1
c) v 0 = v 1 + v 1 – v 1 – v 1
d) v 0 = – v 1 – v 1 + v 1 + v 1
Answer: c
Explanation: The voltages are increased first by the left side of the portion and then are also magnified by the right side of the circuit. There are four inputs given out of which two are magnified twice and the other are magnified only once.
8. You are provided with an ideal op amp and three 10kΩ resistors. Using series and parallel resistor combinations, how many different inverting-amplifier circuit topologies are possible?
a) 2
b) 3
c) 4
d) 5
Answer: c
Explanation: Consider series and parallel combination of the resistances provided and arrange then in the feedback region and as output resistance.
9. The loop gain for an ideal operational amplifier with R 1 = 10kΩ and R 2 = 1MΩ is
a) 20 db
b) 40 db
c) 60 db
d) 80 db
Answer: b
Explanation: Loop gain in this case is given by 20 log .
10. In an inverting op-amp circuit for which the gain is −4 V/V and the total resistance used is 100 kΩ. Then the value of R 1 and R 2
a) R 1 = 20KΩ and R 1 = 80KΩ
b) R 1 = 80KΩ and R 1 = 20KΩ
c) R 1 = 40KΩ and R 1 = 60KΩ
d) R 1 = 50KΩ and R 1 = 50KΩ
Answer = a
Explanation: Solve R 1 + R 2 = 100
R 2 /R 1 = 4 for R 1 and R 2 respectively.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Non Inverting Configuration”.
1. In the non-inverting configuration of operational amplifier
a) The positive terminal is connected to the ground directly
b) The negative terminal is connected to the ground directly
c) The positive terminal is connected to the power source
d) The negative terminal is connected to the power source
Answer: c
Explanation: Non inverting configuration requires a power source connected to the power source.
2. For ideal non-inverting operational amplifier
a) Input and output resistances are infinite
b) Input resistance is infinite and output resistance is zero
c) Input resistance is zero and output resistance is infinite
d) Input and output resistances are zero
Answer: b
Explanation: It is an ideal characteristic of the non-inverting op amp.
3. For an ideal non-inverting operational amplifier having finite gain , the ratio of output voltage (v 0 ) to input voltage (v i ) is (given R 2 is the feedback resistance)
a) //A))
b) //A))
c) //A))
d) //A))
Answer: a
Explanation: It is a standard mathematical expression.
4. The gain for an ideal non-inverting operational amplifier is (given R 2 is the feedback resistance)
a) R 2 /R 1 – 1
b) R 2 /R 1
c) -R 2 /R 1
d) R 2 /R 1 + 1
Answer: d
Explanation: It is a standard mathematical expression.
5. While performing an experiment to determine the gain for an ideal operational amplifier having finite gain, a student mistakenly used the equation 1 + R 2 /R 1 where R 2 is the feedback resistance. What is the percentage error in his result? Given A is the finite voltage gain of the ideal amplifier used.
a) / X 100%
b) / X 100%
c) / X 100%
d) / X 100%
Answer: c
Explanation: The correct formula is //A)).
6. The finite voltage gain of a non-inverting operational amplifier is A and the resistance used is R 1 and R 2 in which R 2 is the feedback resistance. Under what conditions it can one use the expression 1 + R 2 /R 1 to determine the gain of the amplifier?
a) A ~ R 2 /R 1
b) A >> R 2 /R 1
c) A << R 2 /R 1
d) None of the mentioned
Answer: b
Explanation: The formula is valid for the ideal case in which the value of A is infinite, practically it should be very large when compared to R 2 /R 1 .
7. Which of the following is not true for a voltage follower amplifier?
a) Input voltage is equal to output voltage
b) Input resistance is infinite and output resistance is zero
c) It has 100% negative feedback
d) None of the mentioned
Answer: d
Explanation: All the statements are false.
8. For designing a non-inverting amplifier with a gain of 2 at the maximum output voltage of 10 V and the current in the voltage divider is to be 10 μA the resistance required are R 1 and R 2 where R 2 is used to provide negative feedback. Then
a) R 1 = 0.5 MΩ and R 2 = 0.5 MΩ
b) R 1 = 0.5 kΩ and R 2 = 0.5 kΩ
c) R 1 = 5 MΩ and R 2 = 5 MΩ
d) R 1 = 5 kΩ and R 2 = 5 kΩ
Answer: a
Explanation: 1 + R2/R 1 = 2 and 10/(R 1 +R 2 ) = 10 μA. Solve for R 1 and R 2 .
9. It is required to connect a transducer having an open-circuit voltage of 1 V and a source resistance of 1 MΩ to a load of 1-kΩ resistance. Find the load voltage if the connection is done directly and through a unity-gain voltage follower.
a) 1 μV and 1 mV respectively
b) 1 mV and 1 V respectively
c) 0.1 μV and 0.1 mV respectively
d) 0.1 mV and 0.1 V respectively
Answer: b
Explanation: When a unity gain follower is uses then input signal is equal to output signal. When connected directly, output signal is given by 1 X 1kΩ/1MΩ or 1mV.
10. Consider the figure given below. If the resistance R 1 is disconnected from the ground and connected to a third power source v 3 , then expression for the value of
v 0 is
electronic-devices-circuits-questions-answers-non-inverting-configuration-q10
a) 2v 1 + 4v 2 − 3v 3
b) 6v 1 + 8v 2 − 3v 3
c) 6v 1 + 4v 2 − 9v 3
d) 3v 1 + 4v 2 − 3v 3
Answer: c
Explanation: When a third power source is connected to the resistance of 1kΩ, then also the potential between the two input terminals of op amps remains the same. Using this fact the expression c is obtained.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Difference Amplifiers”.
1. For the difference amplifier which of the following is true?
a) It responds to the difference between the two signals and rejects the signal that are common to both the signal
b) It responds to the signal that are common to the two inputs only
c) It has a low value of input resistance
d) The efficacy of the amplifier is measured by the degree of its differential signal to the preference of the common mode signal
Answer: a
Explanation: All the statements are not true except for the fact that it responds only when there is difference between two signals only.
2. If for an amplifier the common mode input signal is v c , the differential signal id v d and A c and A d represent common mode gain and differential gain respectively, then the output voltage v 0 is given by
a) v 0 = A d v d – A c v c
b) v 0 = – A d v d + A c v c
c) v 0 = A d v d + A c v c
d) v 0 = – A d v d – A c v c
Answer: c
Explanation: It is a standard mathematical expression.
3. If for an amplifier v 1 and v 2 are the input signals, v c and v d represent the common mode and differential signals respectively, then the expression for CMRR is
a) 20 log (|A d | / |A c |)
b) -10 log (|A c | / |Ad|) 2
c) 20 log (v 2 – v 1 / 0.5(v 2 + v 1 ))
d) All of the mentioned
Answer: d
Explanation: Note that all the expressions are identical.
4. The problem with the single operational difference amplifier is its
a) High input resistance
b) Low input resistance
c) Low output resistance
d) None of the mentioned
Answer: b
Explanation: Due to low input resistance a large part of the signal is lost to the source’s internal resistance.
5. For the difference amplifier as shown in the figure show that if each resistor has a tolerance of ±100 ε % then the worst-case CMRR is given approximately by (given K = R 2 /R 1 = R 4 /R 3 )
electronic-devices-circuits-questions-answers-difference-amplifiers-q5
a) 20 log [K+1/4ε].
b) 20 log [K+1/2ε].
c) 20 log [K+1/ε].
d) 20 log [2K+2/ε].
Answer: a
Explanation: None.
6. For the circuit given below determine the input common mode resistance.
electronic-devices-circuits-questions-answers-difference-amplifiers-q5
a) (R 1 + R 3 ) || (R 2 ) || + (R 4 )
b) (R 1 + R 4 ) || (R 2 + R 3 )
c) (R 1 + R 2 ) || (R3 + R 4 )
d) (R 1 + R 3 ) || (R 2 + R 4 )
Answer: c
Explanation: Parallel combination of series combination of R 1 & R 3 with the series combination of R 3 and R 4 is the required answer as is visible by the circuit.
7. For the circuit shown below express v 0 as a function of v 1 and v 2 .
electronic-devices-circuits-questions-answers-difference-amplifiers-q7
a) v 0 = v 1 + v 2
b) v 0 = v 2 – v 1
c) v 0 = v 1 – v 2
d) v 0 = -v 1 – v 2
Answer: b
Explanation: Considering the fact that the potential at the input terminals are identical and proceeding we obtain the given result.
8. For the difference amplifier shown below, let all the resistors be 10kΩ ± x%. The expression for the worst-case common-mode gain is
electronic-devices-circuits-questions-answers-difference-amplifiers-q7
a) x / 50
b) x / 100
c) 2x /
d) 2x /
Answer: d
Explanation: None.
9. Determine A d and A c for the given circuit.
electronic-devices-circuits-questions-answers-difference-amplifiers-q9
a) A c = 0 and A d = 1
b) A c ≠ 0 and A d = 1
c) A c = 0 and A d ≠ 1
d) A c ≠ 0 and A d ≠ 1
Answer: a
Explanation: Consider the fact that the potential at the input terminals are identical and obtain the values of V 1 and V 2 . Thus obtain the value of V d and V c .
10. Determine the voltage gain for the given circuit known that R 1 = R 3 = 10kΩ abd R 2 = R 4 = 100kΩ.
electronic-devices-circuits-questions-answers-difference-amplifiers-q5
a) 1
b) 10
c) 100
d) 1000
Answer: b
Explanation: Voltage gain is 100/10.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Integrators and Differentiators”.
1. The other name for Miller Circuit is
a) Non-Inverting Integrator
b) Inverting Integrator
c) Non-Inverting Differentiator
d) Inverting Differentiator
Answer: b
Explanation: Miller Circuit is also called Inverting integrator.
2. The slope of the frequency response of an integrator is
a) Linear with negative slope
b) Linear with positive slope
c) Exponential increase
d) Exponential decrease
Answer: a
Explanation: The slope is linear and negative.
3. The integrating transfer function has the value of
a) jωCR
b) –jωCR
c) 1 / jωCR
d) -1 / jωCR
Answer: d
Explanation: Standard mathematical expression for the transfer function.
4. The expression for the integration frequency is
a) CR
b) 1/CR
c) R/C
d) C/R
Answer: b
Explanation: Standard mathematical expression for the integrator frequency.
5. Determine the expression for the transfer function for the circuit shown below.
electronic-devices-circuits-questions-answers-integrators-differentiators-q5
a) /
b) /
c) – /
d) – /
Answer: c
Explanation: It is a standard expression.
6. The frequency transfer function of a differentiator is given by
a) jωCR
b) 1/jωCR
c) – jωCR
d) – 1/jωCR
Answer: a
Explanation: Standard mathematical expression for the transfer function of a differentiator.
7. The slope of the frequency response of a differentiator is
a) Linear with negative slope
b) Linear with positive slope
c) Exponential increase
d) Exponential decrease
Answer: b
Explanation: The slope is linear with a positive slope.
8. The phase in the integrator and differentiator circuit respectively are
a) +90 degrees and +90 degrees
b) -90 degrees and -90 degrees
c) -90 degrees and +90 degrees
d) +90 degrees and -90 degrees
Answer: d
Explanation: These are the characteristics of the integrators and differentiators circuits respectively.
9. Consider a symmetrical square wave of 20-V peak-to-peak, 0 average, and 2-ms period applied to a Miller integrator. Find the value of the time constant CR such that the triangular waveform at the output has a 20-V peak-to-peak amplitude.
a) 0.25ms
b) 0.50ms
c) 2.5ms
d) 5.0ms
Answer: b
Explanation: According to the question 1/CR = 2.
10. The expression for the differentiator time constant is
a) CR
b) 1/CR
c) R/C
d) C/R
Answer: a
Explanation: Standard mathematical expression for the time constant for the differentiators.
This set of Electronic Devices and Circuits Quiz focuses on “DC Imperfections in Operational Amplifiers”.
1. Consider an inverting amplifier with a nominal gain of 1000 constructed from an op amp with an input offset voltage of 3 mV and with output saturation levels of ±10 V. What is the peak sine-wave input signal that can be applied without output clipping?
a) 7 mV
b) 10 mV
c) 13 mV
d) 9mV
Answer: a
Explanation: The maximum that can be sent without clipping is 10V – 1000 X 3mV or 7V.
Misplaced & Consider an inverting amplifier with a nominal gain of 1000 constructed from an op amp with an input offset voltage of 3 mV and with output saturation levels of ±10 V. If the effect of V Os is nulled at room temperature (25 0 C), how large an input can one now apply if:
2. The circuit is to operate at a constant temperature?
a) 8.5 mV
b) 9 mV
c) 9.5 mV
d) 10 mV
Answer: d
Explanation: Explanation: Maximum signal that will not be clipped is 10mV because 10mV X 1000 = 10V.
3. The circuit is to operate at a temperature in the range 0°C to 75°C and the temperature coefficient of V OS is 10 μV/°C?
a) 8.5 mV
b) 9 mV
c) 9.5 mV
d) 10 mV
Answer: c
Explanation: Since the effect is nullified at 25 o C, the peak that can be sent now is given by 10 – X 0.1 mV.
4. One of the DC imperfections of the amplifiers are dc offset voltage which is
a) Existence of output signal even when the common mode signal is zero
b) Existence of common mode signal causing zero output signal
c) Existence of output signal even when the differential signal is zero
d) Existence of differential signal causing zero output signal
Answer: c
Explanation: DC offset voltage is existence of output signal even when the differential signal is zero.
5. For the amplifier shown determine the value of the bias current (I b ) and input offset current (I o ) respectively.
electronic-devices-circuits-questions-answers-quiz-q5
a) I b = I B1 + I B2 I o = I B1 – I B2
b) I b = I B1 + I B2 I o = | I B1 – I B2 |
c) I b = 0.5(I B1 + I B2 ) I o = | I B1 – I B2 |
d) I b = 0.5(I B1 + I B2 ) I o = I B1 – I B2
Answer: c
Explanation: Standard mathematical expressions are used with the given variables.
6. Consider the circuit shown below which reduces the impact of the input bias current. If I B1 = I B2 = Input bias current, then determine the value of R 3 so that the output voltage (v 0 ) is not impacted by the input bias current.
electronic-devices-circuits-questions-answers-quiz-q6
a) /
b) /
c) R1-/
d) R2- /
Answer: a
Explanation: This will be possible when R 3 has the same value as the net effect of R 1 and R 2 .
7. Consider an inverting amplifier circuit designed using an op amp and two resistors, R 1 = 10 kΩ and R 2 = 1 MΩ. If the op amp is specified to have an input bias current of 100 nA and an input offset current of 10 nA, find the output dc offset voltage resulting.
a) 0.1 mV
b) 1 mV
c) 10 mV
d) 100 mV
Answer: d
Explanation: Use the mathematical definition of bias current and offset current.
Consider a Miller integrator with a time constant of 1ms and an input resistance of 10 kΩ. Let the op amp have V OS = 2 mV and output saturation voltages of ±12 V.
8. Assuming that when the power supply is turned on the capacitor voltage is zero, how long does it take for the amplifier to saturate?
a) 3s
b) 6s
c) 9s
d) 12s
Answer: b
Explanation: Use v O = V OS
(V OS /CR)t.
9. Select the largest possible value for a feedback resistor RF so that at least ±10 V of output signal swing remains available.
a) 10 kΩ
b) 100 kΩ
c) 1 MΩ
d) 10 MΩ
Answer: d
Explanation: Use v O = V OS
(V OS /CR)t.
10. What is the corner frequency of the resulting STC network?
a) 1 Hz
b) 0.16 Hz
c) 0.33 Hz
d) 0.5 Hz
Answer: b
Explanation: The required answer is given by 1/6 Hz.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Effect of Finite Open-Loop gain and Bandwidth on Circuit Performance”.
1. An internally compensated op amp is specified to have an open-loop dc gain of 106 dB and a unity gain bandwidth of 3 MHz. Find f b and the open-loop gain at f b .
a) 15Hz and 103 db
b) 30Hz and 103 db
c) 15 Hz and 51.5 db
d) 30 Hz and 51.5 db
Answer: a
Explanation: Use the equations below.
electronic-devices-circuits-multiple-choice-questions-answers-q1
2. A single-pole model has __________ db/decade roll-off of the gain.
a) -3 db/decade
b) -6 db/decade
c) -10 db/decade
d) -20 db/decade
Answer: d
Explanation: It is a standard characteristic of a single-pole model.
3. Single-pole model is also known as
a) Frequent pole
b) Stable pole
c) Dominant pole
d) Responsive pole
Answer: c
Explanation: Single-pole model is also called dominant pole.
4. An op amp having a 106-dB gain at dc and a single-pole frequency response with f t = 2 MHz is used to design a non-inverting amplifier with nominal dc gain of 100. The 3-dB frequency of the closed-loop gain is
a) 10 kHz
b) 20 kHz
c) 30 kHz
d) 40 kHz
Answer: b
Explanation:
Use the equation below to obtain a frequency response curve and proceed further.
electronic-devices-circuits-multiple-choice-questions-answers-q4
5. An internally compensated op amp has a dc open-loop gain of 106 V/V and an AC open-loop gain of 40 dB at 10 kHz. Estimate its gain–bandwidth product and its expected gain at 1 kHz.
a) 0.1 MHz and -60 db
b) 10 MHz and -60 db
c) 10 MHz and 60 db
d) 1 MHz and 60 db
Answer: d
Explanations: Use the following results.
electronic-devices-circuits-multiple-choice-questions-answers-q5
6. An inverting amplifier with nominal gain of −20 V/V employs an op amp having a dc gain of 104 and a unity-gain frequency of 106 Hz. What is the 3-dB frequency f3dB of the closed-loop amplifier?
a) 2π 23.8 kHz
b) 2π 47.6 kHz
c) 2π 71.4 kHz
d) 2π 95.2 kHz
Answer: b
Explanation:
electronic-devices-circuits-multiple-choice-questions-answers-q6
7. cascading two identical amplifier stages, each having a low-pass STC frequency response with a 3dB frequency f1, results in an overall amplifier with a 3dB frequency given by
a) √ f1
b) √ f1
c) √ f1
d) √ f1
Answer: c
Explanation:
electronic-devices-circuits-multiple-choice-questions-answers-q7
8. Find the ft required for internally compensated op amps to be used in the implementation of the closed loop amplifiers with dc gain of +100 V/V and 3db bandwidth of 100kHz?
a) 1 kHz
b) 10 kHz
c) 100 kHz
d) 1 MHz
Answer: b
Explanation: None
9. A particular op amp, characterized by a gain–bandwidth product of 20 MHz, is operated with a closed-loop gain of +100 V/V. What 3-dB bandwidth results? At what frequency does the closed-loop amplifier exhibit a −6° phase shift?
a) 21 kHz
b) 31.5 kHz
c) 42 kHz
d) 52.5 kHz
Answer: a
Explanation:
electronic-devices-circuits-multiple-choice-questions-answers-q9
10. Find the f t required for internally compensated op amps to be used in the implementation of the closed loop amplifiers with dc gain of -2 V/V and 3db bandwidth of 10 MHz?
a) 7.5 MHz
b) 15 MHz
c) 22.5 MHz
d) 30 MHz
Answer: d
Explanation: -R 2 ⁄ R 1 = -2 V ⁄ V
f 3db = 10 MHZ f t = 10 MHZ = 30 MHZ.
This set of Electronic Devices and Circuits online quiz focuses on “Large Signal Operations on operational Amplifiers”.
1. Slew rate of an amplifier is defined as
a) Minimum rate of change of the output possible in a real operational amplifier
b) Maximum rate of change of the output possible in a real operational amplifier
c) Average rate of change of the output possible in a real operational amplifier
d) Ratio of the maximum and the average rate of change of the output in a real amplifier
Answer: b
Explanation: By definition slew rate is the maximum rate of change of the output possible in a real operational amplifier.
2. Determine the slew rate of the amplifier having full power bandwidth f 0 and the rated output voltage as V 0 . Given that the input signal is of sinusoidal nature.
a) 2πf 0 V 0
b) V 0 / 2πf 0
c) V 0 / f 0
d) f 0 V 0
Answer: a
Explanation: v = V 0 sin wt
dv/dt = wV 0 sin wt
max value of dv/dt = wV 0
max value of w = w 0 = 2πf 0
w 0 V 0 = Slew Rate = 2πf 0 V 0 .
3. The units of the full power bandwidth is
a) Watt
b) Joule
c) Seconds
d) Hertz
Answer: d
Explanation: It has the units of frequency.
4. The full-power bandwidth, f M , is the maximum frequency at which
a) an output sinusoid with an amplitude equal to the op-amp rated output voltage (V o max ) can be produced without distortion
b) it is the range of the frequencies in which the amplitude of output signal is equal to or greater than half of the op-amp rated output voltage
c) it is the range of the frequencies in which the amplitude of output signal is equal to or less than half of the op-amp rated output voltage
d) It is the range of the frequencies in which the power gain is half or more than half of the maximum rated power gain of the op-amp
Answer: a
Explanation: This is the only statement that satisfies the definition of the full-power bandwidth.
5. Which of the following is not limitation of the operational amplifier
a) Output voltage saturation
b) Output current limits
c) Slew rate
d) None of the mentioned
Answer: d
Explanation: None of the mentioned are the limitations of the operational amplifier.
6. A particular op amp using ±15-V supplies operates linearly for outputs in the range −12 V to +12 V. If used in an inverting amplifier configuration of gain –100, what is the rms value of the largest possible sine wave that can be applied at the input without output clipping?
a) 120 mV
b) 60 mV
c) 84.85 mV
d) 42.42 mV
Answer: c
Explanation: Peak value of input wave = 12/100 or 120 mV. Hence the rms value is 120/√2 or 84.85 mV.
7. For operation with 10-V output pulses with the requirement that the sum of the rise and fall times represent only 20% of the pulse width , what is the slew-rate requirement for an op amp to handle pulses 2 µs wide?
a) 10 V/µs
b) 20 V/µs
c) 40 V/µs
d) 80 V/µs
Answer: c
Explanation: None.
8. An op amp having a slew rate of 20 V/µs is to be used in the unity-gain follower configuration, with input pulses that rise from 0 to 3 V. What is the shortest pulse that can be used while ensuring full-amplitude output?
a) 0.10 µs
b) 0.15 µs
c) 0.20 µs
d) 0.30 µs
Answer: b
Explanation: Time taken to reach 3V from 0V with slew rate of 20V/µs is 3/20 µs or 0.15 µs.
In designing with op amps one has to check the limitations on the voltage and frequency ranges of operation of the closed-loop amplifier, imposed by the op-amp finite bandwidth (f t ), slew rate , and output saturation (V o max ). Consider the use of an op amp with ft t = 2 MHz, SR = 1 V/µs, and V 0 max = 10 V in the design of a non-inverting amplifier with a nominal gain of 10. Assume a sine-wave input with peak amplitude Vi.
9. If Vi = 0.5 V, what is the maximum frequency before the output distorts?
a) 31.8 kHz
b) 318 kHz
c) 3.18 kHz
d) 3.18 MHz
Answer: a
Explanation: V i = 0.5v, V 0 = 0.5 X 10 = 5V
2πf V 0 = SR or f = 31.8 kHz.
10. If f = 20 kHz, what is the maximum value of V i before the output distorts?
a) 0.397 V
b) 0.795 V
c) 1.192 V
d) 1.590 V
Explanation: V 0 = 10Vi
2πf V 0 = SR = 20πf V i , here f is 20 kHz, SR is 1 V/µs. Hence the value of V i is 0.795 V.
This set of Electronic Devices and Circuits Questions and Answers for Campus interviews focuses on “MOSFETs Device Strucuture and Physical Operation”.
1. For NMOS transistor which of the following is not true?
a) The substrate is of p-type semiconductor
b) Inversion layer or induced channel is of n type
c) Threshold voltage is negative
d) None of the mentioned
Answer: c
Explanation: The threshold voltage is positive for NMOS.
2. Process transconductance parameter is directly proportional to
a) Electron mobility only
b) -1 only
c) Oxide capacitance only
d) Product of oxide capacitance and electron mobility
Answer: d
Explanation: It is the product of the electronic mobility with the oxide capacitance (F/m 2 ).
3. The SI Units of the Process transconductance Parameter is
a) V 2 /A
b) A/V 2
c) V/A
d) A/V
Answer: b
Explanation: k’ = μ n C ox where μ n is electronic mobility (m 2 /V s ) and C ox is oxide capacitance is (F/m 2 ).
4. Aspect ratio of the MOSFET has the units of
a) No units
b) m
c) m 2
d) m -1
Answer: a
Explanation: It is the ratio of the induced channel width to the induced channel length .
5. The MOSFET transconductance parameter is the product of
a) Process transconductance and inverse of aspect ratio
b) Inverse of Process transconductance and aspect ratio
c) Inverse of Process transconductance and inverse of aspect ratio
d) Process transconductance and aspect ratio
Answer: d
Explanation: This statement only satisfies the mathematical expression.
6. With the potential difference between the source and the drain kept small (V DS is small), the MOSFET behaves as a resistance whose value varies __________ with the overdrive voltage
a) Linearly
b) Inversely
c) Exponentially
d) Logarithmically
Answer: b
Explanation: For small V DS , resistance r is given by
R = 1 / ((μ n C ox )(V OV )).
7. For a p channel MOSFET which of the following is not true?
a) The source and drain are a p type semiconductor
b) The induced channel is p type region which is induced by applying a positive potential to the gate
c) The substrate is a n type semiconductor
d) None of the mentioned
Answer: b
Explanation: The induced channel is p type region which is induced by applying a negative potential to the gate.
8. When the voltage across the drain and the source (V DS ) is increased from a small amount (assuming that the gate voltage, VG with respect to the source is higher than the threshold voltage, V t ), then the width of the induced channel in NMOS (assume that V DS is always small when compared to the V ov )
a) Will remain as was before
b) Will become non uniform and will take a tapered shape with deepest width at the drain
c) Will become non uniform and will take a tapered shape with deepest width at the source
d) Will remain uniform but the width of the channel will increase
Answer: c
Explanation: The voltage across the source will be V OV and the voltage will decrease linearly to V OV – V DS as we reach the drain end. The width of the induced channel is proportional to the voltage.
9. The saturation current of the MOSFET is the value of the current when
a) The voltage between the drain and drain becomes equal to the overdrive voltage
b) The voltage between the drain and drain becomes equal to the threshold voltage
c) The voltage between the drain and drain becomes equal to the voltage applied to the gate
d) The voltage between the drain and drain becomes equal to difference the overdrive voltage and the threshold voltage
Answer: a
Explanation: By definition of the MOSFET saturation current.
10. At channel pinch off
a) The width of the induced channel becomes non linear
b) The width of the induced channel becomes very large
c) width becomes 1/e times the maximum possible width
d) The width of the induced channel becomes zero and the current saturates
Answer: d
Explanation: It is a characteristics of a channel pinch off.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “MOSFETs Current-Voltage Characterisitcs”.
1. If a MOSFET is to be used in the making of an amplifier then it must work in
a) Cut-off region
b) Triode region
c) Saturation region
d) Both cut-off and triode region can be used
Answer: c
Explanation: Only in the saturation region a MOSFET can operate as an amplifier.
2. For MOSFET is to be used as a switch then it must operate in
a) Cut-off region
b) Triode region
c) Saturation region
d) Both cut-off and triode region can be used
Answer: d
Explanation: In both regions it can perform the task of a switch.
Misplaced & Using the circuit shown below,
electronic-devices-circuits-questions-answers-mosfets-current-voltage-characterisitcs-q3
3. Determine the conditions in which the MOSFET is operating in the triode region.
i. V GD > V t
ii. V D S > V OV
iii. I D ∝ (V OV – 0.5V DS )V DS
a) i, ii, and iii are correct
b) i and iii are correct
c) i and ii are correct
d) ii and iii are correct
Answer: b
Explanation: Only the points I and iii are correct and ii is false.
4. Determine the conditions in which the MOSFET is operating in the saturation region
i. V GD > V t
ii. V DS > V OV
iii. I D ∝ (V OV ) 2
a) i, ii, and iii are correct
b) i and iii are correct
c) i and ii are correct
d) ii and iii are correct
Answer: d
Explanation: i is false and ii and iii are true.
5. In the saturation region of the MOSFET the saturation current is
a) Independent of the voltage difference between the source and the drain
b) Depends directly on the voltage difference between the source and the drain
c) Depends directly on the overdriving voltage
d) Depends directly on the voltage supplied to the gate terminal
Answer: a
Explanation: Saturation current does not depends on the voltage difference between the source and the drain in the saturation region of a MOSFET.
6. An n-channel MOSFET operating with V OV =0.5V exhibits a linear resistance = 1 kΩ when V DS is very small. What is the value of the device transconductance parameter k n ?
a) 2 mA/V 2
b) 20 mA/V 2
c) 0.2 A/V 2
d) 2 A/V 2
Answer: a
Explanation: Use the standard mathematical expression to determine the value of k n .
7. An NMOS transistor is operating at the edge of saturation with an overdrive voltage V OV and a drain current I D . If is V OV is doubled, and we must maintain operation at the edge of saturation, what value of drain current results?
a) 0.25I D
b) 0.5I D
c) 2I D
d) 4I D
Answer: c
Explanation: I 0 is directly proportional to V OS .
Using the circuit below answer the question
electronic-devices-circuits-questions-answers-mosfets-current-voltage-characterisitcs-q8
8. Which of the following is true for the triode region?
a. V DG > V tp
b. V SD < V OV
c. I D ∝ V OV
d. None of the mentioned
Answer: d
Explanation: V DG > |V tp | and V SD < |V OV |.
9. Which of the following is true for the saturation region?
a) V DG ≤ |V tp |
b) V SD ≤| V OV |
c) V DG < |V tp |
d) V SD <| V OV |
Answer: a
Explanation: It is a characteristic for the saturation region.
10. The current i D
a) Depends linearly on V OV in the saturation region
b) Depends on the square of V OV in the saturation region
c) Depends inversely on V OV in the triode region
d) None of the mentioned
Answer: b
Explanation: Use the standard mathematical expressions for i 0 in different regions.
This set of Tough Electronic Devices and Circuits Questions and Answers focuses on “MOSFETs Circuits at DC”.
1. The transistor in the circuit shown below has k n = 0.4 mA/V 2 , V t = 0.5 V and λ = 0. Operation at the edge of saturation is obtained when
tough-electronic-devices-circuits-questions-answers-q1
a) R D = 0.5 kΩ
b) R D = 1.0 kΩ
c) R D = 1.5 kΩ
d) R D = 2.0 kΩ
Answer: c
Explanation: Use the standard formula for edge saturation.
2. The PMOS transistor in the circuit shown has V t = −0.7 V, μ p C ox = 60 μA/V 2 , L = 0.8 μm, and λ = 0. Find the value of R in order to establish a drain current of 0.115 mA and a voltage V D of 3.5 V.
tough-electronic-devices-circuits-questions-answers-q2
a) 12.5 KΩ
b) 25 kΩ
c) 37.5 kΩ
d) 50 kΩ
Answer: a
Explanation:
tough-electronic-devices-circuits-questions-answers-q2a
3. The NMOS transistors in the circuit shown have V t = 1 V, μ n C OX = 120 μA/V 2 , λ = 0, and L 1 = L 2 = L 3 = 1μm. Then which of the following is not the value of the width of these MOSFETs shown
a) 2 µm
b) 8µm
c) All of the mentioned
d) None of the mentioned
Answer: d
Explanation:
tough-electronic-devices-circuits-questions-answers-q3
4. The MOSFET shown has Vt = 1V, k n = 100µA/V 2 and λ = 0. Find the required values of W/L and of R so that when v I = V DD = +5 V, r DS = 50 Ω, and V O = 50 mV.
tough-electronic-devices-circuits-questions-answers-q4
a) W/L = 25 and R = 4.95 kΩ
b) W/L = 25 and R = 9.90 kΩ
c) W/L = 50 and R = 4.95 kΩ
d) W/L = 50 and R = 9.90 kΩ
Answer: c
Explanation:
tough-electronic-devices-circuits-questions-answers-q4a
For each of the circuits shown find the labeled voltages. For all transistors, k n = 1 mA/V2, V t = 2V, and λ = 0
5. Find V 3
tough-electronic-devices-circuits-questions-answers-q5
a) 2.41V
b) 3.41V
c) 4.41V
d) 1.41V
Answer: b
Explanation: V 3 = 10- 4 * 2 + 1.4 = 3.4v.
6. Find V 4 and V 5
tough-electronic-devices-circuits-questions-answers-q6
a) 4V and -5V respectively
b) -4V and 5V respectively
c) 4V and 5V respectively
d) -4V and -5V respectively
Answer: a
Explanation: tough-electronic-devices-circuits-questions-answers-q6a
7. Find V 1 and V 2
a) 2V and -4V
b) -2V and 4V
c) 2V and 4V
d) -2V and -4V
Answer: c
Explanation: I D = 1 = 1 ⁄ 2 * 1 * (V GS – 2) 2 => V GS = 3.41v.
V 3 = 3.41v.
Misplaced & For each of the circuits shown find the labeled node voltages. The NMOS transistors have V t = 1 V and k n = 2 mA/V 2 and λ = 0
8. Find V 1 and V 2
tough-electronic-devices-circuits-questions-answers-q8
a) 2.44 and -1.28 V
b) 2.44 and -2.56 V
c) 1.22 and -2.56 V
d) 1.22 and -1.28 V
Answer: b
Explanation:
tough-electronic-devices-circuits-questions-answers-q8a
9. Find V 3 and V 4
tough-electronic-devices-circuits-questions-answers-q9
a) 3.775V and 5V
b) 3.775V and 2.55V
c) 7.55V and 2.55V
d) 7.555V and 5V
Answer: d
Explanation:
tough-electronic-devices-circuits-questions-answers-q9a
10. For the PMOS transistor in the circuit shown k n = 8 µA/V 2 , W/L = 25,|V tp | = 1V and I = 100μA. For what value of R is V SD = V SG ?
tough-electronic-devices-circuits-questions-answers-q10
a) 0 Ω
b) 12.45 kΩ
c) 25.9 kΩ
d) 38.35 kΩ
Answer: a
Explanation: V SG will be equal to V SD only when the resistance shown is zero or in other words there should not be any resistance.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “MOSFET in Amplfier Design”.
Misplaced & Various measurements are made on an NMOS amplifier for which the drain resistor R D is 20 kΩ. First, DC measurements show the voltage across the drain resistor, V RD , to be 2 V and the gate-to-source bias voltage to be 1.2 V. Then, ac measurements with small signals show the voltage gain to be −10 V/V.
1. What is the value of V t for this transistor?
a) 0.7V
b) 0.8V
c) 0.9V
d) 1.0V
Answer: b
Explanation:
electronic-devices-circuits-questions-answers-mosfet-amplifier-design-q1
2. If the process transconductance parameter is 50μA/V 2 , what is the MOSFET’s W/L?
a) 25
b) 50
c) 75
d) 100
Answer: a
Explanation: electronic-devices-circuits-questions-answers-mosfet-amplifier-design-q1
Consider the amplifier below for the case V DD = 5 V, R D = 24 kΩ, = 1 mA/V 2 , and V t = 1 V.
electronic-devices-circuits-questions-answers-mosfet-amplifier-design-q3
3. If the amplifier is biased to operate with an overdrive voltage V OV of 0.5 V, find the incremental gain at the bias point.
a) -3 V/V
b) -6 V/V
c) -9 V/V
d) -12 V/V
Answer: d
Explanation:
electronic-devices-circuits-questions-answers-mosfet-amplifier-design-q3a
4. For amplifier biased to operate with an overdrive voltage of 0.5V, and disregarding the distortion caused by the MOSFET’s square-law characteristic, what is the largest amplitude of a sine-wave voltage signal that can be applied at the input while the transistor remains in saturation?
a) 1.61 V
b) 1.5 V
c) 0.11 V
d) 3.11 V
Answer: b
Explanation: Use the standard mathematical formula to obtain the result.
5. For the input signal of 1.5V what is the value of the gain value obtained?
a) -12.24 V/V
b) -12.44 V/V
c) -12.64 V/V
d) -12.84 V/V
Answer: c
Explanation: The amplitude of the output voltage signal that results is approximately equal to V oq – V OB = 2 – 0.61 = 1.39v.
The gain implied by amplitude is
Gain = -1.39/0.11 = -12.64 V/V.
6. Which of the following is the fastest switching device?
a) JEFT
b) Triode
c) MOSFET
d) BJT
Answer: c
Explanation: MOSFET is the fastest switching device among the given four options.
7. Bias point is also referred by the name
a) DC Operating Point
b) Quiescent Point
c) None of the mentioned
d) All of the mentioned
Answer: d
Explanation: Bias point is called dc operating point as the MOSFET functions best at this point. Also since at the bias point no signal component is present it is called quiescent point
Consider the amplifier circuit shown below. The transistor is specified to have V t = 0.4 V, k n = 0.4 mA/V2, W/L = 10 and λ = 0. Also, let V DD = 1.8V, R D = 17.5kΩ, V GS = 0.6V and v gs = 0V.
electronic-devices-circuits-questions-answers-mosfet-amplifier-design-q8
8. Find I D .
a) 0.08 mA
b) 0.16 mA
c) 0.4 mA
d) 0.8 mA
Answer: a
Explanation: electronic-devices-circuits-questions-answers-mosfet-amplifier-design-q9
9. Find V DS .
a) 0.1V
b) 0.2 V
c) 0.4 V
d) 0.8 V
Answer: c
Explanation: electronic-devices-circuits-questions-answers-mosfet-amplifier-design-q9
10. Find A v .
a) -12 V/V
b) -14 V/V
c) -16 V/V
d) -18 V/V
Solution: b
Explanation: A v = – k n V ov R D
= -0.4 * 10 * 0.2 * 17.5
= – 14.4v
This set of Tricky Electronic Devices and Circuits Questions and Answers focuses on “MOSFET in Small Signal Operation”.
1. An NMOS technology has μ n C ox = 50 μA/V 2 and V t = 0.7 V. For a transistor with L = 1μm, find the value of W that results in g m 1mA/V at I D = 0.5 mA.
a) 10 μm
b) 20 μm
c) 30 μm
d) 40 μm
Answer: b
Explanation:
tricky-electronic-devices-circuits-questions-answers-q1
2. Consider an NMOS transistor having k n = 2 mA/V 2 . Let the transistor be biased at V OV = 1V. For operation in saturation, what dc bias current I D results? If a +0.1-V signal is superimposed on V GS , find the corresponding increment in collector current by evaluating the total collector current I D and subtracting the dc bias current I D .
a) I D = 1mA and Increment = 0.21 mA
b) I D = 1mA and Increment = 0.42 mA
c) I D = 2mA and Increment = 0.21 mA
d) I D = 2mA and Increment = 0.42 mA
Answer: a
Explanation:
tricky-electronic-devices-circuits-questions-answers-q2
3. We know I D =1/2 k n (V GS + v gs – V t ) 2 . Let the signal v gs be a sine wave with amplitude V gs , and substitute v gs = V gs sin ω t in Eq.. Using the trigonometric identity show that the ratio of the signal at frequency 2ω to that at frequency ω , expressed as a percentage is
a) V gs /V ov x 100%
b) 1/2V gs /V ov x 100%
c) 1/4V gs /V ov x 100%
d) 1/8V gs /V ov x 100%
Answer: c
Explanation:
tricky-electronic-devices-circuits-questions-answers-q3
4. If in a particular application V gs is 10 mV, find the minimum overdrive voltage at which the transistor should be operated so that the second-harmonic distortion is kept to less than 1%.
a) 1V
b) 0.75V
c) 0.5V
d) 0.25V
Answer: d
Explanation:
tricky-electronic-devices-circuits-questions-answers-q4
An NMOS amplifier is to be designed to provide a 0.50-V peak output signal across a 50-kΩ load that can be used as a drain resistor.
5. If a gain of at least 5 V/V is needed, what value of g m is required?
a) 0.1 mA/V
b) 0.2 mA/V
c) 0.4 mA/V
d) 0.8 mA/V
Answer: a
Explanation: g m R d = 5 or g m = 5/50 mA/V.
6. Using a dc supply of 3 V, what values of I D and V OV would you choose?
a) 0.34 mA and 0.35 V respectively
b) 0.34 mA and 0.69 V respectively
c) 0.034 mA and 0.35 V respectively
d) 0.034 mA and 0.69 V respectively
Answer: d
Explanation:
tricky-electronic-devices-circuits-questions-answers-q6
7. What W/L ratio is required if μ n C ox = 200 μA/V 2 ?
a) 1.23
b) 1.23
c) 1.43
d) 1.53
Answer: c
Explanation:
tricky-electronic-devices-circuits-questions-answers-q7
For a 0.8-μm CMOS fabrication process: V tn = 0.8 V, V tp = −0.9 V, μ n C ox = 90 μA/V 2 , μ p C ox = 30 μA/V 2 , Cox = 1.9 fF/μm2, V A = 8L , and |V A | = 12L .
8. Find the small-signal model parameters (g m , r o and g mb ) for an NMOS transistor having W/L = 20 μm/2 μm and operating at I D = 100 μA and |V SB | = 1V.
a) g m = 0.42mA/V, r o = 160 kΩ, g mb = 0.084 mA/V
b) g m = 0.21mA/V, r o = 160 kΩ, g mb = 0.042 mA/V
c) g m = 0.42mA/V, r o = 80 kΩ, g mb = 0.042 mA/V
d) g m = 0.24mA/V, r o = 80 kΩ, g mb = 0.084 mA/V
Answer: a
Explanation:
tricky-electronic-devices-circuits-questions-answers-q8
9. Find the small-signal model parameters (g m , r o and g mb ) for a PMOS transistor having W/L = 20 μm/2 μm and operating at I D = 100 μA and |V SB | = 1V.
a) g m = 0.24mA/V, r o = 240 kΩ, g mb = 0.024 mA/V
b) g m = 0.24mA/V, r o = 120 kΩ, g mb = 0.048 mA/V
c) g m = 0.24mA/V, r o =240 kΩ, g mb = 0.048 mA/V
d) g m = 0.12mA/V, r o = 240 kΩ, g mb = 0.048 mA/V
Answer: c
Explanation:
tricky-electronic-devices-circuits-questions-answers-q9
10. The overdrive voltage at which each device must be operating is
a) NMOS = 0.83V and PMOS = 0.48V
b) NMOS = 0.48V and PMOS = 0.83V
c) NMOS = 0.24V and PMOS = 0.41V
d) NMOS = 0.41V and PMOS = 0.24V
Answer: b
Explanation:
NMOS case
tricky-electronic-devices-circuits-questions-answers-q10
PMOS case
tricky-electronic-devices-circuits-questions-answers-q10a
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Basic MOSFET Amplifier Configurations”.
1. In which of the following configuration does a MOSFET works as an amplifier?
a) Common Source
b) Common Gate
c) Common drain
d) All of the mentioned
Answer: d
Explanation: There are three basic configurations for connecting the MOSFET as an amplifier. Each of these configurations is obtained by connecting one of the three MOSFET terminals to ground, thus creating a two-port network with the grounded terminal being common to the input and output ports.
2. The MOSFET in the following circuit is in which configuration?
electronic-devices-circuits-questions-answers-basic-mosfet-amplifier-configurations-q2
a) Common Source
b) Common Gate
c) Common Drain
d) None of the mentioned
Answer: b
Explanation: It is the circuit for Common gate configuration.
3. The MOSFET in the following circuit is in which configuration?
electronic-devices-circuits-questions-answers-basic-mosfet-amplifier-configurations-q3
a) Common Source
b) Common Gate
c) Common Drain
d) None of the mentioned
Answer: c
Explanation: It is the circuit for Common drain configuration.
4. The MOSFET in the following circuit is in which configuration?[/expand] electronic-devices-circuits-questions-answers-basic-mosfet-amplifier-configurations-q4
a) Common Source
b) Common Gate
c) Common Drain
d) None of the mentioned
Answer: a
Explanation: It is the circuit for Common source configuration.
Reference circuit for Q.5-Q.10 The circuit below is the characterization for the amplifier as a functional block.
electronic-devices-circuits-questions-answers-basic-mosfet-amplifier-configurations-q5
5. If the value of R in for the common source configuration is R 1 and that for common source with a source resistance configuration is R 2 ideally. The ratio of R 1 /R 2 will be
a) R 1 /R 2 = 1
b) 0 < R 1 /R 2 < 1
c) R 1 /R 2 > 1
d) R 1 /R 2 = 0
Answer: a
Explanation: Ideally both must have infinite resistance.
6. Which is true for the value of A vo for common source (Represented by A 1 ) and common source with a source resistance (represented by A 2 ).
a) A 1 = A 2
b) A 1 > 2
c) A 1 < A 2
d) |A 1 | < |A 2 |
Answer: c
Explanation: A 1 = -g m R D and A 2 = -g m R D /1+g m R S
Reference circuit for Common source configuration
electronic-devices-circuits-questions-answers-basic-mosfet-amplifier-configurations-q6
Reference circuit for common source with source resistance R S
electronic-devices-circuits-questions-answers-basic-mosfet-amplifier-configurations-q6a
7. Which of the following is true for the voltage gain for the common source configuration and the common gate configuration ?
a) A 1 = A 2
b) |A 1 | = |A 2 | and A1 ≠ A 2
c) |A 1 | > |A 2 |
d) |A 1 | < |A 2 |
Answer: b
Explanation: A 1 = -g m (R L ||R D ) and A 1 = g m (R L ||R D )
Reference figure for common source configuration
electronic-devices-circuits-questions-answers-basic-mosfet-amplifier-configurations-q6
Reference figure for common gate configuration
electronic-devices-circuits-questions-answers-basic-mosfet-amplifier-configurations-q6
8. The value of the voltage gain (A v ) for the common source with source resistance (represented by A 1 ) and common gate configuration (represented by A 2 ) are related to each other by
a) A 1 > A 2
b) |A 1 | > |A 2 |
c) A 1 < A 2
d) A 1 > A 2 and |A 1 | > |A 2 |
Answer: c
Explanation: A 1 = – g m (R L ||R D )/ 1 + g m R S and
A 2 = g m (R L ||R D )
Reference figure for common source with source resistance configuration
electronic-devices-circuits-questions-answers-basic-mosfet-amplifier-configurations-q6a
Reference figure for common gate configuration
electronic-devices-circuits-questions-answers-basic-mosfet-amplifier-configurations-q6
9. In which of the following configuration is the input resistance (R i ) not equal to zero ideally?
a) Common source configuration
b) Common source configuration with source resistance
c) Common gate configuration
d) Source follower configuration
Answer: c
Explanation: Refer to the circuit for the common gate configuration
electronic-devices-circuits-questions-answers-basic-mosfet-amplifier-configurations-q6
10. Which of the following has A VO independent of the circuit elements?
a) Common source configuration
b) Common gate configuration
c) Source follower configuration
d) None of the mentioned
Answer: c
Explanation: A VO = 1 source follower.
This set of Electronic Devices and Circuits Problems focuses on “Biasing in MOS Amplifier Circuit”.
A discrete MOSFET common source amplifier has R in = 2 MΩ, g m m = 4mA/V, r o = 100 kΩ, R D = 10 kΩ, C gs = 2pF and C gd = 0.5pF. The amplifier is fed from a voltage source with an internal resistance of 500 kΩ and Is connected to the a load of 10 kΩ.
1. The value of the overall mixed gain A M is
a) -15.2 V/V
b) -16.2 V/V
c) -17.2 V/V
d) -18.2 V/V
Answer: a
Explanation:
electronic-devices-circuits-problems-q1
2. The upper 3-db frequency f H is
a) 11.1 kHz
b) 22.1 kHz
c) 33.1 kHz
d) 44.1 kHz
Answer: c
Explanation:
electronic-devices-circuits-problems-q2
The amplifier in the figure shown below is biased to operate at I D = 1mA and g m = 1mA/V.
electronic-devices-circuits-problems-q3
3. Find the midband gain.
a) 0.43 V/V
b) 1.43 V/V
c) 2.43 V/V
d) 3.43 V/V
Answer: b
Explanation:
electronic-devices-circuits-problems-q3a
4. Find the value of C S that places F L at 10Hz
a) 6.57 µF
b) 12.57 µF
c) 18.57 µF
d) 24.57 µF
Answer: c
Explanation:
electronic-devices-circuits-problems-q4
)In the NMOS transistor of the circuit shown below is biased to have g m = 1mA/V and r 0 = 100 kΩ.
electronic-devices-circuits-problems-q5
5. Find A M
a) 1.02 V/V
b) 2.04 V/V
c) 3.06 V/V
d) 4.08 V/V
Answer: c
Explanation:
electronic-devices-circuits-problems-q3a
6. Find f H if C gs = 1 pF and C gd = 0.2 pF.
a) 875 kHz
b) 875 kHz
c) 875 kHz
d) 875 kHz
Answer: c
Explanation:
electronic-devices-circuits-problems-q6
For a particular depletion mode NMOS device, V t = -2V, k n W/L = 200 µA/V 2 and λ = 0.02V -1 . For V DS = 2V
7. What is the drain current that flows
a) 104 µA
b) 208 µA
c) 312µA
d) 416 µA
Answer: d
Explanation:
electronic-devices-circuits-problems-q7
8. What is the value of the drain current if both L and W are doubled?
a) 408 µA
b) 412 µA
c) 416 µA
d) 420 µA
Answer: a
Explanation:
electronic-devices-circuits-problems-q8
A depletion type N channel MOSFEt with k n W/L = 2 mA/V 2 and V t = 3V has its source and gate grounded. For V d = 0.1V and neglecting channel length modulating effect
9. Find drain current.
a) 1.18 mA
b) 0.89 mA
c) 0.59 mA
d) 0.3 mA
Answer: b
Explanation:
electronic-devices-circuits-problems-q9
10. In which region is the triode operating?
a) Triode region
b) End of saturation region
c) Saturation region
d) None of the mentioned
Answer: a
Explanation:
electronic-devices-circuits-problems-q10
This set of Advanced Electronic Devices and Circuits Questions and Answers focuses on “Discrete-Circuit MOS Amplifiers”.
1. Calculate the overall voltage gain G v of a common source amplifier for which g m = 2mA/V, R D = 10 kΩ, R 0 = 10 kΩ and R G = 10 MΩ. The amplifier is fed from a signal source of Thevenin resistance of 0.5MΩ and the amplifier is coupled with a load of 10 kΩ.
a) -11.2 V/V
b) -22.4 V/V
c) -33.6 V/V
d) -44.8 V/V
Answer: a
Explanation:
advanced-electronic-devices-circuits-questions-answers-q1
The MOSFEt circuit below has V t = 1V, k n W/L = 0.8 mA/V 2 and V A = 40V
advanced-electronic-devices-circuits-questions-answers-q2
2. Find the value of R G so that i D = 0.1 mA, the largest possible value of R D is used while the maximum signal swing at the drain is of 1V and the input resistance at the gate is 10 MΩ.
a) 1 MΩ
b) 10 MΩ
c) 0.1 MΩ
d) 0.01 MΩ
Answer: b
Explanation:
advanced-electronic-devices-circuits-questions-answers-q2a
3. Find the value of g m at the bias point
a) 0.1 mA/V
b) 0.2 mA/V
c) .0.3 mA/V
d) 0.4 mA/V
Answer: d
Explanation:
advanced-electronic-devices-circuits-questions-answers-q3
4. If terminal Z is grounded, X is connected to a signal source having a resistance of 1 MΩ and terminal Y is connected to a load resistance of 40 kΩ, find the voltage gain from the signal source to the load.
a) 3.25 v/V
b) 6.5 V/V
c) 9.75 V/V
d) 13 V/V
Answer: b
Explanation:
advanced-electronic-devices-circuits-questions-answers-q4
5. If terminal Y is grounded find the voltage gain from X to Z with Z open-circuit.
a) 0.33 V/V
b) 0.66 V/V
c) 0.99 V/V
d) None of the mentioned
Answer: c
Explantion:
advanced-electronic-devices-circuits-questions-answers-q5
6. If terminal X is grounded and terminal Z is connected to a current source delivering a current of 10 µA and having a resistance of 100 kΩ, find the voltage signal that can be measured at Y neglecting the effect of V 0 .
a) 0.34V
b) 0.68 V
c) 3.4 V
d) 6.8 V
Answer: a
Explanation:
advanced-electronic-devices-circuits-questions-answers-q6
The NMOS transistor in source follower circuit shown has g m = 5mA/V and a large r 0 .
advanced-electronic-devices-circuits-questions-answers-q7
7. Find the output resistance.
a) 0.1 kΩ
b) 0.2 kΩ
c) 0.3 kΩ
d) 0.4 kΩ
Answer: b
Explanation:
advanced-electronic-devices-circuits-questions-answers-q7a
8. Find the open-Circuit voltage gain.
a) 1 V/V
b) 2 V/V
c) 3 V/V
d) 4 V/V
Answer: a
Explanation :
advanced-electronic-devices-circuits-questions-answers-q7a
advanced-electronic-devices-circuits-questions-answers-q8
9. The NMOS transistor in the common gate amplifier as shown in the circuit below has g m = 5 mA/V. Find the input resistance and the voltage gain.
advanced-electronic-devices-circuits-questions-answers-q9
a) Input resistance: 0.1 kΩ and Voltage gain: 3.05 V/V
b) Input resistance: 0.1 kΩ and Voltage gain: 3.05 V/V
c) Input resistance: 0.2 kΩ and Voltage gain: 3.05 V/V
d) Input resistance: 0.2 kΩ and Voltage gain: 7.1 V/V
Answer: d
Explanation:
advanced-electronic-devices-circuits-questions-answers-q9a
10. If the output of the source follower in is connected to the input of the common gate amplifier of . Determine the overall voltage gain (V 0 /V i ).
advanced-electronic-devices-circuits-questions-answers-q10
a) 1.55 V/V
b) 3.55 V/V
c) 5.55 V/V
d) 7.55 V/V
Answer: b
Explanation:
advanced-electronic-devices-circuits-questions-answers-q10a
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Body Effect”.
1. The _____________ of a MOSFET is affected by the voltage which is applied to the back contact.
a) Threshold Voltage
b) Output Voltage
c) Both threshold and output voltage
d) Neither of the threshold nor the output voltage
Answer: a
Explanation: The voltage difference between the source and the bulk, VBS changes the width of the depletion layer and therefore also the voltage across the oxide due to the change of the charge in the depletion region. This results in a difference in threshold voltage which equals the difference in charge in the depletion region divided by the oxide capacitance.
2. The variation of the threshold voltage with the applied bulk-to-source voltage is typically observed by plotting the _________________ as a function of the source-to-drain voltage.
a) drain current
b) square root of the drain current
c) square of the drain current
d) natural logarithm of the drain current
Answer: b
Explanation: The change in threshold current is directly proportional to the square root of the drain current. For further assistance check the mathematical expression for the same.
3. The SI units of the body effect parameter is
a) Volt
b) Volt X Volt
c) √Volt
d) It has no units
Answer: c
Explanation: Vt = Vt0 + k[√ – √2φf]. In this expression k is the body effect parameter hence its units can be determined.
4. An NMOS transistor has Vt0 = 0.8 V, 2 φf = 0.7 V, and γ = 0.4 V1/2. Find Vt when VSB = 3 V.
a) 0.12 V
b) 1.23 V
c) 2.34 V
d) 3.45 V
Answer: b
Explanation: Vt = Vt0 + k[√ – √2φf]. use this expression to obtain the desired result.
5. The threshold voltage is
a) Increases on increasing temperature
b) May increase or decrease on increasing temperature depending upon other factors
c) Temperature independent
d) Decreases on increasing temperature
Answer: d
Explanation: The threshold voltage depends only on the temperature and it decreases by roughly 2 mV for every degree Celsius increase in the temperature.
6. As the voltage on the drain is increased, a value is reached at which the pn junction between the drain region and substrate suffers avalanche breakdown known as
a) Weak avalanche
b) Strong avalanche
c) Weak storm
d) Punch-through
Answer: a
Explanation: As the voltage on the drain is increased, a value is reached at which the pn junction between the drain region and substrate suffers avalanche breakdown. This breakdown usually occurs at voltages of 20 V to 150 V and results in a somewhat rapid increase in current .
7. A breakdown effect that occurs in modern devices at low voltages is
a) Weak avalanche
b) Strong avalanche
c) Weak storm
d) Punch-through
Answer: d
Explanation: Punch-through occurs in devices with relatively short channels when the drain voltage is increased to the point that the depletion region surrounding the drain region extends through the channel to the source. The drain current then increases rapidly. Normally, punch-through does not result in permanent damage to the device.
8. At ______________ the drain current is no longer related to the Vgs by square law relationship.
a) When the temperature is high
b) When temperature is very low
c) Velocity saturation
d) None of the mentioned
Answer: c
Explanation: At velocity saturation the current depends linearly on the Vgs.
9. In MOSFETs a breakdown may occur at around 30 V. This is due to
a) Velocity saturation
b) Breakdown of the gate diode
c) Sudden decrease in the depletion region
d) Fall of the threshold voltage due to impurities
Answer: b
Explanation: The breakdown of the oxide at the gate may occur when the voltage is around 30 V. This may also permanently damage the device.
10. Which of the below issues may not be experienced when using MOSFETs?
a) Weak avalanche
b) Velocity saturation
c) Punch-through
d) All of the mentioned
Answer: d
Explanation: All of the mentioned are some of the common issues that one may face while dealing with MOSFETs.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “BJTs Device Strucutres and Physical Operations”.
1. Which of the following is not a part of a BJT?
a) Base
b) Collector
c) Emitter
d) None of the mentioned
Answer: d
Explanation: BJT consists of three semiconductor regions, base region, emitter region and collector region.
2. The number of pn junctions in a BJT is/are
a) 1
b) 2
c) 3
d) 4
Answer: b
Explanation: There are two pn junctions, base-emitter junction and collector-emitter junction respectively.
3. In which of the following modes can a BJT be used?
a) Cut-off mode
b) Active mode
c) Saturation mode
d) All of the mentioned
Answer: d
Explanation: These three are the defined regions in which a BJT operates.
4. If a BJT is to be used as an amplifier, then it must operate in___________
a) Cut-off mode
b) Active mode
c) Saturation mode
d) All of the mentioned
Answer: b
Explanation: A BJT operates as an amplifiers in active mode and as a switch in cut-off or saturation mode.
5. If a BJT is to be used as a switch, it must operate in____________
a) Cut-off mode or active mode
b) Active Mode or saturation mode
c) Cut-off mode or saturation mode
d) Cut-off mode or saturation mode or active mode
Answer: c
Explanation: A BJT operates as an amplifiers in active mode and as a switch in cut-off or saturation mode.
6. In cut off mode
a) The base-emitter junction is forward biased and emitter-collector junction is reversed biased
b) The base-emitter junction is forward biased and emitter-collector junction is forward biased
c) The base-emitter junction is reversed biased and emitter-collector junction is reversed biased
d) The base-emitter junction is reversed biased and emitter-collector junction is forward biased
Answer: c
Explanation: In cut-off mode there is no current flowing through the BJT hence both junctions must be reversed biased else if either of them is forward biased then the current will flow.
7. On which of the following does the scale current not depends upon?
a) Effective width of the base
b) Charge of an electron
c) Electron diffusivity
d) Volume of the base-emitter junction
Answer: d
Explanation: The saturation current does not depends upon the volume of the base-emitter junction. Instead it depends upon the area of the cross section of the base-emitter junction in a direction perpendicular to the flow of current.
8. On which of the following does the collector current not depends upon?
a) Saturation current
b) Thermal voltage
c) Voltage difference between the base and emitter
d) None of the mentioned
Answer: d
Explanation: Collector current depends linearly of the saturation current and exponentially to the ratio of the voltage difference between the base and collector and thermal voltage.
9. The range for the transistor parameter also referred as common-emitter current gain has a value of__________ for common devices.
a) 50-200
b) 400-600
c) 750-1000
d) > 1000
Answer: a
Explanation: Most commonly used transistors have a voltage gain of in the range of 50-200. Only some specially designed transistors have a transistor parameter in the range of 1000.
10. The collector current Ic is related to the emitter current Ie by a factor k. If b is the transistor parameter then the value of k in terms of b is
a) k = b/
b) k = /b
c) b = /k
d) None of the mentioned
Answer: a
Explanation: Ic = k Ie and also Ie = /b Ic . Equating these two results we get k = b/.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “BJTs Current-Voltage Characteristics”.
1. The curve between the collector current versus the potential difference between the base and emitter is
a) A straight line inclined to the axes
b) A straight line parallel to the x-axis
c) An exponentially varying curve
d) A parabolic curve
Answer: c
Explanation: The natural logarithm of the collector current depends directly on the the potential difference between the base and the emitter.
2. The curve between the collector current and the saturation is
a) A straight line inclined to the axes
b) A straight line parallel to the x-axis
c) A straight line parallel to the y-axis
d) An exponential curve
Answer: a
Explanation: The collector current depends directly on the saturation current.
3. The magnitude of the thermal voltage is given by
a) k/Tq
b) kT/q
c) q/Kt
d) Tk/q
Answer: b
Explanation: kT/q is the correct mathematical expression for the thermal voltage.
4. The correct relation between the transistor parameters α and ß are related by
a) ß = 1 – α/α
b) ß = 1 + α/α
c) α = ß + 1/ß
d) α = ß/ß + 1
Answer: d
Explanation: Only expression α = ß/ß + 1 is the correct expression that relates α and ß.
5. The correct expression relating the emitter current Ie to the collector current Ic is
a) Ie = α Ic
b) Ic = α Ic
c) Ie = ß Ic
d) Ic = ß Ic
Answer: b
Explanation: Ie = Ic/α or Ic = α Ie
6. The value of the thermal voltage at room temperature can be approximated as
a) 25 mV
b) 30 mV
c) 35 mV
d) 40 mV
Answer: a
Explanation: Thermal voltage is given by kT/q which at T = 25 degrees Celsius is approximately 25 mV.
7. The correct relation between the emitter current Ie and the base current Ib is given by
a) Ib = Ie
b) Ib = Ie
c) Ie = ß Ib
d) Ie = ß Ib
Answer: d
Explanation: The correct mathematical expression are Ie = ß Ib and Ib = Ie respectively.
8. The Early Effect is also called as
a) Base-width modulation effect
b) Base-width amplification effect
c) Both of the mentioned
d) None of the mentioned
Answer: a
Explanation: At a given value of vBE, increasing vCE increases the reverse-bias voltage on the collector–base junction, and thus increases the width of the depletion region of this junction. This in turn results in a decrease in the effective base width W. Also the saturation current is inversely proportional to the width, the saturation current will increase and also makes collector current increases proportionally. This is the Early Effect. For the reasons mentioned above, it is also known as the base-width modulation effect.
9. For the BJT to operate in active mode Collector-Base junction must be
a) Heavily doped
b) Must reversed bias
c) Must be forward bias
d) Lightly doped
Answer: b
Explanation: The BJT operates in active mode when the collector-Base junction is reversed bias. Also doping cannot prevent saturation of the transistor.
10. Collector current reaches zero when
a) Vce = Vt ln
b) Vt = Vce ln
c) Vce = Vt ln
d) Vce = Vt ln
Answer: a
Explanation: Ic = Is exp – Isc exp. In this expression put ic = 0 and simplify.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “BJT Circuits at DC”.
1. Which of the following condition is true for cut-off mode?
a) The collector current Is zero
b) The collector current is proportional to the base current
c) The base current is non zero
d) All of the mentioned
Answer: a
Explanation: The base current as well as the collector current are zero in cut-off mode.
2. Which of the following is true for the cut-off region in an npn transistor?
a) Potential difference between the emitter and the base is smaller than 0.5V
b) Potential difference between the emitter and the base is smaller than 0.4V
c) The collector current increases with the increase in the base current
d) The collector current is always zero and the base current is always non zero
Answer: b
Explanation: Both collector and emitter current are zero in cut-off region.
3. Which of the following is true for a typical active region of an npn transistor?
a) The potential difference between the emitter and the collector is less than 0.5 V
b) The potential difference between the emitter and the collector is less than 0.4 V
c) The potential difference between the emitter and the collector is less than 0.3 V
d) The potential difference between the emitter and the collector is less than 0.2 V
Answer: c
Explanation: Most commonly used transistors have Vce less than 0.4 V for the active region.
4. Which of the following is true for the active region of an npn transistor?
a) The collector current is directly proportional to the base current
b) The potential difference between the emitter and the collector is less than 0.4 V
c) All of the mentioned
d) None of the mentioned
Answer: c
Explanation: The base current and the collector current are directly proportional to each other and the potential difference between the collector and the base is always less than 0.4 V.
5. Which of the following is true for the saturation region of BJT transistor?
a) The collector current is inversely proportional to the base current
b) The collector current is proportional to the square root of the collector current
c) The natural logarithm of the collector current is directly proportional to the base current
d) None of the mentioned
Answer: b
Explanation: The collector current is directly proportional to the base current in the saturation region of the BJT.
6. Which of the following is true for a npn transistor in the saturation region?
a) The potential difference between the collector and the base is approximately 0.2V
b) The potential difference between the collector and the base is approximately 0.3V
c) The potential difference between the collector and the base is approximately 0.4V
d) The potential difference between the collector and the base is approximately 0.5V
Answer: d
Explanation: The commonly used npn transistors have a potential difference of around 0.5V between he collector and the base.
7. The potential difference between the base and the collector Vcb in a pnp transistor in saturation region is ________
a) -0.2 V
b) -0.5V
c) 0.2 V
d) 0.5 V
Answer: b
Explanation: The value of Vcb is -0.5V for a pnp transistor and 0.5V for an npn transistor.
8. For a pnp transistor in the active region the value of Vce is
a) Less than 0.3V
b) Less than 3V
c) Greater than 0.3V
d) Greater than 3V
Answer: a
Explanation: For a pnp transistor Vce is less than 0.3V, for an npn transistor it is greater than 0.3V.
9. Which of the following is true for a pnp transistor in active region?
a) CB junction is reversed bias and the EB junction is forward bias
b) CB junction is forward bias and the EB junction is forward bias
c) CB junction is forward bias and the EB junction is reverse bias
d) CB junction is reversed bias and the EB junction is reverse bias
Answer: a
Explanation: Whether the transistor in npn or pnp, for it be in active region the EB junction must be reversed bias the CB junction must be forward bias.
10. Which of the following is true for a pnp transistor in saturation region?
a) CB junction is reversed bias and the EB junction is forward bias
b) CB junction is forward bias and the EB junction is forward bias
c) CB junction is forward bias and the EB junction is reverse bias
d) CB junction is reversed bias and the EB junction is reverse bias
Answer: b
Explanation: Whether the transistor in npn or pnp, for it be in saturation region the EB junction must be forward bias the CB junction must be forward bias.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “BJT in Amplifier Design”.
1. Find the maximum allowed output negative swing without the transistor entering saturation, and
a) 1.27 mV
b) 1.47 mV
c) 1.67 mV
d) 1.87 mV
Answer: d
Explanation:
electronic-devices-circuits-questions-answers-bjt-amplifier-design-q1
2. The corresponding maximum input signal permitted is
a) 1.64 mV
b) 1.74 mV
c) 1.84 mV
d) 1.94 mV
Answer: d
Explanation: If we assume linear operation right to saturation we can use the gain A v to calculate the maximum input signal. Thus for an output swing ∆ V o = 0.8 we have
∆ V i = ∆ V o / A v = -0.7 / -360 = 1.94 mV.
For the amplifier circuit in Fig. 6.33 with Vcc = +10 V, Rc = 1 kΩ and the DC collector bias current equal to Ic,
3. Find the voltage gain.
a) 100 Ic
b) 200 Ic
C) 400 Ic
d) 800 Ic
Answer: c
Explanation:
electronic-devices-circuits-questions-answers-bjt-amplifier-design-q3
4. The maximum possible positive output signal swing as determined by the need to keep the transistor in the active region.
a) 9.7 + Ic
b) 9.7 – Ic
c) 10.3 + Ic
d) 10.3 – Ic
Answer: a
Explanation: Assuming the output voltage V o = 0.3v is the lowest V ce to stay out of saturation.
V o = 0.3 = 10 – I c R c
= 10 – I c R c + ∆V o
∆ V o = -10 + 0.3 + I c *1.
5. The maximum possible negative output signal swing as determined by the need to keep the transistor in the active region.
a) 0.1 Ic
b) Ic
c) 10 Ic
d) 100 Ic
Answer: b
Explanation: Maximum output voltage before the Transistor is cutoff.
V ce + ∆V o = V cc
∆V o = V cc – V ce
= 10 – 10 + 10 I c
= 10 I c .
6. The transistor in the circuit below is biased at a dc collector current of 0.5 mA. What is the voltage gain?
electronic-devices-circuits-questions-answers-bjt-amplifier-design-q6
a) -1 V/V
b) -10 V/V
c) -100 V/V
d) -1000 V/V
Answer: c
Explanation:
electronic-devices-circuits-questions-answers-bjt-amplifier-design-q6a
7. For a BJt Vt is 5 V, Rc = 1000 ohm and bias current Ic is 12 mA. The value of the voltage gain is __________
a) -1.2 V/V
b) -2.4 V/V
c) -3.6 V/V
d) -4.8 V/V
Answer: b
Explanation: Voltage gain is / Vt.
For the BJT amplifier circuit with Vcc = +10 V, Rc = 1 kΩ and the DC collector bias current equal to 5 mA,
8. The value of the voltage gain is _______________
a) -2 V/V
b) -4 V/V
c) -10 V/V
d) -20 V/V
Answer: a
Explanation: The voltage is 400 X Ic where Ic is 5 mA.
9. The maximum possible positive output signal swing as determined by the need to keep the transistor in the active region.
a) -1.7 V
b) -2.7 V
c) -3.7 V
d) -4.7 V
Answer: d
Explanation: The maximum voltage swing is given by -10 + 0.3 + . Putting Ic as 5 mA, we get -4.7 mV.
10. The maximum possible negative output signal swing as determined by the need to keep the transistor in the active region.
a) 0.5 V
b) 1 V
c) 5 V
d) 10 V
Answer: c
Explanation: It is given by -10 + 10 + . Putting Ic as 5 mA we get 5V.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Small Signal Operations and Model”.
1. A transistor with ß = 120 is biased to operate at a dc collector current of 1.2 mA. Find the value of gm.
a) 12mA/V
b) 24 mA/V
c) 36 mA/V
d) 48 mA/V
Answer: d
Explanation:
electronic-devices-circuits-questions-answers-small-signal-operations-model-q1
2. A transistor with ß = 120 is biased to operate at a dc collector current of 1.2 mA. Find the value of R?p.
a) 625 ohm
b) 1250 ohm
c) 2500 ohm
d) 5000 ohm
Answer: c
Explanation:
electronic-devices-circuits-questions-answers-small-signal-operations-model-q1
3. A transistor with ß = 120 is biased to operate at a dc collector current of 1.2 mA. Find the value of Re.
a) 2.5 ohm
b) 20.6 ohm
c) 25.2 ohm
d) 30.4 ohm
Answer: b
Explanation:
electronic-devices-circuits-questions-answers-small-signal-operations-model-q1
4. A transistor operating with nominal gm of 60 mA/V has a ß that ranges from 50 to 200. Also, the bias circuit, being less than ideal, allows a 20% variation in Ic. What is the smallest value found of the resistance looking into the base?
a) 347 ohm
b) 694 ohm
c) 1041 ohm
d) 1388 ohm
Answer: b
Explanation:
electronic-devices-circuits-questions-answers-small-signal-operations-model-q4
5. A transistor operating with nominal gm of 60 mA/V has a ß that ranges from 50 to 200. Also, the bias circuit, being less than ideal, allows a 20% variation in Ic. What is the largest value found of the resistance looking into the base?
a) 1050 ohm
b) 21000 ohm
c) 3150 ohm
d) 4200 ohm
Answer: d
Explanation:
electronic-devices-circuits-questions-answers-small-signal-operations-model-q4
6. A designer wishes to create a BJT amplifier with a gm of 50 mA/V and a base input resistance of 2000 O or more. What is the minimum ß he can tolerate for the transistor used?
a) 100
b) 150
c) 200
d) 250
Answer: a
Explanation:
electronic-devices-circuits-questions-answers-small-signal-operations-model-q6
7. A designer wishes to create a BJT amplifier with a gm of 50 mA/V and a base input resistance of 2000 O or more. What emitter bis current should he choose?
a) 1.06 mA
b) 1.16 mA
c) 1.26 mA
d) 1.36 mA
Answer: c
Explanation:
electronic-devices-circuits-questions-answers-small-signal-operations-model-q6
8. Which of the following is true?
a) Ib = ß Ic
b) Ib = ß + 1/ Ic
c) Ib = Ic/ß
d) Ib = Ic/ ß – 1
Answer: c
Explanation: The correct relationship between Ic and Ie is Ib = Ic/ß.
9. The SI units of transconductance is
a) Ampere/ volt
b) Volt/ ampere
c) Ohm
d) Siemens
Answer: a
Explanation: Transcoductance is given by Ic/Vt.
10. Which of the following represents the correct mathematical form of the term denoted by the symbol Rp?
a) ß/gm
b) Vt/Ib
c) All of the mentioned
d) None of the mentioned
Answer: c
Explanation: Both of the expressions are identical.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Basic BJT Amplifier Configuration”.
An amplifier is measured to have an internal resistance of 10 kΩ, voltage gain of 100V/V and output resistance of 100 Ω. Also, when a load resistance of 1 kΩ is connected between the output resistance if found to decrease to 8 kΩ. If the amplifier is fed with the signal source having an internal resistance of 2 kΩ, then
1. Find Gm.
a) 1 A/V
b) 10 A/V
c) 100 A/V
d) 1000 A/V
Answer: a
Explanation: Gm = / or 100/100 A/V.
2. Find Av.
a) 9.09 V/V
b) 10 V/V
c) 90.9 V/V
d) 100 V/V
Answer: c
Explanation: Av = Avo X Rl/Ro+Rl or 100 X 1000/1000+100 or 90.9 V/V.
3. Find Gvo.
a) 53.3 V/V
b) 63.3 V/V
c) 73.3 V/V
d) 83.3 V/V
Answer: d
Explanation: Gvo = / .
4. Find Gv.
a) 53.4 V/V
b) 72.7 V/V
c) 83.3 V/V
d) 90.9 V/V
Answer: b
Explanation: Gv = / Avo or 83.3 X 90.9 / 100 V/V.
5. Find R out.
a) 146 Ω
b) 292 Ω
c) 584 Ω
d) 1168 Ω
Answer: a
Explanation: Rout = Rl . Put in the respective values and solve.
6. Find Ai.
a) 182 A/A
b) 364 A/A
c) 546 A/A
d) 728 A/A
Answer: d
Explanation: / gives the required value of Ai.
The circuit shown below is a small sine wave signal with average zero and transistor ß =
electronic-devices-circuits-questions-answers-basic-bjt-amplifier-configuration-q7
7. Find the value of R to establish a dc emitter current of about 0.5 mA.
a) 28.57 kΩ
b) 57.04 kΩ
c) 114.08 kΩ
d) 228.16 kΩ
Answer: a
Explanation:
electronic-devices-circuits-questions-answers-basic-bjt-amplifier-configuration-q7a
8. Find R to establish a dc collector voltage of about +5V.
a) 5 kΩ
b) 10 kΩ
c) 15 kΩ
d) 20 kΩ
Answer: d
Explanation: V c = 15 – R c .I c
5 = 15 – R c * 0.99 * 0.5m
R c = 20.2kΩ
= 20kΩ.
9. For R = 10 kΩ and transistor Ro = 200 kΩ, determine the overall voltage gain.
a) -21 V/V
b) -42 V/V
c) -86 V/V
d) -123 V/V
Answer: c
Explanation:
electronic-devices-circuits-questions-answers-basic-bjt-amplifier-configuration-q9
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Biasing in BJT Amplifier Circuits”.
consider the figure shown below and answer the questions that proceed.
electronic-devices-circuits-questions-answers-biasing-bjt-amplifier-circuits-q1
1. For Vcc = 15V, R1 = 100 kΩ, R = 3.9 kΩ, R = 6.8 kΩ and ß = 100, determine the dc collector current for each transistor.
a) 0.29 mA
b) 0.48 mA
c) 0.96 mA
d) 1.92 mA
Answer: c
Explanation:
electronic-devices-circuits-questions-answers-biasing-bjt-amplifier-circuits-q1a
2. For Vcc = 15V, R1 = 100 kΩ, R = 3.9 kΩ, R = 6.8 kΩ and ß = 100, determine the dc collector voltage for each transistor.
a) 4.25 V
b) 8.5 V
c) 12.75 V
d) 17 V
Answer: b
Explanation:
electronic-devices-circuits-questions-answers-biasing-bjt-amplifier-circuits-q1a
3. Find R for R = 5 kΩ.
a) 2.4 kΩ
b) 4.8 kΩ
c) 17.3 kΩ
d) 34.6 kΩ
Answer: a
Explanation: It is the parallel combination of the 32 kΩ resistor and 2.6 kΩ resistor respectively.
4. Find Vb1/Vsig for R = 5 kΩ.
a) 0.08 V/V
b) 0.16 V/V
c) 0.24 V/V
d) 0.32 V/V
Answer; d
Explanation:
electronic-devices-circuits-questions-answers-biasing-bjt-amplifier-circuits-q4
5. Find R .
a) 2.4 kΩ
b) 4.8 kΩ
c) 17.3 kΩ
d) 34.6 kΩ
Answer: a
Explanation: It is the parallel combination of the 32 kΩ resistor and 2.6 kΩ resistor respectively.
6. Find Vb2/Vb1.
a) -34 V/V
b) -51 V/V
c) – 68.1 V/V
d) -100 V/V
Answer: c
Explanation:
electronic-devices-circuits-questions-answers-biasing-bjt-amplifier-circuits-q6
7. For Rl = 2 kΩ find Vo/Vb2.
a) -59.3 V/V
b) -29.7 V/V
c) -89.1 V/V
d) None of the mentioned
Answer: a
Explanation:
electronic-devices-circuits-questions-answers-biasing-bjt-amplifier-circuits-q7
8. Find the overall voltage gain.
a) 323 V/V
b) 646 V/V
c) 969 V/V
d) 1292 V/V
Answer: d
Explanation:
electronic-devices-circuits-questions-answers-biasing-bjt-amplifier-circuits-q8
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Spread Spectrum”.
A pseudo-noise sequence is generated using a feedback shift register of length m = 4. The chip rate is 107 chips per second.
1. The PN sequence length is
a) 10
b) 12
c) 15
d) 18
Answer: c
Explanation: The PN sequence length is N = 2 m – 1 = 16 – 1 = 15.
2. The chip duration is
a) 1µs
b) 0.1 µs
c) 0.1 ms
d) 1 ms
Answer: b
Explanation: Tc = 1/(10 7 ) or 0.1 µs.
3. The period of PN sequence is
a) 1.5 µs
b) 15 µs
c) 6.67 ns
d) 0.67 ns
Answer: b
Explanation: The period of the PN sequence is T = NTc = 15 x 0.1 = 1.5 s
4. A slow FH/MFSK system has the following parameters.
Number of bits per MFSK symbol = 4
Number of MFSK symbol per hop = 5
The processing gain of the system is
a) 13.4 dB
b) 37.8 dB
c) 6 dB
d) 26 dB
Answer: d
Explanation: PG = Wc/Rs = 5 x 4 = 20 or 26 db.
5. A fast FH/MFSK system has the following parameters.
Number of bits per MFSK symbol = 4
Number of pops per MFSK symbol = 4
The processing gain of the system is
a) 0 dB
b) 7 dB
c) 9 dB
d) 12 dB
Answer: d
Explanation: PG = 4 x 4 = 16 or 12 db.
A rate 1/2 convolution code with dfrec = 10 is used to encode a data sequence occurring at a rate of 1 kbps. The modulation is binary PSK. The DS spread spectrum sequence has a chip rate of 10 MHz.
6. The coding gain is
a) 7 dB
b) 12 dB
c) 14 dB
d) 24 dB
Answer: a
Explanation: 0.5 x 10 = 5 or 7 db is the coding gain.
7. The processing gain is
a) 14 dB
b) 37 dB
c) 58 dB
d) 104 dB
Answer: b
Explanation: PG = (10 7 )/ = 5000 or 37 db.
An FH binary orthogonal FSK system employs an m 15 stage liner feedback shift register that generates an ML sequence. Each state of the shift register selects one of L non over lapping frequency bands in the hopping pattern. The bit rate is 100 bits/s. The demodulator employ non coherent detection.[/expand]
8. If the hop rate is one per bit, the hopping bandwidth
for this channel is
a) 6.5534 MHz
b) 9.4369 MHz
c) 2.6943 MHz
d) None of the mentioned
Answer: a
Explanation: The length of the shift-register sequence is L = 2 m – 12 15 = 32767 bits
For binary FSK modulation, the minimum frequency separation is 2/T, where 1/T is the symbol rate. The hop rate is 100 hops/sec. Since the shift register has L 32767 states and each state utilizes a bandwidth of 2/T = 200 Hz, then the total bandwidth for the FH signal is 6.5534 MHz.
9. Suppose the hop rate is increased to 2 hops/bit and the receiver uses square law combining the signal over two hops. The hopping bandwidth for this channel is
a) 3.2767 MHz
b) 13.1068 MHz
c) 26.2136 MHz
d) 1.6384 MHz
Answer: b
Explanation: If the hopping rate is 2 hops/bit and the bit rate is 100 bits/sec, then, the hop rate is 200 hops/sec. The minimum frequency separation for orthogonality 2/T 400 Hz. Since there are N 32767 states of the shift register and for each state we select one of two frequencies separated by 400 Hz, the hopping bandwidth is 13.1068 MHz.
10. In a fast FH spread spectrum system, the information is transmitted via FSK with non coherent detection. Suppose there are N = 3 hops/bit with hard decision decoding of the signal in each hop. The channel is AWGN with power spectral density 0.5No and an SNR 20 ~13 dB . The probability of error for this system is
a) 0.013
b) 0.0013
c) 0.049
d) 0.0049
Answer: b
Explanation: The total SNR for three hops is 20 ~ 13 dB. Therefore the SNR per hop is 20/3. The probability of a chip error with non-coherent detection is
electronic-devices-circuits-questions-answers-spread-spectrum-q10
This set of Electronic Devices and Circuits Mcqs focuses on “The AC Analysis of a Small-Signal Low-Frequency Common Emitter Transistor”.
1. The feature of an approximate model of a transistor is
a) it helps in quicker analysis
b) it provides individual analysis for different configurations
c) it helps in dc analysis
d) ac analysis is not possible
Answer: a
Explanation: The small signal model helps in quicker ac analysis of a transistor. The approximate model is applicable for all the configurations. The dc analysis is not obtained by using a small signal model of transistor.
2 A transistor has h fe =100, h ie =2kΩ, h oe =0.005mmhos, h re =0. Find the output impedance if the lad resistance is 5kΩ.
a) 5kΩ
b) 4kΩ
c) 20kΩ
d) 15kΩ
Answer: b
Explanation: R O =I/h oe =1/0.005m
=20kΩ.R O I = R O || R L I =20||5
=4kΩ.
3. A CE amplifier when bypassed with a capacitor at the emitter resistance has
a) increased input resistance and increased voltage gain
b) increased input resistance and decreased voltage gain
c) decreased input resistance and increased voltage gain
d) decreased input resistance and decreased voltage gain
Answer: c
Explanation: When a transistor is bypassed with a capacitor, it short circuits in the small signal analysis of transistor and the resistor too shorts. The input resistance becomes RI=hie. The value of the input resistance is decreased and the gain now will be increasing.
4. A transistor has h ie =2kΩ, h oe =25µmhos and h fe =60 with an unbypassed emitter resistor Re=1kΩ. What will be the input resistance and output resistance?
a) 90kΩ and 50kΩ respectively
b) 33kΩ and 45kΩ respectively
c) 6kΩ and 40kΩ respectively
d) 63kΩ and 40kΩ respectively
Answer: d
Explanation: As the emitter is unbypassed, the input resistance Ri=h ie +(1+h fe )Re
=2+61=63kΩ. The output resistance R O =1/h oe =1/25MΩ=40kΩ.
5. A transistor has h ie =1KΩ and h fe =60 with an bypassed emitter resistor R e =1kΩ. What will be the input resistance and output resistance?
a) 90kΩ and 50kΩ respectively
b) 33kΩ and 45kΩ respectively
c) 6kΩ and 40kΩ respectively
d) 63kΩ and 40kΩ respectively
Answer: d
Explanation: As the emitter is bypassed, the input resistance R i =h ie
=1kΩ. The output resistance R O =1/h oe but the value is not given.
So, h oe =0 and R O =1/0=∞.
6. In the given circuit, find the equivalent resistance between A and B nodes.
electronic-devices-circuits-questions-answers-mcqs-q6
a) 100kΩ
b) 50kΩ
c) 40kΩ
d) 60kΩ
Answer: b
Explanation: R AB =R O ||100Ω
= (R S I +h ie /1+h fe )||100
=9+1/100||100=100||100=50Ω.
7. Which of the following acts as a buffer?
a) CC amplifier
b) CE amplifier
c) CB amplifier
d) cascaded amplifier
Answer: a
Explanation: The voltage gain of a common collector amplifier is unity. It is then used as a buffer. The CC amplifier is also called as an emitter follower. Though there is no amplification done, the output will be stabilised.
8. Which of the following is true?
a) CC amplifier has a large current gain
b) CE amplifier has a large current gain
c) CB amplifier has low voltage gain
d) CC amplifier has low current gain
Answer: b
Explanation: The CE amplifier has high current and voltage gains. The CC amplifier has unity voltage gain which cannot be regarded as high. The common base amplifier has a unity current gain and high voltage gain.
9. In an NPN silicon transistor, α=0.995, I E =10mA and leakage current I CBO =0.5µA. Determine I CEO .
a) 10µA
b) 100µA
c) 90µA
d) 500µA
Answer: b
Explanation: I C =α I E +I CBO =0.995*10mA+0.5µA=9.9505mA.
I B =I E -I C =10-9.9505=0.0495mA. β=α/=0.995/=199
I CEO =9.9505-199*0.0495=0.1mA==100µA.
10. In CB configuration, the value of α=0.98A. A voltage drop of 4.9V is obtained across the resistor of 5KΩ when connected in collector circuit. Find the base current.
a) 0.01mA
b) 0.07mA
c) 0.02mA
d) 0.05mA
Answer: c
Explanation: Here, I C =4.9/5K=0.98mA
α = I C /I E .So,
I E =I C /α=0.98/0.98=1mA.
I B =I E -I C =1-0.98=0.02mA.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Biasing Parameters”.
1. The current gain of BJT is_________
a) g m r o
b) g m /r o
c) g m r i
d) g m /r i
Answer: c
Explanation: We know, current gain A V =h fe . In π model, h fe is referred to β.
We know, r i = β/g m .
From this, β=r i g m .
2. For the amplifier circuit of fig. The transistor has β of 800. The mid band voltage gain VO/VI of the circuit will be_________
electronic-devices-circuits-questions-answers-biasing-parameters-q2
a) 0
b) <1
c) =1
d) 800
Answer: c
Explanation: The circuit is PNP transistor, collector coupled amplifier. The voltage gain is unity for a CC amplifier. Hence on observation, the CC amplifier gives a unity gain.
3. In a bipolar transistor at room temperature, the emitter current is doubled the voltage across its base emitter junction_________
a) doubles
b) halves
c) increase by about 20mV
d) decreases by about 20mV
Answer: c
Explanation: The change in voltage with temperature can be found by, V = 2.3mV O . In a bipolar transistor at room temperature if the emitter current is doubled the voltage across its base emitter junction thereby doubles.
4. A common emitter transistor amplifier has a collector current of 10mA, when its base current is 25µA at the room, temperature. What is input resistance?
a) 3kΩ
b) 5kΩ
c) 1kΩ
d) 7kΩ
Answer: c
Explanation: We know, β/g m =r i
= (I C /I B )/(I C /V T )=V T /I B =25m/25µ=1k.
5. For an NPN transistor connected as shown in below, V BE =0.7V. Give that reverse saturation current of junction at room temperature is 10-13A, the emitter current is_________
electronic-devices-circuits-questions-answers-biasing-parameters-q5
a) 30mA
b) 39mA
c) 29mA
d) 49mA
Answer: d
Explanation: When the collector and base are shorted, the transistor behaves as a normal diode. So, the diode equations imply. I E =I O (e V/V0 -1). We get, I E =49mA.
6. The voltage gain of given circuit below is_________
electronic-devices-circuits-questions-answers-biasing-parameters-q6
a) 100
b) 20
c) 10
d) 30
Answer: c
Explanation: The gain for the given circuit can be found by, A V =R F /R S
=100K/10K=10.
7. A small signal source V=Acos20t+Bsin10000t, is applied to a transistor amplifier as shown. The transistor has β=150 and h ie =3KΩ. What will be the V O ?
electronic-devices-circuits-questions-answers-biasing-parameters-q7
a) 1500
b) -150
c) -1500Bsin10000t
d) -150Bsin10000t
Answer: d
Explanation: A V =-h fe R L I /h ie =3*150/3=-150. So, V O =-150V
But cos20t has low frequency so capacitors are open circuited. Only, the sine component is allowed.
So, V o =-150Bsin10000t.
8. Which of the following statements are correct for basic transistor configurations?
a) CB Amplifiers has low input impedance and low current gain
b) CC Amplifiers has low input impedance and high current gain
c) CE Amplifiers has very poor voltage gain but very high input impedance
d) The current gain of CB Amplifier is higher than the current gain of CC Amplifiers
Answer: a
Explanation: The CE amplifier has moderate input and output impedances. The CC amplifier has unity voltage gain. The common ba se amplifier has a unity current gain and high voltage gain.
9. The collector current is 2.945A and α=0.98. The leakage current is 2µA. What is the emitter current and base current?
a) 3mA and 55µA
b) 2.945mA and 55µA
c) 3.64mA and 33µA
d) 5.89mA and 65µA
Answer: a
Explanation: (I C – I CBO )/α=I E
= /0.98=3mA.
I E =I C +I B . So, I B =3-2.495=0.055mA=55µA.
10. The change in collector emitter voltage from 6V to 9V causes increase in collector current from 6mA to 6.3mA. Determine the dynamic output resistance.
a) 20kΩ
b) 10kΩ
c) 50kΩ
d) 60kΩ
Answer: b
Explanation: r o =∆V CE /∆I C
=3/0.3m=10kΩ.
11. A transistor is connected in CB configuration. The emitter voltage is changed by 200mV, the emitter by 5mA. During this transition the collector base voltage is kept constant. What is the input dynamic resistance?
a) 30Ω
b) 60Ω
c) 40Ω
d) 50Ω
Answer: c
Explanation: The ratio of change in emitter base voltage (∆V EB ) to resulting change in emitter current (∆ IE ) at constant collector base voltage (V CB ) is defined as input resistance. This is denoted by r i .
We know, ∆V EB /∆ IE =r i
=200/5=40Ω.
This set of Electronic Devices and Circuits Assessment Questions and Answers focuses on “Problems on AC and DC Analysis”.
1. In the circuit, transistor has β =60, V BE =0.7V. Find the collector to emitter voltage drop h oe .
electronic-devices-circuits-assessment-questions-answers-q1
a) 5V
b) 3V
c) 8V
d) 6V
Answer: d
Explanation: We know, I C =(V CC -V BE )/R B
By putting the values, we have I C =5.9mA. I E =I C /α. So, I E =5.99mA.
h oe = V CC -R C (I C +I B ). We have h oe =6V.
2. For the circuit shown, find the quiescent point.
electronic-devices-circuits-assessment-questions-answers-q2
a)
b)
c)
d)
Answer: c
Explanation: We know, h oe =12V
(I C ) SAT =V CC /R L =12/6K=2mA. I B =10V/0.5M=20µA. I C = βI B =1mA. I
h oe =V CC -I C R L =12-1*6=6V. So, quiescent point is .
3. For the circuit shown, find the quiescent point.
electronic-devices-circuits-assessment-questions-answers-q3
a)
b)
c)
d)
Answer: c
Explanation: We know, I E =V EE /RE=10/5kΩ=2mA
I C =α I E =I E =2mA
V CB =V CC -I C R L =20-10=10V. So, quiescent point is .
4. In the circuit shown below, β =100 and V BE =0.7V. The Zener diode has a breakdown voltage of 6V. Find the operating point.
electronic-devices-circuits-assessment-questions-answers-q4
a)
b)
c)
d)
Answer; a
Explanation: We know, by KVL -12+(I C +I B )1K+6+V BE =0
We have I E =5.3. I C = αI E =5.24mA. From another loop, -12+I E IK+V BE =0
We have, h oe =12-5.3m*1000=6.7V. Hence the Q point is .
5. 10. When the β value is large for a given transistor, the I C and h oe values are given by_________
a) (V CC -V BE )/R B , V CC -R C I C
b) (V CC +V BE )/R B , V CC -R C (I C +I B )
c) (V CC +V BE )/R B , V CC +R C (I C +I B )
d) (V CC +V BE )/R B ,V CC +R C (I C -I B )
Answer: a
Explanation: The base current I B is zero when β value is large. So, the h oe changes to V CC -R C I C . The collector current I C is changed to (V CC -V BE )/R B from β(V CC -V BE )/R E + R B .
6. For the circuit shown, find the quiescent point.
electronic-devices-circuits-assessment-questions-answers-q6
a)
b)
c)
d)
Answer: c
Explanation: We know, I E =V EE /R E =30/10kΩ=3mA
I C =α I E =I E =3mA
V CB =V CC -I C RL=25-15=10V. So, quiescent point is .
7. The PNP transistor when used for switching the power, then it is called_________
a) sourcing current
b) sinking current
c) forward sourcing
d) reverse sinking
Answer: a
Explanation: Sometimes DC current gain of a bipolar transistor is too low to directly switch the load current or voltage, so multiple switching transistors is used. The load is connected to ground and the transistor switches the power to it.
8. In which of the regions, the small capacitors are open circuited?
a) high frequency
b) medium frequency
c) low frequency
d) the region does not affect he capacitors
Answer: c
Explanation: In low frequency region, the small capacitors are open circuited and large capacitors are in active state. In high frequency region, the large capacitors are short circuited and small capacitors are active.
9. In mid frequency region, the large capacitors are short and small capacitors are open circuited. What happens to the R C coupled circuit?
a) the circuit is now frequency blind
b) it is DC isolated
c) the circuit turns reactive
d) the circuit is AC dependent
Answer: a
Explanation: In the mid frequency region, the whole circuit is resistive because the large capacitors are short and small capacitors are open circuited. The gain is constant in this region. So, the circuit is frequency blind as the gain is constant in this region.
10. In low frequency region, the gain_________
a) increases
b) decreases
c) remains same
d) depends on value of capacitors
Answer: b
Explanation: At very low frequency, the gain decreases due to the coupling capacitor. When the frequency is decreased, the reactance of the circuit increases and the drop across the coupling capacitor increases. The gain is therefore decreased as the output decreased.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Two-Port Devices and Hybrid Model”.
1. How many terminals does a two – port network have?
a) 2
b) 4
c) 6
d) 8
Answer: b
Explanation: A two – port network has four terminals, two input terminals and two output terminals. Each pair of input or output terminals consist of a current and voltage terminal. In a two – port network, port 1 is usually the input terminal while port 2 is the output terminal.
2. What is the most efficient way to calculate voltage or current in a two – port network?
a) Z parameters
b) Y parameters
c) H parameters
d) G parameters
Answer: c
Explanation: We use H parameters or hybrid parameters to calculate the voltage or current of a two – port network. Z parameters stand for impedance parameters, Y parameters stand for admittance parameters and G parameters stand for inverse hybrid parameters.
3. What is the size of a h parameter matrix for a two – port network?
a) 2 × 2
b) 2 × 4
c) 4 × 4
d) 4 × 2
Answer: a
Explanation: The hybrid parameters for a two – port network consist of a 2 × 2 matrix consisting of four elements: h 11 , h 12 , h 21 and h 22 . They are calculated by short circuiting the input / output port and calculating the respective ratios.
4. What are the units of the off – diagonal h parameters?
a) Volt
b) Ampere
c) dimensionless
d) Ohm
Answer: c
Explanation: The off – diagonal elements of the hybrid parameter matrix are dimensionless. One of the diagonal parameters is measured in ohm while the other diagonal parameter is measured in mho.
5. How many types of parameters are used for two port networks?
a) 8
b) 1
c) 2
d) 6
Answer: d
Explanation: There are essentially 6 different parameters involving two – port networks. They are Z parameter , Y parameter , T parameter , T’ parameter , H parameter and G parameter .
6. How do we calculate V 1 in a two – port network?
a) V 1 = h 21 I 2 + h 22 V 2
b) V 1 = h 11 I 1 + h 12 V 2
c) V 1 = h 11 I 2 + h 12 V 2
d) V 1 = h 21 I 1 + h 22 V 2
Answer: b
Explanation: We use the hybrid matrix parameters to calculate the input output voltage and current of the two ports. V 1 is the input voltage from the first port and it is calculated using the formula: V 1 = h 11 I 1 + h 12 V 2 .
7. What is the value of I 2 if h 21 = 2, I 1 = 1A, h 22 = 0.04 and V 2 = 50V?
a) 2A
b) 0.2A
c) 0.4A
d) 4A
Answer: d
Explanation: To find the current I 2 we use the below mentioned formula. Given values are h 21 = 2, I 1 = 1A, h 22 = 0.04 and V 2 = 50V.
I 2 = h 21 I 1 + h 22 V 2
I 2 = × + × = 2 + 2 = 4A.
8. At what frequency range do hybrid models work the best?
a) Very low frequency
b) Low to moderate frequency
c) Moderate to high frequency
d) High frequency
Answer: b
Explanation: Hybrid models with varying current and voltage work best at low to moderate frequencies. At higher frequencies, we use power and energy variables instead of the current and voltage variables.
9. Hybrid parameters were also called “Series – parallel parameters”.
a) True
b) False
Answer: a
Explanation: Hybrid parameters or h parameters were initially called series – parallel parameters. Later they were renamed as hybrid parameters due to their ability to adapt to the physical characteristics of transistors.
10. Why are h parameters used for transistor calculations?
a) Ease of use
b) Ease of use, easy to measure
c) It is not used for transistor calculations
d) Easy to measure
Answer: b
Explanation: H parameters are widely used for transistor calculations for multiple reasons. It is easier to use since in some cases we are unable to calculate the Z and Y parameters. It is also easy to measure since most transistor manufacturers mention the h parameters, else it is easy to calculate from the static characteristic curve.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Transistor Hybrid Model”.
1. The h-parameters analysis gives correct results for __________
a) large signals only
b) small signals only
c) both large and small
d) not large nor small signals
Answer: b
Explanation: Every linear circuit is associated with h –parameters. When this linear circuit is terminated with load r L , we can find input impedance, current gain, voltage gain, etc in terms of h-parameters. For small a.c. signal, transistor behaves as linear device. Under such circumstances the a.c. signal operation of a transistor can be described in terms of h-parameters.
2. For what type of signals does a transistor behaves as linear device?
a) small signals only
b) large signals only
c) both large and small signal
d) no signal
Answer: a
Explanation: The small variation in the total voltage and current due to an application of signal moves the point up and down just by a bit and that whole up and down dynamics of the operating point from its DC value point can be approximated to be along a straight line. Whole analysis can be done with same assumption of linearity with the limit of signal being in the same vicinity of the DC operating point.
That’s how we get all those equations for linear operation and also its small signal equivalent model using h-parameters.
3. How many h-parameters are there for a transistor?
a) two
b) three
c) four
d) five
Answer: c
Explanation: A transistor has four h-parameters –
H 11 = V 1 /i 1
H 21 = i 2 /i 1
H 12 = V 1 /V 2
H 22 = i 2 /V 2 .
4. The dimensions of h ie parameters are _______
a) MHO
b) OHM
c) Farad
d) Ampere
Answer: b
Explanation: h ie = v BE /i b ; common emitter input impedance
For V CE = 0 i.e. output short circuited
Where v BE = Base emitter voltage i.e. input voltage
i b = Base current i.e. input current
We know that V/I = R. Its unit is ohm.
5. The h fe parameter is called _______ in CE arrangement with output short circuited.
a) Voltage Gain
b) Current gain
c) Input impedance
d) Output impedance
Answer: b
Explanation: h fe in CE arrangement is given as
H fe = I c /I b for V CE = 0 short circuited
So, this is current gain as it is then output to input current ratio.
6. What happens to the h parameters of a transistor when the operating point of the transistor changes?
a) It also changes
b) Does not change
c) May or may not change
d) Nothing happens
Answer: a
Explanation: It is very difficult to get exact values of h parameters for a particular transistor. It is because these parameters are subject to considerable variation unit to unit variation, variation due to change in temperature and variation due to the operating point.
7. If temperature changes, h parameters of a transistor _____
a) also change
b) does not change
c) remains same
d) may or may not change
Answer: a
Explanation: It is very difficult to get exact values of h parameters for a particular transistor. It is because these parameters are subject to considerable variation unit to unit variation, variation due to change in temperature and variation due to the operating point.
8. In CE arrangement, the value of input impedance is approximately equal to _____
a) H IE
b) H IB
c) H OE
d) H RE
Answer: a
Explanation: h ie = v BE /i b ; common emitter input impedance
For V CE = 0 i.e. output short circuited
Where v BE =Base emitter voltage i.e. input voltage
i b = Base current i.e. input current.
9. How many h-parameters of a transistor are dimensionless?
a) Four
b) Two
c) Three
d) One
Answer: b
Explanation: H 11 = V 1 /I 1 ; for V 2 = 0
This parameter is input impedance with output short.
Its unit is ohm.
H 21 = i 2 /i 1 ; for V 2 = 0
This parameter is Current gain ratio with output short.
It is unit less or dimensionless.
H 12 = V 1 /V 2 ; for i 1 = 0
This is voltage gain feedback ratio with terminals open.
And it is unit less or dimensionless.
H 22 = i 2 /v 2 ; for i 1 = 0
This is output admittance with input terminals open.
Its unit is ohm -1 or mho.
Thus there are two h-parameters which are unit less or dimensionless.
10. The values of h-parameters of a transistor in CE arrangement are ________ arrangement.
a) same as for CB
b) same as for CC
c) different from that in CB
d) similar to no
Answer: c
Explanation: The values of h-parameter in CE arrangement:
H ie = V be /I b ; for V ce = 0
H fe = i c /i b ; for V ce = 0
H re = V be /V ce ; for i b = 0
H oe = i c /v ce ; for i b = 0
The values of h-parameter in CB arrangement:
H ib = V be /I e ; for V bc = 0
H fb = i c /I e ; for V bc = 0
H rb = V be /V bc ; for i e = 0
H ob = i c /v be ; for i e = 0 .
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Determination of the H-Parameters”.
1. How do we determine the hybrid parameters h 11 and h 21 of a two – port network?
a) Short circuiting the input terminal
b) Open circuiting the input terminal
c) Short circuiting the output terminal
d) Open circuiting the output terminal
Answer: c
Explanation: There are four h parameters, they are: h 11 , h 12 , h 21 and h 22 . To determine the values of h 11 and h 21 we need to short circuit the output terminal of a given two – port network. Short circuiting the output terminal makes the output terminal voltage V 2 equal to zero.
2. In a hybrid model of a two – port network, parameter h 11 is also known as?
a) Input conductance
b) Input resistance
c) Output conductance
d) Output resistance
Answer: b
Explanation: The parameter h 11 can be calculated by short circuiting the output port. By doing so, we can calculate h 11 = V 1 / I 1 . Since this gives h 11 the units of volts / ampere or in other words ohms, it is also known as input resistance.
3. How do we determine the value of the hybrid parameter h 21 in a two – port network?
a) h 21 = I 1 / I 2
b) h 21 = V 1 / V 2
c) h 21 = I 2 / I 1
d) h 21 = V 2 / V 1
Answer: c
Explanation: To calculate the value of h 21 , we short circuit the output terminal so the output voltage V 2 equals zero. Hence the equation for hybrid parameters becomes, I 2 = h 21 I 1 + 0. Therefore, h 21 = I 2 / I 1 .
4. Which hybrid parameter is used to calculate forward current gain in a two – port network?
a) h 11
b) h 12
c) h 21
d) h 22
Answer: c
Explanation: The hybrid parameter h 21 can be calculated by short circuiting the output port. By doing so, we can calculate h 21 = I 2 / I 1 . Since this is a ratio between two current units, h 21 is unitless. It is known as the forward current gain of the circuit.
5. In a hybrid model, which parameter is known as the output conductance?
a) h 11
b) h 12
c) h 21
d) h 22
Answer: d
Explanation: When the input terminal of a two – port device is open circuited, the input current flowing equals to zero. Hence the value of h 22 = I 2 / V 2 . The unit of this parameter is ampere / volt or mho. Hence it is known as output conductance.
6. Which of the following four hybrid parameters is used to calculate reverse voltage gain?
a) h 11
b) h 12
c) h 21
d) h 22
Answer: b
Explanation: The hybrid parameter h 12 can be calculated by open circuiting the input port. By doing so, we can calculate h 12 = V 1 / V 2 . Since this is a ratio between two voltage units, h 12 is unitless. It is known as the reverse voltage gain of the circuit.
7. How do we calculate the value of hybrid parameter h 12 if V 2 = 20V and V 1 = 50V in a two – port device?
a) 0.04
b) 0.25
c) 0.4
d) 2.5
Answer: d
Explanation: To find the hybrid parameter h 12 we use the below mentioned formula. Given values are V 2 = 20V and V 1 = 50V.
h 21 = V 1 / V 2 = 50 / 20 = 2.5.
8. How do we determine the hybrid parameters h 12 and h 22 in a two – port linear device?
a) Short circuiting the input terminal
b) Open circuiting the input terminal
c) Short circuiting the output terminal
d) Open circuiting the output terminal
Answer: b
Explanation: From the four hybrid parameters, to determine the value of h 12 and h 22 we need to open circuit the input terminal. This in turn makes the current from the input terminal I 1 equal to zero.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Measurement of H-Parameters”.
1. For the two – port network shown below, assume R 1 = R 2 = 10kΩ. What is the value of the hybrid parameter h 22 ?
a) 0.1 mho
b) 10 mhos
c) 20 mhos
d) 15 mhos
Answer: a
Explanation: To calculate h 22 , we open circuit the input terminal. Hence the current flowing through resistor R 1 is zero and the voltage V 1 across the input terminal is equal to the voltage V 2 across output terminal. Therefore, to find h 22 , given that R 2 = 10kΩ.
V 1 = V 2
I 2 = V 2 / 10kΩ
h 22 = I 2 / V 2 = 0.1 mho
2. For the two – port network shown below, assume R 1 = R 2 = 10kΩ. What is the value of the hybrid parameter h 11 ?
a) 10kΩ
b) 20kΩ
c) 10kΩ
d) 5kΩ
Answer: c
Explanation: To calculate h 11 , we short circuit the output terminal. Hence the current flowing through resistor R 2 is zero. Therefore, to find h 11 , given that R 1 = 10kΩ.
V 1 = R 1 x I 1
h 11 = V 1 / I 1 = R 1 x I 1 / I 1 = R 1 = 10kΩ.
3. How do we calculate the inverse hybrid parameter g 11 of a two – port network?
a) g 11 = I 1 / V 1
b) g 11 = I 2 / V 1
c) g 11 = I 1 / V 2
d) g 11 = I 2 / V 2
Answer: a
Explanation: The inverse hybrid parameter g 11 for a two – port network is calculated by open circuiting the output terminal. In this case, the current I 2 is equal to zero. The parameter g 11 is calculated by taking the ratio of input terminal current over input terminal voltage.
4. What are the hybrid parameters used to analyze?
a) MOSFET
b) Junction Field Effect Transistor
c) Bipolar Junction Transistor
d) It has no use
Answer: c
Explanation: The four hybrid parameters of a two – port network are used to analyze the workings of a Bipolar Junction Transistor or in other words a BJT. The inverse hybrid parameters are used to analyze the Junction Field Effect Transistor of JFET.
5. How is the h 21 hybrid parameter of a two – port network used?
a) It is used as a voltage source at input port
b) It is used as a current source at input port
c) It is used as a voltage source at output port
d) It is used as a current source at output port
Answer: d
Explanation: While checking the hybrid parameter equivalent network of a two – port network we see that the hybrid parameter h 21 is sued as a current source at the output port. While using Kirchhoff’s current law, the hybrid parameter equations can be represented as a current source of h 21 I 1 .
6. How do we calculate the inverse hybrid parameter g 21 of a two – port network?
a) g 21 = I 1 / I 2
b) g 21 = I 2 / I 1
c) g 21 = V 2 / V 1
d) g 21 = V 1 / V 2
Answer: c
Explanation: The inverse hybrid parameter g 21 for a two – port network is calculated by open circuiting the output terminal. In this case, the current I 2 is equal to zero. The parameter g 21 is calculated by taking the ratio of output terminal voltage over input terminal voltage. It is also known as the open circuit voltage gain.
7. How is the h 12 hybrid parameter of a two – port network used?
a) It is used as a voltage source at output port
b) It is used as a current source at output port
c) It is used as a voltage source at input port
d) It is used as a current source at input port
Answer: c
Explanation: While checking the hybrid parameter equivalent network of a two – port network we see that the hybrid parameter h 12 is sued as a voltage source at the input port. While using Kirchhoff’s voltage law, the hybrid parameter equations can be represented as a fixed voltage source of h 21 V 2 .
8. How do we calculate the inverse hybrid parameter g 12 of a two – port network?
a) g 12 = I 2 / I 1
b) g 12 = I 1 / I 2
c) g 12 = V 1 / V 2
d) g 12 = V 2 / V 1
Answer: b
Explanation: The inverse hybrid parameter g 12 for a two – port network is calculated by short circuiting the input terminal. In this case, the current V 1 is equal to zero. The parameter g 12 is calculated by taking the ratio of input terminal current over output terminal current.
9. How do we calculate the inverse hybrid parameter g 22 of a two – port network?
a) g 22 = V 2 / I 1
b) g 22 = I 1 / V 2
c) g 22 = V 1 / I 2
d) g 22 = V 2 / I 2
Answer: d
Explanation: The inverse hybrid parameter g 22 for a two – port network is calculated by short circuiting the output terminal. In this case, the current V 1 is equal to zero. The parameter g 22 is calculated by taking the ratio of output terminal voltage over output terminal current.
10. We can measure hybrid h – parameters using inverse hybrid g – parameters.
a) True
b) False
Answer: a
Explanation: The hybrid h – parameters can be measured using the inverse hybrid g – parameters. All the parameters used to analyze a two – port network can be used to convert and measure other different parameters.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Conversion Formulas for the Parameters of the Three-Transistor Configurations”.
1. What are the units of hybrid parameter h ie for a transistor?
a) Ohms
b) Siemens
c) Voltage
d) Dimensionless
Answer: a
Explanation: The hybrid parameter h ie is the input impedance in a common emitter circuit. Hence, the units of this hybrid parameter will be measured in Ohms. It is the same unit as the hybrid parameters h ib and h ic .
2. How many hybrid parameters are there in a transistor circuit?
a) 16
b) 8
c) 12
d) 4
Answer: c
Explanation: There are a total of 12 hybrid parameters in a transistor circuit. There are four parameters for each mode of operation. Each mode of operation has only four working hybrid parameters, which are derived from the four basic hybrid parameters: h 11 , h 12 , h 21 and h 22 . The 12 transistor parameters are: h ib , h ie , h ic , h rb , h re , h rc , h fb , h fe , h fc , h ob , h oe and h oc .
3. What is the equivalent hybrid parameter for the transistor hybrid parameter h re ?
a) h 11
b) h 12
c) h 21
d) h 22
Answer: b
Explanation: The transistor hybrid parameter h re is equivalent to the hybrid parameter h 12 . The transistor hybrid parameter is a measure of the open circuit reverse voltage transfer ratio. The transistor equivalent parameter is a ratio of the voltages across two terminals.
4. What is the equivalent hybrid parameter for the transistor hybrid parameter h fe ?
a) h 11
b) h 12
c) h 21
d) h 22
Answer: c
Explanation: The transistor hybrid parameter h fe is equivalent to the hybrid parameter h 21 . The transistor hybrid parameter is a measure of the output short circuit forward current gain. The transistor equivalent parameter is a ratio of the currents across two terminals.
5. What is the equivalent hybrid parameter for the transistor hybrid parameter h oe ?
a) h 11
b) h 12
c) h 21
d) h 22
Answer: d
Explanation: The transistor hybrid parameter h oe is equivalent to the hybrid parameter h 22 . The transistor hybrid parameter is a measure of the input open circuit with output admittance. The transistor equivalent parameter is a ratio of the current across one terminal and the voltage.
6. What is the equivalent hybrid parameter for the transistor hybrid parameter h ib ?
a) h 11
b) h 12
c) h 21
d) h 22
Answer: a
Explanation: The transistor hybrid parameter h ib is equivalent to the hybrid parameter h 11 . The transistor hybrid parameter is a measure of the output short circuit with input impedance. The transistor equivalent parameter is a ratio of the voltage across one terminal and the current.
7. Which external parameters affect the transistor hybrid parameters?
a) Temperature
b) Collector current
c) Temperature and collector current
d) External parameters don’t affect hybrid parameters
Answer: c
Explanation: The transistor hybrid boundaries are not consistent and change with temperature, collector current and collector emitter voltage. Consequently, while planning the circuit the hybrid boundaries ought to be estimated under similar conditions as the actual circuit.
8. Why do we use hybrid parameter models while describing the working of a transistor?
a) Ease of use
b) Characterize bipolar transistors
c) Acts as a switch
d) Makes no difference
Answer: b
Explanation: H-boundaries are one framework for describing bipolar semiconductors. The h-boundaries of a semiconductor will give you a smart thought what it can do, how to utilize it viably in a circuit, and whether it is proper for a specific circuit. Hence, we use them to characterize bipolar transistors.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Analysis of Transistor Amplifier Circuit using h-parameters”.
1. What is the current gain of a transistor amplifier circuit if I 1 = 10mA, I 2 = 20mA, V 1 = 25V, V 2 = 15V?
a) -0.5
b) -2
c) 0.5
d) 2
Answer: b
Explanation: To calculate the current gain of a transistor amplifier circuit, we use the below formula. The current gain of a transistor amplifier is the ratio of output to input currents. Given the values of I 1 = 10mA and I 2 = 20mA.
Current gain = A = – I 2 / I 1 = 20 / 10 = 2.
2. How the current gain of a common emitter transistor amplifier expressed in terms of transistor hybrid parameters?
a) -h fe / (1 + Z L × h re )
b) h fe / (1 + Z L × h re )
c) -h fe / (1 + Z L × h oe )
d) h fe / (1 + Z L × h oe )
Answer: c
Explanation: The current gain of a transistor amplifier circuit is expressed as -I 2 / I 1 . This can be expressed in terms of transistor hybrid parameters. The current gain of a common emitter transistor amplifier in terms of hybrid parameters is expressed as -h fe / (1 + Z L × h oe ), where Z L is the load impedance.
3. How the current gain of a common base transistor amplifier expressed in terms of transistor hybrid parameters?
a) -h fb / (1 + Z L × h ob )
b) h fb / (1 + Z L × h ob )
c) -h fb / (1 + Z L × h rb )
d) h fb / (1 + Z L × h rb )
Answer: a
Explanation: The current gain of a transistor amplifier circuit is expressed as -I 2 / I 1 . This can be expressed in terms of transistor hybrid parameters. The current gain of a common base transistor amplifier in terms of hybrid parameters is expressed as -h fb / (1 + Z L × h ob ), where Z L is the load impedance.
4. How the current gain of a common collector transistor amplifier expressed in terms of transistor hybrid parameters?
a) -h fc / (1 + Z L × h rc )
b) h fc / (1 + Z L × h rc )
c) -h fc / (1 + Z L × h oc )
d) h fc / (1 + Z L × h oc )
Answer: c
Explanation: The current gain of a transistor amplifier circuit is expressed as -I 2 / I 1 . This can be expressed in terms of transistor hybrid parameters. The current gain of a common collector transistor amplifier in terms of hybrid parameters is expressed as -h fc / (1 + Z L × h oc ), where Z L is the load impedance.
5. What is the input impedance of a transistor amplifier circuit if I 1 = 10mA, I 2 = 20mA, V 1 = 25V, V 2 = 15V?
a) 0.5kΩ
b) 2kΩ
c) 1.5kΩ
d) 2.5kΩ
Answer: d
Explanation: To calculate the input impedance of a transistor amplifier circuit, we use the below formula. The input impedance of a transistor amplifier is the ratio of input voltage to input current. Given the values of I 1 = 10mA and V 1 = 25V:
Input impedance = Z i = V 1 / I 1 = 25 / 10 = 2.5kΩ
6. How the input impedance of a common emitter transistor amplifier expressed in terms of transistor hybrid parameters?
a) h ie – (h fe × h oe / Y L + h re )
b) h ie + (h fe × h oe / Y L + h re )
c) h ie – (h fe × h re / Y L + h oe )
d) h ie + (h fe × h re / Y L + h oe )
Answer: c
Explanation: The input impedance of a transistor amplifier circuit is expressed as V 1 / I 1 . This can be expressed in terms of transistor hybrid parameters. The input impedance of a common emitter transistor amplifier in terms of hybrid parameters is expressed as h ie – (h fe × h re / Y L + h oe ), where Y L is the load admittance.
7. How the input impedance of a common base transistor amplifier expressed in terms of transistor hybrid parameters?
a) -h ib + (h fb × h ib / Y L + h ob )
b) h ib – (h fb × h ib / Y L + h ob )
c) -h ib + (h fb × h rb / Y L + h ob )
d) h ib – (h fb × h rb / Y L + h ob )
Answer: d
Explanation: The input impedance of a transistor amplifier circuit is expressed as V 1 / I 1 . This can be expressed in terms of transistor hybrid parameters. The input impedance of a common base transistor amplifier in terms of hybrid parameters is expressed as h ib – (h fb × h rb / Y L + h ob ), where Y L is the load admittance.
8. How the input impedance of a common collector transistor amplifier expressed in terms of transistor hybrid parameters?
a) h ic + (h fc × h rc / Y L + h oc )
b) h ic – (h fc × h rc / Y L + h oc )
c) -h ic + (h fc × h rc / Y L + h oc )
d) -h ic – (h fc × h rc / Y L + h oc )
Answer: b
Explanation: The input impedance of a transistor amplifier circuit is expressed as V 1 / I 1 . This can be expressed in terms of transistor hybrid parameters. The input impedance of a common collector transistor amplifier in terms of hybrid parameters is expressed as h ic – (h fc × h rc / Y L + h oc ), where Y L is the load admittance.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Comparison of Transistor Amplifier”.
1. Which type of amplifiers exhibits the current gain approximately equal to unity without any current amplification?
a) CE
b) CB
c) CC
d) Cascade
Answer: b
Explanation: In common base amplifier, input signal is applied at emitter terminal while the amplified output signal is obtained at the collector terminal with respect to ground.
For the AC signals, the base terminal is specifically connected to ground through the capacitor.
Even, the output resistance is very high & hence, the current gain is approximately equal to unity. Due to this, there is no possibility of current amplification. Consequently, the CB amplifier exhibits high voltage gain.
2. Why is the Darlington configuration not suitable for more than two transistors?
a) Because leakage current increases and voltage gain decreases with multiple numbers of transistors
b) Because leakage current decreases and voltage gain increases with multiple numbers of transistors
c) Because leakage current as well as voltage gain increases with multiple numbers of transistors
d) Because leakage current as well as voltage gain decreases with multiple numbers of transistors
Answer: a
Explanation: As the number of transistors increases, the leakage current also increases. The leakage current gets multiplied by the current gain of Darlington configuration.
Generally, the voltage gain of CC configuration is nearly equal to ‘1’ but the voltage gain of Darlington configuration is very less than ‘1’.
Therefore, if we increase the number of transistors in Darlington configuration the voltage gain will ultimately reduce. But, in order to prevent these likely undesirable conditions, Darlington configuration is completely inappropriate for more than two transistors.
3. What should be the level of input resistance to allow the occurrence of source loading in common base amplifier configuration?
a) low
b) high
c) moderate
d) stale
Answer: a
Explanation: As per the configuration of CB amplifier, it is evident that its input resistance is very low but its output resistance is enormously high.
However, the lower value of input resistance allows the stipulation of source loading in common base amplifier circuit.
Thus, there is no current amplification because of unity current gain. These all reasons eventually add to high level of voltage gain.
4. Which among the below assertions is not a relevant property of CE amplifier?
a) High voltage gain
b) High current gain
c) High input resistance
d) High output resistance
Answer: d
Explanation: The voltage gain, current gain and input resistance CE amplifiers are utterly high but it has low output resistance.
The collector resistor (R c ) performs the purpose of controlling the collector current. Input and emitter resistors are adopted for biasing of transistor in an active region so that it becomes possible for the transistor to function as an amplifier.
Due to high current gain at the output of RC coupled CE amplifier, the resistance level at the output is exceedingly low.
5. What is the phase-shift between input and output voltages of CE amplifier?
a) 90°
b) 120°
c) 180°
d) 270°
Answer: c
Explanation: During the amplification method of RC coupled CE amplifier, there is a phase shift of about 180o between input and output. Basically, the output is said to be reversed version of input.
The magnitude of output voltage becomes higher as compared to that of input signal but the shape is correctly similar to that of an input signal.
But, the input ac signal gets amplified along with the phase-shift of 180o between input and output.
6. Which capacitor is used to block DC portion by allowing to pass only AC portion of the amplified signal to load?
a) Input Coupling Capacitor
b) Bypass Capacitor
c) Output Coupling Capacitor
d) Both coupling and bypass capacitor
Answer: c
Explanation: In RC coupled CE amplifier, the transistor is connected in common-emitter configuration and capacitors C 1 & C 2 are coupling capacitors.
Input coupling capacitor is used for coupling the ac input voltage to the transistor base. Inversely, the output coupling capacitor (C 2 ) is used for coupling an output of an amplifier to the load resistance or to the next stage of an amplifier.
Besides these, input coupling capacitor blocks any DC element present in AC input voltage & couples only AC component of input signal whereas bypass capacitor offers a low reactance to the amplified AC signal.
7. The configuration in which voltage gain of transistor amplifier is lowest is ____________
a) common collector
b) common emitter
c) common base
d) common emitter & base
Answer: a
Explanation: In common collector configuration because the emitter voltage follows that of the base. Offering a high input impedance and a low output impedance it is extensively used as a buffer. The voltage gain is unity, even though current gain is high. The input and output signals are in phase.
8. The configuration in which current gain of transistor amplifier is lowest is ___________
a) common collector
b) common base
c) common emitter
d) common emitter & base
Answer: b
Explanation: In Common base configuration, the input impedance is very low; While offering a high output impedance. Although the voltage is high, the current gain is low and the overall power gain is also low when compared to the other transistor configurations available. Thus, there is no current amplification because of unity current gain.
9. The configuration in which input impedance of transistor amplifier is lowest is ___________
a) common collector
b) common emitter
c) common base
d) common emitter & base
Answer: c
Explanation: In Common base configuration, the input impedance is very low; While offering a high output impedance. Although the voltage is high, the current gain is low and the overall power gain is also low when compared to the other transistor configurations available.
10. The configuration in which output impedance of transistor amplifier is highest is ___________
a) common collector
b) common base
c) common emitter
d) common collector and base
Answer: b
Explanation: In Common base configuration, the input impedance is very low; While offering a high output impedance. Although the voltage is high, the current gain is low and the overall power gain is also low when compared to the other transistor configurations available.
11. Q. What should be the gain of an amplifier at 20 kHz if the half power frequencies are f L = 20 Hz and f H = 15 kHz along with mid band gain = 80?
a) 22.76
b) 45.09
c) 40.08
d) 48.07
Answer: d
Explanation:
f L = 20 Hz
f H = 15 kHz
A v = 80
To determine: Voltage gain at 20 kHz
Formula : A v = A v / √1 + (f/f H ) 2
To determine the voltage gain at 20 kHz, we recognize that,
A v = A v / √1 + (f/f H ) 2
= 80 / √1 + 2
= 48.07.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Linear Analysis of a Transistor Circuit”.
1. What does the static characteristic curve of a transistor define?
a) Steady state relations
b) Current
c) Voltage
d) It is not applicable for transistors
Answer: a
Explanation: The static transistor curve of a transistor defines the steady state relations between the input and output current and voltages. They are calculated and plotted with the help of DC measurements and linear analysis is performed.
2. How many known variables do we need to perform a linear analysis of a transistor circuit?
a) 3
b) 5
c) 6
d) 4
Answer: d
Explanation: There are a total of 6 variables in a transistor circuit. A transistor is a three-terminal device with each terminal having current and voltage measurements. We require any four variables to determine the other two variables of a transistor circuit.
3. Which of the following is an independent variable in a linear analysis of a common base transistor?
a) Base current
b) Collector base voltage
c) Collector current
d) Collector emitter voltage
Answer: b
Explanation: In a linear analysis of a common base transistor, we have two independent variables. The collector base voltage is an independent variable and the collector current is the dependent variable. We can obtain I C using V CB from the characteristic curve.
4. What frequency should be used to check the linear analysis of a transistor?
a) High frequency
b) Medium-high frequency
c) Medium-low frequency
d) Low frequency
Answer: d
Explanation: Low frequency conditions are used to check the linear analysis of transistor models. This is because, under low frequency d-c conditions the value of emitter current is almost equal to the value of collector current.
5. Which of the following is an independent variable in a linear analysis of a common base transistor?
a) Collector emitter voltage
b) Collector current
c) Emitter base voltage
d) Emitter current
Answer: d
Explanation: In a linear analysis of a common base transistor, we have two independent variables. The emitter current is an independent variable and the emitter base voltage is the dependent variable. We can obtain V EB using I E from the characteristic curve.
6. Where is the operating point of transistor located in a linear analysis?
a) Linear region
b) Saturation region
c) Cut-off region
d) It is not located on this graph
Answer: a
Explanation: The operating point of the quiescent operating point is in the linear region of the output of the characteristic curve. In this region the input variations are proportional to the output variations and are easy to determine.
7. Which of the following is an inference made from the linear analysis of a transistor circuit?
a) 99% of the collector current flows to the emitter
b) 99% of the emitter current flows to the collector
c) 0.99% of the collector current flows to the emitter
d) 0.99% of the emitter current flows to the collector
Answer: a
Explanation: While performing the linear analysis of a transistor circuit we observe that 99% of the collector current flows to the emitter. Current always flows from the collector to the emitter and not the other way around since the emitter terminal is usually grounded.
8. What is the function of a capacitor in the linear analysis of a transistor?
a) To find the operating point
b) To block d-c signals
c) To block a-c signals
d) Makes no difference
Answer: b
Explanation: Capacitors are used in the linear analysis of a transistor circuit to block out and isolate the d-c signals while letting the a-c signals pass through. This is useful to calculate the load line in the analysis of the transistor.
9. What is the equation of the load line in the linear analysis of a transistor?
a) V CE = V CC + I B R L
b) V CE = V CC – I B R L
c) V CE = V CC + I C R L
d) V CE = V CC – I C R L
Answer: d
Explanation: Let the collector voltage supply be V CC in series with a load resistor R L . A straight line superimposed on the collector characteristic will provide us with the load line. The equation for the load line is V CE = V CC – I C R L .
10. What do the small signal parameters in a transistor circuit vary with respect to in the linear analysis?
a) Load line
b) Bias point
c) Temperature
d) Makes no difference
Answer: b
Explanation: The small signal parameters in a transistor circuit vary with respect to the bias point in the linear analysis model. This occurs even in the linear region of the characteristic curve of the transistor.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Physical Model of a CB Transistor”.
1. Where is the input measured in a common base transistor physical model?
a) Collector terminal
b) Emitter terminal
c) Base terminal
d) Ground
Answer: b
Explanation: In the physical model of a common base transistor amplifier the input is measured at the emitter terminal of the BJT biased device. Whereas, the output is measured across the collector terminal of the biased BJT device.
2. Which parameter of the physical model is varied while measuring the input characteristics of a common-base transistor?
a) Emitter current
b) Emitter voltage
c) Collector current
d) Emitter base voltage
Answer: d
Explanation: To determine the input characteristics, the collector-base voltage is kept constant at zero volts and the emitter base voltage is increased from zero volts to different voltage levels. For each voltage level of the input voltage, the input current is recorded.
3. Where is the output measured in a common base transistor physical model?
a) Collector terminal
b) Emitter terminal
c) Base terminal
d) Ground
Answer: a
Explanation: In the physical model of a common base transistor amplifier the output is measured at the collector terminal of the BJT biased device. Whereas, the input is measured across the emitter terminal of the biased BJT device.
4. Which parameter of the physical model is varied while measuring the output characteristics of a common-base transistor?
a) Emitter current
b) Emitter voltage
c) Collector current
d) Collector base voltage
Answer: d
Explanation: To determine the output characteristics, the emitter current is kept constant at zero and the collector base voltage is increased from zero volts to varying voltage levels. For each voltage level of the output voltage, the collector current is recorded.
5. What is the path of the output characteristic of a CB transistor physical model in the active region?
a) Linear
b) Exponential decrease
c) Exponential increase
d) Constant
Answer: d
Explanation: In a physical model of a common base transistor, the output characteristic curve first increases exponentially and then remains constant. In the active region of working the output characteristics show a constant value collector current with respect to collector base voltage.
6. What is the path of the output characteristic of a CB transistor physical model in the saturation region?
a) Linear
b) Exponential decrease
c) Exponential increase
d) Constant
Answer: c
Explanation: In a physical model of a common base transistor, the output characteristic curve first increases exponentially and then remains constant. In the saturation region of working the output characteristics show an exponential increase in the collector current with respect to collector base voltage.
7. What is the path of the input characteristic of a CB transistor physical model?
a) Linear
b) Exponential decrease
c) Exponential increase
d) Constant
Answer: a
Explanation: In a physical model of a common base transistor, the input characteristic curve increases exponentially. The emitter current increases exponentially with respect to emitter base voltage for fixed values of collector base voltage.
8. How do you calculate the dynamic input resistance of a CB transistor?
a) ΔV BE / ΔI C
b) ΔV BE / ΔI E
c) ΔV CB / ΔI C
d) ΔV CB / ΔI E
Answer: b
Explanation: Dynamic input resistance is defined as the ratio of change in emitter base voltage to the corresponding change in the emitter current. While the collector voltage is kept at a constant value. Therefore, r i = ΔV BE / ΔI E .
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Approximate Hybrid Model and its use in CE, CB, CC”.
1. What is the approximate hybrid parameter current gain of a CE amplifier?
a) -h fe
b) h fe
c) -h fc
d) h fc
Answer: a
Explanation: The current gain of a common emitter amplifier can be expressed in terms of approximate hybrid parameters. The current gain h fe is the forward transfer characteristics and is used to calculate the current gain in common emitter amplifiers.
2. What is the approximate hybrid parameter input resistance of a CE amplifier?
a) h oe
b) h ib
c) h ic
d) h ie
Answer: d
Explanation: The input resistance of a common emitter amplifier can be expressed in terms of approximate hybrid parameters. The input resistance is universally expressed as h ie . This is also the same for approximate as well as exact hybrid models.
3. What is the approximate hybrid parameter voltage gain of a CE amplifier?
a) h fe × R L / h ie
b) -h fe × R L / h ie
c) h fe × R L / h oe
d) -h fe × R L / h oe
Answer: b
Explanation: The voltage gain of a common emitter amplifier can be expressed in terms of approximate hybrid parameters. The voltage gain is equal to -h fe × R L / h ie where, R L is load resistance, h ie is input resistance and -h fe is the current gain.
4. What is the approximate hybrid parameter current gain of a CB amplifier?
a) -h fe / 1 + h fe
b) -h fe / 1 – h fe
c) h fe / 1 + h fe
d) h fe / 1 – h fe
Answer: c
Explanation: The current gain of a common base amplifier can be expressed in terms of approximate hybrid parameters. The current gain h fe is the forward transfer characteristics. The current gain of a common base amplifier is expressed as h fe / 1 + h fe .
5. What is the approximate hybrid parameter input resistance of a CB amplifier?
a) h fe / 1 – h fe
b) h fe / 1 + h fe
c) h ie / 1 – h fe
d) h ie / 1 + h fe
Answer: d
Explanation: The input resistance of a common base amplifier can be expressed in terms of approximate hybrid parameters. The input resistance is h ie / 1 + h fe where, h ie and h fe are the input resistance and current gain of CE amplifier respectively.
6. What is the approximate hybrid parameter voltage gain of a CB amplifier?
a) h fe × R L / h ie
b) -h fe × R L / h ie
c) h fe × R L / h oe
d) -h fe × R L / h oe
Answer: a
Explanation: The voltage gain of a common base amplifier can be expressed in terms of approximate hybrid parameters. The voltage gain is equal to h fe × R L / h ie where, R L is load resistance, h ie is input resistance of CE amplifier and -h fe is the current gain of CE amplifier.
7. What is the approximate hybrid parameter current gain of a CC amplifier?
a) 1 + h oe
b) 1 – h oe
c) 1 + h fe
d) 1 – h fe
Answer: c
Explanation: The current gain of a common collector amplifier can be expressed in terms of approximate hybrid parameters. The current gain h fe is the forward transfer characteristics. The current gain of a common collector amplifier is expressed as 1 + h fe .
8. What is the approximate hybrid parameter input resistance of a CC amplifier?
a) h ie – (1 + h fe ) × R L
b) h ie + (1 + h fe ) × R L
c) h ie – (1 + h fe ) / R L
d) h ie + (1 + h fe ) / R L
Answer: b
Explanation: The input resistance of a common collector amplifier can be expressed in terms of approximate hybrid parameters. The input resistance is h ie + (1 + h fe ) × R L where, h ie and h fe are the input resistance and current gain of CE amplifier respectively and R L is the load resistance.
9. What is the approximate hybrid parameter voltage gain of a CC amplifier?
a) 1 + h ie × R i
b) 1 – h ie × R i
c) 1 + h ie / R i
d) 1 – h ie / R i
Answer: d
Explanation: The voltage gain of a common collector amplifier can be expressed in terms of approximate hybrid parameters. The voltage gain is equal to 1 – h ie / R i where, R i is equal to h ie + (1 + h fe ) × R L .
10. We use approximate hybrid models for transistors because they introduce less than 10% error.
a) True
b) False
Answer: a
Explanation: For the analysis of transistor amplifiers we use approximate hybrid models instead of exact hybrid models. They introduce less than 10% error which can be ignored since hybrid parameters themselves are not steady but vary considerably for the same type of transistor.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Transistors Millers Theorem”.
1. What technique is used in Millers Theorem?
a) Two-port network
b) Hybrid parameters
c) Grounding
d) Short circuiting
Answer: a
Explanation: In Millers Theorem, we use the equivalent two-port network of the given electrical circuit. We divide the circuit into two parts, each part representing a different port for easier analysis of the circuit.
2. Which law is the Millers Theorem based on?
a) Ohm’s Law
b) Moore’s Law
c) Coulomb’s Law
d) Kirchhoff’s Current and Voltage Law
Answer: d
Explanation: The Millers Theorem deals with the impedance supplied by two current/voltage sources connected in parallel. These two versions were derived by Kirchhoff’s two laws: Voltage and Current laws. The dual Millers Theorem is based mainly on the current laws while the other theorem focuses on the voltage law.
3. According to Millers Theorem, what should be the configuration of voltages?
a) Both dependent voltages
b) Both independent voltages
c) One dependent voltage and one independent voltage
d) No specification necessary
Answer: c
Explanation: Miller theorem proposes that an impedance segment is provided by two irregular voltage sources that are associated in arrangement through the common ground. For all intents and purposes, one of them goes about as an autonomous voltage source and different demonstrations a directly reliant voltage.
4. What is theimpedance from the input port according to Millers Theorem?
a) Z in1 = Z × K / 1 – K
b) Z in1 = Z × K / 1 + K
c) Z in1 = Z / 1 – K
d) Z in1 = Z / 1 + K
Answer: c
Explanation: According to Millers Theorem, to calculate the input impedance from the input port we use: Z in1 = Z / 1 – K. Where, Z is the original circuit impedance and K is the ratio of the two nodal voltages V 2 / V 1 .
5. What is the impedance from the output port according to Millers Theorem?
a) Z in2 = Z × K / K – 1
b) Z in2 = Z × K / K + 1
c) Z in2 = Z / K – 1
d) Z in2 = Z / K +1
Answer: a
Explanation: According to Millers Theorem, to calculate the input impedance from the output port we use: Z in2 = Z × K / K – 1. Where, Z is the original circuit impedance and K is the ratio of the two nodal voltages V 2 / V 1 .
6. According to Millers Theorem, what is the impedance in input port if I 1 = 20 mA, I 2 = 30 mA and Z = 2.3 kΩ?
a) 0.38 kΩ
b) 3.83 kΩ
c) 0.57 kΩ
d) 5.75 kΩ
Answer: d
Explanation: To calculate theimpedance in input port, given I 1 = 20 mA, I 2 = 30 mA and Z = 2.3 kΩ:
Z in1 = × Z
α = I 2 / I 1 = 1.5
Z in1 = × Z = × 2.3 = 5.75 kΩ.
7. According to Millers Theorem, what is the impedance in output port if I 1 = 20 mA, I 2 = 30 mA and Z = 2.3 kΩ?
a) 0.38 kΩ
b) 3.83 kΩ
c) 0.57 kΩ
d) 5.75 kΩ
Answer: b
Explanation: To calculate theimpedance in output port, given I 1 = 20 mA, I 2 = 30 mA and Z = 2.3 kΩ:
Z in2 = × Z / α
α = I 2 / I 1 = 1.5
Z in2 = × Z / α = × 2.3 / 1.5 = 3.83 kΩ.
8. What are the applications of Millers Theorem?
a) Lower power consumption
b) Additional voltage source
c) Additional current source
d) Circuit optimization
Answer: b
Explanation: Millers Theorem is applied in an amplifier setting known as Millers amplifier. The amplifier can be used as an additional voltage source which converts the actual impedance into a virtual impedance. The virtual impedance can be thought of as a component connected in parallel to the amplifier input.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Junction Field-Effect Transistor – 1”.
1. FET is a voltage controlled device.
a) True
b) False
Answer: a
Explanation: Field Effect Transistors are voltage controlled devices, by applying some voltage between the gate and source, the drain current can be controlled. In order to control the operation of FET the gate to drain voltage is varied to operate the FET in different regions of operation.
2. Which of the following statement is true about FET?
a) It has high output impedance
b) It has high input impedance
c) It has low input impedance
d) It does not offer any resistance
Answer: b
Explanation: Because of the Sio2 insulator, doped between drain and source at the top, the resistance offered by this is very high. The insulator will stop the flow of electron from one part to another which acts as an open circuit.
3. Comparing the size of BJT and FET, choose the correct statement?
a) BJT is larger than the FET
b) BJT is smaller than the FET
c) Both are of same size
d) Depends on application
Answer: a
Explanation: BJT usually are built with a thickness of up to 1cm whereas the FET uses a fabrication technique which makes its size in mm.
4. What is the main advantage of FET which makes it more useful in industrial applications?
a) Voltage controlled operation
b) Less cost
c) Small size
d) Semiconductor device
Answer: c
Explanation: Because of its small size, the IC chips can be made even smaller which reduces the wear and tear. The process technology used with process technology constant a which is the ratio of Width and Length, the FET is made more advantageous.
5. For a FET when will maximum current flows?
a) V gs = 0V
b) V gs = 0v and V ds >= |V p |
c) V DS >= |V p |
d) V p = 0
Answer: b
Explanation: For a FET the current reaches maximum that is IDSS occurs when V gs = 0V and V DS >= |V p |
6. What is the value of current when the gate to source voltage is less than the pinch off voltage?
a) 1A
b) 5A
c) 100A
d) 0
Answer: d
Explanation: When the gate to source voltage is less than pinch off, both of the junctions will be reverse biased and hence no current flows.
7. The Shockley equation is __________________
a) I D = (1 – V gs /V p ) 2
b) I D = I DSS (1 – V gs /V p ) 2
c) I D = I DSS (1 – V gs /V p ) 1
d) I D = I DSS (1 + V gs /V p ) 2
Answer: b
Explanation: By using the diode equation Shockley derived the equation between the drain current and voltage V gs .
8. What is the value of drain current when V gs =pinch off voltage?
a) 0A
b) 1A
c) 2A
d) Cannot be determined
Answer: a
Explanation: I D = I DSS (1-V gs /V p ) 2
If V gs = V p ,then
I D = I DSS =0.
9. To use FET as a voltage controlled resistor, in which region it should operate?
a) Ohmic region
b) cut off
c) Saturation
d) cut off and saturation
Answer: a
Explanation: By varying the gate to source voltage, Resistance can be varied as follows r d = r o /(1-V gs /V p ) 2
10. For an n-channel FET, What is the direction of current flow?
a) Source to drain
b) Drain to source
c) Gate to source
d) Gate to drain
Answer: b
Explanation: When a voltage greater than pinch off is applied, the current starts flowing from Drain to source.
11. For a p-channel FET, What is the direction of current flow?
a) Source to drain
b) Drain to source
c) Gate to source
d) Gate to drain
Answer: a
Explanation: When the voltage is lesser than pinch off, the current flows from Source to Drain.
The forward bias drain and gate is the reason for the flow of electron from Drain to source, as the conventional current flows opposite to the electron flow, the current will flow from Source to Drain.
This set of Electronic Devices and Circuits Questions & Answers for Exams focuses on “The Junction Field-Effect Transistor – 2”.
1. Which of the following can be considered to be an advantage of FET amplifier as compared to BJT amplifier?
A – Higher input impedance
B – Good bias stability
C – Higher gain-bandwidth product
D – Lower noise figure
Select the correct answer using the codes given below
Codes:
a) A, B and C
b) A, B and D
c) B, C and D
d) A, C and D
Answer: b
Explanation: Advantage of FET over BJT are
i) Higher input impedance
ii) Good bias stability
iii) Lower noise figure.
2. Two identical FETs, each characterized by the parameters g_m and r_d are connected in parallel .The composite FET is then characterized by the parameters_____________
a) g m /2 and 2r d
b) g m /2 and 2r d
c) 2g m and r d /2
d) 2g m and r d /2
Answer: c
Explanation: By converting FET into ac equivalent circuits and connecting them in parallel,
g n = 2g m and r n = r d /2.
3. The pinch off voltage of JFET is 5v. What is its cut off voltage?
a) 2.5V
b) 3V
c) 4V
d) 5V
Answer: d
Explanation: Pinch off voltage =5V
At cut off the gate to source voltage of JFET is equal to pinch off voltage
V gs = V P => V gs = 5V.
4. The action of JFET in its equivalent circuit can be represented as which of the following?
a) Current controlled current source
b) Current controlled voltage source
c) Voltage controlled current source
d) Voltage controlled Voltage source
Answer: c
Explanation: In JFET equivalent circuit, the output current is controlled by the gate to source voltage and hence we can say it is a Voltage controlled current source.
5. Which of the following is the main advantage of Self bias?
a) Eliminates the need of two power supply
b) Maximum stability
c) Minimum stability
d) Maximum & Minimum stability
Answer: a
Explanation: Self bias eliminates the need of 2 power supply by connecting gate resistance to the supply voltage.
6. FET amplifier does not obey the law of conservation of energy.
a) True
b) False
Answer: b
Explanation: Amplifier obeys the law of conservation of energy.
7. Which of the following is the necessary condition to design an amplifier?
a) V ce ≤ 1 ⁄ 10 of RC
b) |V ce | ≤ 1 ⁄ 10 of RC
c) |V ce | ≤ 1 ⁄ 1000 of RC
d) |V ce | ≤ 1 ⁄ 100 of RC
Answer: b
Explanation: In order to design an amplifier the time constant and RC and voltage across specified terminals.
8. At higher frequency, the capacitance of an amplifier circuit is mainly because of which capacitance?
a) Coupling capacitors
b) Stray capacitance
c) Resistors
d) Inductors
Answer: b
Explanation: During High frequency, the stray capacitance is the only source of capacitance in an amplifier circuit, the stray capacitance and junction capacitance.
9. Which of the following is the equation for stray capacitance frequency?
a) F H = 1/27R
b) F H = 1/R
c) F H = 54/27R
d) F H = 11/27R
Answer: a
Explanation: During high frequency the capacitance is due to combination of transmission capacitance and junction capacitance. The equation is given by F H = 1/27R.
10. What is the maximum value of gain of an amplifier?
a) 140dB
b) 130dB
c) 120db
d) 100dB
Answer: c
Explanation: The maximum gain of amplifier is 1000000. When we consider this in decibel scale, it gives 120dB => gain in dB = 20log10 = 120dB.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Pinch off Voltage Vp”.
1. What is pinch off voltage?
a) The minimum voltage required to turn on the FET
b) The maximum voltage a FET can withstand
c) Current amplification factor/voltage gain
d) The value of voltage at which the current gets pinched to zero
Answer: d
Explanation: Once the voltage difference between the gate and source goes near to the pinch off voltage, the channel will get pinched off resulting in off state of the FET, which makes no conventional current flow.
2. A p-channel Ge JFET has max-half channel width 5µm and channel conductivity of 2/Ωcm, if E r = 2000cm 2 /Vsec, What is the value of pinch off voltage?
a) 8.21V
b) 82.1V
c) 88.21V
d) 5.2V
Answer: c
Explanation: Vp=qN A a 2 /2E
= (5×10 -4 ) (5×10 -4 )×2/(2×16×8.854×2000×10 -14 )
= 88.21V.
3. What will happen if gate voltage applied is positive to pinch off voltage?
a) Device burns
b) More current flows
c) Nothing happens
d) Current remains the same
Answer: a
Explanation: When the voltage applied across the gate terminal becomes more positive, all of the current will start flowing from drain to gate terminal. This results in breaking of insulator layer, resulting in device destruction.
4. How does a FET behave when the v-I characteristics are to the left of pinch off for an n channel FET?
a) Voltage controlled resistor
b) Amplifier
c) Switch
d) Diode
Answer: a
Explanation: When the voltage V gs is less than pinch off voltage, there will be no current flow, resulting in Ohmic region, on controlling the V gs and V Ds values, the FET acts as Voltage controlled resistor.
5. What is the relation between the drain current and source current once the voltage crosses pinch off?
a) I D = I S
b) I D = I S + 1.5
c) I D = 1/I S
d) I S – 2I D = 0
Answer: a
Explanation: For a FET, once the voltage exceeds the pinch off voltage, the electrons will start flowing from source to drain or vice versa, since no current flows through the gate terminal, the current transfer takes place only between the drain and source. Hence I D = I S .
6. If a FET has a pinch off voltage =-1V and I D =1mA, If V gs =0V, What is the value of I DSS ?
a) 1A
b) 1mA
c) 0A
d) 100mA
Answer: b
Explanation: I D =I DSS (1-V gs /V p ) 2
V gs =0V => : I D =I DSS =1mA.
7. For what value of V gs , the drain current will be 1/4th of its maximum current?
a) 0
b) 1
c) V p
d) V p /2
Answer: d
Explanation: I D =I DSS (1-V gs /V p ) 2
When I D =I DSS /4 => (1-V gs /V p ) 2 = 1/4
=>(1-V gs /V p ) 2 =1/2
=>V gs =V p /2.
8. For a n-channel FET, what is the condition of V gs for which the current becomes zero?
a) 0
b) 100V
c) V p
d) Infinite
Answer: c
Explanation: I D =I DSS (1-V gs /V p ) 2
When ID=0, (1-V gs /V p ) 2 =0
1=V gs /V p => V gs =V p .
9. For a FET having I DSS =2mA V gs =2V and V p =-1V, What is the value of source current?
a) 9mA
b) 18mA
c) 3mA
d) 1mA
Answer: b
Explanation: I D =I DSS (1-V gs /V p ) 2
I D =I S =2×=3 2 ×2=18mA
I S =18mA.
10. Find the current through gate if the FET was given with gate to source voltage =10V and drain to source voltage =20V, the pinch off voltage was -2V and I D =2mA.
a) 10mA
b) 20mA
c) 0mA
d) 2mA
Answer: c
Explanation: We know that for a FET, the gate terminal is heavily doped with an insulator, resulting in infinite resistance. Hence no current flows through the gate terminal. It has no relationship with the other current parameters. It is constant for all FETs.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The JFET Volt-Ampere Characteristics”.
1. For low value of V DS , the JFET behaves like a __________
a) Voltage Variable Resistor
b) Constant Voltage Device
c) Amplifier
d) Switch
Answer: a
Explanation: When V DS voltage is very less, there will not be much current flow, since as the V DS changes, the current value changes very little. Hence we can say that FET works as a voltage controlled Resistor.
2. If a JFET with length L=10µm, a=2µm, W=8µm, V p =-4V.What is the value of r ds at V gs = 0V?
a) 2KΩ
b) 5.2KΩ
c) 10KΩ
d) 9.8KΩ
Answer: d
Explanation: r ds =L/(2aqN D µ nW ) = Nd = 2V/qa 2 =1.33×10 21 atoms/m 3
µn=0.15m 2 /v-sec
On substituting the values, we get r ds =9.8KΩ.
3. A JFET has I D =10mA, I DSS =1A, V p =-1v, what is the value of V gs ?
a) -0.9V
b) 1V
c) -1V
d) 0.5V
Answer: a
Explanation: electronic-devices-circuits-questions-answers-jfet-volt-ampere-characteristics-q3
4. Where does the transfer curve lie for a p- channel FET?
a) First quadrant
b) Second quadrant
c) Third quadrant
d) Fourth quadrant
Answer: a
Explanation: For a P- channel FET, since the pinch off lies on right side of the origin, the current Id will increase from V P and rise to its left until it reaches I DSS . Usually I DSS lies on positive Y axis, therefore we can say transfer curve for the p-channel FET lies on first quadrant.
5. For an n-channel FET, the current v at V gs =-2V and V ds =5V was found to be 2mA, what will be the value of I D at V gs =0V and V ds =5V?
a) 0A
b) 2mA
c) Lesser than 0A
d) Greater than 2mA
Answer: d
Explanation: We know that for a n channel FET, the current will flow only if V gs is greater than V p , since in this case current is flowing, which indicates V gs is greater than pinch off. But as V gs increases the flow of current also increases, hence it will be greater than 2mA.
6. For a p-channel FET, the current I D at V gs = 2V and V ds = 5V and V gs = 3V was found to be 2mA, what will be the value of I D at V gs = 0V and V ds = 5V?
a) 0A
b) 2mA
c) Lesser than 0A
d) Greater than 2mA
Answer: a
Explanation: We know that for a p channel FET, the current will flow only if V gs is lesser than V gs , but here pinch off voltage is 3V, when V gs =5V which is greater than pinch off voltage the channel will be destroyed resulting in no conventional current flow.
7. A FET is biased at I D = I DSS /4, at this point the value of trans conductance is __________________ (I D =10mA, V p =-5V).
a) 1mA/V
b) 2mA/V
c) 3mA/V
d) 4mA/V
Answer: b
Explanation: electronic-devices-circuits-questions-answers-jfet-volt-ampere-characteristics-q7
8. What is the value of resultant g m if two non identical FETs are connected in parallel?
a) (µ 1 r d1 + µ 2 r d2 )/ (r d1 +r d2 )
b) (µ 1 r d2 + µ 2 r d1 )/ (r d1 +r d2 )
c) 0
d) µ 1 r d1 + µ 2 r d2
Answer: b
Explanation: we know that µ=g m r d
But g m’ =g m1 +g m2
And r d’ =r d1 ×r d2 /r d1 + r d2
From the above two equations, g m’ =(µ 1 r d2 + µ 2 r d1 )/ (r d1 +r d2 ).
9. What will be the value of r d , if two identical FETs are connected in parallel?
a) Doubles
b) Reduces to half
c) 0
d) Infinite
Answer: b
Explanation: r d’ =r d1 ×/r d1 + r d2
If r d1 =r d1 =r d => then r d’ =r d 2 /2r d =r d /2.
10. What will be the value of trans conductance if two Identical FETs are connected in parallel?
a) Doubles
b) Reduces to half
c) 0
d) Infinite
Answer: a
Explanation: g m’ =I DS /V gs at V DS =0V, if two FET are identical then,
g m1 =g m2 =g m
But g m’ =g m1 +g m2
Therefore g m’ =g m +g m =2g m .
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The FET Small-Signal Model”.
1. What is trans-conductance?
a) Ratio of change in drain current to change in collector current
b) Ratio of change in drain current to change in gate to source voltage
c) Ratio of change in collector current to change in drain current
d) Ratio of change in collector current to change in gate to source voltage
Answer: b
Explanation: The change in drain current which is resulted due to change in gate to source voltage in a FET is measured by trans-conductance. This is termed as trans because it provides relationship between input and output quantity.
2. For a FET, graph is drawn by taking voltage V GS in X axis and drain current in Y axis, if the value of X changes from 10 to 20 results in change in value of Y axis from 2 to 3. What is the value of trans conductance?
a) 1
b) 2
c) 0.1
d) 0.01
Answer: c
Explanation: Trans-conductance= change in drain current/ change in gate to source voltage
trans conductance=3-2/20-10=1/10
trans conductance=0.1.
3. The slope obtained in V GS vs I D was 0.002. What is the value ofg m ?
a) 1
b) 2
c) 0.002
d) 0
Answer: c
Explanation: g m = change in drain current/ change in gate to source voltage g m = slope of V GS vs I D g m = 0.002.
4. Which of the following is an expression for g m0 ?
a) g m0 = I DSS /V p
b) g m0 = 2I DSS /|V p |
c) g m0 = I DSS /5V p
d) g m0 = I DSS /2V p
Answer: b
Explanation: g m0 is the value of g m when V GS =0, we have seen that trans conductance is the ratio of drain current to change in gate to source voltage, but this is an exceptional case, here the transistor will be working in cut off region.
5. Find the maximum value of g m for FET with I DSS =10mA, V p =-2V, V GS =5V?
a) 10mS
b) 20mS
c) 1mS
d) 0
Answer: a
Explanation: g m0 =2I DSS /|V p | g m0 =2×10mA/2V g m0 =10mS.
6. Find the value of g m for FET with I DSS =8mA, V p =4V, V GS =-0.5V?
a) 1mS
b) 2mS
c) 3mS
d) 3.5mS
Answer: d
Explanation: electronic-devices-circuits-questions-answers-fet-small-signal-model-q6
7. A FET has I DSS =4I D and g m0 = 10mS then g m = _________________________
a) 10mS
b) 20mS
c) 5mS
d) 14mS
Answer: c
Explanation: electronic-devices-circuits-questions-answers-fet-small-signal-model-q7
8. Determine the value of output impedance for JFET, if the value of g m =1mS?
a) 1Kohm
b) 0
c) 100Kohm
d) 5Kohm
Answer: a
Explanation: Output impedance=inverse of trans conductance
Output impedance=1/1mS
Output impedance=1Kohm.
9. In a small signal equivalent model of an FET, What does g m V GS stand for?
a) A pure resistor
b) Voltage controlled current source
c) Current controlled current source
d) Voltage controlled voltage source
Answer: b
Explanation: For FET, the voltage is applied across the gate and source to control the drain current, hence while writing small signal model of an FET, on the output side g m V GS represents a current source which can be controlled by the input voltage V GS .
10. Given y fs = 3.6mS and y os = 0.02mS, determine r 0 ?
a) 100Kohm
b) 50Mohm
c) 50Kohm
d) 20Kohm
Answer: c
Explanation: r 0 =1/y os
It is independent of y fs. => r 0 =1/20mS
r 0 =50Kohm.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Insulated-Gate FET – 1”.
1. Which of the following is true about MOSFET?
a) There is no direction between channel and gate terminal
b) There exists a channel and gate short connection
c) Channel is not present and cannot be created
d) They have low input impedance
Answer: a
Explanation: According to the physical structure of MOSFET, there is a Silicon dioxide layer between channel and gate terminal. Since Silicon dioxide acts as a resistor with infinite resistance, there is no connection between the channel and gate terminal of MOSFET.
2. What will happen if a metal is used instead of Silicon dioxide in the fabrication of MOSFET?
a) Device burns
b) No changes in the behaviour of MOSFET
c) Input impedance increases
d) Current cannot be generated in MOSFET
Answer: a
Explanation: If a metal is used instead of Silicon dioxide, once the biasing condition is achieved, as the drain to source voltage current starts floating to gate terminal resulting in burning of the device.
3. Necessary condition to create a channel in n-channel enhancement MOSFET is ____________
a) V gs > V t
b) V gs < V t
c) V gs < 2V t
d) 2V gs > V t
Answer: a
Explanation: When gate to source voltage exceeds the threshold voltage, the source and substrate and drain and substrate forms a forward bias resulting in the formation of depletion region which acts as a channel.
4. Determine the least voltage V DS required to operate E-MOSFET on active region. (V gs =2V,V t =0.7V)
a) 1.2V
b) 2V
c) 0V
d) 1.3V
Answer: d
Explanation: V DS =V gs -V t V DS =2-0.7V V DS =1.3V.
5. Determine the value of K for a MOSFET with V gs =8V,V t =2V and I D =10A?
a) 0.001A/v 2
b) 0.278 A/v 2
c) 0.5761A/v 2
d) 0.0021A/v 2
Answer: b
Explanation: K=I D /(V gs -V t ) 2
K=10/36
K=0.278.
6. Which of the following is the necessary to hold drain current in 0 for n channel E-MOSFET?
a) V gs > V t
b) V gs < V t
c) V gs < 2V t
d) 2V gs > V t
Answer: b
Explanation: For V gs < V t , the channel is not created and hence there will be no electron flow from source to drain. Therefore the drain current will be zero.
7. Which of the following is the necessary to hold drain current in 0 for p channel E-MOSFET?
a) V gs > V t
b) V gs < V t
c) V gs < 2V t
d) 2V gs > V t
Answer: a
Explanation: For V gs > V t , both of the drain-substrate and source-substrate will be reverse biased and hence the depletion region is narrower. Hence the channel is not created and hence there will be no electron flow from source to drain. Therefore the drain current will be zero.
8. In a p-channel E-MOSFET, the current becomes constant in saturation region. Which of the following condition depicts this?
a) V DS = 2V gs -V t
b) V DS > V gs -V t
c) V DS ≤ V gs -V t
d) V DS = V gs
Answer: c
Explanation: The p-channel E-MOSFET operates in the saturation region when V DS > V gs -V t , since we will be fixing the gate to source voltage and threshold is a system constant, whatever may be the value of V DS which is lesser than V gs -V t , the current remains the same.
9. In an n-channel E-MOSFET, the current becomes constant in saturation region. Which of the following condition depicts this?
a) V DS = 2V gs -V t
b) V DS > V gs -V t
c) V DS ≤ V gs -V t
d) V DS = V gs
Answer: b
Explanation: The n-channel E-MOSFET operates in the saturation region when V DS > V gs -V t , since we will be fixing the gate to source voltage and threshold is a system constant, whatever may be the value of V DS which is greater than V gs -V t , the current remains the same.
10. Which of the following is true about n-channel E-MOSFET?
a) Electrons are the majority charge carries
b) Holes are the majority charge carries
c) Both holes and electrons are present in same ratio
d) Neutrons are the charge carriers
Answer: a
Explanation: n-channel E-MOSFET is designed in such a way that the source and drain are formed by the n-type impurity which contains electrons as the majority charge carriers. The doping concentration of source and drain is slightly more than the gate and body, and thus the channel created is by electrons of a source and drain.
This set of Electronic Devices and Circuits Objective Questions & Answers focuses on “The Insulated-Gate FET – 2”.
1. The threshold voltage of an n-channel MOSFET can be controlled by which of the following parameter?
a) Increasing the channel dopant concentration
b) Reducing the channel dopant concentration
c) Reducing the gate-oxide thickness
d) Reducing the channel
Answer: b
Explanation: The threshold voltage of n-channel MOSFET can be increased by reducing the channel dopant concentration or by increasing the oxide thickness.
2. MOSFET can be used as ________
a) Voltage controlled capacitor
b) Current controlled capacitor
c) Voltage controlled inductor
d) Current controlled inductor
Answer: a
Explanation: By using proper techniques, MOSFET can be used as Voltage controlled capacitor.
3. The effective channel length of a MOSFET in saturation decreases with increase in which of the following parameter?
a) Gate voltage
b) Drain voltage
c) Source voltage
d) Body voltage
Answer: b
Explanation: The channel length of a MOSFET in saturation decreases with increase in drain voltage of the MOSFET.
4. In a MOSFET operating in a saturation region, the channel length modulation effect causes
a) An increase in gate-source capacitance
b) Decrease in Trans conductance
c) Decrease in the unity gain cut off
d) Decrease in the output impedance
Answer: d
Explanation: Under channel length modulation, 1/r ds =dI ds /dV ds =dI Dsat =1/r o
Hence it decreases from ∞ to 1/r o .
5. Which of the following effects can be caused by decrease in temperature?
a) Increase in MOSFET current
b) Increase in BJT current
c) Decrease in MOSFET current
d) Decrease in BJT current
Answer: c
Explanation: MOSFET has positive temperature coefficient and drain resistance increase with decrease in temperature .Hence current decreases with decrease in temperature.
6. At room temperature, what is the possible value for the mobility of electrons in the inversion layer of a silicon n-channel MOSFET?
a) 450 cm 2 /v-s
b) 1350 cm 2 /v-s
c) 1800 cm 2 /v-s
d) 3600cm 2 /v-s
Answer: b
Explanation: The mobility of electron in s-type Si semiconductor is 1350 cm 2 /v-s.
In Inversion layer mobility of electron is 1350 cm 2 /v-s.
7. The drain current of a MOSFET in saturation is given by I D =K(V GS -V P ) (V GS -V P )
where k is a constant. Determine the magnitude of g m .
a) K(V GS -V T ) (V GS -V T )/ V DS
b) 2K(V GS -V T )
c) I D /V GS V dS
d) K(V GS -V T ) (V GS -V T )/ V GS
Answer: b
Explanation: g m =∂I D /∂V GS |V DS =constant
=>g m = 2K(V GS -V T ).
8. The depletion type MOSFET is equivalent to normally closed switch.
a) True
b) False
Answer: a
Explanation: The depletion type has its gate short circuited with source terminal and hence always in On condition .Therefore we can say it as closed switch.
9. Which of the following statement is true about enhancement MOSFET?
a) It acts as closed switch
b) It acts as open switch
c) It acts as resistor with small resistance
d) Capacitor
Answer: b
Explanation: Because of an absence of channel in enhancement MOSFET, no current flows and hence acts as open switch.
10. For a transistor in its circuit symbol, the line between drain and source was broken, what does this indicate?
a) BJT
b) JFET
c) Depletion type MOSFET
d) Enhancement type MOSFET
Answer: d
Explanation: While representing a Transistor, if the solid line is broken then it represents an open switch which is Enhancement type MOSFET.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Common Source Amplifier”.
1. An amplifier is designed using fixed bias configuration, what is its output impedance ?
a) R D +r d
b) R G
c) R D || r d
d) 0
Answer: c
Explanation: By using ac equivalent of FET, g m V gs is made 0. On observing the circuit from output side, Z O =R D || r d , where R D is the drain resistance. In general, r d >>R D , Z O =R D .
2. An amplifier is designed using fixed bias configuration, what is its input impedance ?
a) R D +r d
b) R G
c) R D || r d
d) 0
Answer: b
Explanation: From the ac small signal model, the input impedance of FET is open circuit.
Gate resistance R G is in parallel with open circuit resulting in Z in =∞||R G =R G .
3. Which of the following is true about the common Source amplifier?
a) It has low input impedance
b) It has high output impedance
c) Infinite gain
d) Phase reversal voltage output
Answer: d
Explanation: If we consider a fixed bias based amplifier, the voltage gain is A v =-g m R D ,
The negative sign indicates that the voltage waveform is phase reversed by 180 degrees.
4. A Self bias configuration contains R D =3.3, R s =1 KΩ, R G =1MΩ and g m =1.5mS. Determine A v ?
a) -2
b) 3
c) -4
d) 5
Answer: a
Explanation: A v =-g m R D / (1+g m R S ) because R S is given and it is not bypassed.
A v =-1.5×3.3KΩ/
A v =-2.
5. What is the reason for connecting a capacitor in parallel with R s ?
a) It blocks the noise
b) For ac signal it acts a short circuit resulting in grounding source terminal
c) It blocks the noise & for ac signal it acts a short circuit resulting in grounding source terminal
d) To increase impedance
Answer: c
Explanation: The capacitor connected across Source blocks all the DC noise signals, when AC input is applied, at high frequency the capacitive reactance will be almost zero resulting in short circuit making the source to be directly connected to ground.
6. A sinusoidal signal has 1.5V at t=2S, If this signal is given to Common source amplifier with |A v |=5V, what will be its output at t=2S?
a) 7.5V
b) -7.5V
c) 0
d) 15V
Answer: b
Explanation: Input at t=2S is 1.5V, |A v |=5V=> output=5×1.5V=7.5V,
But amplifier produces 180 degree phase-shift => output=-7.5V.
7. For a self bias amplifier, R S =R D , the output will be_________________
a) Same as Input
b) Same as input but phase reversed
c) 0
d) Infinite
Answer: b
Explanation: A v =-g m R D / (1+g m R S ),
If R S =R D , A v =-g m R D / (1+g m R D ) => A v is nearly equal to -1 A v =V o /V i => V o =-V i .
8. Which of the following statement is true about FET common source amplifier compared to BJT amplifier?
a) It has High input impedance
b) It has low input impedance
c) No input Voltage is needed
d) Input Voltage is needed
Answer: a
Explanation: FET due to its physical nature has very high input impedance; hence the voltage drop in input side will be very less making the entire signal to pass through the input port.
9. If gain is need to be stabilized, which of the following element is used for a Common source amplifier?
a) Capacitive
b) Inductive
c) Resistive
d) LC tank circuit
Answer: c
Explanation: The negative feedback is given through a resistor which stabilizes the gain.
Hence a resistor is connected across source terminal. Any temperature changes will not result in fluctuation of the output because of the negative feedback.
10. What is the input impedance of voltage divider configuration?
a) R 1 +R 2
b) R G
c) electronic-devices-circuits-questions-answers-common-source-amplifier-q10
d) 0
Answer: c
Explanation: From Ac equivalent, Z in =R 1 || R 2 ||∞
Z in = R 1 || R 2
=> electronic-devices-circuits-questions-answers-common-source-amplifier-q10 .
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Common Drain Amplifier”.
1. Output Impedance of Common Drain Amplifier is______________
a) R S +r d
b) R S
c) R S || r d
d) R S ||1/g m
Answer: d
Explanation: By using ac equivalent of FET, g m V gs is made 0. On observing the circuit from output side, Z O =V o /I o
But I o = V o ((1/r d )+(1/R S )+(g m )),
=> Z O =V o /V o ((1/r d )+(1/R S )+(g m ))= r d |(|R S |)|1/g m
Z O =R S ||1/g m .
2. A Common drain amplifier is designed using fixed bias configuration, what is its input impedance?
a) R D + r d
b) R G
c) R D || r d
d) 0
Answer: b
Explanation: From the ac small signal model, the input impedance of FET is open circuit.
Gate resistance R G is in parallel with open circuit resulting in Z in =∞||R G =R G .
3. Which of the following is true about Common Drain amplifier?
a) It has low input impedance
b) It has high output impedance
c) Infinite gain
d) Output will be same as input
Answer: d
Explanation: The voltage gain for a Common Drain Amplifier, A v =1, hence the output at the source terminal follows the input provided to the gate terminal.
4. A sinusoidal signal has 1.5V at t=2S, If this signal is given to Common drain amplifier, what will be its output at t=2S?
a) 7.5V
b) -7.5V
c) 0
d) 1.5V
Answer: d
Explanation: Input at t=2S is 1.5V, |A v |=1
=>output=1×1.5V=1.5V.
5. A sine signal with period 2S has a peak of 2V at 0.5S and at 1.5 it was -2V, if this signal is applied to Common drain configured amplifier, What will be the output at t=1S?
a) 2V
b) -2V
c) 0V
d) 1V
Answer: c
Explanation: For the given sine signal, since the period is 2S, the value of voltage at t=1S will be 0V and this signal is provided as input to Common drain configured amplifier which follows the input signal. Therefore the output will be 0 at t=1S.
6. Which of the following is true about the effects of r d in Drain amplifier?
a) r d Increases output impedance
b) r d Decreases output impedance
c) Output impedance remains the same
d) r d Increases input impedance
Answer: b
Explanation: Typically Z O =c||1/g m
But with r d , Z O =R S ||1/g m ||r d ,
Hence output impedance will decrease.
7. Which of the following is the other name for Common Drain Amplifier?
a) Source Follower
b) Current Booster
c) Voltage booster
d) Voltage limiter
Answer: a
Explanation: The voltage gain for a Common Drain Amplifier, A v =1, hence the output at the source terminal follows the input provided to the gate terminal. Hence we call a Common Drain Amplifier as Source follower as the name itself says it the source terminal follows the input signal.
8. While choosing the operating conditions, the V gs value was set lesser to V t . What will the output if the FET is made to work as Source Follower?
a) 0
b) Same as the input
c) Same as input but phase reversed
d) Cannot be determined
Answer: a
Explanation: If FET is needed to work as an amplifier, then the bias point must lie on Saturation, but in the above case the FET is operating in cut-off resulting in off condition.
Hence output will be 0.
9. What will be the gain value of FET if g mo =0?
a) 0
b) 1
c) 2
d) Infinite
Answer: a
Explanation: A v =g m R S /(1+g m R S ) but g m is having linear relationship with g mo .
Since g mo =0, voltage gain will be zero resulting in 0 Output.
10. If g m =0.5mS, R S =2KΩ, determine Z O for source follower?
a) 2KΩ
b) 1KΩ
c) 3KΩ
d) 1.5KΩ
Answer: b
Explanation: Z O =R S ||1/g m
=>1/g m =1/0.5mS=>2KΩ
2KΩ||2KΩ=1KΩ,
=>Z O =1KΩ.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Biasing the FET”.
1. Which of the following relation is true about gate current?
a) I G =I D +I S
b) I D =I G
c) I S = I G
d) I G =0
Answer: d
Explanation: The FET physical structure which contains silicon dioxide provides infinite resistance. Hence no current will flow through the gate terminal.
2. Which of the following equations gives the relation between I D and V gs ?
a) I D =I DSS (1-V gs /V p ) 2
b) I D =I DSS (1-V gs /V p ) 1
c) I D =I DSS (1-V gs /V p ) 3
d) I D =I DSS (1-V gs /V p ) 4
Answer: a
Explanation: The above equation called as Shockley’s equation depicts the relation between I D and V gs . When V gs becomes equal toV p , the current will become zero, which clearly satisfies the physical nature of FET.
3. For a fixed bias circuit the drain current was 1mA, what is the value of source current?
a) 0mA
b) 1mA
c) 2mA
d) 3mA
Answer: c
Explanation: We know that for an FET same current flows through the gate and source terminal, Hence source current=1mA.
4. For a fixed bias circuit the drain current was 1mA, V DD =12V, determine drain resistance required if V DS =10V?
a) 1KΩ
b) 1.5KΩ
c) 2KΩ
d) 4KΩ
Answer: c
Explanation: V DS =V DD -I D R D
=>10=12-R D ×1mA
=>R D =2/1mA=2 KΩ.
5. Which of the following equation brings the relation between gate to source voltage and drain current in Self Bias?
a) V gs =V DD
b) V gs =-I D R s
c) V gs =0
d) V gs =1+I D R s
Answer: b
Explanation: V R s =I D R s
ButV R s +V gs =0
V gs =-I D R s .
6. For a self-bias circuit, find drain to source voltage if V DD =12V, I D =1mA, R s =R D =1KΩ?
a) 1V
b) 2V
c) 10V
d) 5V
Answer: c
Explanation: V DS =V DD -I D (R D +R s )
=>V DS =12-1mA
=>V DS =10V.
7. Find the gate voltage for voltage divider having R 1 =R 2 =1KΩ and V DD =5V?
a) 1V
b) 5V
c) 3V
d) 2.5V
Answer: d
Explanation: V G = R 2 ×V DD /R 1 +R 2
=>V G =1×5/2
=> V G = 2.5V.
8. Find the gate to source voltage for voltage divider having R 1 =R 2 =2KΩ and V DD =12V, I D =1mA and R S =4KΩ?
a) 3V
b) 2V
c) 0V
d) 1V
Answer: b
Explanation: V G = R 2 ×V DD /R 1 +R 2
=>V G =2×12/4
=>V G= 6V
=>V GS =V G -I D R s
=>V GS =2V.
9. What will happen if values of R s increase?
a) V gs Increases
b) V gs Decreases
c) V gs Remains the same
d) V gs =0
Answer: b
Explanation: Increasing values of R s result in lower quiescent values of I D and more negative values of V gs .
10. What is the current flowing through the R 1 resistor for voltage divider (R 1 =R 2 =1KΩ, V DD =10V)?
a) 5mA
b) 3mA
c) 1mA
d) 2mA
Answer: a
Explanation: I R 1 =I R 2 =V DD /R 1 +R 2
=>I R 1 = 10/2KΩ
=>I R 1 = 5mA.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “A Generalized FET Amplifier”.
1. Ideal maximum voltage for common drain amplifier is _________
a) 0
b) 1
c) 0.5
d) 2
Answer: b
Explanation: Similar to the transistor emitter follower, the source follower configuration itself provides a high level of buffering and a high input impedance. The actual input resistance of the FET itself is very high as it is a field effect device. This means that the source follower circuit is able to provide excellent performance as a buffer. The voltage gain is unity, although current gain is high. The input and output signals are in phase.
2. If a certain drain JFET has a transconductance of 4ms. And has external drain resistance of 1.5 ohm than ideal voltage gain will be _________
a) 4
b) 5
c) 6
d) 8
Answer: c
Explanation: The transconductance, g m is defined as
g m = ΔI D / ΔV GS
so g m = Voltage gain / R D
Therefore, voltage gain = g m * R D
=4 * 1.5
=6.
3. Input resistance of common gate of the amplifier is __________
a) zero
b) infinity
c) extremely low
d) extremely high
Answer: c
Explanation: For a Common gate amplifier, Current gain is about unity, input resistance is low, output resistance is high a CG stage is a current “buffer”. It takes a current at the input that may have a relatively small Norton equivalent resistance and replicates it at the output port, which is a good current source due to the high output resistance.
4. A FET circuit has a transconductance of 2500 µ seconds and drain resistance equals to 10Kohms than voltage gain will be __________
a) 20
b) 25
c) 30
d) 35
Answer: b
Explanation: The transconductance, g m is defined as
g m = ΔI D / ΔV GS
so g m = Voltage gain / R D
Therefore, voltage gain = g m * R D
=2500*10 -6 * 10 * 10 3
= 25.
5. Voltage gain of common drain amplifier is always slightly less than _____
a) 0.5
b) 1
c) 1.5
d) 2
Answer: b
Explanation: In common drain amplifier
Writing KCL at the source node ;
G m (v in – v out ) – g mbs v out – g ds v out = 0
v out v in = G m / G m + G mbs + g ds
Therefore gain is less than one.
6. A common gate amplifier has _______
a) low input impedance
b) high input impedance
c) infinite input impedance
d) no impedance
Answer: a
Explanation: Common gate amplifier just like common base amplifier had a very large voltage gain but input impedance is very low. Also, it gives unity current gain. Hence, the power of the amplified signal will be less.
7. D-MOSFET in case of common source amplifier can operate with gate to source voltage zero at ______
a) Peak positive point
b) Peak negative point
c) Q point
d) Origin
Answer: c
Explanation: Q-point” needs to be found for the correct biasing of the JFET amplifier circuit with single amplifier configurations of Common-source, Common-drain or Source-follower and the Common-gate available for most FET devices.
8. A common source amplifier has _______
a) no source resistance
b) no drain resistance
c) no gate resistance
d) low input impedance
Answer: a
Explanation: In a common source amplifier the source resistance is connected to the ground because of which its source resistance is kept zero.
9. The drain of FET is analogous to BJT
a) collector
b) emitter
c) base
d) drain
Answer: a
Explanation: A common collector amplifier is one of three basic single-stage bipolar junction transistor amplifier, typically used as a voltage buffer.
In this circuit the base terminal of the transistor serves as the input, the emitter is the output, and the collector is common to both. The analogous field-effect transistor circuit is the common drain amplifier and the analogous tube circuit is the cathode follower.
10. Input signal of common drain amplifier is applied to the gate through ________
a) input resistor
b) coupling capacitor
c) output capacitor
d) transformer
Answer: b
Explanation: A common-drain amplifier is also called a source-follower. Self-biasing is used in this particular circuit. The input signal is applied to the gate through a coupling capacitor, and the output signal is coupled to the load resistor through the other capacitor.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Unijunction Transistors”.
1. How many terminals are there in a unijunction transistor?
a) 1
b) 2
c) 3
d) 4
Answer: c
Explanation: A unijunction transistor consists of three terminals. It is a semiconductor device that displays negative resistance and switching characteristics. It is used as a relaxation oscillator in phase control applications.
2. What are unijunction transistors used for?
a) Amplifying a circuit
b) Circuit breaker
c) Splitting device
d) On-Off switching device
Answer: d
Explanation: Unlike bipolar transistors or field effect transistors these unijunction transistors cannot be used to amplify a circuit. The unijunction transistors are used as an on-off switching transistor. They have unidirectional conductivity.
3. Unijunction transistors have unidirectional conductivity and positive impedance characteristics.
a) True
b) False
Answer: b
Explanation: Unijunction transistors have unidirectional conductivity and negative impedance characteristics. During circuit breakdown, they function more as a variable voltage divider. They are used in gate pulse, timing circuits and trigger generation.
4. What is the generalized value for the voltage across resistor R B1 in a unijunction transistor?
a) V RB1 = (R B2 / R B1 + R B2 ) × V BB
b) V RB1 = (R B2 / R B1 – R B2 ) × V BB
c) V RB1 = (R B1 / R B1 + R B2 ) × V BB
d) V RB1 = (R B1 / R B1 – R B2 ) × V BB
Answer: c
Explanation: The generalized value for the voltage across resistor R B1 is (R B1 / R B1 + R B2 ) × V BB . Where, R B1 and R B2 are the resistance across two terminals B 1 and B 2 . The voltage V BB is the voltage across the two terminals.
5. What is the intrinsic stand-off ratio of a unijunction transistor when R B1 = 10kΩ and R BB = 15kΩ?
a) 0.67
b) 0.55
c) 0.80
d) 0.44
Answer: a
Explanation: The intrinsic stand-off ratio is equal to the ratio R B1 / R BB . Standard values of η range from 0.5 to 0.8. Given R B1 = 10kΩ and R BB = 15kΩ:
η = R B1 / R BB = 10 / 15 = 0.67
6. Which type of material is the channel of a unijunction transistor made up of?
a) PN type
b) It doesn’t affect the working
c) P type
d) N type
Answer: d
Explanation: The core conducting component is the N-type channel of the transistor. Unijunction transistor consists of a single solid piece of N-type semiconductor material forming the main current carrying channel.
7. What are the terminals of a unijunction transistor?
a) Collector, Base and Emitter
b) Emitter, Base 1 and Base 2
c) Gate, Drain and Source
d) Gate, Drain, Body and Source
Answer: b
Explanation: The unijunction transistor consists of a single solid piece of N-type channel with its two outer connections marked as Base 2 and Base 1. The third connection is the Emitter and it is located along the channel. The emitter lies closer to base 2 as compared to base 1.
8. What are the working regions of a unijunction transistor?
a) Linear region
b) Negative Resistance region
c) Saturation region
d) Cut-off region
Answer: b
Explanation: As soon as the transistor reaches the triggering voltage the unijunction transistor will turn on. If the applied voltage increases to the emitter lead, the peak voltage is achieved. The voltage drops from peak voltage to Valley Point. This happens in spite of the current increasing; this is because of the negative resistance.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Class A Large Signal Amplifiers”.
1. Why do we use CE amplifier as a large signal class a amplifier?
a) It has very high output impedance
b) It has very high input impedance
c) It has very high voltage gain
d) It is very much stable
Answer: c
Explanation: Since CE amplifier has reasonably high voltage gain and hence can work with high voltage or large signals.
2. What does class A amplifier do?
a) Delivers KV of voltage to load
b) Delivers KW of power
c) Delivers Kilo Pascal pressure
d) Delivers more resistance
Answer: b
Explanation: Class A always delivers more power to load because of a larger current delivered from the transistor.
3. What is the efficiency of Class A amplifiers?
a) 30 or less
b) 50 or less
c) 100
d) 75
Answer: a
Explanation: Since in Class A amplifiers distortion is more and as it requires tuned circuit as load, the efficiency is very low.
4. An ideal large signal amplifier delivers 100% DC power to its load.
a) True
b) False
Answer: a
Explanation: In an ideal case, no distortion occurs, that is noise effects are completely ignored, hence it gives 100% efficiency.
5. Which of the following statement is true about class A amplifiers?
a) They are weak against distortions
b) They supress noise signals
c) More efficient
d) Delivers 100% power to load
Answer: a
Explanation: Since class A amplifiers use basic transistor, even the noise signal will get amplified when the noise signal exists in input.
6. Why there is a need for heat sinks in Class A amplifier?
a) To control the external temperature
b) To avoid temperature changes affecting the transistor
c) To control heat dissipation
d) To increase output resistance
Answer: c
Explanation: When even no input is given to class A amplifiers, it produces some current to load, and hence heat sinks are required to avoid this.
7. Basic operation of Class A amplifiers____________
a) Crossover distortion
b) Law of Conservation of energy
c) Millers law
d) Switching transistor theory
Answer: d
Explanation: The class A amplifier is the simplest form of power amplifier that uses the switching transistor in the standard common emitter configuration.
8. If DC power for a Class A amplifier is 500W and AC power is 150W, what is its efficiency?
a) 50%
b) 75%
c) 20%
d) 30%
Answer: d
Explanation: efficiency=AC POWER/DC POWER
Efficiency=150/500=3/10
In percentage 3*100/10=30%.
9. Which of the following technique is used to increase the efficiency of class A amplifier?
a) By using FET
b) By using PNP transistor
c) By using matched transformers as load
d) By using potentiometers as load
Answer: c
Explanation: We can increase the efficiency of class A amplifier by using load matching concept, hence a pair of matched transformer can be used as load which in turn increases the efficiency.
10. What is the conduction angle of class A amplifier?
a) 90
b) 180
c) 270
d) 360
Answer: d
Explanation: Since the amplifier is always On, it conducts all the current waveform which can be used to state that it has 360 degree conduction angle.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Second Harmonic Distortion”.
1. What is the cause of harmonic distortion in a power amplifier?
a) Exact replication of output signal
b) Perfectly synced harmonics
c) Non-linearity of transistors
d) Linearity of transistors
Answer: c
Explanation: Harmonic distortion in a power amplifier is caused due to the non-linearity of transistor devices. Transistors are the active elements in the circuit and are used for amplification. However, they do not equally amplify each point of the input wave form which then creates distortion.
2. What is the second harmonics of a wave form with fundamental frequency of 50Hz?
a) 50Hz
b) 150Hz
c) 25Hz
d) 100Hz
Answer: d
Explanation: The harmonics of a wave are the integer multiples of the fundamental frequency. Hence, the second harmonics of a wave with fundamental frequency 50Hz is 100Hz.
Second harmonics = 2 × fundamental frequency = 2 × 50Hz = 100Hz
3. The second order harmonics is the most prominent even harmonic.
a) True
b) False
Answer: a
Explanation: All audio amplifiers produce a certain level of distortion usually in the form of even harmonics. Out of all the even harmonics, the most prominent one is the second order harmonics. Second order harmonic distortion is the amount of 2nd order harmonic content present in the output signal with respect to the fundamental frequency.
4. What is the total harmonic distortion if the amplitude of first harmonic distortion is 20dB and second harmonic distortion 25dB?
a) 1.2
b) 1.52
c) 1.25
d) 1.5
Answer: c
Explanation: The total harmonic distortion is calculated by the formula given below. Given the first harmonic distortion V 1 = 20dB and V 2 = 25dB:
THD = (V 2 2 + V 3 2 +…+ V n 2 ) 1/2 / V 1 = (V 2 2 ) 1/2 / V 1 = V 2 / V 1 = 25 / 20 = 1.25
5. What is the signal to noise ratio of a second harmonic wave if the input signal amplitude is 20dB and the output signal amplitude is 25dB?
a) 0.8
b) 0.5
c) 1.25
d) 1.5
Answer: a
Explanation: To calculate the SNR we take the ratio of the input signal amplitude to the output signal amplitude.
SNR = input wave / output wave = 20dB / 25dB = 0.8
6. What reduces the second harmonic distortion?
a) Amplifiers
b) Demodulators
c) Modulators
d) Transistor switch arrays
Answer: d
Explanation: The second harmonic distortion can be reduced by transistor switch arrays. The switch arrays are implemented in a pseudo-differential configuration. This design of the transistor switch arrays as a side effect reduces the second harmonic distortion.
7. What is the unit of second harmonic distortion?
a) Ampere
b) dBc
c) Volts
d) Watts
Answer: b
Explanation: The second order harmonic distortion is the ratio of second-order harmonic to the input signal or the carrier signal. It is measured as dBc. dBc is the power ratio of a signal to a carrier signal, expressed in decibels.
8. Which of the following configuration classes has the lowest second order harmonics?
a) Class A
b) Class AB
c) Class B
d) Class C
Answer: a
Explanation: Class A configuration has the highest linearity due to which is produces the least amount of distortion, which in turn produces even lower second order harmonic distortion. It is then followed by Class AB, Class B and Class C.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Higher-Order Harmonic Distortion”.
1. What does THD+N represent in higher order harmonic distortion?
a) Noise
b) Higher order distortion
c) Higher order distortion with noise component
d) Second order distortion with noise component
Answer: c
Explanation: THD+N represents the addition of all noise components to the fundamental frequency. Subsequently noise is also accounted for in this scale, hence it is better than the THD scale. The cause for noise being present in the amplifier output includes the power supply interference, RF interference, switching noises and many more.
2. What is the higher order harmonics of order 5 of a wave form with fundamental frequency of 50Hz?
a) 50Hz
b) 150Hz
c) 250Hz
d) 100Hz
Answer: c
Explanation: The harmonics of a wave are the integer multiples of the fundamental frequency. Hence, the fifth harmonics of a wave with fundamental frequency 50Hz is 250Hz.
Fifth harmonics = 5 × fundamental frequency = 5 × 50Hz = 250Hz.
3. Lower order harmonics are harder to hear compared to the higher order harmonics.
a) True
b) False
Answer: a
Explanation: All audio amplifiers produce a certain level of distortion usually in the form of even harmonics. Out of all the even harmonics, lower order harmonics are harder to hear compared to the higher order harmonics.
4. What is the total harmonic distortion if the amplitude of first harmonic distortion is 20dB, second harmonic distortion 25dB and external noise is 10dB?
a) 11.2dB
b) 11.52dB
c) 11.25dB
d) 11.5dB
Answer: c
Explanation: The total harmonic distortion is calculated by the formula given below. Given the first harmonic distortion V 1 = 20dB, V 2 = 25dB and N = 10dB:
THD = (V 2 2 + V 3 2 + … + V n 2 ) 1/2 / V 1 = (V 2 2 ) 1/2 / V 1 = V 2 / V 1 = 25 / 20 = 1.25
THD + N = 1.25 + 10 = 11.25dB.
5. What is the value of a THD F for a square wave?
a) 48.3%
b) 80.3%
c) 12.1%
d) 37.0%
Answer: a
Explanation: The THD F values of certain standard waveforms are calculated diagnostically. To compute the THD F for a square wave, we use the formula given below:
THD F = ((π 2 / 8) – 1) 1/2 = 0.483 = 48.3%.
6. What happens to the gap between THD F and THD R for higher order harmonic distortion?
a)Increases
b) Decreases
c) Remains unchanged
d) Not related
Answer: a
Explanation: THD F represents the total harmonic distortion for fundamental frequency and THD R represents the total harmonic distortion for root mean square. For lower harmonics, the gap between these two values is negligible. However, for higher order harmonics the gap increases.
7. What is the value of a THD F for a sawtooth wave?
a) 48.3%
b) 80.3%
c) 12.1%
d) 37.0%
Answer: b
Explanation: The THD F values of certain standard waveforms are calculated diagnostically. To compute the THD F for a sawtooth wave, we use the formula given below:
THD F = ((π 2 / 6) – 1) 1/2 = 0.803 = 80.3%.
8. What is the value of a THD F for a symmetrical triangle wave?
a) 48.3%
b) 80.3%
c) 12.1%
d) 37.0%
Answer: c
Explanation: The THD F values of certain standard waveforms are calculated diagnostically. To compute the THD F for a symmetrical triangle wave, we use the formula given below:
THD F = ((π 4 / 96) – 1) 1/2 = 0.121 = 12.1%.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Cascading Transistor Amplifiers”.
1. What is the advantage of using a cascading transistor amplifier?
a) High gain and high bandwidth
b) High gain and low bandwidth
c) Low gain and high bandwidth
d) Low gain and low bandwidth
Answer: a
Explanation: A circuit having a single transistor amplifier does not provide suitable bandwidth or gain. To overcome this difficulty, we combine several amplification stages. The cascading transistor amplifier theory is used for high gain as well as high bandwidth.
2. What is the advantage of using CB amplifier configuration in a cascading transistor amplifier?
a) High efficiency
b) Low distortion
c) Good high frequency operation
d) Good low frequency operation
Answer: c
Explanation: The cascading transistor amplifier circuit can be constructed with two configurations of a transistor that is CE and CB . The CB configuration delivers a good high-frequency operation.
3. What is the total voltage gain of a cascading transistor amplifier if the gain of first stage is 4 and the gain of the second stage is 10?
a) 0.4
b) 4
c) 40
d) 2.5
Answer: c
Explanation: The total gain of a cascading transistor amplifier can be calculated by the formula given below. Where, A V = overall gain, A V1 = voltage gain of first stage and A V2 = voltage gain of second stage.
A V = A V1 × A V2 = 4 × 10 = 40
4. What is the total voltage gain of a cascading transistor amplifier if input of first stage is 10V and the output of the second stage is 25V?
a) 0.4
b) 4
c) 40
d) 2.5
Answer: d
Explanation: The total gain of a cascading transistor amplifier is the ration of the output of second stage amplifier to the input of first stage amplifier. Where, V output = 25V and V input = 10V.
A V = V output / V input = 25V / 10V = 2.5
5. What does cascading of two transistor amplifiers imply?
a) Output of first stage sent to input of second stage
b) Output of first stage sent to coupling device
c) Input of first stage sent to input of second stage
d) Not related
Answer: b
Explanation: In multi-stage amplifiers, the output of first stage is coupled to the input of second stage using a coupling device. The coupling devices is typically a capacitor or a transformer. Cascading is known as the process of combining two amplifier stages using a coupling device.
6. How many methods of coupling are used for a cascading transistor amplifier?
a) 3
b) 4
c) 2
d) 5
Answer: b
Explanation: There are four types of coupling mechanisms used. They are Resistance-capacitance coupling, thermal coupling, impedance coupling and transformer coupling. Resistance-capacitance coupling is the most vastly used mechanism.
7. What is the total voltage gain of a cascading transistor amplifier if the gain of first stage is 14dB, gain of the second stage is 12dB and gain of third stage is 24dB?
a) 40dB
b) 50.32dB
c) 50dB
d) 40.32dB
Answer: c
Explanation: The total gain of a cascading transistor amplifier is the sum of the multiple stages when the gain is calculated is decibels. Given, A V1 = 14dB, A V2 = 12dB and A V3 = 24dB
A V = A V1 + A V2 + A V3 = 14dB + 12dB + 24dB = 50dB
8. The coupling device is essential for a cascading transistor amplifier.
a) True
b) False
Answer: a
Explanation: One of the main functions of the coupling device in a cascading transistor amplifier is to block the direct current from passing through to the input of next stage from the output of the first stage. Another function of the coupling device is to transfer the current from the output of first stage to the input of the second stage.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “N-Stage Cascading Amplifiers”.
1. What is the purpose of using ann-stage cascading transistor amplifier?
a) Increase voltage gain
b) Decrease voltage gain
c) Increase current gain
d) Decrease current gain
Answer: a
Explanation: A circuit having a single transistor amplifier does not provide suitable bandwidth or gain. The purpose of an n-stage cascading amplifier is to provide an increase in the voltage gain. The total gain of ann-stage cascading amplifier is the product of the voltage gains of the discrete stages.
2. Which two terminals of the transistors are connected in an n-stage cascading amplifier?
a) Collector
b) Base
c) Emitter
d) Collector, base and emitter
Answer: c
Explanation: The n-stage cascading transistor amplifier circuit can be constructed by connecting the emitters of two consecutive transistors. A resistor is placed between each stage to act as a coupling device.
3. What is the total voltage gain of a cascading transistor amplifier if the individual gains are: 1.5, 2.3, 3.46 and 2?
a) 18.52
b) 9.26
c) 23.87
d) 11.26
Answer: c
Explanation: The total gain of a cascading transistor amplifier can be calculated by taking the product of the individual stages. Where,
A V = A V1 × A V2 × A V3 × A V4 = 1.5 × 2.3 × 3.46 × 2 = 23.87
4. What is the total voltage gain of the nth stage in an n-stage cascading amplifier?
a) A VK = A IK × R LK × R iK
b) A VK = A IK × R iK / R LK
c) A VK = A IK / R LK × R iK
d) A VK = A IK × R LK / R iK
Answer: d
Explanation: The gain for any intermediate stage in an n-stage cascading amplifier can be calculated using A VK = A IK × R LK / R iK . Where, A IK , R LK and R iK is the estimated with respect to the values of the previous stages. A IK represents the voltage gain while R LK and R iK represent the effective load impedance.
5. What are n-stage cascading amplifiers also known as?
a) Common collector amplifier
b) Multistage amplifiers
c) Common base amplifier
d) Common emitter amplifier
Answer: b
Explanation: A single stage of amplifier provides an insufficient current gain or voltage gain. Several amplifier stages connected in cascade are used instead. Hence it is known as an n-stage cascading amplifier of multistage amplifier.
6. How many ports are there in an n-stage cascading amplifier?
a) 3
b) 4
c) 2
d) 5
Answer: c
Explanation: An n-stage cascading amplifier is a two-port network assembled from a series of amplifiers. Each amplifier directs its output to the input of the subsequent amplifier. The performance obtained from a single-stage amplifier is inadequate, there fore numerous stages are combined to form a multistage amplifier.
7. What is the total voltage gain of an n-stage cascading amplifier if the gain of stages are 5dB, 14dB, 20dB, 15dB and 24dB?
a) 76dB
b) 46dB
c) 78dB
d) 79dB
Answer: c
Explanation: The total gain of ann-stage cascading amplifier is the sum of the gain measured at each stage only when the gain is calculated is decibels. Given, A V1 = 5dB, A V2 = 14dB, A V3 = 20dB, A V4 = 15dB and A V5 = 24dB
A V = A V1 + A V2 + A V3 + A V4 + A V5 = 5dB + 14dB + 20dB + 15dB + 24dB = 78dB
8. The middle stage of an n-stage cascading amplifier provides most gain.
a) True
b) False
Answer: a
Explanation: The middle stage of an n-stage cascading amplifier provides most gain. It provides the maximum voltage gain along with fluctuating the DC level of the signal. It also converts the signal from one mode to another while processing.
9. At what frequencies does the gain of the n-stage cascading amplifier get compromised?
a) Mid-range frequencies
b) High-range frequencies
c) Low-range frequencies
d) It does not affect the gain
Answer: c
Explanation: Multistage cascading amplifiers or n-stage cascading amplifiers have a low cut off frequency. The gain of the system is compromised at lower frequencies. To calculate the cut off frequency f c = f 1 / √(2 1/n – 1), where f 1 is initial frequency.
10. How is the voltage gain measured in an n-stage cascading amplifier?
a) Volts
b) Ampere
c) Decibels
d) Dimensionless
Answer: c
Explanation: The voltage gain is measured in terms of decibels. The total voltage gain is calculated as the product of individual stages or as a sum of all the stages if the gain calculated at each intermediate stage was measured in decibels as well.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Decibel”.
1. What is the symbol of a decibel?
a) dB
b) Db
c) DB
d) db
Answer: a
Explanation: The decibel is a comparative unit of measurement equivalent to one tenth of a Bel unit. A Bel unit is expressed as B and hence the decibel is subsequently expressed by the symbol dB. They are a unit of a logarithmic scale to measure power or intensity levels.
2. What is the corresponding change in level for a factor of 10 change in power?
a) 100dB
b) 20dB
c) 10dB
d) 0.1dB
Answer: c
Explanation: A decibel is ten times the value of power in its logarithm in base 10. Therefore a 10dB change is resultant of a change in power by a factor of 10. However, when calculating in terms of amplitude the change is due to a factor on 20.
3. What is the corresponding change in level for a factor of 10 change in amplitude?
a) 100dB
b) 20dB
c) 10dB
d) 0.1dB
Answer: b
Explanation: A decibel is ten times the value of power in its logarithm in base 10. Therefore a 20dB change is resultant of a change in amplitude by a factor of 10. However, when calculating in terms of power the change is due to a factor on 10.
4. What is the value of voltage gain in decibels if V output = 20V and V input = 15V?
a) 1.75dB
b) 2.28dB
c) 1.37dB
d) 2.48dB
Answer: d
Explanation: The voltage gain in terms of decibels is expressed as 20 × log (V output / V input ) . Given, V output = 20V and V input = 15V.
Voltage gain = 20 × log (V output / V input ) = 20 × log = 20 × 0.1239 = 2.48dB.
5. What is the value of power gain in decibels if P output = 250W and P input = 100W?
a) 2.48dB
b) 3.98dB
c) 1.78dB
d) 3.54dB
Answer: b
Explanation: The power gain in terms of decibels is expressed as 10 × log (P output / P input ) . Given, P output = 250W and P input = 100W
Power gain = 10 × log (P output / P input ) = 10 × log = 10 × 0.3979 = 3.98dB.
6. How is the power ratio corresponding to 12dB change in level?
a) 16.38
b) 14.24
c) 15.85
d) 12.88
Answer: c
Explanation: To calculate the power ratio corresponding to 12dB change in level:
G = 10 × 1 = 10 1.2 = 15.85.
7. What is dBFS used to measure in terms of decibels?
a) Peak amplitude of a signal
b) Power relative to 1mW
c) Amplitude of a signal
d) Volume unit
Answer: c
Explanation: We use dBFS to measure the amplitude of a signal in comparison to the maximum value a device could handle prior to clipping. FS stands for full scale and is used to define the power level sinusoid or a full scale square wave.
8. The symbols dBA, dBB and dBC are used for to estimated human ear’s reaction to sound.
a) True
b) False
Answer: a
Explanation: The symbols dBA, dBB and dBC are used for to estimated human ear’s reaction to sound. They all fall under the subcategory of dB SPL which stands for sound pressure level. They are all weighted measures and are used to refer to the effects of noise on human and animals’ ears.
9. What is dBTP used to measure in terms of decibels?
a) Peak amplitude of a signal
b) Power relative to 1mW
c) Amplitude of a signal
d) Volume unit
Answer: a
Explanation: We use dBTP to measure the peak amplitude of a signal in comparison to the maximum value a device could handle prior to clipping. TP stands for true peak and the values measured are always negative or zero since they are less compared to the full scale values.
10. What is dBm used to measure in terms of decibels?
a) Peak amplitude of a signal
b) Power relative to 1mW
c) Amplitude of a signal
d) Volume unit
Answer: c
Explanation: We use dBm to measure the power relative to 1mW. It is generally compared relative to a 600Ω impedance, which equates to a voltage level of 0.775V. The m stands for milliwatts. The derivate dBm0 stands for dBm measured at zero transmission level point.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Simplified CE Hybrid Model”.
1. What is the value of input resistance in the simplified CE hybrid model?
a) h oe
b) h ib
c) h ic
d) h ie
Answer: d
Explanation: The input resistance of a common emitter amplifier can be expressed in terms of hybrid parameters in the simplified CE hybrid model. The input resistance is universally expressed as h ie . This is also the same for approximate as well as exact hybrid models.
2. How do we calculate the value of I C in the simplified CE hybrid model?
a) I C = h re × I B
b) I C = -h re × I B
c) I C = h fe × I B
d) I C = -h fe × I B
Answer: c
Explanation: The collector current of a common emitter amplifier can be expressed in terms of hybrid parameters in the simplified CE hybrid model. It is expressed as the product of h fe × I B . This is the same for approximate hybrid models as well.
3. What is the value of current gain in the simplified CE hybrid model?
a) h fe
b) -h fe
c) h re
d) –h re
Answer: b
Explanation: The current gain of a common emitter amplifier can be expressed in terms of hybrid parameters in the simplified CE hybrid model. The current gain h fe is the forward transfer characteristics and is used to calculate the current gain in common emitter amplifiers.
4. What is the value of voltage gain in the simplified CE hybrid model?
a) h fe × R L / h re
b) –h fe × R L / h re
c) h fe × R L / h ie
d) –h fe × R L / h ie
Answer: d
Explanation: The voltage gain of a common emitter amplifier can be expressed in terms of hybrid parameters in the simplified CE hybrid model. The voltage gain is equal to -h fe × R L / h ie where, R L is load resistance, h ie is input resistance and -h fe is the current gain.
5. What is the magnitude of voltage generated at emitter circuit of a simplified CE hybrid model?
a) h re × h fe / I b × R L
b) h re × h fe × I b × R L
c) h re × h fe / I C × R L
d) h re × h fe × I C × R L
Answer: b
Explanation: The magnitude of voltage generated at emitter circuit of a common emitter amplifier can be expressed in terms of hybrid parameters in the simplified CE hybrid model. The magnitude of voltage generated is equal to h re × h fe × I b × R L where, R L is load resistance and I b is the base current.
6. What is the value of output resistance in the simplified CE hybrid model?
a) ∞
b) 0
c) h ic
d) h ie
Answer: a
Explanation: The output resistance in a simplified common emitter hybrid model is calculated as the ratio of collector voltage over collector current while V S is equal to zero and R L is omitted. Due to this configuration I C is equal to zero and hence output resistance is infinite.
7. What is the value power gain in the simplified CE hybrid model?
a) A V – A I
b) A V + A I
c) A V × A I
d) A V / A I
Answer: c
Explanation: The power gain in a simplified common emitter hybrid model is calculated as the product of voltage gain and power gain. It is expressed as A V × A I . It can also be further simplified as A I 2 × R L / R i , where R L is load resistance and R i is input resistance.
8. What is the value of overall current gain in the simplified CE hybrid model while considering R S ?
a) A I × I B / I S
b) A I × I C / I S
c) A I × I S / I C
d) A I × I S / I B
Answer: a
Explanation: The current gain in a simplified common emitter hybrid model while considering R S is calculated as the product of current gain without external parameters and the ratio of base current and external applied current. It is expressed as A I × I B / I S . Where, I S is the external applied current on the amplifier.
9. What is the value of overall voltage gain in the simplified CE hybrid model while considering R S ?
a) V B / V S
b) V C / V S
c) V B × V S
d) V C × V S
Answer: a
Explanation: The overall voltage gain in a simplified common emitter hybrid model while considering R S is calculated as ratio of base voltage over the applied external voltage. It is expressed as V B / V S . Where, V S is the external applied voltage on the amplifier.
10. We use simplified CE hybrid models to obtain approximate values.
a) True
b) False
Answer: a
Explanation: It is more practical to calculate the approximate values of voltage and current gains instead of computing the exact values of a common emitter amplifier circuit. We use a simplified common emitter hybrid model to obtain these approximate values without reducing the accuracy.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Simplified Calculations for the CC Configuration”.
1. How do we computethe current gain of a simplified CC amplifier model using hybrid parameters?
a) 1 + h oe
b) 1 – h oe
c) 1 + h fe
d) 1 – h fe
Answer: c
Explanation: The current gain of a simplified common collector amplifier can be expressed in terms of approximate hybrid parameters. The current gain h fe is the forward transfer characteristics. The current gain of a common collector amplifier is expressed as 1 + h fe .
2. How do we compute the input resistance of a simplified CC amplifier model using hybrid parameters?
a) h ie – (1 + h fe ) × R L
b) h ie + (1 + h fe ) × R L
c) h ie – (1 + h fe ) / R L
d) h ie + (1 + h fe ) / R L
Answer: b
Explanation: The input resistance of a simplified common collector amplifier can be expressed in terms of approximate hybrid parameters. The input resistance is h ie + (1 + h fe ) × R L where, h ie and h fe are the input resistance and current gain of CE amplifier respectively and R L is the load resistance.
3. How do we compute the voltage gain of a simplified CC amplifier model using hybrid parameters?
a) 1 + h ie × R i
b) 1 – h ie × R i
c) 1 + h ie / R i
d) 1 – h ie / R i
Answer: d
Explanation: The voltage gain of as implified common collector amplifier can be expressed in terms of approximate hybrid parameters. The voltage gain is equal to 1 – h ie / R i where, R i is equal to h ie + (1 + h fe ) × R L .
4. What is the value of voltage V B of a CC amplifier configuration if V CC = 12V, R 1 = 5kΩ and R 2 = 8kΩ?
a) 6.54V
b) 0.7V
c) 4.6V
d) 7.38V
Answer: d
Explanation: The voltage V B of a simplified common collector amplifier configuration can be calculated using V B = I × R 2 . Given, V CC = 12V and R 1 = 5kΩ and R 2 = 8kΩ:
I = V CC / R 1 + R 2 = 12 / = 0.92mA
V B = I × R 2 = 0.92mA × 8kΩ = 7.38V
5. How do we compute the base impedance of a simplified CC amplifier?
a) β × (R e – r e ‘)
b) β × (R e + r e ‘)
c) R B || β × (R e – r e ‘)
d) R B || β × (R e + r e ‘)
Answer: b
Explanation: The base impedance of a simplified common collector amplifier can be expressed in terms circuit resistor values. The base impedance is β × (R e + r e ‘) . Where β is the transistor amplifiers current gain, R e is the equivalent emitter resistance and r e ‘ is the alternating current resistance of the emitter-base diode.
6. How do we compute the input impedance of a simplified CC amplifier?
a) β × (R e – r e ‘)
b) β × (R e + r e ‘)
c) R B || β × (R e – r e ‘)
d) R B || β × (R e + r e ‘)
Answer: d
Explanation: The input impedance of a simplified common collector amplifier can be expressed in terms circuit resistor values. The base impedance is R B || β × (R e + r e ‘) . Where β is the transistor amplifiers current gain, Reis the equivalent emitter resistance, R B is the equivalent base resistance and r e ‘ is the alternating current resistance of the emitter-base diode.
7. How do we compute the output impedance of a simplified CC amplifier?
a) R E || (r e ‘ – (R 1 || R 2 ) / )
b) R E || (r e ‘ – (R 1 || R 2 ) / )
c) R E || (r e ‘ + (R 1 || R 2 ) / )
d) R E || (r e ‘ + (R 1 || R 2 ) / )
Answer: c
Explanation: The output impedance of a simplified common collector amplifier can be expressed in terms circuit resistor values. The base impedance is R E || (r e ‘ + (R 1 || R 2 ) / ) . Where β is the transistor amplifiers current gain, R e is the equivalent emitter resistance and r e ‘ is the alternating current resistance of the emitter-base diode.
8. What is the voltage gain of a simplified CC configuration amplifier if R E = 25kΩ and r e ‘ = 12kΩ?
a) 0.68
b) 0.59
c) 0.48
d) 0.086
Answer: a
Explanation: The voltage gain A V of a simplified common collector amplifier configuration can be calculated using A V = R E / (R E + r e ‘) . Given, R E = 25kΩ and r e ‘ = 12kΩ:
A V = R E / (R E + r e ‘) = 25 / 25 + 12 = 25 / 37 = 0.675 ≈ 0.68
9. What is the approximate value of the voltage gain of a simplified CC amplifier configuration?
a) 1
b) β
c) ∞
d) 0
Answer: a
Explanation: The voltage gain of a simplified CC amplifier configuration is approximately equal to unity (A V ≈ 1) and its current gain, A i is approximately equal to β(A i ≈ β) which depends on the value of each individual transistors.
10. There is no collector resistance in a simplified CC amplifier configuration.
a) True
b) False
Answer: a
Explanation: The simplified common collector amplifier configuration does not have a collector resistance R C . The collector terminal of the bipolar junction transistor is common to both the input and output circuit terminals.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Simplified Calculations for the CB Configuration”.
1. How do we compute the current gain of a simplified CB amplifier model using hybrid parameters?
a) -h fe / 1 + h fe
b) -h fe / 1 – h fe
c) h fe / 1 + h fe
d) h fe / 1 – h fe
Answer: c
Explanation: The current gain of a simplified common base amplifier can be expressed in terms of approximate hybrid parameters. The current gain h fe is the forward transfer characteristics. The current gain of a common base amplifier is expressed as h fe / 1 + h fe .
2. How do we compute the input resistance of a simplified CB amplifier model using hybrid parameters?
a) h fe / 1 – h fe
b) h fe / 1 + h fe
c) h ie / 1 – h fe
d) h ie / 1 + h fe
Answer: d
Explanation: The input resistance of a simplified common base amplifier can be expressed in terms of approximate hybrid parameters. The input resistance is h ie / 1 + h fe where, h ie and h fe are the input resistance and current gain of CE amplifier respectively.
3. How do we compute the voltage gain of a simplified CB amplifier model using hybrid parameters?
a) h fe × R L / h ie
b) -h fe × R L / h ie
c) h fe × R L / h oe
d) -h fe × R L / h oe
Answer: a
Explanation: The voltage gain of a simplified common base amplifier can be expressed in terms of approximate hybrid parameters. The voltage gain is equal to h fe × R L / h ie where, R L is load resistance, h ie is input resistance of CE amplifier and -h fe is the current gain of CE amplifier.
4. What is the value of voltage V B of a CB amplifier configuration if V CC = 12V, R 1 = 25kΩ and R 2 = 6kΩ?
a) 2.84V
b) 13.03V
c) 2.45V
d) 2.32V
Answer: d
Explanation: The voltage V B of a simplified common base amplifier configuration can be calculated using V B = I × R 2 . Given, V CC = 12V and R 1 = 25kΩ and R 2 = 6kΩ:
I = V CC / R 1 + R 2 = 12 / = 0.387mA
V B = I × R 2 = 0.387mA × 6kΩ = 2.32V
5. What is the bias of a simplified common base amplifier configuration?
a) Reverse biased
b) Forward biased
c) Forward and reverse biased
d) It is independent of the bias
Answer: b
Explanation: The common base design amplifier works to such an extent that the input signal is applied to the emitter terminal and the yield is taken from the collector terminal. The emitter current is likewise the input current, and the collector current is additionally the yield current. The base-emitter intersection is forward-biased.
6. How do we compute the voltage gain of a simplified CB amplifier?
a) R B || r e ‘
b) R C || r e ‘
c) R B / r e ‘
d) R C / r e ‘
Answer: d
Explanation: The voltage gain of a simplified common base amplifier can be expressed in terms circuit resistor values. The base impedance is R C / r e ‘. Where R C is the equivalent current resistance and r e ‘ is the alternating current resistance of the emitter-base diode.
7. How do we compute the output impedance of a simplified CB amplifier?
a) R C + R L
b) R B + R L
c) R C || R L
d) R B || R L
Answer: c
Explanation: The output impedance of a simplified common base amplifier can be expressed in terms circuit resistor values. The base impedance is R C || R L . Where R C is the equivalent collector resistance and R L is the equivalent load resistance.
8. What is the value of β in a simplified CB configuration amplifier if I C = 124mA and I B = 1.26mA?
a) 98.41
b) 99.24
c) 100
d) 97.33
Answer: a
Explanation: The value of β in a simplified common base amplifier configuration can be calculated using β = I C / I B . Given, I C = 124mA and I B = 1.26mA:
β = I C / I B = 124 / 1.26 = 98.41
9. What is the approximate value of the current gain of a simplified CB amplifier configuration?
a) 1
b) β
c) ∞
d) 0
Answer: a
Explanation: The current gain of a simplified CB amplifier configuration is approximately equal to unity (A i ≈ 1). A i is approximately equal to β / β + 1. The value of β depends on the value of each individual transistors.
10. The simplified common base amplifier configuration is useful in audio and radio frequency applications.
a) True
b) False
Answer: a
Explanation: The common base amplifier plan is very helpful in sound and radio recurrence applications because of its input yield impedance attributes. The current buffer is utilized to coordinate a low-impedance source to a high-impedance.
This set of Electronic Devices and Circuits Puzzles focuses on “CE Amplifier with an Emitter Resistance”.
1. For a common-emitter amplifier, the purpose of swamping is ________
a) to minimize gain
b) to reduce the effects of r’ e
c) to maximize gain
d) no purpose
Answer: b
Explanation: A swamping resistor is an unbiased resistance in the emitter circuit of a common-emitter amplifier. The swamping resistor stabilizes the voltage gain and reduces distortion.
2. What is the use of coupling capacitors in CE amplifier?
a) blocks dc
b) pass ac
c) reduce distortion
d) pass ac & blocks dc
Answer: d
Explanation: In analog circuits, a coupling capacitor is used to connect two circuits such that only the AC signal from the first circuit can pass through to the next while DC is blocked. This technique helps to isolate the DC bias settings of the two coupled circuits.
3. CE amplifier is mostly preferred in amplifier circuits because _________
a) it provides better voltage and current gain
b) of low output impedance
c) of high output impedance
d) it has better Q-point
Answer: a
Explanation: The common emitter configuration provides maximum voltage and current gain. The other configurations provide either high current gain or voltage gain but not both for a BJT. That’s why CE amplifier is always preferred and hence does the choice of CE becomes obvious.
4. Which type of biasing is used in CE amplifier?
a) Fixed bias
b) Collector to base bias
c) Voltage divider bias
d) Emitter bias
Answer: c
Explanation: The single stage common emitter amplifier uses biasing commonly called “Voltage Divider Biasing”. This type of biasing arrangement uses two resistors as a potential divider network across the supply with their center point supplying the required Base bias voltage to the transistor.
5. What is the correct phase shift in CE amplifier?
a) 45 degrees
b) 180 degrees
c) 60 degrees
d) 120 degrees
Answer: b
Explanation: The negative alternation of an AC signal will cause a decrease in the base current. This action then causes a corresponding decrease in emitter current through R L . The output signal of a common- emitter amplifier is therefore 180 degrees out of phase with the input signal.
6. The use of by-pass capacitor in CE amplifier is _________
a) to increase voltage gain
b) to increase negative feedback
c) for decreasing the frequency
d) to block dc
Answer: a
Explanation: When an emitter resistance is added in a CE amplifier, its voltage gain is reduced, but the input impedance increases. Whenever bypass capacitor is linked in parallel with an emitter resistance, the voltage gain of CE amplifier increases.
7. In analog circuits, a coupling capacitor is used to connect two circuits such that only the AC signal from the first circuit can pass through to the next while DC is blocked.
a) True
b) False
Answer: a
Explanation: A coupling capacitor allows ac signal to pass from one circuit to the other. This technique helps to isolate the DC bias settings of the two coupled circuits.
8. When the signal is applied, the ratio of change of collector current to the ratio of change of base current is called_________
a) dc current gain
b) ac current gain
c) base current amplification factor
d) emitter current amplification factor
Answer: b
Explanation: The ac current gain is given by β=∆I C /∆I B . When the signal is applied, the ratio of change of collector current to the ratio of change of base current is called ac current gain.
9. In CE configuration, if the voltage drop across 6kΩ resistor connected in the collector circuit is 6V. Find the value of IB when β=50.
a) 0.01mA
b) 0.25mA
c) 0.03mA
d) 0.02mA
Answer: d
Explanation: I C =V across R L /R L =5V/5KΩ=1mA
I B =I C /β=1/50=0.02MA.
10. The___________ configuration is used frequently for impedance matching.
a) fixed bias
b) voltage- divider
c) emitter follower
d) collector feedback
Answer: c
Explanation: The emitter follower configuration is mostly used as a voltage buffer. These configurations are extensively used in impedance matching applications because of their high input impedance.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Emitter Follower”.
1. What type of amplifier is an emitter follower amplifier?
a) Voltage amplifier
b) Wideband amplifier
c) Feedback amplifier
d) Power amplifier
Answer: c
Explanation: The emitter follower amplifier is one of the most noticeable feedback amplifiers. It provides a negative current feedback to the circuit. These type of amplifiers are usually used in the end stage or the last stage of a series of amplifiers.
2. How is the input impedance of an emitter follower amplifier?
a) Irrelevant
b) Moderate
c) Low
d) High
Answer: d
Explanation: The construction of an emitter follower circuit are approximately similar to a normal amplifier. One of the most important differentiating feature of an emitter follower circuit is its high input impedance.
3. How is the output impedance of an emitter follower amplifier?
a) Irrelevant
b) Moderate
c) Low
d) High
Answer: c
Explanation: The construction of an emitter follower circuit are approximately similar to a normal amplifier. One of the most important differentiating feature of an emitter follower circuit is its low output impedance.
4. How is the current gain of an emitter follower amplifier?
a) Irrelevant
b) Moderate
c) Low
d) High
Answer: d
Explanation: The construction of an emitter follower circuit are approximately similar to a normal amplifier. One of the most important differentiating feature of an emitter follower circuit is its high current gain.
5. How is the power gain of an emitter follower amplifier?
a) Irrelevant
b) Moderate
c) Low
d) High
Answer: d
Explanation: The construction of an emitter follower circuit are approximately similar to a normal amplifier. One of the most important differentiating feature of an emitter follower circuit is its high power gain.
6. What is the application of an emitter follower amplifier?
a) Positive feedback
b) Voltage gain
c) Power gain
d) Impedance matching
Answer: d
Explanation: The high input impedance along with low output impedance makes the emitter follower amplifier ideal for impedance matching. Along with that, its key characteristics include a comparatively high current and power gain.
7. What type of negative feedback does the emitter follower amplifier provide?
a) Voltage, current and power
b) Voltage
c) Current
d) Power
Answer: c
Explanation: The emitter follower configuration is prominently known for its feedback amplifier qualities. It is a negative feedback network that provides a negative feedback of current to the circuit. It is generally used in the last stage of an amplifying network.
8. What is the value of β in an emitter follower amplifier if V f = 12V and V O = 12V?
a) 1
b) 124
c) 1.24
d) 0.124
Answer: a
Explanation: The value of β in an emitter follower amplifier configuration can be calculated using β = V f / V O . Given, V f = 12V and V O = 12V:
β = V f / V O = 12 / 12 = 1
9. What is the voltage gain of an emitter follower amplifier?
a) 1
b) β
c) ∞
d) 0
Answer: a
Explanation: The voltage gain of an emitter follower amplifier configuration is roughly equal to unity. AV is approximately equal to 1. This is due to the relatively high input impedance and low output impedance.
10. What is the other name for an emitter follower amplifier configuration?
a) Common collector
b) Common base
c) Common emitter
d) Amplifier circuit
Answer: a
Explanation: The emitter follower circuit configuration is also known as the common collector configuration as it provides a high input impedance and a low output impedance. The emitter follower circuit acts as a buffer stage, and in result is used in several circuits where there is a need to prevent the loading of a circuit.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “High Input Resistance Transistor Circuit”.
1. Which of the following have high input resistance transistor circuits?
a) Common collector
b) Common base
c) Feedback amplifier
d) Power amplifier
Answer: a
Explanation: The common collector amplifier circuit is one of the most noticeable high input resistance transistor circuits. In most applications, along with CC configuration the common emitter also generally has a high input resistance.
2. What is the other name for a high input resistance transistor amplifier circuit amplifier configuration?
a) Common collector
b) Common base
c) Common emitter
d) Amplifier circuit
Answer: a
Explanation: The high input resistance transistor amplifier circuit configuration is also known as the common collector configuration as it provides a high input impedance and a low output impedance. The high input resistance transistor amplifier circuit acts as a middle stage.
3. How is the current gain of a high input resistance transistor amplifier circuit amplifier?
a) Irrelevant
b) Moderate
c) Low
d) High
Answer: d
Explanation: The construction of a high input resistance transistor amplifier circuit is approximately similar to a normal amplifier. One of the most important differentiating feature of a high input resistance transistor amplifier circuit is its high current gain.
4. How is the output impedance of a high input resistance transistor amplifier circuit?
a) Irrelevant
b) Moderate
c) Low
d) High
Answer: c
Explanation: The construction of a high input resistance transistor amplifier circuit is approximately similar to a normal amplifier. One of the most important differentiating feature of a high input resistance transistor amplifier circuit is its low output impedance.
5. What is the application of a high input resistance transistor amplifier circuit amplifier?
a) Positive feedback
b) Voltage gain
c) Power gain
d) Impedance matching
Answer: d
Explanation: The high input impedance and resistance along with low output impedance makes the transistor amplifier circuit amplifier ideal for impedance matching. Along with that, its key characteristics include a comparatively high current and power gain.
6. What type of negative feedback does a high input resistance transistor amplifier circuit amplifier provide?
a) Voltage, current and power
b) Voltage
c) Current
d) Power
Answer: c
Explanation: A high input resistance transistor amplifier circuit configuration is prominently known for its feedback amplifier qualities. It is a negative feedback network that provides a negative feedback of current to the circuit. It is generally used in the last stage of an amplifying network.
7. How is the power gain of a high input resistance transistor amplifier circuit amplifier?
a) Irrelevant
b) Moderate
c) Low
d) High
Answer: d
Explanation: The construction of a high input resistance transistor amplifier circuit is approximately similar to a normal amplifier. One of the most important differentiating feature of a high input resistance transistor amplifier circuit is its high power gain.
8. What is the value of β in a high input resistance transistor amplifier circuit amplifier if V f = 7V and V O = 6.89V?
a) 1
b) 1.12
c) 1.4
d) 0.124
Answer: a
Explanation: The value of β in a high input resistance transistor amplifier circuit amplifier configuration can be calculated using β = V f / V O . Given, V f = 7V and V O = 6.89V:
β = V f / V O = 7 / 6.89 = 1.01 ≈ 1.
9. What is the voltage gain of a high input resistance transistor amplifier circuit amplifier?
a) 1
b) β
c) ∞
d) 0
Answer: a
Explanation: The voltage gain of a high input resistance transistor amplifier circuit amplifier configuration is roughly equal to unity. A V is approximately equal to 1. This is due to the relatively high input impedance and low output impedance.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Cascode Transistor Configuration”.
1. What is the advantage of using a cascading transistor configuration?
a) High gain and high bandwidth
b) High gain and low bandwidth
c) Low gain and high bandwidth
d) Low gain and low bandwidth
Answer: a
Explanation: A circuit having a single transistor configuration does not provide suitable bandwidth or gain. To overcome this difficulty, we combine several amplification stages. The cascading transistor configuration theory is used for high gain as well as high bandwidth.
2. What is the advantage of using CB amplifier configuration in a cascading transistor configuration?
a) High efficiency
b) Low distortion
c) Good high frequency operation
d) Good low frequency operation
Answer: c
Explanation: The cascading transistor configuration circuit can be constructed with two configurations of a transistor that is CE and CB . The CB configuration delivers a good high-frequency operation.
3. What is the total voltage gain of a cascading transistor configuration if the gain of first stage is 48 and the gain of the second stage is 12?
a) 6.75
b) 6.55
c) 5.76
d) 5.65
Answer: c
Explanation: The total gain of a cascading transistor configuration can be calculated by the formula given below. Where, A V = overall gain, A V1 = voltage gain of first stage and A V2 = voltage gain of second stage.
A V = A V1 × A V2 = 4.8 × 1.2 = 5.76.
4. What is the total voltage gain of a cascading transistor configuration if input of first stage is 50V and the output of the second stage is 100V?
a) 20
b) 0.2
c) 2.5
d) 2
Answer: d
Explanation: The total gain of a cascading transistor configuration is the ration of the output of second stage amplifier to the input of first stage amplifier. Where, V output = 100V and V input = 50V.
A V = V output / V input = 100V / 50V = 2.
5. What does cascading of two transistor amplifiers imply?
a) Output of first stage sent to input of second stage
b) Output of first stage sent to coupling device
c) Input of first stage sent to input of second stage
d) Not related
Answer: b
Explanation: In multi-stage amplifiers, the output of first stage is coupled to the input of second stage using a coupling device. The coupling devices is typically a capacitor or a transformer. Cascading is known as the process of combining two amplifier stages using a coupling device.
6. How many methods of coupling are used for a cascading transistor configuration?
a) 3
b) 4
c) 2
d) 5
Answer: b
Explanation: There are four types of coupling mechanisms used. They are Resistance-capacitance coupling, thermal coupling, impedance coupling and transformer coupling. Resistance-capacitance coupling is the most vastly used mechanism.
7. What is the total voltage gain of a cascading transistor configuration if the gain of first stage is 10dB, gain of the second stage is 20dB and gain of third stage is 30dB?
a) 60.32dB
b) 50dB
c) 60dB
d) 50.32dB
Answer: c
Explanation: The total gain of a cascading transistor configuration is the sum of the multiple stages when the gain is calculated is decibels. Given, A V1 = 10dB, A V2 = 20dB and A V3 = 30dB
A V = A V1 + A V2 + A V3 = 10dB + 20dB + 30dB = 60dB.
8. The coupling device is essential for a cascading transistor configuration.
a) True
b) False
Answer: a
Explanation: One of the main functions of the coupling device in a cascading transistor configuration is to block the direct current from passing through to the input of next stage from the output of the first stage. Another function of the coupling device is to transfer the current from the output of first stage to the input of the second stage.
9. What is the purpose of using a cascading transistor configuration?
a) Increase voltage gain
b) Decrease voltage gain
c) Increase current gain
d) Decrease current gain
Answer: a
Explanation: A circuit having a single transistor configuration does not provide suitable bandwidth or gain. The purpose of a cascading transistor configuration is to provide an increase in the voltage gain. The total gain of a cascading transistor configuration is the product of the voltage gains of the discrete stages.
10. How is the voltage gain measured in a cascading transistor configuration?
a) Volts
b) Ampere
c) Decibels
d) Dimensionless
Answer: c
Explanation: The voltage gain is measured in terms of decibels. The total voltage gain is calculated as the product of individual stages or as a sum of all the stages if the gain calculated at each intermediate stage was measured in decibels as well.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Transistor Difference Amplifiers”.
1. What is the voltage gain of a differential amplifier is all amplifier values are equal?
a) 1
b) 0
c) ∞
d) Cannot be determined
Answer: a
Explanation: If all the resistor values are of the equal value, then the circuit will have a voltage gain equal to exactly one or unity. Hence it is also called as a Unity Gain Differential Amplifier. The output voltage expression would be the difference between the two voltages.
2. What is the problem with a single operational difference amplifier?
a) High input resistance
b) Low input resistance
c) Low output resistance
d) High output resistance
Answer: b
Explanation: The problem with a single operational difference amplifier is its low input resistance. This low input resistance results in a loss of a major component of the signal. Operational Amplifier is internally a Differential Amplifier with features like High Input Impedance, Low Output Impedance.
3. What is the total voltage gain of a difference amplifier if the gain of first stage is 4.8 and the gain of the second stage is 1.2?
a) 6.75
b) 6.55
c) 5.76
d) 5.65
Answer: c
Explanation: The total gain of a difference amplifier can be calculated by the formula given below. Where, A V = overall gain, A V1 = voltage gain of first stage and A V2 = voltage gain of second stage.
A V = A V1 × A V2 = 4.8 × 1.2 = 5.76
4. What is the total voltage gain of a difference amplifier if input of first stage is 50V and the output of the second stage is 100V?
a) 20
b) 0.2
c) 2.5
d) 2
Answer: d
Explanation: The total gain of a difference amplifier is the ratio of the output of second stage amplifier to the input of first stage amplifier. Where, V output = 100V and V input = 50V.
A V = V output / V input = 100V / 50V = 2
5. How is the voltage gain measured in a difference amplifier?
a) Volts
b) Ampere
c) Decibels
d) Dimensionless
Answer: c
Explanation: The voltage gain is measured in terms of decibels. The total voltage gain is calculated as the product of individual stages or as a sum of all the stages if the gain calculated at each intermediate stage was measured in decibels as well.
6. How do we calculate the output voltage of a difference amplifier?
a) V O = -A d × V d + A C × V C
b) V O = -A d × V d – A C × V C
c) V O = A d × V d + A C × V C
d) V O = A d × V d – A C × V C
Answer: c
Explanation: The differential gain of a difference amplifier is defined as the gain obtained at the output signal with respect to the difference in the input signals applied. To calculate the output voltage of a difference amplifier we use A d × V d + A C × V C . Where, A d is the differential gain and A C and V C represent the common mode gain.
7. What is the purpose of using a difference amplifier?
a) Increase voltage gain
b) Decrease voltage gain
c) Increase current gain
d) Decrease current gain
Answer: a
Explanation: A circuit having a single transistor configuration does not provide suitable bandwidth or gain. The purpose of a difference amplifier is to provide an increase in the voltage gain. The total gain of a difference amplifier is the product of the voltage gains of the discrete stages.
8. How many methods of coupling are used for a difference amplifier?
a) 3
b) 4
c) 2
d) 5
Answer: b
Explanation: There are four types of coupling mechanisms used. They are Resistance – capacitance coupling, thermal coupling, impedance coupling and transformer coupling. Resistance – capacitance coupling is the most vastly used mechanism.
9. What is the total voltage gain of a difference amplifier if the gain of first stage is 10dB, gain of the second stage is 20dB and gain of third stage is 30dB?
a) 60.32dB
b) 50dB
c) 60dB
d) 50.32dB
Answer: c
Explanation: The total gain of a difference amplifier is the sum of the multiple stages when the gain is calculated is decibels. Given, A V1 = 10dB, A V2 = 20dB and A V3 = 30dB
A V = A V1 + A V2 + A V3 = 10dB + 20dB + 30dB = 60dB
10. The coupling device is essential for a difference amplifier.
a) True
b) False
Answer: a
Explanation: One of the main functions of the coupling device in a difference amplifier is to block the direct current from passing through to the input of next stage from the output of the first stage. Another function of the coupling device is to transfer the current from the output of first stage to the input of the second stage.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The High Frequency T Model”.
1. What is the T represent in high frequency T model?
a) Transmission
b) Transistor
c) Tailgate
d) Thorough
Answer: b
Explanation: The T in high frequency T model represents transistor models. This demonstrates the study of high frequency input operations on multiple basic transistor models to observe the variation in the output characteristics.
2. Which model is similar to the high frequency T model?
a) G model
b) Hybrid-pi model
c) Inverse G model
d) Z model
Answer: b
Explanation: The Hybrid-pi model is similar to the high frequency T model. The T model can use either a voltage or a current as the variable that controls the current source. In the T model, the current source’s expression is either g m V BE or αI E .
3. How do we determine the hybrid parameters h 11 and h 21 in a high frequency T model?
a) Short circuiting the input terminal
b) Open circuiting the input terminal
c) Short circuiting the output terminal
d) Open circuiting the output terminal
Answer: c
Explanation: There are four h parameters, they are: h 11 , h 12 , h 21 and h 22 . To determine the values of h 11 and h 21 we need to short circuit the output terminal of a given two-port network. Short circuiting the output terminal makes the output terminal voltage V 2 equal to zero.
4. Which hybrid parameter is used to calculate forward current gain in a high frequency T model?
a) h 11
b) h 12
c) h 21
d) h 22
Answer: c
Explanation: The hybrid parameter h 21 can be calculated by short circuiting the output port in a high frequency T model. By doing so, we can calculate h 21 = I 2 / I 1 . Since this is a ratio between two current units, h 21 is unit less. It is known as the forward current gain of the circuit.
5. What is h 11 also known as in a high frequency T model?
a) Input conductance
b) Input resistance
c) Output conductance
d) Output resistance
Answer: b
Explanation: The parameter h 11 can be calculated by short circuiting the output port. By doing so, we can calculate h 11 = V 1 / I 1 . Since this gives h 11 the units of volts / ampere or in other words ohms, it is also known as input resistance in a high frequency T model.
6. How do we calculate the value of hybrid parameter h 12 if V 2 = 20V and V 1 = 50V in a high frequency T model?
a) 0.04
b) 0.25
c) 0.4
d) 2.5
Answer: d
Explanation: To find the hybrid parameter h 12 in a high frequency T model we use the below mentioned formula. Given values are V 2 = 20V and V 1 = 50V.
h 21 = V 1 / V 2 = 50 / 20 = 2.5
7. How do we determine the hybrid parameters h 12 and h 22 in a high frequency T model?
a) Short circuiting the input terminal
b) Open circuiting the input terminal
c) Short circuiting the output terminal
d) Open circuiting the output terminal
Answer: b
Explanation: From the four hybrid parameters, to determine the value of h 12 and h 22 we need to open circuit the input terminal. This in turn makes the current from the input terminal I 1 equal to zero. It is the same procedure in a high frequency T model.
8. Which of the following four hybrid parameters is used to calculate reverse voltage gain in a high frequency T model?
a) h 11
b) h 12
c) h 21
d) h 22
Answer: b
Explanation: The hybrid parameter h 12 can be calculated by open circuiting the input port in a high frequency T model. By doing so, we can calculate h 12 = V 1 / V 2 . Since this is a ratio between two voltage units, h 12 is unitless. It is known as the reverse voltage gain of the circuit.
9. In a high frequency T model, which parameter is known as the output conductance?
a) h 11
b) h 12
c) h 21
d) h 22
Answer: d
Explanation: When the input terminal of a two-port device is open circuited in a high frequency T model, the input current flowing equals to zero. Hence the value of h 22 = I 2 / V 2 . The unit of this parameter is ampere / volt or mho. Hence it is known as output conductance.
10. How do we determine the value of the hybrid parameter h 21 in a high frequency T model?
a) h 21 = I 1 / I 2
b) h 21 = V 1 / V 2
c) h 21 = I 2 / I 1
d) h 21 = V 2 / V 1
Answer: c
Explanation: To calculate the value of h 21 high frequency T model, we short circuit the output terminal so the output voltage V 2 equals zero. Hence the equation for hybrid parameters becomes, I 2 = h 21 I 1 + 0. Therefore, h 21 = I 2 / I 1 .
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The CB Short Circuit Current Frequency Response”.
1. Where is the input measured in a CB short circuit current frequency response?
a) Collector terminal
b) Emitter terminal
c) Base terminal
d) Ground
Answer: b
Explanation: In the physical model of a CB short circuit current frequency response the input is measured at the emitter terminal of the BJT biased device. Whereas, the output is measured across the collector terminal of the biased BJT device.
2. Which parameter of the physical model is varied while measuring the input characteristics of a CB short circuit current frequency response?
a) Emitter current
b) Emitter voltage
c) Collector current
d) Emitter base voltage
Answer: d
Explanation: To determine the input characteristics, the collector – base voltage is kept constant at zero volts and the emitter base voltage is increased from zero volts to different voltage levels. For each voltage level of the input voltage, the input current is recorded.
3. Where is the output measured in a CB short circuit current frequency response?
a) Collector terminal
b) Emitter terminal
c) Base terminal
d) Ground
Answer: a
Explanation: In the physical model of a CB short circuit current frequency response the output is measured at the collector terminal of the BJT biased device. Whereas, the input is measured across the emitter terminal of the biased BJT device.
4. Which parameter of the physical model is varied while measuring the output characteristics of a CB short circuit current frequency response?
a) Emitter current
b) Emitter voltage
c) Collector current
d) Collector base voltage
Answer: d
Explanation: To determine the output characteristics, the emitter current is kept constant at zero and the collector base voltage is increased from zero volts to varying voltage levels. For each voltage level of the output voltage, the collector current is recorded.
5. What is the path of the output characteristic of a CB short circuit current frequency response in the active region?
a) Linear
b) Exponential decrease
c) Exponential increase
d) Constant
Answer: d
Explanation: In a CB short circuit current frequency response model, the output characteristic curve first increases exponentially and then remains constant. In the active region of working the output characteristics show a constant value collector current with respect to collector base voltage.
6. What is the path of the output characteristic of a CB short circuit current frequency response in the saturation region?
a) Linear
b) Exponential decrease
c) Exponential increase
d) Constant
Answer: c
Explanation: In a physical model of a CB short circuit current frequency response, the output characteristic curve first increases exponentially and then remains constant. In the saturation region of working the output characteristics show an exponential increase in the collector current with respect to collector base voltage.
7. What is the path of the input characteristic of a CB short circuit current frequency response?
a) Linear
b) Exponential decrease
c) Exponential increase
d) Constant
Answer: a
Explanation: In a CB short circuit current frequency response, the input characteristic curve increases exponentially. The emitter current increases exponentially with respect to emitter base voltage for fixed values of collector base voltage.
8. How do you calculate the dynamic input resistance of a CB short circuit current frequency response?
a) ΔV BE / ΔI C
b) ΔV BE / ΔI E
c) ΔV CB / ΔI C
d) ΔV CB / ΔI E
Answer: b
Explanation: Dynamic input resistance is defined as the ratio of change in emitter base voltage to the corresponding change in the emitter current. While the collector voltage is kept at a constant value. Therefore, r i = ΔV BE / ΔI E .
9. What is the value of voltage V B of a CB short circuit current frequency response configuration if V CC = 12V, R 1 = 25kΩ and R 2 = 6kΩ?
a) 2.84V
b) 13.03V
c) 2.45V
d) 2.32V
Answer: d
Explanation: The voltage V B of a simplified CB short circuit current frequency response configuration can be calculated using V B = I × R 2 . Given, V CC = 12V and R 1 = 25kΩ and R 2 = 6kΩ:
I = V CC / R 1 + R 2 = 12 / = 0.387mA
V B = I × R 2 = 0.387mA × 6kΩ = 2.32V
10. What is the bias of a simplified CB short circuit current frequency response configuration?
a) Reverse biased
b) Forward biased
c) Forward and reverse biased
d) It is independent of the bias
Answer: b
Explanation: The CB short circuit current frequency response operates such that the input signal is applied to the emitter terminal and the output is taken from the collector terminal. The emitter current is also the input current, and the collector current is also the output current. The base – emitter junction is forward – biased.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Hybrid PI CE Transistor Model”.
1. What is the value of input resistance in the hybrid-pi CE transistor model?
a) h oe
b) h ib
c) h ic
d) h ie
Answer: d
Explanation: The input resistance of a common emitter amplifier can be expressed in terms of hybrid parameters in the hybrid-pi CE transistor model. The input resistance is universally expressed as h ie . This is also the same for approximate as well as exact hybrid models.
2. How do we calculate the value of I C in the hybrid-pi CE transistor model?
a) I C = h re × I B
b) I C = -h re × I B
c) I C = h fe × I B
d) I C = -h fe × I B
Answer: c
Explanation: The collector current of a common emitter amplifier can be expressed in terms of hybrid parameters in the hybrid-pi CE transistor model. It is expressed as the product of h fe × I B . This is the same for approximate hybrid models as well.
3. What is the value of current gain in the hybrid-pi CE transistor model?
a) h fe
b) -h fe
c) h re
d) –h re
Answer: b
Explanation:The current gain of a common emitter amplifier can be expressed in terms of hybrid parameters in the hybrid-pi CE transistor model. The current gain h fe is the forward transfer characteristics and is used to calculate the current gain in common emitter amplifiers.
4. What is the value of voltage gain in the hybrid-pi CE transistor model?
a) h fe × R L / h re
b) –h fe × R L / h re
c) h fe × R L / h ie
d) –h fe × R L / h ie
Answer: d
Explanation: The voltage gain of a common emitter amplifier can be expressed in terms of hybrid parameters in the hybrid-pi CE transistor model. The voltage gain is equal to -h fe × R L / h ie where, R L is load resistance, h ie is input resistance and -h fe is the current gain.
5. What is the magnitude of voltage generated at emitter circuit of a hybrid-pi CE transistor model?
a) h re × h fe / I b × R L
b) h re × h fe × I b × R L
c) h re × h fe / I C × R L
d) h re × h fe × I C × R L
Answer: b
Explanation: The magnitude of voltage generated at emitter circuit of a common emitter amplifier can be expressed in terms of hybrid parameters in the hybrid-pi CE transistor model. The magnitude of voltage generated is equal to h re × h fe × I b × R L where, R L is load resistance and I b is the base current.
6. What is the value of output resistance in the hybrid-pi CE transistor model?
a) ∞
b) 0
c) h ic
d) h ie
Answer: a
Explanation: The output resistance in a simplified common emitter hybrid model is calculated as the ratio of collector voltage over collector current while V S is equal to zero and R L is omitted. Due to this configuration I C is equal to zero and hence output resistance is infinite.
7. What is the value power gain in the hybrid-pi CE transistor model?
a) A V – A I
b) A V + A I
c) A V × A I
d) A V / A I
Answer: c
Explanation: The power gain in a simplified common emitter hybrid model is calculated as the product of voltage gain and power gain. It is expressed as A V × A I . It can also be further simplified as A I 2 × R L / R i , where R L is load resistance and R i is input resistance.
8. What is the value of overall current gain in the hybrid-pi CE transistor model while considering R S ?
a) A I × I B / I S
b) A I × I C / I S
c) A I × I S / I C
d) A I × I S / I B
Answer: a
Explanation: The current gain in a simplified common emitter hybrid model while considering R S is calculated as the product of current gain without external parameters and the ratio of base current and external applied current. It is expressed as A I × I B / I S . Where, I S is the external applied current on the amplifier.
9. What is the value of overall voltage gain in the hybrid-pi CE transistor model while considering R S ?
a) V B / V S
b) V C / V S
c) V B × V S
d) V C × V S
Answer: a
Explanation: The overall voltage gain in a simplified common emitter hybrid model while considering R S is calculated as ratio of base voltage over the applied external voltage. It is expressed as V B / V S . Where, V S is the external applied voltage on the amplifier.
10. We use hybrid-pi CE transistor model to obtain approximate values.
a) True
b) False
Answer: a
Explanation: It is more practical to calculate the approximate values of voltage and current gains instead of computing the exact values of a common emitter amplifier circuit. We use a simplified common emitter hybrid model to obtain these approximate values without reducing the accuracy.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Hybrid PI Conductances in Low Frequency H Parameters”.
1. For the two-port network shown below, assume R 1 = R 2 = 10kΩ. What is the value of the hybrid-pi conductance in low frequency H-parameters h 22 ?
a) 0.1 mho
b) 10 mhos
c) 20 mhos
d) 15 mhos
Answer: a
Explanation: To calculate h 22 , we open circuit the input terminal. Hence the current flowing through resistor R 1 is zero and the voltage V 1 across the input terminal is equal to the voltage V 2 across output terminal. Therefore, to find h 22 , given that R 2 = 10kΩ.
V 1 = V 2
I 2 = V 2 / 10 kΩ
h 22 = I 2 / V 2 = 0.1 mho
2. For the two-port network shown below, assume R 1 = R 2 = 10kΩ. What is the value of the hybrid-pi in low frequency H-parameters h 11 ?
a) 10kΩ
b) 20kΩ
c) 10kΩ
d) 5kΩ
Answer: c
Explanation: To calculate h 11 , we short circuit the output terminal. Hence the current flowing through resistor R 2 is zero. Therefore, to find h 11 , given that R 1 = 10kΩ.
V 1 = R 1 x I 1
h 11 = V 1 / I 1 = R 1 x I 1 / I 1 = R 1 = 10kΩ.
3. How do we calculate the inverse hybrid-pi in low frequency H-parameters g 11 of a two-port network?
a) g 11 = I 1 / V 1
b) g 11 = I 2 / V 1
c) g 11 = I 1 / V 2
d) g 11 = I 2 / V 2
Answer: a
Explanation: The inverse hybrid-pi conductance in low frequency H-parameters g 11 for a two-port network is calculated by open circuiting the output terminal. In this case, the current I 2 is equal to zero. The parameter g 11 is calculated by taking the ratio of input terminal current over input terminal voltage.
4. What are the hybrid-pi conductance in low frequency H-parameters used to analyze?
a) MOSFET
b) Junction Field Effect Transistor
c) Bipolar Junction Transistor
d) It has no use
Answer: c
Explanation: The four hybrid-pi conductance in low frequency H-parameters of a two-port network are used to analyze the workings of a Bipolar Junction Transistor or in other words a BJT. The inverse hybrid-pi conductance in low frequency H-parameters are used to analyze the Junction Field Effect Transistor of JFET.
5. How is the h 21 hybrid-pi conductance in low frequency H-parameters of a two-port network used?
a) It is used as a voltage source at input port
b) It is used as a current source at input port
c) It is used as a voltage source at output port
d) It is used as a current source at output port
Answer: d
Explanation: While checking the hybrid-pi conductance in low frequency H-parameters equivalent network of a two-port network we see that the hybrid-pi conductance in low frequency H-parameters h 21 is sued as a current source at the output port. While using Kirchhoff’s current law, the hybrid-pi conductance in low frequency H-parameters equations can be represented as a current source of h 21 I 1 .
6. How do we calculate the inverse hybrid-pi conductance in low frequency H-parameters g 21 of a two-port network?
a) g 21 = I 1 / I 2
b) g 21 = I 2 / I 1
c) g 21 = V 2 / V 1
d) g 21 = V 1 / V 2
Answer: c
Explanation: The inverse hybrid-pi conductance in low frequency H-parameters g 21 for a two-port network is calculated by open circuiting the output terminal. In this case, the current I 2 is equal to zero. The parameter g 21 is calculated by taking the ratio of output terminal voltage over input terminal voltage. It is also known as the open circuit voltage gain.
7. How is the h 12 hybrid-pi conductance in low frequency H-parameters of a two-port network used?
a) It is used as a voltage source at output port
b) It is used as a current source at output port
c) It is used as a voltage source at input port
d) It is used as a current source at input port
Answer: c
Explanation: While checking the hybrid-pi conductance in low frequency H-parameters equivalent network of a two-port network we see that the hybrid-pi conductance in low frequency H-parameters h 12 is sued as a voltage source at the input port. While using Kirchhoff’s voltage law, the hybrid-pi conductance in low frequency H-parameters equations can be represented as a fixed voltage source of h 21 V 2 .
8. How do we calculate the inverse hybrid-pi conductance in low frequency H-parameters g 12 of a two-port network?
a) g 12 = I 2 / I 1
b) g 12 = I 1 / I 2
c) g 12 = V 1 / V 2
d) g 12 = V 2 / V 1
Answer: b
Explanation: The inverse hybrid-pi conductance in low frequency H-parameters g 12 for a two-port network is calculated by short circuiting the input terminal. In this case, the current V 1 is equal to zero. The parameter g 12 is calculated by taking the ratio of input terminal current over output terminal current.
9. How do we calculate the inverse hybrid-pi conductance in low frequency H-parameters g 22 of a two-port network?
a) g 22 = V 2 / I 1
b) g 22 = I 1 / V 2
c) g 22 = V 1 / I 2
d) g 22 = V 2 / I 2
Answer: d
Explanation: The inverse hybrid-pi conductance in low frequency H-parameters g 22 for a two-port network is calculated by short circuiting the output terminal. In this case, the current V 1 is equal to zero. The parameter g 22 is calculated by taking the ratio of output terminal voltage over output terminal current.
10. We can measure hybrid h-parameters using inverse hybrid g-parameters.
a) True
b) False
Answer: a
Explanation: The hybrid h-parameters can be measured using the inverse hybrid g-parameters. All the parameters used to analyze a two-port network can be used to convert and measure other different parameters.
11. What is h 11 also known as in hybrid-pi conductance in low frequency H-parameters?
a) Input conductance
b) Input resistance
c) Output conductance
d) Output resistance
Answer: b
Explanation: The parameter h 11 can be calculated by short circuiting the output port. By doing so, we can calculate h 11 = V 1 / I 1 . Since this gives h 11 the units of volts/ampere or in other words ohms, it is also known as input resistance.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The CE Short Circuit Gain Obtained with the Hybrid PI Model”.
1. Why do we need CE short circuit gain obtained with hybrid pi model?
a) To provide a non – linear output
b) To maintain transistor in active region
c) To maintain transistor in saturation region
d) To maintain transistor in cut – off region
Answer: b
Explanation: We use a CE short circuit gain obtained with hybrid pi model to collect negative feedback and to maintain the transistor in an active region. To maintain stability in active region, the DC base biased voltage is resultant from the collector voltage V C .
2. What is the function of R E in the CE short circuit gain obtained with hybrid pi model?
a) To improve stability and decrease positive feedback
b) To improve stability and increase positive feedback
c) To improve stability and decrease negative feedback
d) To improve stability and increase negative feedback
Answer: d
Explanation: In a CE short circuit gain obtained with hybrid pi model circuit, the emitter resistor provides additional stability along with increasing the negative feedback sent to the collector. The addition of the emitter resistance enables the transistor’s emitter to no longer be grounded to zero – volt potential.
3. In the circuit given below, assume V CC = 5V, V BE = 0.7V, R E = 10kΩ, R B = 20kΩ and β = 50. How much is the current I E ?
a) 0.561mA
b) 0.335mA
c) 0.413mA
d) 0.513mA
Answer: c
Explanation: To find the value of I E , we substitute the values in the below equation. Given values are V CC = 5V, V BE = 0.7V, R E = 10kΩ, R B = 20kΩ and β = 50.
I E = (V CC – V BE ) / (R E + (R B / β))
I E = / )
I E = 4.3 / = 0.413mA.
4. What happens if collector current increases in a CE short circuit gain obtained with hybrid pi model?
a) Emitter voltage increases therefore base voltage increases
b) Emitter voltage decreases therefore base voltage decreases
c) Emitter voltage increases therefore base voltage decreases
d) Emitter voltage decreases therefore base voltage increases
Answer: a
Explanation: If the collector current I C is increased, the corresponding emitter current also increases. Which in turn causes the voltage across R E to increase. This in turn causes a proportional rise in the base voltage since V B = V E + V BE .
5. Why is CE short circuit gain obtained with hybrid pi model better for linear circuits?
a) Independent of β
b) Dependent on β
c) Highly predictable
d) Not stable
Answer: b
Explanation: CE short circuit gain obtained with hybrid pi model is better for linear circuits as compared to self – bias circuits as it is dependent on β. Voltage divider bias circuits are highly predictable whereas self – bias circuits are independent of β. Therefore, for a CE short circuit gain obtained with hybrid pi model linear circuits are preferred.
6. How does emitter resistor R E provide stability in CE short circuit gain obtained with hybrid pi model?
a) Consumes less power
b) Has an easier circuit design
c) Automatically biases the circuit
d) It does not provide stability
Answer: c
Explanation: In a CE short circuit gain obtained with hybrid pi model, the emitter resistor provides stability by automatically biasing the circuit using negative feedback. The negative feedback negates any change due to the collector current with an opposing change provided by the base bias voltage and thus helps maintain circuit stability.
7. In the circuit given below, assume V CC = 12V, V BE = 0.7V, R B = 330kΩ, R C = 3.3kΩ, R E = 2.7kΩ and β = 50. What is the base current I B ?
a) 16.432µA
b) 17.856µA
c) 20.542µA
d) 17.936µA
Answer: d
Explanation: To find the value of I B , we substitute the values in the below equation. Given values are V CC = 12V, V BE = 0.7V, R B = 330kΩ, R C = 3.3kΩ, R E = 2.7kΩ and β = 50.
I B = (V CC – V BE ) / (R B + β × (R C + R E ))
I B = ) = 11.3 / = 11.3 / 630 = 17.936µA.
8. How do you calculate the value of V CE in CE short circuit gain obtained with hybrid pi model?
a) V CE = V CC + I C (R C + R E )
b) V CE = V CC – I C (R C + R E )
c) V CE = V CC – I C (R B + R E )
d) V CE = V CC + I C (R B + R E )
Answer: b
Explanation: The value of V CE can be calculated using this equation: V CE = V CC – I C (R C + R E ). It is the voltage between the collector and emitter terminal of the transistor and is measured as the output of the transistor.
9. How does the negative feedback help a CE short circuit gain obtained with hybrid pi model circuit?
a) Helps make it more predictable
b) Provides opposing change in base voltage
c) Helps make it more predictable, provides opposing change in base voltage
d) It doesn’t affect
Answer: c
Explanation: The negative feedback in a CE short circuit gain obtained with hybrid pi model circuit provides a negative feedback which in turn helps make the circuit more predictable as it provides opposing change in the base voltage which cancels out any change in the collector current.
10. What are the disadvantages of CE short circuit gain obtained with hybrid pi model circuits?
a) Requires few resistors
b) Provides a lot of stability
c) Provides negative feedback
d) Provides positive feedback
Answer: c
Explanation: A collector emitter bias circuit provides negative feedback as well as requires multiples resistors for a small change. The negative feedback limits the frequency range it will work in. Higher frequencies will provide poor performance. It also requires a greater number of resistors just to provide a stability against a small parameter.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The Alpha Cutoff Frequency”.
1. What is the function of an alpha cutoff frequency?
a) To simplify the circuit
b) To provide a non – linear output
c) To optimize the power
d) To provide steady current or voltage
Answer: d
Explanation: The Q – point of a device is the direct current or voltage of a device when no input is applied. The alpha cutoff frequency is a part of the device with provides the steady current or voltage. It is designed by determining the necessary voltage and current levels across each resistor.
2. How many main types of alpha cutoff frequency are there in bipolar transistors?
a) 3
b) 4
c) 5
d) 6
Answer: c
Explanation: There are five main type of biasing circuits. They are Fixed Bias, Collector – to – Base Bias, Voltage Divider Bias, Fixed Bias with Emitter Resistor and Emitter Bias. Each one of these have a different alpha cutoff frequency value.
3. Why do we require R E for a good stable alpha cutoff frequency?
a) To obtain a current I E sensitive to α and V BE
b) To obtain a current I B sensitive to α and V BE
c) To obtain a current I E insensitive to α and V BE
d) To obtain a current I B insensitive to α and V BE
Answer: c
Explanation: The aim of having an alpha cutoff frequency is to maintain a stable current or voltage, regardless of other parameters changing. Hence, we need R E to get a stable I E current. Without R E the value of I E changes drastically with a small change in α.
4. In the circuit given below, assume V CC = 4V, V BB = 5V, V BE = 0.7V, R E = 3.3kΩ, R C = 10kΩ, R B = 40kΩ and α = 100. What is the current I E ?
a) 1.388mA
b) 0.012mA
c) 10.47mA
d) 1.165mA
Answer: a
Explanation: Given the values V CC = 4V, V BB = 5V, V BE = 0.7V, R E = 3.3kΩ, R C = 10kΩ, R B = 40kΩ and α = 100. To find the value of I E , we substitute the values in this equation:
I E = (V BB – V BE ) / (R E + (R B / α + 1))
I E = / )
I E = 4.3 / = 1.388mA.
5. Why is it essential to stabilize the operating point of a circuit?
a) Thermal runaway, power efficiency, individual variations
b) Temperature dependency of I B , individual variations, power efficiency
c) Thermal runaway, Temperature dependency of I C , individual variations
d) Temperature dependency of I C , thermal runaway, cost efficiency
Answer: c
Explanation: It is essential to stabilize the operating point of a circuit to prevent thermal runaway, temperature dependency of I C and individual variations. Thermal runaway occurs because of the self – destruction of an unstable transistor. Temperature dependency on I C causes major variations in I E . Individual variations of parameters such as α and V BE cause huge fluctuations in I E .
6. In an ideal stable self – alpha cutoff frequency, what should be the current I B with respect to I C ?
a) I B should be 20% of I C
b) I B should be 5% of I C
c) I B should be 10% of I C
d) I B should be 50% of I C
Answer: c
Explanation: Ideally, the resistor values are designed in a way that the voltage drop across R E is approximately 10% of V cc and I B should be 10% of I c . Hence, stable self – alpha cutoff frequency work best at low power supply voltages.
7. What values of V BE and α would provide maximum stability?
a) V BE = 5V, α = 50
b) V BE = 4V, α = 100
c) V BE = 4.5V, α = 50
d) It is irrelevant
Answer: d
Explanation: In a stable alpha cutoff frequency, the variations in the value of V BE and α do not affect the stability of the system. The current through the emitter remains unchanged. Hence, the values can vary without hampering the system.
8. What is the stability factor if R E = 8kΩ and R TH = 11kΩ during alpha cutoff frequency?
a) 1.335
b) 2.375
c) 1.727
d) 0.272
Answer: b
Explanation: Stability factor is calculated by the following equation. On plugging in the given values R E = 8kΩ and R TH = 11kΩ, we get:
S = 1 + R TH / R E
S = 1 + 11 / 8 = 2.375.
9. Collector base feedback and emitter feedback combined together provide stability for self – alpha cutoff frequency.
a) True
b) False
Answer: a
Explanation: Collector base feedback and emitter feedback combined provide stability for self – alpha cutoff frequency. This is because the emitter base junction is forward biased due to the voltage drop across R E .
10. Where does degeneration take place in a self – alpha cutoff frequency?
a) Across R B
b) Across R E
c) No degeneration occurs
d) Across R C
Answer: b
Explanation: Degeneration or negative feedback occurs across R E which in turn stabilizes the fluctuations of current I E due to temperature changes and variations in V BE and α. In a stable alpha cutoff frequency, the variations in the value of V BE and α do not affect the stability of the system.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “The CE Short Circuit Current Frequency Response”.
1. What is the value of input resistance in the CE short circuit current frequency response model?
a) h oe
b) h ib
c) h ic
d) h ie
Answer: d
Explanation: The input resistance of a common emitter amplifier can be expressed in terms of hybrid parameters in the CE short circuit current frequency response model. The input resistance is universally expressed as h ie . This is also the same for approximate as well as exact hybrid models.
2. How do we calculate the value of I C in the CE short circuit current frequency response model?
a) I C = h re × I B
b) I C = -h re × I B
c) I C = h fe × I B
d) I C = -h fe × I B
Answer: c
Explanation: The collector current of a common emitter amplifier can be expressed in terms of hybrid parameters in the CE short circuit current frequency response model. It is expressed as the product of h fe × I B . This is the same for approximate hybrid models as well.
3. What is the value of current gain in the CE short circuit current frequency response model?
a) h fe
b) -h fe
c) h re
d) –h re
Answer: b
Explanation: The current gain of a common emitter amplifier can be expressed in terms of hybrid parameters in the CE short circuit current frequency response model. The current gain h fe is the forward transfer characteristics and is used to calculate the current gain in common emitter amplifiers.
4. What is the value of voltage gain in the CE short circuit current frequency response model?
a) h fe × R L / h re
b) –h fe × R L / h re
c) h fe × R L / h ie
d) –h fe × R L / h ie
Answer: d
Explanation: The voltage gain of a common emitter amplifier can be expressed in terms of hybrid parameters in the CE short circuit current frequency response model. The voltage gain is equal to -h fe × R L / h ie where, R L is load resistance, h ie is input resistance and -h fe is the current gain.
5. What is the magnitude of voltage generated at emitter circuit of a CE short circuit current frequency response model?
a) h re × h fe / I b × R L
b) h re × h fe × I b × R L
c) h re × h fe / I C × R L
d) h re × h fe × I C × R L
Answer: b
Explanation: The magnitude of voltage generated at emitter circuit of a common emitter amplifier can be expressed in terms of hybrid parameters in the CE short circuit current frequency response model. The magnitude of voltage generated is equal to h re × h fe × I b × R L where, R L is load resistance and I b is the base current.
6. What is the value of output resistance in the CE short circuit current frequency response model?
a) ∞
b) 0
c) h ic
d) h ie
Answer: a
Explanation: The output resistance in a simplified common emitter hybrid model is calculated as the ratio of collector voltage over collector current while V S is equal to zero and R L is omitted. Due to this configuration I C is equal to zero and hence output resistance is infinite.
7. What is the value power gain in the CE short circuit current frequency response model?
a) A V – A I
b) A V + A I
c) A V × A I
d) A V / A I
Answer: c
Explanation: The power gain in a simplified common emitter hybrid model is calculated as the product of voltage gain and power gain. It is expressed as A V × A I . It can also be further simplified as A I 2 × R L / R i , where R L is load resistance and R i is input resistance.
8. What is the value of overall current gain in the CE short circuit current frequency response model while considering R S ?
a) A I × I B / I S
b) A I × I C / I S
c) A I × I S / I C
d) A I × I S / I B
Answer: a
Explanation: The current gain in a simplified common emitter hybrid model while considering R S is calculated as the product of current gain without external parameters and the ratio of base current and external applied current. It is expressed as A I × I B / I S . Where, I S is the external applied current on the amplifier.
9. What is the value of overall voltage gain in the CE short circuit current frequency response model while considering R S ?
a) V B / V S
b) V C / V S
c) V B × V S
d) V C × V S
Answer: a
Explanation: The overall voltage gain in a simplified common emitter hybrid model while considering R S is calculated as ratio of base voltage over the applied external voltage. It is expressed as V B / V S . Where, V S is the external applied voltage on the amplifier.
10. We use CE short circuit current frequency response models to obtain approximate values.
a) True
b) False
Answer: a
Explanation: It is more practical to calculate the approximate values of voltage and current gains instead of computing the exact values of a common emitter amplifier circuit. We use a simplified common emitter hybrid model to obtain these approximate values without reducing the accuracy.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Current Gain with Resistive Load”.
1. What type of amplifier is a current gain with resistive load amplifier?
a) Voltage amplifier
b) Wideband amplifier
c) Feedback amplifier
d) Power amplifier
Answer: c
Explanation: The current gain with resistive load amplifier is one of the most noticeable feedback amplifiers. It provides a negative current feedback to the circuit. These type of amplifiers are usually used in the end stage or the last stage of a series of amplifiers.
2. How is the input impedance of a current gain with resistive load amplifier?
a) Irrelevant
b) Moderate
c) Low
d) High
Answer: d
Explanation: The construction of a current gain with resistive load circuit are approximately similar to a normal amplifier. One of the most important differentiating feature of a current gain with resistive load circuit is its high input impedance.
3. How is the output impedance of a current gain with resistive load amplifier?
a) Irrelevant
b) Moderate
c) Low
d) High
Answer: c
Explanation: The construction of a current gain with resistive load circuit are approximately similar to a normal amplifier. The low output impedance is one of the most prominent differentiating feature of a current gain with resistive load amplifier circuit.
4. How is the current gain of a current gain with resistive load amplifier?
a) Irrelevant
b) Moderate
c) Low
d) High
Answer: d
Explanation: The construction of a current gain with resistive load circuit are approximately similar to a regular feedback amplifier. Nonetheless, one of the most striking differentiating feature of a current gain with resistive load circuit is its high current gain.
5. How is the power gain of a current gain with resistive load amplifier?
a) Irrelevant
b) Moderate
c) Low
d) High
Answer: d
Explanation: One of the most significant difference between a normal amplifier and of a current gain with resistive load circuit is its high power gain. The construction of a current gain with resistive load circuit are approximately similar to a normal amplifier, however the resistive load provides high power gain.
6. What is the application of a current gain with resistive load amplifier?
a) Positive feedback
b) Voltage gain
c) Power gain
d) Impedance matching
Answer: d
Explanation: The high input impedance along with low output impedance makes the current gain with resistive load amplifier ideal for impedance matching. Along with that, its key characteristics include a comparatively high current and power gain.
7. What type of negative feedback does the current gain with resistive load amplifier provide?
a) Voltage, current and power
b) Voltage
c) Current
d) Power
Answer: c
Explanation: The current gain with resistive load configuration is prominently known for its feedback amplifier qualities. It is a negative feedback network that provides a negative feedback of current to the circuit. It is generally used in the last stage of an amplifying network.
8. What is the value of β in a current gain with resistive load amplifier if V f = 12V and V O = 12V?
a) 1
b) 124
c) 1.24
d) 0.124
Answer: a
Explanation: The value of β in a current gain with resistive load amplifier configuration can be calculated using β = V f / V O . Given, V f = 12V and V O = 12V:
β = V f / V O = 12 / 12 = 1.
9. What is the voltage gain of a current gain with resistive load amplifier?
a) 1
b) β
c) ∞
d) 0
Answer: a
Explanation: The voltage gain of a current gain with resistive load amplifier configuration is roughly equal to unity. A V is approximately equal to 1. Since this type of amplifier focuses more towards the current gain, the voltage gain is usually maintained at unity.
10. What is the other name for a current gain with resistive load amplifier configuration?
a) Common collector
b) Common base
c) Common emitter
d) Amplifier circuit
Answer: a
Explanation: The current gain with resistive load circuit configuration is also known as the common collector configuration as it provides a high input impedance and a low output impedance. The current gain with resistive load circuit acts as a buffer stage, and in result is used in several circuits where there is a need to prevent the loading of a circuit.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Transistor Amplifier Response, Taking Source Resistance into Account”.
1. What does the static characteristic curve of a transistor define?
a) Steady state relations
b) Current
c) Voltage
d) It is not applicable for transistors
Answer: a
Explanation: The static transistor curve of a transistor defines the steady state relations between the input and output current and voltages. They are calculated and plotted with the help of DC measurements and linear analysis is performed.
2. How many known variables do we need to perform a linear analysis of a transistor circuit taking source resistance into account?
a) 3
b) 5
c) 6
d) 4
Answer: d
Explanation: There are a total of 6 variables in a transistor circuit taking source resistance into account. A transistor is a three-terminal device with each terminal having current and voltage measurements. We require any four variables to determine the other two variables of a transistor circuit taking source resistance into account.
3. Which of the following is an independent variable in a linear analysis of a common base transistor?
a) Base current
b) Collector base voltage
c) Collector current
d) Collector emitter voltage
Answer: b
Explanation: In a linear analysis of a common base transistor, we have two independent variables. The collector base voltage is an independent variable and the collector current is the dependent variable. We can obtain I C using V CB from the characteristic curve.
4. What frequency should be used to check the linear analysis of a transistor?
a) High frequency
b) Medium-high frequency
c) Medium-low frequency
d) Low frequency
Answer: d
Explanation: Low frequency conditions are used to check the linear analysis of transistor models. This is because, under low frequency d-c conditions the value of emitter current is almost equal to the value of collector current.
5. Which of the following is an independent variable in a linear analysis of a common base transistor?
a) Collector emitter voltage
b) Collector current
c) Emitter base voltage
d) Emitter current
Answer: d
Explanation: In a linear analysis of a common base transistor, we have two independent variables. The emitter current is an independent variable and the emitter base voltage is the dependent variable. We can obtain V EB using I E from the characteristic curve.
6. Where is the operating point of transistor located in a linear analysis?
a) Linear region
b) Saturation region
c) Cut-off region
d) It is not located on this graph
Answer: a
Explanation: The operating point of the quiescent operating point is in the linear region of the output of the characteristic curve. In this region the input variations are proportional to the output variations and are easy to determine.
7. Which of the following is an inference made from the linear analysis of a transistor circuit taking source resistance into account?
a) 99% of the collector current flows to the emitter
b) 99% of the emitter current flows to the collector
c) 0.99% of the collector current flows to the emitter
d) 0.99% of the emitter current flows to the collector
Answer: a
Explanation: While performing the linear analysis of a transistor circuit taking source resistance into account we observe that 99% of the collector current flows to the emitter. Current always flows from the collector to the emitter and not the other way around since the emitter terminal is usually grounded.
8. What is the function of a capacitor in the linear analysis of a transistor?
a) To find the operating point
b) To block d-c signals
c) To block a-c signals
d) Makes no difference
Answer: b
Explanation: Capacitors are used in the linear analysis of a transistor circuit taking source resistance into account to block out and isolate the d-c signals while letting the a-c signals pass through. This is useful to calculate the load line in the analysis of the transistor.
9. What is the equation of the load line in the linear analysis of a transistor?
a) V CE = V CC + I B R L
b) V CE = V CC – I B R L
c) V CE = V CC + I C R L
d) V CE = V CC – I C R L
Answer: d
Explanation: Let the collector voltage supply be V CC in series with a load resistor R L . A straight line superimposed on the collector characteristic will provide us with the load line. The equation for the load line is V CE = V CC – I C R L .
10. What do the small signal parameters in a transistor circuit taking source resistance into account vary with respect to in the linear analysis?
a) Load line
b) Bias point
c) Temperature
d) Makes no difference
Answer: b
Explanation: The small signal parameters in a transistor circuit taking source resistance into account vary with respect to the bias point in the linear analysis model. This occurs even in the linear region of the characteristic curve of the transistor.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Frequency Response – Millers Theorem”.
1. What technique is used in Millers Theorem?
a) Two – port network
b) Hybrid parameters
c) Grounding
d) Short circuiting
Answer: a
Explanation: In Millers Theorem, we use the equivalent two – port network of the given electrical circuit. We divide the circuit into two parts, each part representing a different port for easier analysis of the circuit.
2. Which law is the Millers Theorem based on?
a) Ohm’s Law
b) Moore’s Law
c) Coulomb’s Law
d) Kirchhoff’s Current and Voltage Law
Answer: d
Explanation: The Millers Theorem deals with the impedance supplied by two current / voltage sources connected in parallel. These two versions were derived by Kirchhoff’s two laws: Voltage and Current laws. The dual Millers Theorem is based mainly on the current laws while the other theorem focuses on the voltage law.
3. According to Millers Theorem, what should be the configuration of voltages?
a) Both dependent voltages
b) Both independent voltages
c) One dependent voltage and one independent voltage
d) No specification necessary
Answer: c
Explanation: Miller theorem recommends that an impedance segment is provided by two irregular voltage sources that are associated in arrangement through the common ground. For all intents and purposes, one of them goes about as a free voltage source and the other acts a linearly dependent voltage.
4. What is theimpedance from the input port according to Millers Theorem?
a) Z in1 = Z × K / 1 – K
b) Z in1 = Z × K / 1 + K
c) Z in1 = Z / 1 – K
d) Z in1 = Z / 1 + K
Answer: c
Explanation: According to Millers Theorem, to calculate the input impedance from the input port we use: Z in1 = Z / 1 – K. Where, Z is the original circuit impedance and K is the ratio of the two nodal voltages V 2 / V 1 .
5. What is the impedance from the output port according to Millers Theorem?
a) Z in2 = Z × K / K – 1
b) Z in2 = Z × K / K + 1
c) Z in2 = Z / K – 1
d) Z in2 = Z / K + 1
Answer: a
Explanation: According to Millers Theorem, to calculate the input impedance from the output port we use: Z in2 = Z × K / K – 1. Where, Z is the original circuit impedance and K is the ratio of the two nodal voltages V 2 / V 1 .
6. According to Millers Theorem, what is the impedance in input port if I 1 = 20 mA, I 2 = 30 mA and Z = 2.3 kΩ?
a) 0.38 kΩ
b) 3.83 kΩ
c) 0.57 kΩ
d) 5.75 kΩ
Answer: d
Explanation: To calculate theimpedance in input port, given I 1 = 20 mA, I 2 = 30 mA and Z = 2.3 kΩ:
Z in1 = × Z
α = I 2 / I 1 = 1.5
Z in1 = × Z = × 2.3 = 5.75 kΩ
7. According to Millers Theorem, what is the impedance in output port if I 1 = 20 mA, I 2 = 30 mA and Z = 2.3 kΩ?
a) 0.38 kΩ
b) 3.83 kΩ
c) 0.57 kΩ
d) 5.75 kΩ
Answer: b
Explanation: To calculate theimpedance in output port, given I 1 = 20 mA, I 2 = 30 mA and Z = 2.3 kΩ:
Z in2 = × Z / α
α = I 2 / I 1 = 1.5
Z in2 = × Z / α = × 2.3 / 1.5 = 3.83 kΩ
8. What are the applications of Millers Theorem?
a) Lower power consumption
b) Additional voltage source
c) Additional current source
d) Circuit optimization
Answer: b
Explanation: Millers Theorem is applied in an amplifier setting known as Millers amplifier. The amplifier can be utilized as an extra voltage source which changes over the real impedance into a virtual impedance. The virtual impedance can be considered as a part associated in parallel to the amplifier input.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “High-Frequency Response of the Emitter Followers”.
1. What type of amplifier is a high frequency response of emitter follower amplifier?
a) Voltage amplifier
b) Wideband amplifier
c) Feedback amplifier
d) Power amplifier
Answer: c
Explanation: The high frequency response of emitter follower amplifier is one of the most noticeable feedback amplifiers. It provides a negative current feedback to the circuit. These type of amplifiers are usually used in the end stage or the last stage of a series of amplifiers.
2. How is the input impedance of a high frequency response of emitter follower amplifier?
a) Irrelevant
b) Moderate
c) Low
d) High
Answer: d
Explanation: The construction of a high frequency response of emitter follower circuit is approximately similar to a normal amplifier. One of the most important differentiating feature of a high frequency response of emitter follower circuit is its high input impedance.
3. How is the output impedance of a high frequency response of emitter follower amplifier?
a) Irrelevant
b) Moderate
c) Low
d) High
Answer: c
Explanation: The structure of a high frequency response of emitter follower circuit is roughly comparable to a normal amplifier. The low output impedance is one of the most prominent differentiating feature of a high frequency response of emitter follower circuit.
4. How is the current gain of a high frequency response of emitter follower amplifier?
a) Irrelevant
b) Moderate
c) Low
d) High
Answer: d
Explanation: The construction of a high frequency response of emitter follower amplifier circuit is almost similar to an ordered feedback amplifier. Nonetheless, one of the most striking differentiating feature of a current gain with resistive load circuit is its high current gain.
5. How is the power gain of a high frequency response of emitter follower amplifier?
a) Irrelevant
b) Moderate
c) Low
d) High
Answer: d
Explanation: One of the most significant difference between a normal amplifier and of a high frequency response of emitter follower amplifier circuit is its high power gain. The construction of a high frequency response of emitter follower amplifier circuit is approximately similar to a normal amplifier, however the resistive load provides high power gain.
6. What is the application of a high frequency response of emitter follower amplifier?
a) Positive feedback
b) Voltage gain
c) Power gain
d) Impedance matching
Answer: d
Explanation: The high input impedance along with low output impedance makes the high frequency response of emitter follower amplifier ideal for impedance matching. Along with that, its key characteristics include a comparatively high current and power gain.
7. What type of negative feedback does the high frequency response of emitter follower amplifier provide?
a) Voltage, current and power
b) Voltage
c) Current
d) Power
Answer: c
Explanation: The high frequency response of emitter follower configuration is prominently known for its feedback amplifier qualities. It is a negative feedback network that provides a negative feedback of current to the circuit. It is generally used in the last stage of an amplifying network.
8. What is the value of β in a high frequency response of emitter follower amplifier if V f = 12V and V O = 12V?
a) 1
b) 124
c) 1.24
d) 0.124
Answer: a
Explanation: The value of β in a high frequency response of emitter follower amplifier configuration can be calculated using β = V f / V O . Given, V f = 12V and V O = 12V:
β = V f / V O = 12 / 12 = 1.
9. What is the voltage gain of a high frequency response of emitter follower amplifier?
a) 1
b) β
c) ∞
d) 0
Answer: a
Explanation: The voltage gain of a high frequency response of emitter follower amplifier configuration is roughly equal to unity. AV is approximately equal to 1. This is due to the relatively high input impedance and low output impedance.
10. What is the other name for a high frequency response of emitter follower amplifier configuration?
a) Common collector
b) Common base
c) Common emitter
d) Amplifier circuit
Answer: a
Explanation: The high frequency response of emitter follower circuit configuration is also known as the common collector configuration as it provides a high input impedance and a low output impedance. The high frequency response of emitter follower circuit acts as a buffer stage, and in result is used in several circuits where there is a need to prevent the loading of a circuit.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “High Frequency Response of the Differential Amplifiers”.
1. What is the voltage gain of a high frequency response of differential amplifier?
a) 1
b) 0
c) ∞
d) Cannot be determined
Answer: a
Explanation: If all the resistor values are of the equal value, then the circuit will have a voltage gain equal to exactly one or unity. Hence it is also called as a Unity Gain High frequency response of differential amplifier. The output voltage expression would be the difference between the two voltages.
2. What is the problem with a single operational high frequency response of differential amplifier?
a) High input resistance
b) Low input resistance
c) Low output resistance
d) High output resistance
Answer: b
Explanation: The problem with a single operational high frequency response of differential amplifier is its low input resistance. This low input resistance results in a loss of a major component of the signal. Operational Amplifier is internally a High frequency response of differential amplifier with features like High Input Impedance, Low Output Impedance.
3. What is the total voltage gain of a high frequency response of differential amplifier if the gain of first stage is 4.8 and the gain of the second stage is 1.2?
a) 6.75
b) 6.55
c) 5.76
d) 5.65
Answer: c
Explanation: The total gain of a high frequency response of differential amplifier can be calculated by the formula given below. Where, A V = overall gain, A V1 = voltage gain of first stage and A V2 = voltage gain of second stage.
Given A V1 = 4.8, A V2 = 1.2
A V = A V1 × A V2 = 4.8 × 1.2 = 5.76.
4. What is the total voltage gain of a high frequency response of differential amplifier if input of first stage is 50V and the output of the second stage is 100V?
a) 20
b) 0.2
c) 2.5
d) 2
Answer: d
Explanation: The total gain of a high frequency response of differential amplifier is the ratio of the output of second stage amplifier to the input of first stage amplifier. Where, V output = 100V and V input = 50V.
A V = V output / V input = 100V / 50V = 2.
5. How is the voltage gain measured in a high frequency response of differential amplifier?
a) Volts
b) Ampere
c) Decibels
d) Dimensionless
Answer: c
Explanation: The voltage gain is measured in terms of decibels. The total voltage gain is calculated as the product of individual stages or as a sum of all the stages if the gain calculated at each intermediate stage was measured in decibels as well.
6. How do we calculate the output voltage of a high frequency response of differential amplifier?
a) V O = -A d × V d + A C × V C
b) V O = -A d × V d – A C × V C
c) V O = A d × V d + A C × V C
d) V O = A d × V d – A C × V C
Answer: c
Explanation: The differential gain of a high frequency response of differential amplifier is defined as the gain obtained at the output signal with respect to the difference in the input signals applied. To calculate the output voltage of a high frequency response of differential amplifier we use A d × V d + A C × V C . Where, A d is the differential gain and A C and V C represent the common mode gain.
7. What is the purpose of using a high frequency response of differential amplifier?
a) Increase voltage gain
b) Decrease voltage gain
c) Increase current gain
d) Decrease current gain
Answer: a
Explanation: A circuit having a single transistor configuration does not provide suitable bandwidth or gain. The purpose of a high frequency response of differential amplifier is to provide an increase in the voltage gain. The total gain of a high frequency response of differential amplifier is the product of the voltage gains of the discrete stages.
8. How many methods of coupling are used for a high frequency response of differential amplifier?
a) 3
b) 4
c) 2
d) 5
Answer: b
Explanation: There are four types of coupling mechanisms used. They are Resistance-capacitance coupling, thermal coupling, impedance coupling and transformer coupling. Resistance-capacitance coupling is the most vastly used mechanism.
9. What is the total voltage gain of a high frequency response of differential amplifier if the gain of first stage is 10dB, gain of the second stage is 20dB and gain of third stage is 30dB?
a) 60.32dB
b) 50dB
c) 60dB
d) 50.32dB
Answer: c
Explanation: The total gain of a high frequency response of differential amplifier is the sum of the multiple stages when the gain is calculated is decibels. Given, A V1 = 10dB, A V2 = 20dB and A V3 = 30dB
A V = A V1 + A V2 + A V3 = 10dB + 20dB + 30dB = 60dB.
10. The coupling device is essential for a high frequency response of differential amplifier.
a) True
b) False
Answer: a
Explanation: One of the main functions of the coupling device in a high frequency response of differential amplifier is to block the direct current from passing through to the input of next stage from the output of the first stage. Another function of the coupling device is to transfer the current from the output of first stage to the input of the second stage.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “High Frequency Response of Multistage Amplifiers”.
1. What is the advantage of using a high frequency response of multistage amplifier?
a) High gain and high bandwidth
b) High gain and low bandwidth
c) Low gain and high bandwidth
d) Low gain and low bandwidth
Answer: a
Explanation: A circuit having a single transistor amplifier does not provide suitable bandwidth or gain. To overcome this difficulty, we combine several amplification stages. The high frequency response of multistage amplifiers theory is used for high gain as well as high bandwidth.
2. What is the advantage of using CB amplifier configuration in a high frequency response of multistage amplifier?
a) High efficiency
b) Low distortion
c) Good high frequency operation
d) Good low frequency operation
Answer: c
Explanation: The high frequency response of multistage amplifiers circuit can be constructed with two configurations of a transistor that is CE and CB . The CB configuration delivers a good high – frequency operation.
3. What is the total voltage gain of a high frequency response of multistage amplifier if the gain of first stage is 4 and the gain of the second stage is 10?
a) 0.4
b) 4
c) 40
d) 2.5
Answer: c
Explanation: The total gain of a high frequency response of multistage amplifier can be calculated by the formula given below. Where, A V = overall gain, A V1 = voltage gain of first stage and A V2 = voltage gain of second stage.
Given, A V1 = 4 & A V2 = 10
A V = A V1 × A V2 = 4 × 10 = 40.
4. What is the total voltage gain of a high frequency response of multistage amplifier if input of first stage is 10V and the output of the second stage is 25V?
a) 0.4
b) 4
c) 40
d) 2.5
Answer: d
Explanation: The total gain of a high frequency response of multistage amplifier is the ration of the output of second stage amplifier to the input of first stage amplifier. Where, V output = 25V and V input = 10V.
A V = V output / V input = 25V / 10V = 2.5.
5. What does cascading of two transistor amplifiers imply?
a) Output of first stage sent to input of second stage
b) Output of first stage sent to coupling device
c) Input of first stage sent to input of second stage
d) Not related
Answer: b
Explanation: In multi – stage amplifiers, the output of first stage is coupled to the input of second stage using a coupling device. The coupling devices is typically a capacitor or a transformer. Cascading is known as the process of combining two amplifier stages using a coupling device.
6. How many methods of coupling are used for a high frequency response of multistage amplifier?
a) 3
b) 4
c) 2
d) 5
Answer: b
Explanation: There are four types of coupling mechanisms used. They are Resistance – capacitance coupling, thermal coupling, impedance coupling and transformer coupling. Resistance – capacitance coupling is the most vastly used mechanism.
7. What is the total voltage gain of a high frequency response of multistage amplifier if the gain of first stage is 14dB, gain of the second stage is 12dB and gain of third stage is 24dB?
a) 40dB
b) 50.32dB
c) 50dB
d) 40.32dB
Answer: c
Explanation: The total gain of a high frequency response of multistage amplifier is the sum of the multiple stages when the gain is calculated is decibels. Given, A V1 = 14dB, A V2 = 12dB and A V3 = 24dB
A V = A V1 + A V2 + A V3 = 14dB + 12dB + 24dB = 50dB.
8. The coupling device is essential for a high frequency response of multistage amplifier.
a) True
b) False
Answer: a
Explanation: One of the main functions of the coupling device in a high frequency response of multistage amplifier is to block the direct current from passing through to the input of next stage from the output of the first stage. Another function of the coupling device is to transfer the current from the output of first stage to the input of the second stage.
9. Which two terminals of the transistors are connected in high frequency response of multistage amplifier?
a) Collector
b) Base
c) Emitter
d) Collector, base and emitter
Answer: c
Explanation: The high frequency response of multistage amplifier circuit can be constructed by connecting the emitters of two consecutive transistors. A resistor is placed between each stage to act as a coupling device.
10. What are multistage amplifiers also known as?
a) Common collector amplifier
b) N – stage amplifier
c) Common base amplifier
d) Common emitter amplifier
Answer: b
Explanation: A single stage of amplifier provides an insufficient current gain or voltage gain. Several amplifier stages connected in cascade are used instead. Hence it is known as an n – stage cascading amplifier or multistage amplifier.
11. How many ports are there in a high frequency response of multistage amplifier?
a) 3
b) 4
c) 2
d) 5
Answer: c
Explanation: A high frequency response of multistage amplifier is a two – port network assembled from a series of amplifiers. Each amplifier directs its output to the input of the subsequent amplifier. The performance obtained from a single – stage amplifier is inadequate, therefore numerous stages are combined to form a multistage amplifier.
12. How is the voltage gain measured in a high frequency response of multistage amplifier?
a) Volts
b) Ampere
c) Decibels
d) Dimensionless
Answer: c
Explanation: The voltage gain is measured in terms of decibels. The total voltage gain is calculated as the product of individual stages or as a sum of all the stages if the gain calculated at each intermediate stage was measured in decibels as well.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Frequency Compensation”.
1. If the ratio R B / R E tends to infinity, what should be the value of stability factor in voltage divider circuit?
a) Unity
b) Zero
c) 1 – β dc
d) 1 + β dc
Answer: d
Explanation: Stability factor is fundamentally dependent on the ratio of base resistance to emitter resistance(R B / R E ). Hence, if this ratio is estimated to be small, then the stability factor in such case is equal to unity.
On the contrary, if this ratio R B / R E tends to infinite value, then the value of stability factor in voltage divider circuit becomes nearly equivalent to 1+ β dc .
Therefore, if the ratio is constant then it leads to an increase the stability factor by increasing the value of current gain (β dc ). Besides this, smaller values of base resistance also yield improved stabilization.
2. What are the consequences of diode compensation for the change in base-to-emitter voltage (V BE ) due to temperature?
a) Temperature compensation takes place by variation in forward voltage (V F )
b) Collector current become sensitive to the change in base-to-emitter voltage (V BE )
c) Depends
d) Nothing happens
Answer: a
Explanation: In some of the negative feedback circuits, the amplification level of AC signals gets ablated suddenly. Since it becomes difficult for the circuits to tolerate the loss of signals, it is essential to reduce the drift in an operating point by means of compensation and stabilization.
Compensation techniques include diode compensation, bias compensation using thermistor, sensistor & so on. They play a major role in providing maximum bias in addition to the thermal stabilization to the circuits.
3. Generally, the resistance of thermistor decreases _______
a) Linearly with an increase in temperature
b) Linearly with the decrease in temperature
c) Exponentially with an increase in temperature
d) Exponentially with the decrease in temperature
Answer: c
Explanation: Thermistor is a device or resistor whose resistance is totally dependent on the temperature. Thermistor is widely applicable for several compensation techniques.
It exhibits variation in its resistance with respect to change in temperature. As the resistance of thermistor decreases with an increase in temperature, this property of thermistor is also regarded as negative temperature coefficient of resistivity.
4. Fidelity is nothing but an ability of amplifier to reproduce ________
a) Input signal without any distortion
b) Output signal without any distortion
c) Phase shift signal
d) Amplitude shift signal
Answer: a
Explanation: Fidelity is the property of amplifier which is usually taken into consideration during the analysis of an amplifier’s frequency response. The ability or a potential of an amplifier in order to reproduce the distortion-less nature of input signal is known as ‘Fidelity’.
Fidelity of an amplifier is dependent on the frequency response and the bandwidth of an amplifier to the greater extent. Fidelity gives an implication of satisfying the condition of distortion less output.
This condition for distortion-less output implies that gain of an amplifier should be independent of frequency while the phase shift introduced by an amplifier should reveal proportionality with respect to frequency or zero.
5. Which among the following is an output provided by transresistance amplifier?
a) Output current proportional to signal voltage
b) Output voltage proportional to signal current
c) Output voltage proportional to input voltage
d) Output current proportional to signal current
Answer: b
Explanation: An amplifier which produces the output voltage in proportion to the signal current where the proportionality factor is independent of source and load resistances, is known as ‘Transresistance amplifier’. For a practical transresistance amplifier, input resistance must be very less than source resistance and the output resistance must be very less than load resistance.
Generally, it is the ratio of voltage to signal current. Conversely, the transconductance amplifier exhibits output current proportional to signal voltage. Voltage amplifier exhibits the output voltage proportional to input voltage & current amplifier exhibits the output current proportional to the signal current.
6. Which among the below specified conditions is applicable to prevent the occurrence of thermal runaway in voltage divider bias circuit?
a) V CE < V CC / 2
b) V CE = V CC / 2
c) V CE > V CC / 2
d) None of the mentioned
Answer: a
Explanation: Thermal runaway is a growing process which can ultimately harm the transistor due to unnecessary internal heating or increase in ambient temperature. It occurs due to increase in collector current ahead of the maximum specified value. Additionally, it also occurs due to excessive internal power dissipation above the maximum allowed value. Thermal runaway can be disallowed by maintaining the provision of thermal stability in addition to the treatment of heat sink.
According to thermal stability condition, V CE should be less than V cc /2 results in the Q-point operation at a safe level. On the contrary, if the location of Q-point is at V CE > V cc /2, then the transistor is more likely to get damaged due to thermal runaway.
7. Which of the following process plays a crucial role in devising the independency of operating point over the variations in temperature or transistor parameters?
a) Bias stabilization
b) Bias compensation
c) Bias stabilization & compensation
d) No process
Answer: a
Explanation: Bias stabilization is a process which makes the operating point independent of change in temperature or any change in transistor parameters.
As Q-point exhibits its independent nature over the several variations, it plays a major role in providing the stability to greater extent especially for self-bias, fixed-bias and collector to base bias circuits. Since the stability factor of circuit gets unnatural due to several variations, bias stabilization has a provision of maintaining the Q-point condition irrespective of the changes in temperature or transistor parameters and in that way ensuring the confined level of bias stability to circuits.
8. What should be the level of input resistance to allow the occurrence of source loading in common base amplifier configuration?
a) Low
b) High
c) Moderate
d) Stable
Answer: a
Explanation: As per the configuration of CB amplifier, it is obvious that its input resistance is very low whereas its output resistance is extremely high. However, the lower value of input resistance allows the provision of source loading in common base amplifier circuit. Thus, there is no current amplification due to unity current gain. These all reasons ultimately contribute to high level of voltage gain.
9. The value of dBm in power measurement is predictable by assuming the reference, which is equal to ________
a) 1mW
b) 10mW
c) 1/10 mW
d) 1/100 mW
Answer: a
Explanation: Basically, the Decibel is a logarithmic unit which is used to state the gain of audio frequency amplifiers. This is so for the reason that the response of a human ear to the sound intensity exhibits logarithmic nature.
Apart from this, dBm is a unit which is often used in power measurement. dBm refers to the dB value which is estimated by assumption of a reference whose value is equal to 1mW.
10. The cut-off frequency (f β ) is basically the frequency at which the short circuit __________
a) CB gain of transistor drops by 3 dB from its value at low frequency
b) CE gain of transistor drops by 3 dB from its value at low frequency
c) CC gain of transistor drops by 3 dB from its value at low frequency
d) CC gain of transistor drops by 3 dB from its value at high frequency
Answer: b
Explanation: Cut-off frequency (f β ) is basically the frequency at which short circuit CE gain of transistor drops by 3dB from its value at low frequency.
Thus, the frequency range upto the cut-off frequency(f β ) is regarded as the bandwidth of circuit. Also, the current gain reduces to 70.7% of the low frequency current gain.
Likewise, the frequency at which short circuit CB gain of transistor drops by 3dB from its value next to low frequency is known as cut-off frequency (f α ).
The relationship between these cut-off frequencies indicates that the 3dB frequency of CB configuration is (1 + h fe ) times greater or higher as compared to that of CE configuration.
11. Which among the following will possess a higher bandwidth, if two transistors are provided with unity gain frequency?
a) Transistor with lower h fe
b) Transistor with higher h fe
c) Transistor with lower h re
d) Transistor with higher h re
Answer: a
Explanation: In general, the short circuit CE gain of a transistor gets reduced to unity at f = f T . This implies that the transistor loses its capability of current amplification at f = f T .
As the gain-bandwidth product always remains constant, the transistor with lower h fe will definitely possesses higher bandwidth if the two transistors are made available with equal unity gain frequency.
12. What is an angle of phase shift for each designed RC network in the Phase Shift Oscillator circuit?
a) 60 o
b) 90 o
c) 180 o
d) 45 o
Answer: a
Explanation: R-C phase shift oscillator consists of three identical basic R-C phase shifting kind of networks approved in the cascade configuration.
Thus, the total phase shift introduced by three stage R-C network is equal to 180 o , which implies that the output of network leads its input by 180 o . As a result, it infers that each phase-shift stage acquires the phase shift of about 60 o .
RC network is also renowned as ‘Ladder network’. The phase-shift of about 180 o is generated particularly at one particular frequency, which is measured to be the frequency of operation of the oscillator.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Miller Compensation and Pole Splitting”.
1. What technique is used in Millers compensation?
a) Two – port network
b) Hybrid parameters
c) Grounding
d) Short circuiting
Answer: a
Explanation: In Millers compensation, we use the equivalent two – port network of the given electrical circuit. We divide the circuit into two parts, each part representing a different port for easier analysis of the circuit.
2. Which law is the Miller compensation and pole splitting based on?
a) Ohm’s Law
b) Moore’s Law
c) Coulomb’s Law
d) Kirchhoff’s Current and Voltage Law
Answer: d
Explanation: The Miller compensation and pole splitting deals with the impedance supplied by two current / voltage sources connected in parallel. These two versions were derived by Kirchhoff’s two laws: Voltage and Current laws. The dual Miller compensation and pole splitting is based mainly on the current laws while the other theorem focuses on the voltage law.
3. According to Millers compensation, what should be the configuration of voltages?
a) Both dependent voltages
b) Both independent voltages
c) One dependent voltage and one independent voltage
d) No specification necessary
Answer: c
Explanation: Miller compensation suggests that an impedance component is supplied by two random voltage sources that are connected in series through the common ground. Practically, one of them acts as an independent voltage source and the other acts a linearly dependent voltage.
4. What is the impedance from the input port according to Millers compensation and pole splitting?
a) Z in1 = Z × K / 1 – K
b) Z in1 = Z × K / 1 + K
c) Z in1 = Z / 1 – K
d) Z in1 = Z / 1 + K
Answer: c
Explanation: According to Millers compensation, to calculate the input impedance from the input port we use: Z in1 = Z / 1 – K. Where, Z is the original circuit impedance and K is the ratio of the two nodal voltages V 2 / V 1 .
5. What is the impedance from the output port according to Millers compensation and pole splitting?
a) Z in2 = Z × K / K – 1
b) Z in2 = Z × K / K + 1
c) Z in2 = Z / K – 1
d) Z in2 = Z / K + 1
Answer: a
Explanation: According to Millers compensation and pole splitting, to calculate the input impedance from the output port we use: Z in2 = Z × K / K – 1. Where, Z is the original circuit impedance and K is the ratio of the two nodal voltages V 2 / V 1 .
6. According to Millers compensation and pole splitting, what is the impedance in input port if I 1 = 20 mA, I 2 = 30 mA and Z = 2.3 kΩ?
a) 0.38 kΩ
b) 3.83 kΩ
c) 0.57 kΩ
d) 5.75 kΩ
Answer: d
Explanation: To calculate the impedance in input port, given I 1 = 20 mA, I 2 = 30 mA and Z = 2.3 kΩ:
Z in1 = × Z
α = I 2 / I 1 = 1.5
Z in1 = × Z = × 2.3 = 5.75 kΩ.
7. According to Millers compensation and pole splitting, what is the impedance in output port if I 1 = 20 mA, I 2 = 30 mA and Z = 2.3 kΩ?
a) 0.38 kΩ
b) 3.83 kΩ
c) 0.57 kΩ
d) 5.75 kΩ
Answer: b
Explanation: To calculate the impedance in output port, given I 1 = 20 mA, I 2 = 30 mA and Z = 2.3 kΩ:
Z in2 = × Z / α
α = I 2 / I 1 = 1.5
Z in2 = × Z / α = × 2.3 / 1.5 = 3.83 kΩ.
8. What are the applications of Millers compensation and pole splitting?
a) Lower power consumption
b) Additional voltage source
c) Additional current source
d) Circuit optimization
Answer: b
Explanation: Miller compensation and pole splitting is applied in an amplifier setting known as Millers amplifier. The amplifier can be used as an additional voltage source which converts the actual impedance into a virtual impedance. The virtual impedance can be thought of as a component connected in parallel to the amplifier input.
9. Millers compensation and pole splitting is used in the high – frequency analysis of BJTs and FETs.
a) True
b) False
Answer: a
Explanation: Miller compensation is a method used for steadying operational amplifiers through the usage of a capacitance namely called C f which is to be connected in a negative – feedback approach across one of the second stage internal gain.
10. How do we calculate Millers capacitance when a V = 250 and C = 9.90pF?
a) 2.485pF
b) 2.485nF
c) 24.85pF
d) 24.85nF
Answer: b
Explanation: To calculate the Millers capacitance, we use C M = (1 + a V ) × C. Where, C is the feedback capacitance and a V is the inverting voltage amplifier gain.
C M = (1 + a V ) × C = × 9.90 = 2.485nF.
This set of Electronic Devices and Circuits Question Paper focuses on “Input Bias and Offset Currents of the Bipolar Differential Amplifier”.
1. Which transistor bias circuit provides good Q-point stability with a single-polarity supply voltage?
a) Base bias
b) Collector feedback bias
c) Voltage divider bias
d) Emitter bias
Answer: c
Explanation: When the transistor starts operating, temperature at the junction increases. Hence IC increases. As a result of which, Ie increases. Due to this increase in Ie, voltage drop across Re increases. This reduces the forward voltage across the emitter. Thus Ib reduces.
We know that, for any value of I CO , I C = β * I b + I CO
where I CO =reverse saturation current
β = gain
Therefore, β and I CO increases and at the same time there is decrease in I b . Hence above equation confirms that I C can be maintained within limits. Thus the circuit is more thermally stable and the operating point is more stable.
2. Ideally, for non linear operation, a transistor should be biased so that the Q-point is ________
a) near saturation
b) near cut off
c) where I c is maximum
d) halfway between cut off and saturation
Answer: d
Explanation: If Q-point is near to saturation then positive clipping of input signal, and to cutoff then negative clipping of input signal, if IC is maximum then Q-point is in saturation region. Linear operation means output varies according to input without any distortion.
3. The most stable biasing technique used is the ____________
a) voltage-divider bias.
b) base bias
c) emitter bias
d) collector bias
Answer: a
Explanation: Voltage divider biasing is commonly used because of the main reason that the transistor under this biasing always remains in the active region.
In voltage divider biasing, the voltages at the transistor’s base, emitter and collector all depend upon the external circuit i.e. the biasing resistors R1 and R2 whose valued are fixed thus variation with beta is not present here.
However, voltage divider biasing has the advantage that its stability factor is greater than that of collector feedback biasing, that’s why it is used.
4. What is the Q-point for a fixed-bias transistor with I B = 75 µ A , β DC = 100, V CC = 20 V, and R C = 1.5 K Ohm?
a) V C = 0 V
b) V C =12.25V
c) V C = 8.75 V
d) V C = 20V
Answer: c
Explanation: V CE =V cc -I C R c (V CE = V C , β = I C /I B => I C = β *I B )
V c = V cc – β * I B * R c
=20-100x75x10 -6 x1.5x 10 -3
=8.75V.
5. Emitter bias requires _______
a) only a positive supply voltage
b) only a negative supply voltage
c) no supply voltage
d) both positive and negative supply voltage
Answer: d
Explanation: If a dual power supply is there, then it is the most useful, and provides zero bias voltage at the emitter or collector for load. The negative supply is used to forward-bias the emitter junction through emitter resistor. The positive supply is used to reverse-bias the collector junction. Also has good stability.
6. Which transistor bias circuit arrangement has poor stability because its Q-point varies widely with β DC ?
a) base bias
b) voltage-divider bias
c) emitter bias
d) collector bias
Answer: a
Explanation: In base bias IB, RB and IC are fixed therefore it is also called fixed bias transistor. When the temperature is vary then the Q point is also varied. Therefore base bias has no stabilization.
7. What is the most common bias circuit?
a) Base
b) Collector
c) Emitter
d) Voltage-divider
Answer: d
Explanation: Due to the best stabilization, voltage divider circuit is commonly used. Under this biasing technique, the transistor always remains in the active region. In voltage divider biasing, the voltages at the transistor’s base, emitter and collector all depend upon the external circuit i.e. the biasing resistors R1 and R2 whose valued are fixed thus variation with beta is not present here.
8. Voltage-divider bias has a relatively stable Q-point, as does_________
a) base bias
b) collector-feedback bias
c) emitter bias
d) both base and emitter bias
Answer: b
Explanation: The collector feedback biasing is also beta-independent. It has the same circuit as the voltage divider biasing except there is only one resistor used other than load resistor Rc. However, voltage divider biasing has the advantage that its stability factor is greater than that of collector feedback biasing.
9. The linear operating region of a transistor lies along the load line below ________ and above ___________
a) cut off, saturation
b) saturation, cut off
c) active, saturation
d) cut off, active
Answer: b
Explanation: Q-point is generally taken to be the intersection point of load line with the output characteristics of the transistor. That’s why; the linear operating region of a transistor must lie along the load line below saturation and above the cut off.
10. The input resistance of the base of a voltage-divider biased transistor can be neglected _________
a) at all times
b) only if the base current is much smaller than the current through R2
c) at no time
d) only if the base current is much larger than the current through R2 (the lower bias resistor
Answer: b
Explanation: The input resistance can be neglected if the values of R1 and R2 are very large. Therefore the base current becomes very small.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Differential Amplifier with Active Load”.
1. What is the voltage gain of a differential amplifier with active load?
a) 1
b) 0
c) ∞
d) Cannot be determined
Answer: a
Explanation: If all the resistor values are of the equal value, then the circuit will have a voltage gain equal to exactly one or unity. Hence it is also called as a Unity Gain Differential amplifier with active load. The output voltage expression would be the difference between the two voltages.
2. What is the problem with a single operational differential amplifier with active load?
a) High input resistance
b) Low input resistance
c) Low output resistance
d) High output resistance
Answer: b
Explanation: The problem with a single operational differential amplifier with active load is its low input resistance. This low input resistance results in a loss of a major component of the signal. Operational Amplifier is internally a Differential amplifier with active load with features like High Input Impedance, Low Output Impedance.
3. What is the total voltage gain of a differential amplifier with active load if the gain of first stage is 4.8 and the gain of the second stage is 1.2?
a) 6.75
b) 6.55
c) 5.76
d) 5.65
Answer: c
Explanation: The total gain of a differential amplifier with active load can be calculated by the formula given below. Where, A V = overall gain, A V1 = voltage gain of first stage and A V2 = voltage gain of second stage.
A V = A V1 × A V2 = 4.8 × 1.2 = 5.76
4. What is the total voltage gain of a differential amplifier with active load if input of first stage is 50V and the output of the second stage is 100V?
a) 20
b) 0.2
c) 2.5
d) 2
Answer: d
Explanation: The total gain of a differential amplifier with active load is the ratio of the output of second stage amplifier to the input of first stage amplifier. Where, V output = 100V and V input = 50V.
A V = V output / V input = 100V / 50V = 2
5. How is the voltage gain measured in a differential amplifier with active load?
a) Volts
b) Ampere
c) Decibels
d) Dimensionless
Answer: c
Explanation: The voltage gain is measured in terms of decibels. The total voltage gain is calculated as the product of individual stages or as a sum of all the stages if the gain calculated at each intermediate stage was measured in decibels as well.
6. How do we calculate the output voltage of a differential amplifier with active load?
a) V O = -A d × V d + A C × V C
b) V O = -A d × V d – A C × V C
c) V O = A d × V d + A C × V C
d) V O = A d × V d – A C × V C
Answer: c
Explanation: The differential gain of a differential amplifier with active load is defined as the gain obtained at the output signal with respect to the difference in the input signals applied. To calculate the output voltage of a differential amplifier with active load we use A d × V d + A C × V C . Where, A d is the differential gain and A C and V C represent the common mode gain.
7. What is the purpose of using a differential amplifier with active load?
a) Increase voltage gain
b) Decrease voltage gain
c) Increase current gain
d) Decrease current gain
Answer: a
Explanation: A circuit having a single transistor configuration does not provide suitable bandwidth or gain. The purpose of a differential amplifier with active load is to provide an increase in the voltage gain. The total gain of a differential amplifier with active load is the product of the voltage gains of the discrete stages.
8. How many methods of coupling are used for a differential amplifier with active load?
a) 3
b) 4
c) 2
d) 5
Answer: b
Explanation: There are four types of coupling mechanisms used. They are Resistance – capacitance coupling, thermal coupling, impedance coupling and transformer coupling. Resistance – capacitance coupling is the most vastly used mechanism.
9. What is the total voltage gain of a differential amplifier with active load if the gain of first stage is 10dB, gain of the second stage is 20dB and gain of third stage is 30dB?
a) 60.32dB
b) 50dB
c) 60dB
d) 50.32dB
Answer: c
Explanation: The total gain of a differential amplifier with active load is the sum of the multiple stages when the gain is calculated is decibels. Given, A V1 = 10dB, A V2 = 20dB and A V3 = 30dB
A V = A V1 + A V2 + A V3 = 10dB + 20dB + 30dB = 60dB.
10. The coupling device is essential for a differential amplifier with active load.
a) True
b) False
Answer: a
Explanation: One of the main functions of the coupling device in a differential amplifier with active load is to block the direct current from passing through to the input of next stage from the output of the first stage. Another function of the coupling device is to transfer the current from the output of first stage to the input of the second stage.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Multistage Amplifier”.
1. The frequency response of transformer coupling is ________
a) Good
b) Very Good
c) Excellent
d) Poor
Answer: d
Explanation: The transformer coupling has a poor frequency response. The gain varies considerably with frequency. The gain is constant only over small range of frequencies. Thus transformer coupling introduces frequency distortion; due to which its frequency response is poor.
2. What is the purpose of RC or transformer coupling?
a) To block a.c.
b) To separate bias of one stage from another
c) Increase thermal stability
d) Increase Efficiency
Answer: b
Explanation: In RC or transformer coupling, a capacitor / transformer is used as coupling device which connects output of first stage with input of second stage. Its function is to pass the a.c signal and blocks d.c. bias voltage.
3. Why is RC coupling confined to low power applications?
a) Due to large value of coupling capacitor
b) Low efficiency
c) Large number of components
d) Due to is frequency response
Answer: b
Explanation: RC coupled amplifiers have low voltage and power gain. It is because the low resistance presented by the input of each stage to the preceding stage decreases the effective load resistance and hence the gain. Thus its efficiency is reduced.
4. A radio receiver has how many stages of amplification?
a) One
b) Two
c) Three
d) More than one
Answer: d
Explanation: A multistage amplifier circuit affects the high input impedance of a common source stage combined with the input to output isolation of a common gate stage. A radio receiver has more than one stage of amplification because it is required to restore the characteristics of a radio signal over various channels.
5. Which of the following is an advantage of RC coupling scheme?
a) Good impedance matching
b) Economy
c) High efficiency
d) Frequency response
Answer: b
Explanation: It uses the resistor and capacitor which are not expensive so the cost is low. But it has poor impedance matching because its output impedance is several times larger than the device; at its terminal end. It is unsuitable for low frequency application.
6. The voltage gain is practically expressed in _______
a) db
b) volts
c) as a number
d) ampere
Answer: a
Explanation: Db scale is logarithmic. Voltage gain increases exponentially with frequency so using a linear scale means that we need to work with large values of gain, corresponding to small values of frequency.
7. If a three stage amplifier has individual stage gains of 10db, 6db and 15db; then the total gain in db is ______
a) 600db
b) 24db
c) 14db
d) 31db
Answer: d
Explanation: The overall gain of a multistage amplifier is given as the product of the gain of the individual stages.
Gain = A 1 * A 2 * A 3 ……* A n
Alternately, if the gain of each stage is given in db
The overall gain of the amplifier is the sum of gain of each stage
Gain in db = A 1 + A 2 + A 3
= 10+6+15 = 31db.
8. For extremely low frequencies, RC coupling is not used because of ___________
a) There is considerable power loss
b) There is a hum in the output
c) Electrical size of the coupling capacitor
d) Low efficiency
Answer: c
Explanation: X c = the electrical size of coupling capacitor
Relation of Xc with the signal frequency :
X c = 1 \X c is inversely proportional to f.
At low frequencies, X c becomes very large; output reactance of capacitor increases.
The voltage across load resistance also reduces because some voltage drop takes place across X c .
Thus output voltage reduces. Therefore gain is very low.
9. Which transformer is used for impedance matching in transistor coupled amplifier?
a) step-up
b) step-down
c) same turn ratio
d) different turn ratio
Answer: b
Explanation: Usually the impedance of an output device is a few ohms whereas output impedance of the transistor is several 100 ohms. In order to match the impedance a step down transformer of proper turn’s ratio is used. The impedance of secondary of the transformer is made equal to the load impedance and primary impedance equal to the output impedance of the transistor.
10. Gain of an amplifier usually expressed in db because _______________
a) It is a small unit
b) Calculations become easy
c) Human ear response is logarithmic
d) Gain is reduced
Answer: c
Explanation: The human hearing scale is logarithmic in nature. For doubling perceived intensity of sound, the sound power must be increased by 10 times. That means the gain of amplifier which controls sound intensity must have gain of 10 for doubling perceived intensity of sound which is in a bell and in 10 decibel scale.
11. The total gain of a multistage amplifier is less than the product of the gains of individual stages due to ___________
a) Power loss in the coupling device
b) Loading effect of the next stage
c) The use of many transistors
d) The use of many capacitors
Answer: b
Explanation: The output of first amplifier stage is the input to next stage. In this way overall voltage gain can be increased, when number of amplifier stage is used in succession, it is called a multistage amplifier. The load of first amplifier is the input resistance of the second amplifier. Thus overall gain is reduced.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Properties of Negative Feedback”.
1. What happens to the non-linear distortion due to the initiation of the negative feedback?
a) level of non- linear distortion increases
b) level of non- linear distortion decreases
c) level of non- linear distortion remains stable
d) level of non-linear distortion changes sinusoidal
Answer: b
Explanation: Non- linear distortion generally happens because of essence of non linearity in the exchange qualities of transistor.
If a large amplitude signal is connected to an amplifier to extend the operation of a device beyond the linear operational range, then it also generates the non – linear distortion.
The effective input to an amplifier starts to reduce as the negative feedback is introduced; which eventually helps it to operate in the linear region.
This ability of negative feedback turns out to be beneficial in minimizing the level of non-linear distortion.
2. What would be the value of feedback voltage in a negative feedback amplifier with A=100; β =0.03 and input signal voltage = 40mv?
a) 0.03V
b) 0.06V
c) 0.09V
d) 0.12V
Answer: d
Explanation: Given: A =100; β = 0.03
V i = 40mV
To determine: feedback voltage {V f } for negative feedback amplifier;
Formula Used: V f = β * Vo
V f = 0.03*
V f = 0.12V
Therefore, the feedback voltage is 0.12V for given parameters of negative feedback amplifier.
3. Which of the following conditions is responsible to drive a low resistance load by the current amplifier circuit?
a) R o << R L
b) R o >> R L
c) R s >> R i
d) R s << R i
Answer: b
Explanation: As we know, a current amplifier is an amplifier which produces an output current proportional to the signal current, where the proportionality factor is independent of the source or load resistances.
Ideally, current amplifier must have zero value of input resistance and infinite output resistance.
However it indicates that the input resistance should be very small while output resistance should be very large.
Since, Rs>>Ri; High resistance source drives the current amplifier.
Similarly, as R o >> R L so the current amplifier is driven by low resistance load.
4. For an ideal voltage amplifier circuit, what should be the value of input resistance?
a) Zero
b) Infinity
c) Unity
d) Unpredictable
Answer: b
Explanation: Ideally the value of input resistance must be infinite while the value of output resistance must be zero in order to yield maximum amplified voltage gain.
But practically, it can be implied that the voltage must have large value of input resistance and small value of output resistance.
5. Negative feedback in amplifier _____________
a) Improves the signal-to-noise ratio at input
b) Improves the signal-to-noise ratio at output
c) Does not improve the signal-to-noise ratio at I/O
d) Reduce Distortion
Answer: d
Explanation: Since the negative feedback to any amplifier reduces its overall gain; hence any noise and distortion in the amplifier is also reduced.
6. Two non-inverting amplifiers, with gain =1 and gain=20 are made using identical operational amplifiers. As compared to the unity gain amplifier, the amplifier with gain=20 has ___________
a) More input impedance
b) Less negative feedback
c) Less Bandwidth
d) Less input impedance
Answer: c
Explanation: Gain bandwidth product remains constant for any device. So a device with larger gain will have a narrow bandwidth and vice versa. Gain decreases for the high frequency is because of the same phenomenon.
7. An ideal OP-Amp Is an ideal ______________
a) Current controlled current source
b) Current controlled voltage source
c) Voltage controlled voltage source
d) Voltage controlled Current source
Answer: c
Explanation: The term op-amp refers to a voltage –controlled voltage source. The input impedance is very high and the output impedance is very low. An op-amp without feedback measures the input voltage and puts out an output voltage proportional to the input. The ideal operational amplifier voltage is maintained constant. It is controlled by input voltage.
8. A feedback circuit usually employs which type of circuit?
a) Resistive
b) Inductive
c) Capacitive
d) Shunt
Answer: a
Explanation: Using a resistor in the negative feedback loop we can control the gain in two ways-
For inverting amplifier mode; V out = – (R f /R) V in
For non- inverting amplifier mode; V out = (1+R f /R) V in .
9. A system has a gain of 80 db without feedback, if the negative feedback is 1/50th. What is the closed loop gain of the system in db with the addition of the negative feedback?
a) 34db
b) 40db
c) 30db
d) 42db
Answer: a
Explanation: 80db = 20log
Therefore, G= antilog 10
=10,000
Gv = G/ = 10,000/ = 49.75
Gv = 20log = 34db.
10. Regarding the negative feedback amplifier, which of the following statements is wrong?
a) Widens the separation between 3db frequency
b) Improves gain stability
c) Increases gain – bandwidth product
d) Reduces distortion
Answer: c
Explanation: The gain-bandwidth product in a negative amplifier remains constant. Reducing the closed loop gain increases the feedback factor and increases the bandwidth. Thus the gain bandwidth product remains constant.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Voltage Amplifiers”.
1. The emitter of a swamped amplifier __________
a) is grounded
b) has no dc voltage
c) has an ac voltage
d) has no ac voltage
Answer: c
Explanation: The resistance of the emitter diode r’ e equals 25mV / I E and depends on the temperature. Any change in r’ e will change the voltage gain in CE amplifier. In some applications, a change in voltage is acceptable. But in many applications a stable voltage gain is required.
To make it stable, a resistance rE is inserted in series with the emitter and therefore emitter is no longer ac grounded.
Because of this the ac emitter current flows through r E and produces an ac voltage at the emitter. If r E is much greater than r’ e almost the entire ac input signal appears at the emitter, and the emitter is bootstrapped to the base for ac as well as for dc.
2. A swamped amplifier has ________ input impedance as compared to CE stage amplifier.
a) smaller
b) equal
c) larger
d) zero
Answer: c
Explanation: A swamped amplifier reduces variations in voltage gain by increasing the ac resistance of the emitter circuit to get a stable voltage gain. It is also referred as gain-stabilized amplifier. That’s why the input impedance is larger.
3. We can increase _______ to reduce distortion of an amplified signal.
a) Collector resistance
b) Emitter feedback resistance
c) Generator resistance
d) Load resistance
Answer: b
Explanation: In a negative feedback amplifier, when emitter feedback resistance is increased, the magnitude of the gain of the amplifier reduces; but remains stable. Since the gain reduces the distortion and the noise in the signal is also reduced; as the overall signal we get at output is the input signal multiplied by the gain of the amplifier.
4. A swamped amplifier uses________
a) Base bias
b) Positive feedback
c) Negative feedback
d) A grounded emitter
Answer: c
Explanation: A swamped amplifier reduces variations in voltage gain by increasing the ac resistance of the emitter circuit to get a stable voltage gain. As it involves reduction in gain, it is a negative feedback.
5. In a swamped amplifier, the effects of the emitter diode become _________
a) Important to the voltage gain
b) Critical to input impedance
c) Significant to analysis
d) Unimportant
Answer: d
Explanation: To make the voltage gain stable, a resistance r E is inserted in series with the emitter and therefore emitter is no longer ac grounded.
Because of this the ac emitter current flows through r E and produces an ac voltage at the emitter. If r E is much greater than r’ e almost the entire ac input signal appears at the emitter, and the emitter is bootstrapped to the base for ac as well as for dc.
Thus the effects of the emitter diode become unimportant in swamping amplifier.
6. What is the function of feedback resistor?
a) To increase voltage gain
b) Reduce distortion
c) Decrease collector resistance
d) Decrease input impedance
Answer: b
Explanation: In a negative feedback amplifier, when emitter feedback resistance is increased, the magnitude of the gain of the amplifier reduces; but remains stable. Since the gain reduces the distortion and the noise in the signal is also reduced; as the overall signal we get at output is the input signal multiplied by the gain of the amplifier. Therefore, feedback in an amplifier helps in reducing distortion.
7. If the emitter bypass capacitor is open , the ac output voltage will ____________
a) decrease
b) increase
c) remains the same
d) equals zero
Answer: a
Explanation: The emitter bypass capacitor is connected in parallel to the emitter resistance. The use of this connection is to provide an ac ground at the emitter terminal of the transistor. This has the effect of increasing the circuit voltage gain.
The bypass capacitor is normally a high-value component that provides little reactance at the lowest circuit operating frequency. Therefore, in the absence of this bypass capacitor, the circuit voltage gain will reduce.
8. If any capacitor is open, the ac output voltage wills ____________
a) Decrease
b) Increase
c) Remains the same
d) Equals zero
Answer: a
Explanation: Coupling capacitors are use to decouple ac and dc signals so as not to disturb the Q-point of the circuit when ac signals are provided at the input. Bypass capacitors are used to force signal currents around elements by providing a low impedance path at the frequency. The voltage drop across the resistor provides a negative bias to the emitter terminal, thereby reducing its DC gain, while the AC gain will also reduce as the bypass capacitor which was providing low impedance path is left open.
9. If the emitter resistor is open, the ac input voltage will __________
a) Decrease
b) Increase
c) Remains the same
d) Equals zero
Answer: b
Explanation: An AC signal amplifier circuit is mainly used to stabilize the DC biased input voltage to the amplifier and thus only amplify the required AC signal. This stabilization is achieved by the use of an Emitter Resistance which provides the required amount of automatic biasing needed for a common emitter amplifier. If the emitter resistor is open, then the input ac voltage will eventually increase.
10. The input impedance of the base increases when ______
a) Beta increases
b) Supply voltage increases
c) Beta decreases
d) AC collector resistance increases.
Answer: a
Explanation: Input impedance of the base is given as = beta * r E .
Therefore, beta is directly proportional input impedance of the base.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Transconductance Amplifiers”.
1. A transconductance amplifier is also called ___________
a) current to voltage convertor
b) voltage to current convertor
c) resistor
d) inductor
Answer: a
Explanation: A transconductance amplifier converts an input of voltage to an output of current. It is also called a current to voltage converter or I to V converter. There is usually an additional input for a current to control the amplifier’s transconductance.
2. The transconductance of a JFET ranges from ____________
a) 100 to 500 mA/V
b) 500 to 1000 mA/V
c) 0.5 to 30 mA/V
d) above 1000 mA/V
Answer: c
Explanation: It is very often denoted as a conductance, g m , with a subscript, m, for mutual. Transconductance is defined as follows:
G m = I OUT /V IN
The transconductance of a JFET ranges from 0.5 to 30 mA/V.
3. The constant-current region of a JFET lies between__________
a) Cut off and saturation
b) Cut off and pinch off
c) 0 and I DSS
d) Cut off and breakdown
Answer: d
Explanation: Value of drain-source voltage, V DS for breakdown with the increase in negative bias voltage is reduced simply due to the fact that gate-source voltage, V GS keeps adding to the reverse bias at the junction produced by current flow. Thus the current flow remains constant between the cut off and breakdown region of JFET.
4. The source terminal of a JEFT corresponds to ____________ of a vacuum tube
a) plate
b) cathode
c) grid
d) anode
Answer: b
Explanation: Electron-flow from the source terminal towards the drain terminal is influenced by an applied voltage. In vacuum tube, the source terminal of JFET corresponds to the cathode of vacuum tube.
5. The output characteristics of a JFET closely resemble the output characteristics of a __________ Valve
a) diode
b) pentode
c) triode
d) tetrode
Answer: b
Explanation: The pentode consists of an evacuated glass envelope containing five electrodes in this order: a cathode heated by a filament, a control grid, a screen grid, a suppressor grid, and a plate .
Similar to a pentode valve, with which the junction FET greatly resembles in its operational characteristics, the simplest way of calculating stage gain is by the relationship –
A = g fs . R L
6. If the cross-sectional area of the channel in n-channel JFET increases, the drain current__________
a) is increased
b) is decreased
c) remains the same
d) decreases exponentially then increase
Answer: a
Explanation: The gate source junctions are reverse biased as a result depletion regions from which extend to the bar by changing gate to source voltage; effective cross sectional area decreases with the function of the gate to source voltage. With increasing the cross section area of the channel, the drain current is increased.
7. The channel of a JFET is between the ________
a) gate and drain
b) gate and source
c) drain and source
d) input and output
Answer: c
Explanation: The field effect transistor is a three terminal device that is constructed with no PN-junctions within the main current carrying path between the Drain and the Source terminals. The current path between these two terminals is called the “channel” which may be made of whichever a P-type or an N-type semiconductor material.
8. A certain common-source JFET has a voltage gain of 10. If the source bypass capacitor is removed, _____________
a) the voltage gain will increase
b) transconductance will increase
c) voltage gain will decrease
d) the Q-point will shift
Answer: c
Explanation: In common source JFET, The main purpose of a source bypass capacitor is to provide additional gain at AC. Here, the common-source JFET has a voltage gain of 10; which includes the additional AC gain provided. If the source bypass capacitor is removed, then the voltage gain will eventually decrease.
9. At cut-off, the JFET channel is ___________
a) at its widest point
b) completely closed by the depletion region
c) extremely narrow
d) reverse baised
Answer: b
Explanation: At cut off, the JFET channel is completely isolated by the depletion region. Gate connection is completely isolated from the main current carrying channel. At cut-off, V GS is sufficient to cause the JFET to act as an open circuit as the channel resistance is at maximum.
10. MOSFET from a JFET differs mainly because _____________
a) of power rating
b) of the output
c) the JFET has a pn junction
d) the MOSFET has two gates
Answer: c
Explanation: A MOSFET has no pn junction, instead at the gate of the constituent there is an insulated silicon dioxide layer that insulates from the channel. A JFET is a depletion mode transistor. A JFET has diode junction between the gate and the channel.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Classification of Output Stages”.
1. A class A amplifier with R C = 3.3 k and R E = 1.2 k has a V CC = 20 V. Find I C .
a) 4.4mA
b) 6.1mA
c) 20mA
d) 16.7mA
Answer: a
Explanation: Since, at saturation region, V cc = 0 V, so V cc will drop across R c and R E only.
here, R c + R E = 3.3 k ohm + 1.2 k ohm,
= 4.5 k ohm,
I c = V cc /,
= 20/ = 4.44 mA.
2. Which of the power amplifiers has the lowest overall efficiency?
a) Class C
b) Class A
c) Class B or AB
d) Class D
Answer: b
Explanation: The efficiency of this type of circuit is very low and delivers small power outputs for a large drain on the DC power supply. A Class A amplifier stage passes the same load current even when no input signal is applied so large heat sinks are needed for the output transistors.
3. Calculate the efficiency of a class B amplifier for a supply voltage of V CC = 20 V with peak output voltage of V L = 18 V. Assume R L = 16 .
a) 78.54%
b) 75%
c) 70.69%
d) 50%
Answer: c
Explanation: Using Class B efficiency:
Formula:
%n = 78.54 (V pout / V cc ).
Given
V cc = 20V p out .
V L = 18V,
R L = 16V,
%= 78.54,
Efficiency= 70.69%.
4. For BJT power transistors, the collector terminal is always connected to the transistor’s case________
a) for easy circuit connection
b) to prevent shorts
c) because the collector terminal is the critical terminal for heat dissipation
d) because the collector terminal is located nearest the case
Answer: c
Explanation: Transistor case is used for power transistors as the power dissipated at their collector junction is large. If heat dissipation is not done, this will cause large increases in junction temperature.
5. Which operation class is generally used in radio or communications?
a) A
b) B
c) C
d) D
Answer: c
Explanation: In radio or communications, generally class C is used in case of high level modulation. High level modulation uses class C amplifiers in a broadcast AM transmitter and only the final stage or final two stages are modulated, and all the earlier stages can be driven at a constant level.
6. Which of the push-pull amplifiers is presently the most popular form of the class B power amplifier?
a) Quasi-complementary
b) Transformer-coupled
c) Complementary-symmetry
d) Symmetry
Answer: a
Explanation: In class B operation, a push-pull connection is obtained using either a transformer coupling or by using complementary operation with npn and pnp transistors to provide operation on opposite polarity cycles. While transformer operation can provide opposite cycle signals, the transformer itself is quite large in many application.
7. Calculate the harmonic distortion component for an output signal having a fundamental amplitude of 3 V and a second harmonic amplitude of 0.25 V.
a) 3.83%
b) 38.3%
c) 83.3%
d) 8.33%
Answer: d
Explanation: 3 volt is input
0.25 is distortion in input for 3 volt
then *100 = 8.33%.
8. Which of the power amplifiers is not suitable as audio amplifiers?
a) Class A
b) Class B
c) Class C
d) Class D
Answer: c
Explanation: The transistor biasing gives a much improved efficiency of around 80% to the class C amplifier, it introduces a very heavy distortion of the output signal. Therefore, class C amplifiers are not suitable for use as audio amplifiers.
9. Which type of amplifier uses pulse signals in its operation?
a) Class A
b) Class B or AB
c) Class C
d) Class D
Answer: d
Explanation: The class D amplifier is nonlinear switching amplifiers or pulse width modulation amplifiers. These amplifiers are called as digital amplifiers.
10. What is the maximum efficiency of a class B circuit?
a) 90%
b) 78.5%
c) 40%
d) 50%
Answer: b
Explanation: The conduction angle is 180° for a Class B amplifier. Since the active device is switched off for half the input cycle, the active device dissipates less power and hence the efficiency is enhanced. Theoretical maximum efficiency of Class B power amplifier is 78.5%.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Class A Output Stage”.
1. A class A power amplifier uses ______________
a) Two transistors
b) One transistor
c) Three transistors
d) Four transistors
Answer: b
Explanation: In Class A amplifier, if the collector current flows all times during the full cycle of the input signal, the power amplifier is known as class A power amplifier. It is less used for higher power output stages, as it has poor efficiency. So it consists of only one transistor.
2. The maximum efficiency of resistance loaded class A power amplifier is ____________
a) 5%
b) 35%
c) 25%
d) 50%
Answer: c
Explanation: The theoretical maximum efficiency of a Class A power amplifier is 50%. In practice, with the capacitive coupling and inductive loads , the efficiency can decrease as low as 25%. This means 75% of power drawn by the amplifier from the supply line is wasted.
3. The maximum efficiency of transformer coupled class A power amplifier is ____________
a) 50%
b) 25%
c) 30%
d) 5%
Answer: a
Explanation: Using the transformer coupling technique, the efficiency of an amplifier can be enhanced to a great extent. The coupling transformer provides good impedance matching between the load and output, and it is the main reason behind the improved efficiency. Therefore, Its efficiency is 50%.
4. Power amplifiers handle ____________ signals compare to voltage amplifiers
a) small
b) large
c) very small
d) equal
Answer: b
Explanation: The Small Signal Amplifier is generally referred to as a “Voltage” amplifier because they usually convert a small input voltage into a much larger output voltage. Although the amplification is high the efficiency of the conversion from the DC power supply input to the AC voltage signal output is usually poor.
5. In class A operation, the operating point is generally located ____________ of the d.c. load line.
a) At cut off point
b) At the middle
c) At saturation point
d) In active region
Answer: b
Explanation: By referring to the output characteristics of the class A amplifier operation, the Q-point is placed exactly at the centre of the AC load line and the transistor conducts for every point in the input waveform. Therefore, it lies in the middle of d.c. load line.
6. A power amplifier has comparatively ____________ β.
a) Small
b) Large
c) Very large
d) Same
Answer: a
Explanation: A power amplifier has relatively small gain β, As compared to voltage amplifier. Both current and voltage amplifiers have greater gain as compared to power amplifiers. Power amplifies handle large signals as compared to voltage amplifiers. Power amplifiers have the following stages- A, B, C, D, AB.
7. A class A power amplifier is sometimes called ____________ amplifier
a) Reciprocating
b) Single – ended
c) Symmetrical
d) Differential
Answer: b
Explanation: Class-A designs can be simpler than other classes such as class -AB and -B designs require two connected devices in the circuit , each to handle one half of the waveform whereas class A can use a single device that’s why it is considered as a single ended amplifier.
8. When no signal is applied, the approximate collector efficiency of class A power amplifier is ____________
a) 25%
b) 10%
c) 0%
d) 50%
Answer: c
Explanation: The efficiency of a class A amplifier is dependent upon the input signal. The relation is direct. Therefore, with no input signal applied; the efficiency will be equal to zero.
9. The maximum a.c. power output from a class A power amplifier is 10 W. What should be the minimum power rating of the transistor used?
a) 20W
b) 5 W
c) 25W
d) 50W
Answer: a
Explanation: Maximum AC power is given as;
P 0 = (V CEQ . I CEQ )/2
So the minimum power rating used should be,
10W * 2 = 20W.
10. The most costly coupling is ____________ coupling
a) Transformer
b) Impedance
c) RC
d) Direct
Answer: a
Explanation: In Transformer coupling, audio transformers are used. This requires a larger core and more turns, making a more expensive transformer coupling. Transformer coupling is very widely used at radio frequencies, but it is also practical at audio frequencies, say 30 to 30,000 Hz.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Class B Output Stage”.
1. The maximum collector efficiency of class B operation is ________
a) 50%
b) 25%
c) 30%
d) 5%
Answer: d
Explanation: Since the active device is switched off for half the input cycle, the active device dissipates less power and hence the efficiency is improved. Theoretical maximum efficiency of Class B power amplifier is 78.5%.
2. A 2-transistor class B power amplifier is commonly called ________ amplifier
a) Dual
b) Single ended
c) Push-pull
d) Differential
Answer: c
Explanation: When Class B amplifier is to be used with a pair of active devices ; it is arranged in a push-pull mode where one transistor conducts one half cycle and the other transistor conducts the other half cycle.
3. When a transistor is in cut off________
a) Maximum voltage appears across transistor
b) Maximum voltage appears across load
c) Maximum current flows
d) Least current flows
Answer: a
Explanation: A transistor in cutoff mode is off – there is no collector current, and therefore no emitter current. It almost looks like an open circuit. Thus Maximum voltage appears across the transistor.
4. The output stage of a multistage amplifier usually employs ________
a) Push-pull amplifier
b) Preamplifier
c) Class A power amplifier
d) Class B power amplifier
Answer: a
Explanation: In push-pull mode; one transistor conducts one half cycle and the other transistor conducts the other half cycle. The output from both transistors are then combined together to get a scaled replica of the input.
5. The size of a power transistor is made considerably large to ________
a) Provide easy handling
b) Dissipate heat
c) Facilitate connections
d) Larger Gain
Answer: b
Explanation: Power amplifiers handle large signals as compared to voltage amplifiers. The size of power transistors is made considerably large to dissipate heat. Low efficiency of power amplifiers results in more battery consumption.
6. The maximum a.c. power output from a class B power amplifier is 10 W.
What should be the minimum power rating of the transistor used?
a) 10W
b) 15W
c) 4W
d) 20W
Answer: c
Explanation: maximum ac power output voltage = 10W
In class b amplifier, only one half cycle is considered. So the minimum power rating is 4W.
7. The push-pull circuit must use ________ operation.
a) Class A
b) Class B
c) Class C
d) Class AB
Answer: b
Explanation: The basic class B amplifier uses two complimentary transistors for each half of the waveform, they are arranged in a push-pull form. No DC base bias current in the class B power amplifier.
8. The disadvantage of impedance matching is that it ________
a) Gives distorted output
b) Gives low power output
c) Requires a transformer
d) None of the mentioned
Answer: a
Explanation: The disadvantage of impedance matching is that it gives distorted output. The Power amplifiers generally use transformer coupling because the transformer permits impedance matching.
9. The Power amplifiers generally use transformer coupling because the transformer permits ________
a) Cooling of the circuit
b) Impedance Matching
c) Distortion less Output
d) Good Frequency Response
Answer: b
Explanation: The Power amplifiers generally use transformer coupling because the transformer permits impedance matching. The disadvantage of impedance e matching is that it gives distorted output.
10. The most important consideration in power amplifier is ________
a) Biasing the circuit
b) Impedance Matching
c) Collector Efficiency
d) To keep the transformer cool
Answer: c
Explanation: The most important consideration in power amplifier is Collector Efficiency. The pulsating DC applied to power amplifier causes hum in the circuit. For every stage of power amplifier, the collector efficiency is considered most important.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Power BJTs”.
1. A power transistor is a ____________
a) three layer, three junction device
b) three layer, two junction device
c) two layer, one junction device
d) four layer, three junction device
Answer: b
Explanation: A power BJT has three layers, p-n-p or n-p-n forming two junctions.
p-n-p: two positive layers and one negative layers in between them.
n-p-n: two negative layers and one positive layers.
2. For a power transistor, if base current I B is increased keeping V ce constant; then______
a) I C increases
b) I C decreases
c) I C remains constant
d) I C changes sinusoidal
Answer: a
Explanation: I B is directly proportional to I c . The I C curve is linearly distributed when using I B as a parameter, and exponentially distributed using V BE as a parameter. So when V BE is constant, the transistor current I C is almost linear with respect to I B .
3. Which one is the most suitable power device for high frequency switching application?
a) BJT
b) Power MOSFET
c) Schottkey diode
d) Microwave transistor
Answer: b
Explanation: Power MOSFET has a low turn off time. So it can be operated in a frequency range of 1 to 10 MHz.
4. Insulated-gate bipolar transistor has combinational advantages of ______
a) BJTs and SITs
b) BJTs and MOSFETs
c) SITs and MOSFETs
d) FETs and BJTs
Answer: b
Explanation: IGBT combines advantages of BJTs and MOSFETs. IGBT process high input impedance like a MOSFET and has low on state power loss as in BJTs.
5. A Gate Turn Off can be turned on by applying _______
a) positive gate signal
b) positive drain signal
c) positive source signal
d) negative source signal
Answer: a
Explanation: A Gate Turn Off like and SCR is a device four layer, three junction semiconductor device with three external terminal . GTO can be turned ON and OFF by positive pulse or signal respectively, to the gate terminal.
6. SITH is also known as ________
a) Field controlled diode
b) Field controlled rectifier
c) Silicon controlled diode
d) Silicon controlled rectifier
Answer: a
Explanation: The static induction thyristor is a thyristor with a buried gate structure in which the gate electrodes are placed in n-base region. They are normally on-state, gate electrodes must be negatively biased to hold-off state. It is a self controlled GTO like device. Hence it is sometimes called the field controlled diode.
7. The turn on time of an SCR with inductive load is 20 µs. The pulse train frequency is 2.5 KHz with a mark/space ratio of 1/10, and then SCR will ______
a) Turn On
b) Not turn on
c)Turn on if inductance is removed.
d) Turn on if pulse frequency is increased to two times
Answer: a
Explanation: Pulse repetition rate = 1/(2.5 – 10 3 ) = 0.4 ms = 400 µs.
Mark/space ratio = 1/10.
Pulse width = 400/11 = 36.4 µs.
The SCR will ‘turn on’ as the pulse width is greater than SCR turn on time.
8. What are the three terminals of a power MOSFET called?
a) Collector, emitter, Gate
b) Drain, source, gate
c) Collector, emitter, base
d) Drain, emitter, base
Answer: b
Explanation: A power MOSFET has three terminals called Drain, source and gate in place corresponding to the three terminals Collector, emitter and base for BJT.
9. A thyristor can be termed as ______
a) AC switch
b) DC switch
c) Wave switch
d) Square wave switch
Answer: b
Explanation: Thyristor is a unidirectional device, that is it will only conduct current in one direction only, but unlike a diode, a thyristor can be operate as either ran open circuit switch or as a rectifying diode depending on how the thyristor gate is triggered. In other words, the thyristor can operate only in switching mode.
10. If anode current is 400 A, then the amount of current required to turn off the GTO is about ________
a) 20A
b) 200A
c) 400A
d) 100A
Answer: d
Explanation: Generally, anode current required of GTO is four times of turn off current. So the amount of current required to turn off GTO is 400/4 = 100A.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Transistor Case and Heat Sinks”.
1. Why does the heat sink has fins?
a) to provide cooling to the processor
b) to provide airflow to the radiator
c) to preserve the energy
d) to provide radiator to the airflow
Answer: a
Explanation: The heat sink has a thermal conductor that carries heat away from the CPU into fins that provide a large surface area for the heat to dissipate throughout the rest of the computer, thus cooling both the heat sink and processor.
2. Heat sinks work through the process of _______
a) resistive heat transfer
b) conductive and convection heat transfer
c) active process
d) no heat transfer
Answer: b
Explanation: The main purpose of a heat sink is to expel heat from a generating source. Heat sinks work through the process of conductive and convection heat transfer. Heat sinks are a passive form of cooling, as they have no moving parts and require no power. In most cases, heat sinks are used in conjunction with fans.
3. Copper has around twice the thermal conductivity of aluminum.
a) True
b) False
Answer: a
Explanation: Copper has around twice the thermal conductivity of aluminum, around 400 W/M K for pure copper. Its main applications are in industrial facilities, power plants, solar thermal water systems, HVAC systems, gas water heaters, forced air heating and cooling systems, geothermal heating and cooling, and electronic systems.
4. Fin efficiency is increased by_______
a) Using insulating material
b) Decreasing the fin aspect ratio
c) Using more conductive material
d) Decreasing the fin aspect ratio & Using more conductive material
Answer: d
Explanation: Fin efficiency is one of the parameters which formulate a higher thermal conductivity material significant. A fin of a heat sink may be measured to be a flat plate with heat flow in one end and being dissipated into the surrounding fluid as it travels to the other. As heat flows through the fin, the grouping of the thermal resistance of the heat sink impeding the flow and the heat lost due to convection, the temperature of the fin and, consequently, the heat transfer to the fluid, will decrease from the bottom to the end of the fin. Fin efficiency is defined as the actual heat transferred by the fin, separated by the heat transfer were the fin to be isothermal.
5. How can one decrease the spreading resistance in the base of the heat sink?
a) Increase the base thickness.
b) Decrease the base thickness.
c) Choosing a material with less conductivity.
d) There is no possible way.
Answer: a
Explanation: Spreading resistance occurs when thermal energy is transferred from a minute area to a larger area in a substance with finite thermal conductivity. In a heat sink, this means that heat does not distribute uniformly through the heat sink base. The spreading resistance occurrence is shown by how the heat travels from the heat source location and causes a large temperature gradient between the heat source and the edges of the heat sink. This means that some fins are at a lower temperature than if the heat source were uniform from corner to corner of the base of the heat sink.
6. How can a pin fin heat sink be classified on the basis of fin arrangements?
a) Pin, straight pin, flared pin
b) Pin, circular pin, cylindrical pin
c) Straight pin, circular pin, isotopic pin
d) No classification
Answer: a
Explanation: In general pin fins are classified as straight pin and flared pin. For the straight fin it was 44 °C or 6 °C better than the pin fin. Pin fin heat sink performance is significantly better than straight fins when used in their intended application where the fluid flows axially along the pins rather than only tangentially across the pins.
7. _______are usually utilized to extract heat from a variety of heat generating bodies to a heat sink.
a) square fins
b) cylindrical fins
c) cavities
d) no option
Answer: c
Explanation: Cavities embedded in a heat source are the regions formed between adjacent fins that stand for the essential promoters of nucleate boiling or condensation. These cavities are usually utilized to extract heat from a variety of heat generating bodies to a heat sink.
8. Which of the following is not an application of transistor heat sinks?
a) Soldering
b) Light Emitting Diode Lamps
c) Microprocessor Cooling
d) Environment Monitoring Systems
Answer: d
Explanation: Heat dissipation is an unavoidable by-product of electronic devices and circuits. LED performance and lifetime are strong functions of their temperature. Effective cooling is therefore essential. A case study of a LED based down lighter shows an example of the calculations done in order to calculate the required heat sink necessary for the effective cooling of lighting system. Temporary heat sinks are sometimes used while soldering circuit boards, preventing excessive heat from damaging sensitive nearby electronics.
9. The epoxy bond between the heat sink and component is __________
a) Temporary
b) Loosened
c) Permanent/semi-permanent
d) Highend
Answer: c
Explanation: The epoxy bond between the heat sink and component is semi-permanent/permanent. This is done to make the re-work very difficult and at times impossible. The most typical damage caused by rework is the separation of the component die heat spreader from its package.
10. Which of the following characteristics makes passive heat sinks differently from active heat sinks?
a) It possess mechanical components
b) It possess electrical components
c) Does not possess mechanical components
d) It possess electrical & mechanical components
Answer: c
Explanation: Unlike active heat sink passive heat sink do not possess any mechanical component and are made of aluminum finned radiator.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Class B Amplifiers”.
1. Where does the Q point lie for class B amplifier?
a) Active
b) Cut off
c) Saturation
d) Between saturation and active
Answer: b
Explanation: Class B amplifier are designed by fixing the Q point in cut off region of the transfer characteristic.
2. Class B amplifier Produces output even if the input is zero.
a) True
b) False
Answer: b
Explanation: When input provided to the class B amplifier is zero, no output will be achieved, because it is excellent against noise.
3. What happens when class B amplifier is in a quiescent state?
a) No current flows through the transistor
b) Maximum current flows through the transistor
c) Half of the maximum current flows through the transistor
d) Quarter of the maximum current flows
Answer: a
Explanation: When the transistor is in a quiescent state, no input is applied across the base terminal of the transistor and hence no current flows through the transistor.
4. What is the value of the maximum efficiency of the class B amplifier?
a) 25%
b) 35%
c) 35% to 50%
d) 50% to 70%
Answer: d
Explanation: Class B amplifiers are more efficient compare to the class A amplifier because of good protection against noise effects.
5. Which is the main disadvantage of class B amplifiers?
a) Expensive
b) Less efficient
c) More power dissipation
d) More heat dissipation
Answer: a
Explanation: Since class B amplifier uses a balanced centre-tapped transformer in its design, making it expensive to construct.
6. What kind of design is used to avoid transformer usage?
a) High resistance
b) Matched load
c) Complementary symmetry
d) Capacitive Model
Answer: c
Explanation: To avoid transformer usage in Class B amplifier, pair of transistor is connected together in complemented manner.
7. What is cross over distortion?
a) Effect occurred during switching of transistor after every half cycle
b) Distortion occurred due to resistors
c) Distortion occurred due to Capacitors
d) Distortion occurred due to Inductors
Answer: a
Explanation: Transistor takes 0.7V to turn on when during the end of half cycles, the input gets below 0.7V and it is not possible to reproduce these signals. This is called as Cross over distortion.
8. How to avoid cross over distortion?
a) By using more resistance
b) By using more capacitance
c) By using more Inductance
d) By shifting the Q point above cut off
Answer: d
Explanation: By using two more voltage sources and thus by shifting the Q point slightly above the cut off, we can remove noise over distortion.
9. For a Class B amplifier, the utilized load power is 300W and the Dc power is 500W, find efficiency.
a) 30%
b) 60%
c) 90%
d) 100%
Answer: b
Explanation: Efficiency=ac power/dc power
Efficiency=300/500 = 3/5 = 0.6.
10. What is the conduction angle for Class B push-pull amplifier?
a) 0
b) 90
c) 180
d) 270
Answer: c
Explanation: For class B push-pull amplifier, the conduction angle is 180 degree that is it amplifies only one half cycle of the input in one time period.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Class D Amplifiers”.
1. Why a class D amplifier is called as switching amplifier?
a) Because it consists of a transistor
b) Because of high power
c) Because of less noise effect
d) Because of binary switch
Answer: d
Explanation: A class D amplifier or switching amplifier is an electronic amplifier where all power devices are operated as binary switches.
2. For an ideal class D amplifier, what time is required in transitioning between states?
a) 0
b) 10
c) 100
d) Infinite
Answer: a
Explanation: For an Ideal class D amplifier, zero time is spent between transitioning from on to off or vice versa.
3. Which of the following is an example for the application of class D amplifier?
a) Switch
b) Torch
c) Pulse generator
d) Bulb
Answer: c
Explanation: The output stages of class D amplifier are used as pulse generator, because of its switching ability.
4. Which of the following is true about class D amplifier?
a) Its efficiency is very less
b) It can be made to work as bulb
c) It is big in size
d) It operates on lesser frequently
Answer: d
Explanation: class D amplifier work by generating a square wave of which the low frequency portions of the spectrum is essentially the wanted output signal.
5. Efficiency of class D amplifier____________
a) 100%
b) 59%
c) 75%
d) 60%
Answer: a
Explanation: Theoretically the efficiency of the class D power amplifier is 100% because of its switching ability.
6. MOSFET class D amplifier is more advantageous than BJT class D amplifier.
a) True
b) False
Answer: a
Explanation: MOSFET is smaller and is controlled by voltage, hence it is more advantageous compared to BJT.
7. Which of the following is true about MOSFET class D amplifier?
a) It has infinite gain
b) It is less stable
c) It is less advantageous
d) It dissipates less heat
Answer: d
Explanation: When MOSFET is off, no current flows and hence no power dissipation occurs.
8. Why the cost of MOSFET based power amplifier is less?
a) High impedance
b) Low capacitance
c) Low power dissipation
d) Lesser in size
Answer: c
Explanation: Since power dissipation in MOSFET is less, this results in smaller heat sinks and thus the cost of the amplifier will be reduced.
9. What is the conduction angle of class D amplifier?
a) 180
b) 270
c) 0
d) 90
Answer: 0
Explanation: The conduction angle of class D amplifier is 0 because of its switching ability, it turns on and off non linearly.
10. Why filters are required in class D amplifier?
a) Output is completely dependent on the lower frequency spectrum
b) To increase the complexity
c) To obtain only lower frequency
d) To nullify noise
Answer: a
Explanation: Output is completely dependent on the lower frequency spectrum of the output signal and hence to properly obtain lower frequency components, filters are required.
This set of Electronic Devices and Circuits Multiple Choice Questions & Answers focuses on “Oscillators – 1”.
1. What are oscillators?
a) Switching circuits
b) Converts dc to ac
c) Converts ac to dc
d) Filter circuits
Answer: b
Explanation: Oscillator circuits are the one which uses dc power supply and converts it to alternating sinusoidal signals.
2. Give the relation between output and input voltage of an oscillator?
a) A v = V i /V o
b) V i = V o A v
c) V o = A v /V i
d) A v = V o /V i
Answer: d
Explanation: For an oscillator circuit the output voltage is the product of input voltage and gain of an amplifier.
3. What is the value of self-oscillating circuits?
a) 0
b) 1
c) 2
d) 3
Answer: b
Explanation: For self-oscillating circuits, the loop gain Aβ must be equal to 1, otherwise there will be the effect of noise.
4. What is the value of input voltage, if Aβ=1 and input voltage is 5V?
a) 0
b) 2
c) 5
d) 10
Answer: c
Explanation: since Aβ =1, it is a self oscillating circuits and hence output always has the peak amplitude of 5V.
5. For an oscillator, input voltage is 5V, loop gain is 2, find output voltage?
a) 10
b) 20
c) 5
d) 15
Answer: a
Explanation: For an oscillator, the output voltage is the product of loop gain and input voltage, hence output will be twice the input, that is 5*2 =10V.
6. For practical oscillators, which law has to be obeyed?
a) Faraday law
b) Hertz law
c) Fleming law
d) Barkhausen law
Answer: d
Explanation: For an oscillator circuit, it has to obey the Barkhausen criteria, otherwise the output f the oscillator will tend to zero as the time advances.
7. Which of the following expression depicts Barkhausen criteria?
a) Aβ = 1
b) Aβ = 0
c) Aβ < 1 < Aβ
d) Aβ < 1
Answer: a
Explanation: Barkhausen criteria states that loop gain of the closed loop circuit must be equal to one in order to repeat the generation.
8. Barkhausen criteria states phase of loop gain must be 0 for a self sustaining oscillator.
a) True
b) False
Answer: a
Explanation: For an oscillator circuit, the phase of loop gain must be equal to zero or integral multiple of 2nπ.
9. What is the average value of saturation factor in a practical circuit?
a) 1
b) 0
c) 2
d) 3
Answer: a
Explanation: The average value of saturation factor is equal to loop gain of the particular circuit, usually it will be nearly equal to 1.
10. An oscillator requires an input voltage of high amplitude.
a) True
b) False
Answer: b
Explanation: An oscillator circuit does not require an input voltage, an oscillator can be designed without giving an input signal.
This set of Electronic Devices and Circuits written test Questions & Answers focuses on “Oscillators – 2”.
1. For a phase-shift oscillator, the gain of the amplifier stage must be greater than ________
a) 19
b) 30
c) 29
d) 1
Answer: c
Explanation: Since the resistor-capacitor combination in the RC Oscillator circuit also acts as an attenuator producing an attenuation of -1/29th per stage, the gain of the amplifier must be sufficient to overcome the circuit losses.
Therefore, in our three stage RC network above the amplifier gain must be greater than 29.
2. What is the minimum frequency at which a crystal will oscillate?
a) fundamental
b) seventh harmonic
c) second harmonic
d) third harmonic
Answer: a
Explanation: The physical size of the quartz crystal affects the final or fundamental frequency of oscillations. The fundamental frequency is generally called the crystals “characteristic frequency”. The crystals characteristic or characteristic frequency is inversely proportional to its physical thickness between the two metalized surfaces.
3. A circuit that can change the frequency of oscillation with an application of a dc voltage is sometimes called___________
a) a crystal oscillator
b) a voltage-controlled oscillator
c) an astable multivibrator
d) a Hartley oscillator
Answer: b
Explanation: A Voltage controlled oscillator is an oscillator with an output signal whose output can be varied over a range, which is controlled by the input DC voltage. It is an oscillator whose output frequency is directly related to the voltage at its input.
The oscillation frequency varies from few hertz to hundreds of GHz.
4. In order to start up, a feedback oscillator requires______
a) unity feedback equal to 1
b) negative feedback less than 1
c) positive feedback greater than 1
d) no feedback
Answer: c
Explanation: In oscillators using positive feedback it is important that amplitude of the oscillator output remains stable. Therefore the closed loop gain must be 1 . In other words, the gain within the loop; provided by the amplifier, should exactly match the losses within the loop.
5. The Nyquist plot combines the two Bode plots of gain versus frequency and phase shift versus frequency on a single plot.
a) True
b) False
Answer: a
Explanation: Bode plots show the frequency response of a system. There are two Bode plots one for gain and one for phase.
The Nyquist plot combines gain and phase into one plot in the complex plane. It is drawn by plotting the complex gain g for all frequencies ω. That is, the plot is a curve in the plane parameterized by ω.
6. An input signal is needed for an oscillator to start.
a) True
b) False
Answer: b
Explanation: The oscillator is a device which gives an AC output without any input. Oscillator needs an input signal for start. But once the capacitor is charged then, Oscillator can effectively start work without an input signal.
7. The lead-lag circuit in the Wien-bridge oscillator has a resonant frequency at which the attenuation is_________
a) 1/2
b) 1/4
c) 1/5
d) 1/3
Answer: d
Explanation: The response curve for the lead-lag circuit indicates that the output voltage peaks at a frequency called the resonant frequency, fr. At this point, the attenuation (V out /V in ) of the circuit is 1/3.
V out /V in = 1/3
The lead-lag circuit in the Wien-bridge oscillator has a resonant frequency fr, at which the phase shift through the circuit is and the attenuation is 1/3.
8. At series resonance, the impedance of a crystal is________
a) minimum
b) maximum
c) equal
d) Zero
Answer: a
Explanation: The slope of the crystals impedance suggests that as the frequency increases across its terminals. At a particular frequency, the interaction between the series capacitor and the inductor creates a series resonance circuit reducing the crystals impedance to a minimum and equal to Resistance.
9. The twin-T oscillator produces a ________ response.
a) low-pass
b) high-pass
c) band -pass
d) band-stop
Answer: d
Explanation: Twin T-oscillator is an RC oscillator consists of Twin T-network and an op-amp. Twin-T is actually a combination of low pass and high pass filter combined parallel response provides a band-stop filter with a center frequency equal to
fr=1/2πRC
This is resonant frequency, Oscillations can occur only at this frequency as the filter provides significant negative feedback at frequencies below or above the resonant frequency which is not good for oscillation.
10. Which of the following improvements is a result of the negative feedback in a circuit?
a) Higher input impedance
b) Better stabilized voltage gain
c) Lowered frequency response
d) Higher input impedance & Better stabilized voltage gain
Answer: d
Explanation: The applied negative feedback can improve its performance and reduces sensitivity to parameter variations due to manufacturing or environment. Because of these advantages, many amplifiers and control systems use negative feedback.
