Analog Circuits GATE

Institute of Engineering Studies (IES,Bangalore) Analog Electronics Old GATE ECE

1 Institute of Engineering Studies,2nd

floor, 22nd

main, Marenahalli, JP Nagar,2nd

Phase,Bangalore-560 078. www.InstituteofEngineeringStudies.com

Blog: www.OnlineIES.blogspot.com Email: [email protected]

#97419 00225/080 3249 1693.

GATE-1999 One Mark Questions

1. The first dominant pole encountered in the frequency response of a

compensated op-amp is approximately at

a. 5 Hz b. 10 kHz

c. 1 MHz d. 100 MHz

2. Negative feedback in an amplifier

a. reduce gain

b. increase frequency and phase distortions

c. reduces bandwidth

d. increases noise

3. In the cascade amplifier shown in the figure, if the common-emitter stage (Q1)

has a transconductange gm1, and the common base stage (Q2) has a

transcodnuctance gm2 then the overall transconductance g(=i0/vi) of the cascade

amplifier is

Q

i

iV

R

2Q

1

←

V

o

o

L

a. gm1

b. gm2

c. gm1/2



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d. gm2/2

4. Crossover distortion behaviour is characteristic of

a. Class A output stage b. Class B output stage

c. Class AB output stage d. Common-base output stage

GATE-1999 Two Marks Questions

5. An npn transistor (with C = 0.3 pF) has a unit-gain cutoff frequency fT of 400

MHz at a dc bias current Ic = 1mA. The value of its Cµ (in pF) is approximately

(VT = 26 mV)

a. 15 b. 30

c. 50 d. 96

6. An amplifier has an open-loop gain of 100, an input impedance of 1kΩ and an

output impedance of 100Ω. A feedback network with a feedback factor of 0.99 is

connected to the amplifier in a voltage series feedback mode. The new input and

output impedances, respectively are.

a. 10 Ω and 1 Ω b. 10 Ω and 10 Ω

c. 100 k Ω and 1 Ω d. 100 kΩ and 1 k Ω

7. A dc power supply has a no-load voltage of 30 V, and a full-load voltage of 25

V at a full-load current of 1A. Its output resistance and load regulation,

respectively are

a. 5 Ω and 20% b. 25 Ω and 20%

c. 5 Ω and 16.7% d. 25 Ω and 16.7%



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8. An amplifier is assumed to have a single-pole high-frequency transfer function.

The rise time of its output response to a step function input is 35 nsec. The upper –

3 dB frequency (in MHz) for the amplifier to a sinusoidal input is approximately at

a. 4.55 b. 10

c. 20 d. 28.6


9. In the differential amplifier of the figure, if the source resistance of the current

source IEE is infinite, then the common-mode gain is

a. zero

b. infinite

c. indeterminate

d.

EE

EEE

-V

↓

V

RR

in1Vin2

I

10. In the circuit of the figure V0 is



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R

R

0

+15V

+ 1V

V

-15V

a. -1V b. 2V

c. + 1V d. + 15V

11. Introducing a resistor in the emitter of a common amplifier stabilizes the dc

operating point against variations in

a. only the temperature b. only the β of the transistor

c. both temperature and β d. none of the above

12. The current gain of a bipolar transistor drops at high frequencies because of

a. transistor capacitances

b. high current effects in the base

c. parasitic inductive elements

d. the Early effect

13. If the op-amp in the figure is ideal, the v0 is



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CSin

C

C

t

Sin t

a. zero b. (V1-V2) sinωt

c. – (V1 + V2 ) sinωt d. (V1 + V2) sinωt

14. The configuration of the figure is

R

C

CR

a. precision integrator b. Hartely oscillator

c. Butterworth high pass filter d. Wien-bridge oscillator

15. Assume that the op-amp of the figure is ideal. If Vi is a triangular wave then v0

will be



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R

C

a. square wave b. triangular wave

c. parabolic wave d. sine wave

16. The most commonly used amplifier in sample and hold circuits is

a. a unity gain inverting amplifier

b. a unity gain non-inverting amplifier

c. an inverting amplifier with a gain of 10

d. an inverting amplifier with a gain of 100


17. In the circuit of the figure, assume that the transistor is in the active region. It

has a large β and its base emitter voltage is 0.7V. the value of Ic is

I

R

↓10K

5K

15V

c

c

430

a. Indeterminate since Rc is not given



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b. 1mA

c. 5 mA

d. 10 mA

18. If the op-amp in the figure has an input offset voltage of 5mV and an open-

loop voltage gain of 10,000 then v0 will be

+15V

-15V

a. 0V b. 5mV

c. +15V or – 15V d. +50V or – 50V


19. The current gain of a BJT is

a. gmro b. gm/r0

c. gmrπ d. gm/rπ

20. The ideal OP-AMP has the following characteristics.

a. Ri = ∞, A= ∞, R0 = 0 b. Ri = 0, A= ∞, R0 = 0

c. R1 = ∞, A= ∞, R0 = ∞ d. Ri = 0, A= ∞, R0 = ∞

21. Consider the following two statements:

Statement 1:

A stable multivibrator can be used for generating square wave.



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Statement 2:

Bistable multivibrator can be used for storing binary information.

a. Only statement 1 is correct

b. Only statement 2 is correct

c. Both the statements 1 and 2 are correct

d. Both the statements 1 and 2 are incorrect


22. An npn BJT has gm = 38 mA/V, Cµ = 10-14

F, Cπ = 10-13

F, and DC current gain

β0 = 0-. For this transistor fT and fβ are

a. fT = 1.64 x 108 Hz and fβ = 1.47 x 10

10 Hz

b. fT = 1.47 x 1010

Hz and fβ = 1.64 x 108 Hz

c. fT = 1.33 x 1012

Hz and fβ = 1.47 x 1010

Hz

d. fT = 1.47 x 1010

Hz and fβ = 1.33 x 1012

Hz

23. The transistor shunt regulator shown in the figure has a regulated output

voltage of 10V, when the input varies from 20V to 30V. the relevant parameters

for the zener diode and the transistor are: Vz = 9.5, VBE = 0.3V β = 99. neglect the

current through RB. then the maximum power dissipated in the zener diode (Pz)

and the transistor (PT) are

↓→

V

↓

V

20

I

20-30 V

I

R

=10V

B -

z

c

BE

z

+

a. Pz = 75 mW, PT = 7.9 W



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b. Pz = 85 mW, PT = 8.9 W

c. Pz = 95 mW, PT = 9.9 W

d. Pz = 115 mW, PT = 11.9 W

24. The oscillator circuit shown in the figure is

Cc

L

-

C

c

1=2pF

e

2

L

C C

H=10

=2pF

a. Hartley oscillator with foscillation = 79.6 MHz

b. Colpitts oscillator with foscillation = 50.3 MHz

c. Hartley oscillator with foscillation = 159.2 MHz

d. Colpitts oscillator with foscillation = 159.2 MHz

25. The inverting OP-AMP shwn in the figure has an open-loop gain of 100. the

closed-loop gain V0/Vs is



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+

=1K

=10K

-

a. – 8 b. -9

c. -10 d. -11

26. In the figure assume the OP-AMPs to be ideal. The output v0 of the circuit is:

+-

10mH

10010

=10cos (100t)

10 F

3

1

2

+-=10cos (100t)

10

10mH

100

10

1

2 3

F

a. 10 cos (100t) b.

c. d.

GATE-2002 One Mark questions

27. In a negative feedback amplifier using voltage series (i.e. voltage smapling,

series mixing) feedback,



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a. Ri decreases and R0 decreases

b. Ri decreases and R0 increases

c. Ri increases and R0 decreases

d. Ri increases and R0 increases

(Ri and R0 denote the input and output resistance respectively)

28. A 741-type opamp has a gain-bandwidth product of 1 MHz. A non inverting

amplifier using this opamp and having a voltage gain of 20dB will exhibit a 3-dB

bandwidth of

a. 50 KHz b. 100 KHz

c. 1000/17 KHz d/ 1000/7.07 KHz

29. Three identical RC-coupled transistor amplifiers are cascaded. If each of the

amplifiers has a frequency responses as shown in the figure, the overall frequency

response is as given in



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30. An amplifier using an opam with a slew-rate SR=1 V/µsec has a gain of 40dB.

if this amplifier has to faithfully amplifiy sinusoidal signals from dc to 20 KHz

without introducing any slew-rate induced distortion, then the input signal level

must not exceed.

a. 795 mV b. 395 mV

c. 79.5 mV d. 39.5 mV



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31. The circuit in the figure employs positive feedback and is intended to generate

sinusoidal osicallation. If at a frequency f0 B(f)= then to sustain

oscillation at this frequency.

B(f)Network

(f)

↑

↓

(f)

- -+

+

a. R2 = 5R1 b. R2 = 6R1

c. R2 = R1/6 d. R2 = R1/5

32. A Zener diode regulator in the figure is to be designed to meet the

specifications: IL = 10mA, V0 = 10V and Vin varires from 30 V to 50V. The zener

diode has Vz = 10V and Izk (knee current)= 1mA For satisfactory operation

↓↓

I

↑

↓

↑

↓

Dz

z

IL=10mA

+

-

a. R ≤ 1800 Ω b. 2000 Ω R ≤ 22000 Ω

c. 3700 Ω ≤ R ≤ 4000 Ω d. R > 4000 Ω

33. The voltage gain Av = v0/vt of the JFET amplifier shown in the figure is



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+

(1MRsG

= +10VV

↓

+

-

I

DD

R

D=1mA

1

2

R

C

C

s

)

CD

)(3K

)

VDD

= +10V

↓ ID=1mA

RD

(3K )

C

C

C

1

2

s

+

+

-

R RG s

(1M ) (2.5K )

→

(2.5K

→

IDss = 10 mA

(Assume C1, C2 and Cs to be very large)

a. +18 b. -18

c. + 6 d. -6

34. Consider the following statements in connection with the CMOS inverter in

the figure, where both the MOSFETs are of enhancement type and both have a

thresh old voltage of 2V.

Statement 1: T1 conducts when V1 ≥ 2V.

Statement 2: T1 is always in saturation when V0 = 0V.

→

←

+

T

1T

2

5V

Which of the following is correct?



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a. Only statement 1 is TRUE

b. Only Statement 2 is TRUE

c. Both the statements are TRUE

d. Both the Statements are FALSE.


35. Choose the correct match for input resistance of various amplifier

configuration shown below.

Configuration Input resistance

CB: Common Base LO: Low

CC: Common Collector MO: Moderate

CE: Common Emitter HI: High

a. CB-LO, CC-MO, CE-HI b. CB-LO, CC-HI, CE-MO

c. CB-MO, CC-HI, CE-LO d. CB-HI, CC-LO, CE-MO

36. The circuit shown in the figure is best described as a

~ Output

a. bridge rectifier b. ring modulator

c. frequency discriminatory d. voltage doubler



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37. If the input to the ideal comparator shown in the figure is a sinusoidal signal of

8V (peak to peak) without any DC component, then the output of the comparator

has a duty cycle of

=2VVref

InputOutput

a. 1/2 b. 1/3

c. 1/6 d. 1/12

38. If the differential voltage gain and the common mode voltage gain of a

differential amplifier are 48 dB and 2 dB respectively, then its common mode

rejection ratio is

a. 23 dB b. 25 dB

c. 46 dB d. 50 dB

39. Generally, the gain of a transistor amplifier falls at high frequencies due to the

a. internal capacitances of the device

b. coupling capacitor at the input

c. skin effect

d. coupling capacitor at the output


40. An amplifier without feedback has a voltage gain of 50, input resistance of

1KΩ and output resistance of 2.5 KΩ. The input resistance of the current-shunt

negative feedback amplifier using the above amplifier with a feedback factor of

0.2 is

a. 1/11KΩ b. 1/5 KΩ

c. 5 KΩ d. 11K Ω



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41. In the amplifier circuit shown in the figure, the values of R1 and R2 are such

that the transistor is operating at VCE = 3V and IC = 1.5mA when its β is 150. for a

transistor with β of 200, the operating point (VCE, IC) is

a. (2V, 2mA) b. (3V, 2mA)

c. (4V, 2mA) d. (4V, 1mA)

42. The oscillator circuit shown in the figure has an ideal inverting amplifier. Its

frequency of oscillation (in Hz) is

C C C

RR R

a. b.

c. d.

43. The output voltage of the regulated power supply shown in the figure is



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40K

Power source

15V DC

1K

+

Regulated

-

V

20K

z

Unregulated

DC Output

=3V

a. 3V b. 6V

c. 9V d. 12 V

44. The action of a JFET in its equivalent circuit can best be represented as a

a. Current Controlled Current Source

b. Current Controlled Voltage Source

c. Voltage Controlled Voltage Source

d. Voltage Controlled Current Source

45. If the op-amp in the figure is ideal, the output voltage Vout will be equal to



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1K3V

5K

1K

8K

2V

a. 1V b. 6V

c. 14 V d. 17V

46. Three identical amplifiers with each one having a voltage gain of 50, input

resistance of 1KΩ and output resistance of 250 Ω, are cascaded. The open circuit

voltage gain of the combined amplifier is

a. 49 dB b. 51 dB

c. 98 dB d. 102 dB

47. An ideal saw tooth voltage waveform of frequency 500 Hz and amplifier 3 V

is generated by charging a capacitor of 2µ F in every cycle. The charging requires

a. constant voltage source of 3 V for 1 ms

b. constant voltage source of 3 V for 2 ms

c. constant current source of mA for 1 ms

d. constant current source of 3mA for 2 ms

GATE- 2004 One Mark Questions

48. An ideal op-amp is an ideal

a. voltage controlled current source

b. voltage controlled voltage source

c. current controlled current source



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d. current controlled voltage source

49. Voltage series feedback (also called series-shunt feedback) results in

a. increase in both input and output impedances

b. decrease in both input and output impedances

c. increase in input impedance and decrease in output impedance

d. decrease in input impedance and increase in output impedance

50. The circuit in the figure is a

5K

RR

a. low-pass filter b. high-pass filter

c. band-pass filter d. band-reject filter

51. Assuming VCEsat= 0.2V and β = 50, the minimum base current (IB) required to

drive the transistor in the figure to saturation is

3V

→BI

↓1K

Ic

a. 56 µA b.140 mA

c. 60 mA d. 3 mA



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52. A bipolar transistor is operating in the active region with a collector current of

1 MA. Assuming that the β of the transistor is 100 and the thermal voltage (VT) is

25 mV, the transconductance (gm) and the input resistance (rπ) of the transistor in

the common emitter configuration are

a. gm = 25mA/V and rπ = 15.625kΩ

b. gm = 40/V and rπ = 4.0kΩ

c. gm = 25mA/V and rπ = 2.5kΩ

d. gm = 40 mA/V and rπ = 2.5kΩ

53. The value of C required for sinusoidal oscillations of frequency 1 kHz in the

circuit of the figure is

C

C

1K 2.1K

1K

1K

a. b. 2πµ F

c. d. 2 π √6 µ F

54. In the op-amp circuit given in the figure the load current iL is



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Li

↓

a. b.

c. d.

55. In the voltage regulator shown in the figure the load current can vary from

100mA to 500 mA. Assuming that the zener diode is ideal (i.e. the Zener knee

current is negligibly small and zener resistance is zero in the breakdown region),

the value of R is

-

+

5VVariable Load

100 to 500 mA

12V

R

a. 7Ω b. 70 Ω

c. 70/3 Ω d. 14 Ω



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56. In a full-wave rectifier using two ideal diodes Vdc and Vm are the dc and peak

values of the voltage respectively across a resistive load. If PIV is the peak inverse

voltage of the diode, then the appropriate relationships for the rectifier are

a. b.

c. d.


57. The effect of current shunt feedback in an amplifier is to

a. increase the input resistance and decrease the output resistance

b. increase both input and output resistances

c. decrease both input and output resistance

d. decrease the input resistance and increase the output resistance

58. the input resistance of Ri of the amplifier shown in the figure is

a. 30/4 kΩ b. 10 K kΩ

d. 40 kΩ d. infinite

59. The cascade amplifier is a multistage configuration of

a. CC-CB b. CE-CB



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c. CB-CC d. CE-CC

GATE- 2005 Two Marks Questions

60. For an npn transistor connected as shown in the figure VBE = 0.7 volts. Given

that reverse saturation current of the junction at room temperature 3000

K is 10-13

A, the emitter current is

↑

I↓ c

VBE

a. 30 mA b. 39 mA

c. 49 mA d. 20 mA

61. The voltage e0 indicated in the figure has been measured by an ideal voltmeter.

Which of the following can be calculated?

1 M

e0

1 M

a. Bias current of the inverting input only

b. Bias current of the inverting and non-inverting inputs only

c. Input offset current only

d. Both the bias-current and the input offset current



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62. The OP-amp circuit shown in the figure is a filter. The type of filter and its

cut-off frequency are respectively

1K

10 K

F

10 K

1

a. high pass, 1000 rad/sec. b. low pass, 1000 rad/sec.

c. high pass, 10000 rad/sec. d. low pass, 10000 rad/sec.

63. In an ideal differential amplifier shown in the figure, a large value of (RE)

EE-V

a. increases both the differential and common mode gains.

b. increase the common-mode gain only.

c. decrease the differential-mode gain only

d. decrease the common-mode gain only



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64. For an n-channel MOSFET and its transfer curve shown in the figure, the

threshold voltage is

D

characteristics

↑

1V

||

GS

D

→

I

V

=

→V 3V

G

V

S

D

G

= 5V

Transfer

= 1VVs

a. 1V and the device is in active region

b. -1V and the device is in saturation region

c. 1 V and the device is in saturation region

d. -1 V and the device is in active region

65. The circuit using a BJT with β = 50 and VBE = 0.7V is shown in the figure.

The base current IB and collector voltage Vc are respectively

10

40

2k

20 V

1K

430K

F

F

a. 43 µA and 11.4 Volts b. 40µA and 16 Volts

c. 45µA and 11 Volts d. 50µA and 10 Volts



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66. The zener diode in the regulator circuit shown in the figure has a Zener voltage

of 5.8 volts and a Zener knee current of 0.5mA. the maximum load current drawn

from this circuit ensuring proper functioning over the input voltage range between

20 and 30 volts, is

↑

20-30=5.8V

1k

Load

a. 23.7 mA b. 14.2 mA

c. 13.7 mA d. 24.2 mA

67. Given the ideal operational amplifier circuit shown in the figure indicate the

correct transfer characteristics assuming ideal diodes with zero cut-in voltage.

0.5K

-10V2K

+10V

2K



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→

→

↑

→←

-5V-8V

+10V(a)

←-10V

→

→

↑

→←

-5V +8V

+10V(b)

←-10V

→

→

↑

→←

-5V +5V

+5V(c)

←-10V

→

→

↑

→←

-5V +5V

+10V(d)

←-5V

Common Data Questions 68, 69, 70

Given rd = 20 kΩ, IDSS = 10 mA, VP = - 8V



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68. Zi and Z0 of the circuit are respectively

a. 2M Ω and 2k Ω b. 2M Ω and 20/11 k Ω

c. infinity and 2 M Ω d. infinity and 20/11 k Ω

69. ID and VDS under DC conditions respectively

a. 5.625mA and 8.75 V b. 7500 mA and 5.00V

c. 4.500 mA and 11.00 V d. 6.250 mA and 7.50V

70. Transconductance in milli-Siemens (ms) and voltage gain of the amplifier are

respectively

a. 1.875 ms and 3.41 b. 1.875 ms and -3.41

c. 3.3 mS and -6 d. 3.3 mS and 6

GATE- 2006 One Mark Questions

71. The input impedance (Zi) and the output impedance (Z0) of an ideal

transconductance (voltage controlled current source) amplifier are

a. Zi = 0, Z0 = 0 b. Zi = 0, Z0 = ∞

c. Zi =∞, Z0 = 0 d. Zi =∞, Z0 = ∞



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72. An n-channel depletion MOSFET has following two points on its ID – VGS

curve.

(i) VGS = 0 at ID = 12 mA and

(ii) VGS = - 6 Volts at Z0 = ∞

Which of the following Q-points will give the highest trans-conductance gain for

small signals?

a. VGS = -6 Volts b. VGS = - 3 Volts

c. VGS = 0 Volts d. VGS = 3 Volts


73. For the circuit shown in the following figure, the capacitor C is initially

uncharged. At t = 0, the switch S is closed. The voltage VC across the capacitor at t

= 1 millisecond is

1K

10V

FC=1

+

S

-

In the figure shown above, the OP-AMP is supplied with ± 15 V.

a. 0 Volt b. 6.3 Volts

c. 9.45 Volts d. 10 Volts

74. For the circuit shown below, assume that the zener diode is ideal with a break

down voltage of volts. The waveform observed across R is



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~ R12sin t

+

6V

RV

-

(b)6V

(a)

(c)

-12V

(d)

12V

-6V

-6V

Common Data for Questions 75, 76 & 77

In the transistor amplifier circuit shown in the figure below, the transistor has the

following parameters.

βDC = 60, VBE = 0.7, hie →∞, hfe →∞

The capacitance Cc can be assumed to be infinite.

5.3K

53K

1K

C cc

V

-

12v

+

~

In the figure above, the ground has been shown by the symbol ▼



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75. Under the DC conditions, the collect to-emitter voltage drop is

a. 4.8 Volts b. 5.3 volts

c. 6.0 volts d. 6.6 volts

76. If βDC is increased by 10% the collector to emitter voltage drop

a. increases by less than or equal to 10%

b. decreases by less than or equal to 10%

c. increases by more than 10%

d. decreases by more than 10%

77. the small-signal gain of the amplifier vc/vs is

a. -10 b. – 5.3

c. 5.3 d. 10

Common Data for Questions 78 & 79.

\a regulated power supply shown in figure below, has an unregulated input (UR)

of 15 volts and generates a regulated output Vout. Use the component values shown

in the figure

6V

1K12

Q1

24

15V (UR)

+

-

12

In the figure above, the ground has been shown by the symbol of ▼



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78. The power dissipation across the transistor Q1 shown in the figure is

a. 4.8 Watts b. 5.0 Watts

c. 5.4 Watts d. 6.0 Watts

79. If the unregulated voltage increases by 20% the power dissipation across the

transistor Q1

a. increases by 20% b. increases by 50%

c. remains unchanged d. decreases by 20%


80. The correct full wave rectifier circuit is

81. In a transconductance amplifier it is desirable to have

a. a large input resistance and a large output resistance

b. a large input resistance and a small output resistance

c. a small input resistance and a large output resistance

d. a small input resistance and a small output resistance

GATE- 2007 Two Marks Questions

82. The DC current gain (β) of a BJT is 50. Assuming that the emitter injection

efficiency is 0.995, the base transport factor is



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a. 0.980 b. 0.985

c. 0.990 d. 0.995

83. For the Op-Amp circuit shown in the figure, V0 is

1 V

1K

1K

2K

1K

a. -2V b. -1V

c. -0.5 V d. 0.5V

84. For the BJT circuit shown, assume that the β of the transistor is very large and

VBE = 0.7 V. The mode of operation of the BJT is

2V

10V

+-

+-

1K

10K

a. cut-off b. saturation

c. normal active d. reverse active



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85. In the OP-Amp circuit shown, assume that the diode current follow the

equation I=Is exp (V/VT). for Vi = 2V, V0 = v01, and for Vi = 4 V, V0 = V02. the

relationship between V01 and V02 is

2K

a. V02 = √2 V01 b. V02 = e2 V01

c. V02 = V01 In 2 d. V01 – V02 = VT In 2

86. In the CMOS inverter circuit shown, if the transconductance parameters of the

NMOS and PMOS transistors are kn = kp = µn Cox Wn/Ln = µn Cox Wp/Lp = 40 µ

A/V2 and their threshold voltages are VTHn = |VTHp| = 1V, the current is

5 V

PMOS

2.5 VI

NMOS

↓

a. 0A b. 25µA

c. 45 µA d. 90 µA

87. 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 10 Ω. If the input

voltage (Vi) range is from 10 to 16 V, the output voltage (V0) range from



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200

a. 7.00 to 7.29 V b. 7.14 to 7. 29V

c. 7.14 to 7. 43 V d. 7. 29 to 7. 43 V

Common Data Questions 88, 89, 90

The figure shows the high-frequency capacitance voltage (C-V) characteristics of

a Metal/SiO2/silicon (MOS) capacitor having an area of 1 x 10-4

cm2. assume that

the permitivities (ε0 εr) of silicon and SiO2 are 1 x 10-12

F/cm and 3.5 x 10-13

F/cm

respectively.

88. The gate oxide thickness in the MOS capacitor is

a. 50 mm b. 143 mm

c. 350 mm d. 1µm

89. The maximum depletion layer width in silicon is

a. 0.143 µm b. 0.857 µm

c. 1 µm d. 1.143 µm

90. Consider the following statements about the C-V characteristics plot;



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S1: The MOS capacitor has as n-type substrate.

S2: If positive charges are introduced in the oxide, the C-V plot will shift to the

left.

Then which of the following is true?

a. Both S1 and S2 are true b. S1 is true and S2 is false

c. S1 is false and S2 is true d. Both S1 and S2 are false

Statement for Linked Answer Questions 91 & 92.

Consider the Op-Amp circuit shown in the figure.

C

R

91. The transfer function V0 (s)/Vi(s) is

a. b.

c. d.

92. If Vi = V1 sin (ωt) and V0 = V2 sin (ωt + φ), then the minimum and maximum

values of φ (in radians) are respectively

a. – π/2 and π/2 b. 0 and π/2

c. – π and 0 d. – π/2 and 0



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93. In the following limiter circuit, an input voltage Vi = 10 sin 100 πt is applied.

Assume that the diode drop is 0.7V when it is forward biased. The zener

breakdown voltage is 6.8 V.

D11K

Z 6.8V

D2

The maximum and minimum values of the output voltage respectively are

a. 6.1 V, - 0.7 V b. 0.7 V, -7.5 V

c. 7.5 V, - 0.7 V d. 7.5 V, - 7.5 V


94. Consider the following circuit using an ideal OPAMP. The I-V characteristics

of the diode is described by the relation

Where VT = 25m V, I0 = 1µA and V is the voltage across the diode (taken as

positive for forward bias).

=-1V

100 K

D 4 K

For an input voltage Vi = -1V, the output voltage V0 is

a. 0 V b. 0.1 V

c. 0.7 V d. 1.1 V



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95. The OPAMP circuit shown above represents a

C

L

a. high pass filter b. low pass filter

c. band pass filter d. band reject filter

96. Two identical NMOS transistors M1 and M2 are connected as shown below.

Vbias is chosen so that both transistors are in saturation. The equivalent gm of the

pair is defined to be at constant Vout.

The equivalent gm of the pair is

a. the sum of individual gm’s of the transistors

b. the product of individual gm’s of the transistors

c. nearly equal to the gm of M1

d. nearly equal to gm/g0 of M2

97. Consider the Schmidt trigger circuit shown below



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1

2

Iout

M

←

Mbias

V

A triangular wave which goes from – 12V to 12V is applied to the inverting input

of the OPAMP. Assume that the output of the OPAMP swings from +15V to -

15V. the voltage at the non-inverting input switches between

a. -12V and +12v b. -7.5 and +7.5 V

c. -5V and + 5V d. OV and 5 V

Statement for linked Answer Questions 98 and 99.

In the following transistor circuit, VBE = 0.7 V, re = 25 mV/IE and β and all

=9V

C

c2

20K

C

C10K

3K

E

2.3KE

c1

3K

I↓

98. The value of DC current IE is

a. 1mA b. 2 mA



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c. 5 mA d. 10 mA

GATE-2009 Two marks Questions

100. In the circuit below, the diode is ideal. The voltage V is given by

+ V-

↓

1

-+

1

1A

a. min (Vi 1) b. max (Vi 1)

c. min (-Vi 1) d. max (-Vi 1)

101. A small signal source vi (t) = A cos 20t + B sin 106t is applied to a transistor

amplifier as shown below. The transistor has β = 150 and hie = 3KΩ. Which

expression best approximates v0 (t)?

(t)

100 K 3 K

12V

100 nF

900K10

(t)

20 K F

a. v0 (t) = -1500 (A cos 20t + B sin 106t)

b. v0 (t) = -150 (A cos 20t +B sin 106t)

c. v0 (t) = -1500 B sin 106t

d. v0 (t) = -150 B sin 106t

Common Data for Questions 102 and 103.



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Consider a silicon p-n junction at room temperature having the following

parameters:

Doping on the n-side = 1 x 1017

cm-3

Depletion width on the n-side = 0.1 µm

Depletion width one the p-side = 1.0 µm

Intrinsic carrier concentration = 1.4 x 1010

cm-3

Thermal voltage =26 mV

Permittivity of free space = 8.85 x 10-14

F.cm-1

Dielectric constant of silicon = 12

102. The built-in potential of the junction

a. is 0.70 V

b. is 0.76 V

c. is 0.82 V

d. cannot be estimated from the data given

103. The peak electric filed in the device is

a. 0.15 MV.cm-1

, directed from p-region to n-region

b. 0.15MV. cm-1

, directed from n-region to p-region

c. 1.80 MV.cm-1

directed from p-region to n-region

d. 1.80 MV. cm-1

directed from n-region to p-region

Statement for Linked Answer Question 104 and 105.

Consider the CMOS circuit shown, where the gate voltage VG of the n-MOSFET

is increased from zero, while the gate voltage of the p-MOSFET is kept constant at

3V. Assume that for both transistors, the magnitude of the threshold voltage is 1 V



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and the product of the transconductance parameter and the (W/L) ratio, i.e. the

quantity µCox(W/L). is 1mA. V-2

.

←

V

←

→G

3V

5V

104. For small increase in VG beyond 1V, which of the following gives the correct

description of the region of operation of each MOSFET?

a. Both the MOSFETs are in saturation region

b. Both the MOSFETs are in triode region

c. n-MOSFET is in triode and p-MOSFET is in saturation region

n-MOSFET is in saturation and p-MOSFET is in triode region

105. Estimate the output voltage V0 for VG = 1.5 V, [Hints: use the appropriate

current-voltage equation for each MOSFET, based on the answer to Q.57]

a. b.

c. d.



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ANSWERS & EXPLANATIONS

1. (a)

2. (a)

3. (a)

IC1 = βIB, IE2 = IC1

io ≈ βIB, Vi = IB, rπ

4. (b)

5. (a)

6. (c)

A = 100,

B= 0.99

1 + AB = 100

For voltage series Ri ↑ & R0 ↓ by 1 + AB

-- Ri = 100 x 1K = 100 K



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7. (b)

Regulation

O/P Resistance =

8. (b)

tr x B.ω = 0.35

B.ω =

9. (a)

Common mode gain,

VC = AC Vi (Vi1 = Vi2 = Vi)

If Re is infinite then because of symmetry of fig., Vc becomes zero.

ie1 = ie2 = 0

ib2 < < ic2

So ic2 ~ ie2

10. (d)

In positive feedback op-amp act in its satuation region ± Vsat. Here applied voltage

is positive.

V0 = + Vsat = + 15 V

11. (c)



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12. (a)

At low frequency, Ai = - hfe and Ai decreases as frequency increases.

13. (c)

Here

14. (d)

15. (a)

This circuit acts as a differentiator and differentiation of triangular wave gives

square wave.

16. (b)



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ControlGate

C

17. (d)

Vth

R

R

430

10/3

c

th

15V

5V

Since β is large IB ≈ 0, Rth = 5 || 10

18. (c)

V00 = Vi0.A = 5 mV x 10,000 = 50 V

But V00 = ± 15V,

V00 can never be greater than ± Vsat

19. (c)



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hfe = gm .rπ

22. (b)

(β0 = hfe)

23. (C)

V

I

I

I

I1

I

→

→↓z

↓

C

E-

+20-30V

BBE

20

B

↓

R

=10V

(i.e. when Iz = 0)

IE = IC +Iz

IB = Iz (as no current flows in RB)

From (i) IE = βIz + Iz = (99 +1) Iz

IE = 100 Iz



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I1 = IE = 100 Iz

Iz =

Pz = VzIz = 9.5 x 0.01 = 95mW

Ic = 99Iz = 99 x 0.01 = 0.99A

Pc = VCIC = 10 x 0.99 = 9.9 W

24. (b)

Fig shown is Colpitts oscillator.

25. (b)

Avf = - 9

26. (a)



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+-

10mH

10010

=10cos (100t)

10 F

3

1

2

+-=10cos (100t)

10

10mH

100

10

1

2 3

F

KCL at node 1,

V0 = -10V2 = -10(-cos 100t)

V0 = 10cos 100t

27. (c)

Ri increases by factor of 1 + Aβ and R0 decreases by 1+ Aβ.

28. (b)

Gain X B. ω = 1x106

20 log x = 20 dB

X = 10

29. (a)

fL = 20 Hz fH = 1 KHz for single stage.



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For cascaded stage

30. (c)

Slew rate = A. 2π fVm

V = A.Vm sin ωt

20 log X = 40

X = 100 = A

Vm = 79. 5 mV

31. (a)

1

KCL at node 1



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32. (a)

Z

1→

L

↓

z I

LR

↓↓+

LI↑

I

-

(Iz + IL = I1)

When Vin = 30 V

When Vin = 50 V



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R ≤ 3636

33. (d)

VG = 0, Vs = ID .Rs = 1 mA x 2.5 K = 2.5 V

VGS = VG – Vs = - 2.5 V

AV = -gm RD ( because rd is not given, it is taken as ∞).

= - 2ms x 3K = -6

34. (c)

T1 is N-MOSFET which conduct when Vi > Vth When V0 = 0, CMOS inverter has

I/P = 1 i.e. 5 V So T1 is in saturation and conducts.

35. (b)

36. (d)

37. (b)

8Vp-p

So, Vi = 4 sin ω t



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At Vi = 2

Another crossover at

Therefore Duty cycle =

38. (c)

39. (a)

A= 50, β = 0.2

D =1 + A β = 1 + 50 x 0.2 = 11

Current shunt: R0 increases & R1 decreases by D.

41. (a)

VCE = VCC – IC R2

3 = 6 – 1.5 mA x R2

1.5 mA x R2 = 3

R2 = 2 KΩ



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When β = 200,

IC = βIB (as R1 is same IB remains same)

= 0.01 mA x 200

IC = 2 mA

VCE = VCC – IC R2

= 6-2 mA x 2 kΩ

VCE = 2V

42. (a)

Frequency of oscillation for RC phase shift oscillator is

43. (c)

As volt at non inverting terminal is 3V due to zener diode, voltage at inverting

terminal will be 3V because of virtual ground.

So current in 20K is

44. (d)

45. (b)

V0 = 6V

46. (c)



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4

250 250

1K -

1K+

-+

50

+ -

1K V

50 50

250

Volt. Across 1 K after 1st stage =

Similarly

Therefore AV = 40 x 40 x 50 = 8 x 104

AV in dB = 20 log (8 x 104) = 98 dB

47. (d)

I = 3mA

50. (a)

At ω = ∞, & at ω = 0

51. (a)



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VCE = VCC -ICRC

0.2 = 3-IC x 1 K

IC = 2.8 mA

52. (d)

hfe = gm. rπ, hfe = β

53. (a)

←

C

1K

↓↓

2.1K

1K

1K

C

R= 1K



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For oscillation imaginary part is zero.

i.e.

ω2 C

2 R

2 – 1 = 0

54. (a)

1K

↓i

V

L



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…..(i)

….. (ii)

Putting V0 from (1)

55. (d)

When IL = 100 mA,

When IL = 500mA,

Therefore R= 14 Ω (choosing minimum one)

56. (b)

57. (d)



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58. (b)

59 (b)

60. (c)

When two terminals of a transistor are shorted it acts as diode.

61. (c)

B1 1M

↓

I

I

1M

B2

↑

0

+-

e

V1 = - IB1 x 1M, V2 = V1 = - IB1 x 1M (due to virtual ground)

Drop in feedback resistor 1M = IB2 x 1M

e0 = V2 + IB2 x 1M

e0 = - IB1 x 1M + IB2 x 1M

e0 = (IB2 – IB1) x 1M

where (IB2- IB1) is offset current

62. (a)

Since O/P is taken across 10K it is a high pass filter. I/P is at non-inverting point.

So,



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63. (d)

Only common mode gain depends on RE and differential mode gain is independent

of RE.

64. (c)

From the graph its clear that Vth = 1V

VGs = 3-1 = 2V

VDS = 5-1= 4V

Since VDs ≥ VGS – VT S. MOSFET is in saturation region.

65. (b)

IE = IC + IB = βIB +IB = (β+ 1)IB

KVL in I/P loop gives,

VCC – VBE = IBRB + IE RE = IB RB + (β+1) IB RE

IB = 40 µA

IC = β = 50 x 40 µA = 2000 µA = 2 mA

VC = VCC- ICRC = 20-2 mA X 2K

VC = 16 V

66. (a)

→ →

20-30V5.8V

I

IL

↓

I1K

Load

1

z



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VL= ‘5.8 V

Maximum load current will be when V1 = Vmax

24.2mA = IL + Iz

IL = 24.2 mA – 0.5 mA = 23.7 mA

67. (b)

Vut = βu Vsat

When lower diode is ON,

Vιt = - βι Vsat (when upper diode is ON, βι =

Vut & Vιt are upper and lower transition voltage.

68. (b)

Zin = 2 MΩ

Z0 = rd || RD = 20 K || 2 K

69. (a)



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ID = 10 mA x

VDS = VDD – ID RD = 20- 5.625 mA x 2 K

VDS = 8. 75 V

70. (b)

Gm = 1.875 ms

AV = - gm (rd || RD) (gm Z0)

AV = - 3.41

72. (d)

From the graph it is clear that as VGS increase conductance i.e. slope of graph

increase.

)GS

0

D

=

→

↑ I

V-6V

12mA

(Z

Transfer character of n-channel D-MOSFET

73. (d)



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-

10V

+

10V

1

F

1K

1

Applying KCL at node 1

74. (b)

Zener diode works as normal diode in FB. So, when Vin < 0, VR = Vin When 0 <

Vin < 6, Diode is OFF and VR = 0.

When Vin > 6, Diode conducts and VR = Vin

75. (c)

Applying KVL in base-emitter loop,

12-IERC – IB . Rf – 0.7 = 0

12-0.7 = IE . 1K +

VCE = VCC – IE . RC = 12-6 mA x 1 K = 6 V

76. (b)

When β increases by 10%, new β = 66



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VCE = VCC – IE . RC = 12- 6.31 mA x 1K = 5.7

% change in

77. (a)

78. (d)

Volt across 24K = 6V due to virtual ground concept. So volt across 12K is 3V.

Vout = V12K + V24K = 3+6

Vout = 9 V

VCE = 15- Vout = 15-9 = 6V

Therefore P= VCE . IC = 6 V x 1A

P= 6 Watts

79. (b)

6V

1K 12K

Q1

24K

15V

+

- 6v

12K (3v)

(6V)



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New unregulated voltage = 18V

Therefore VCE = 18-9 = 9 V

IC = 1A

Therefore P = 9 x 1 = 9 watts

80. (c)

81. (a)

82. (b)

PE

PP n→I

→nEI

-

PC1

-+ +

EB CBV

I

V

Transport factor

Current in emitter is both due to holes and electrons Neglect current due to

electrons



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83. (c)

1K

1k

1k

2K

1V

yV

X

1

X = 1 volt y = 0.5 (using voltage division rule)

84. (b)

Given β is large so IB = 0 & IE = IC

Assuming BJT is in active

Applying KVL in Base. Emitter loop

2 – 0.7 = 1 K Ω x IE

IE = 1.3 mA

Now ICsat =

As ICactive > ICsat

So BJT is in saturation

85. (d)

Applying KCL



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IR = ID

V0 = - VT

Now, V01 = - VT

V01 – V02 = VT / n 2

86. (d)

VGS for each MOS is 2.5V

VT = 1 volt device parameter K = 40 µ A/v2

So ID = K (VGS – VT)2

= 40

ID = 90 µA

87. (c)

VZ = 7 volt IK = 0, R z = 10 Ω

Range of Vi = 10 to 16V

Range of voltage across 200 Ω = Vi- Vz = 3 to 9 volt

Range of current through 200 Ω = 15 to 45 mA

Range of variation in output voltage



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= (15 to 45mA) x Rz = 0.15 to 0.45 Volt

So range of output voltage = 7 + (0.15 to 0.45 volt)

= 7.15 volt to 7. 45 volt

88. (a)

89. (c)

90. (d)

91. (a)

From the figure given in the question



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92. (c)

θ = - tan -1

ωRC – tan-1

ωRC

θ = - 2tan -1

ωRC

minimum value of θ = - π (at ω → ∞)

maximum value of θ = 0 (at ω = 0)

93. (c)

For the positive half of Vi

D1 is forward biased and Zener diode is in breakdown stage

V0 = 0.7 + 6.8 = 7.5 V

For the negative half of Vi D2 is forward biased.

V0 = - 0.7 V

94. (b)



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95. (b)

Which is equivalent to standard form of transfer function of low pass filter, i.e.

97. (c)

Le the voltage at the non-inverting input be V1

Applying KCL at non-inverting input end

15-V1 + V0 – V1 = V1 + 15

Since, V0 swings from – 15V to + 15 V,

Therefore V1 switches between – 5V & + 5V.

98. (a)

The given circuit can be redesigned as shown below



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↓

R

I

3K

E

Th

9V=

2.3KVTh

= 6.67 k

Since β is large IB can be ignored

99. (d)

Mid-band voltage gain,

= - 60

100. (a)

101. (b)

The best approx answer for output voltage v0 is

V0 = Ar. vi



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≈ - 150 (A cos 20t + B sin 106 )

Note: Magnitude of gain is taken by total approx.

102. (b)

N side is heavily doped

103. (*)

(directed from N to P side)

104. (d)

105. (*?)

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