Analog Circuits GATE

May 15, 2019 | Author: Prateek Khare | Category: Amplifier, Operational Amplifier, Bipolar Junction Transistor, Mosfet, Transistor
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Institute of Engineering Studies (IES,Bangalore)

Analog Electronics Old GATE ECE

GATE-1999 One Mark Questions

1. The first dominant pole encountered 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, fig ure, if the common-emitter stage (Q 1) has a transconductange g m1, and the common base stage (Q 2) has a transcodnuctance transcodnuctance gm2 then the overall transconductance g(=i 0 /vi) of the cascade amplifier is Q2

← io Vo RL

Vi

Q1

a. gm1 b. gm2 c. gm1 /2

1

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Analog Electronics Old GATE ECE

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 f T of 400 MHz at a dc bias current I c = 1mA. The value of its C µ  (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%

2

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Institute of Engineering Studies (IES,Bangalore)

Analog Electronics Old GATE ECE

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

GATE-2000 One Mark Questions

9. In the differential amplifier of the figure, if the source resistance of the current source IEE is infinite, then the common-mode common-mode gain is a. zero b. infinite c. indeterminate indeterminate d.

R

R

V in1

Vin2



IEE E -VEE

10. In the circuit of the figure V 0 is

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Analog Electronics Old GATE ECE

+15V

V0 + 1V R -15V

R

a. -1V

b. 2V

c. + 1V

d. + 15V

11. Introducing a resistor in the emitter of a common a mplifier 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 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 v 0 is

4

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Institute of Engineering Studies (IES,Bangalore)

Analog Electronics Old GATE ECE

C

Sin

t

Sin

t

C

C

a. zero

b. (V1-V2) sinωt

c. – (V1 + V2 ) sinωt

d. (V1 + V2) sinωt

14. The configuration of the figure is

R

C

R

C

a. precision integrator 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 V i is a triangular wave then v 0 will be

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Analog Electronics Old GATE ECE

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

GATE-2000 Two Marks Questions

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 I c is 15V

Rc 10K

↓I

c

5K 430

a. Indeterminate since R c is not given

6

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Analog Electronics Old GATE ECE

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 openloop voltage gain of 10,000 then v 0 will be +15V

-15V

a. 0V

b. 5mV

c. +15V or – 15V

d. +50V or – 50V

GATE-2001 One Mark Questions

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= ∞, R 0 = 0

b. Ri = 0, A= ∞, R 0 = 0

c. R1 = ∞, A= ∞, R 0 = ∞

d. Ri = 0, A= ∞, R 0 = ∞

21. Consider the following two statements: Statement 1: A stable multivibrator can be used for generating square wave.

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Analog Electronics Old GATE ECE

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

GATE-2001 Two Marks Questions -14

22. An npn BJT has g m = 38 mA/V, C µ  = 10

-13

F, Cπ = 10

F, and DC current gain

β0 = 0-. For this transistor f T and f β are 8

a. f T = 1.64 x 10 Hz and f β = 1.47 x 10

Hz

10

Hz and f β = 1.64 x 10 Hz

12

Hz and f β = 1.47 x 10

10

Hz and f β = 1.33 x 10

b. f T = 1.47 x 10 c. f T = 1.33 x 10

10

d. f T = 1.47 x 10

8

10

Hz

12

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

neglect the

current through R B. then the maximum power dissipated in the zener diode (P z) and the transistor (P T) are 20 → ↓I z



Ic

Vz 20-30 V

RB

=10V

+ VBE

-

a. Pz = 75 mW, P T = 7.9 W

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Analog Electronics Old GATE ECE

b. Pz = 85 mW, PT = 8.9 W c. Pz = 95 mW, P T = 9.9 W d. Pz = 115 mW, PT = 11.9 W

24. The oscillator circuit shown in the figure is Lc

L=10 H

Cc

C1 =2pF C2 =2pF

Ce

a. Hartley oscillator with f oscillation = 79.6 MHz b. Colpitts oscillator with f oscillation oscillation = 50.3 MHz c. Hartley oscillator with f oscillation = 159.2 MHz d. Colpitts oscillator with f oscillation oscillation = 159.2 MHz

25. The inverting OP-AMP shwn in the figure has an open-loop gain of 100. the closed-loop gain V 0 /Vs is

9

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Analog Electronics Old GATE ECE

=10K

=1K

+ -

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:

10

10mH

F

10 =10cos =10cos (100t) ( 100t)

1

100

+ 2

a. 10 cos (100t)

b.

c.

d.

3

GATE-2002 One Mark questions

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

10

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Analog Electronics Old GATE ECE

a. Ri decreases and R 0 decreases b. Ri decreases and R 0 increases c. Ri increases and R 0 decreases d. Ri increases and R 0 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 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

11

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Analog Electronics Old GATE ECE

GATE-2002 Two Marks Questions

30. An amplifier using an opam with a slew-rate SR=1 V/µsec V/µ sec 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

12

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Analog Electronics Old GATE ECE

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

sinusoidal osicallation. If at a frequency f 0 B(f)=

then to sustain

oscillation at this frequency.

+↑ (f) Network B(f)

+ - (f)

-↓

a. R2 = 5R1

b. R2 = 6R1

c. R2 = R1 

/6

2

= R1 /5d. R

32. A Zener diode regulator in the figure is to be designed to meet the specifications: I L = 10mA, V 0 = 10V and Vin varires from 30 V to 50V. The zener diode has Vz = 10V and Izk  (knee current)= 1mA For satisfactory operation ↑

+↑ Iz

IL = 10mA ↓



Dz

-





a. R ≤ 1800 Ω

 b. 2000 Ω R ≤ 22000 Ω

c. 3700 Ω ≤ R ≤ 4000 Ω

d. R > 4000 Ω

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

13

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Analog Electronics Old GATE ECE

VDD = +10V

↓ ID=1mA

RD (3K )

C2

+

C1 →

+ RG (1M (1 M

-

)

Rs (2.5K

Cs

)

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: T 1 conducts when V1 ≥ 2V. Statement 2: T 1 is always in saturation when V 0 = 0V. + 5V →

T2



T1

Which of the following is correct?

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Analog Electronics Old GATE ECE

a. Only statement 1 is TRUE b. Only Statement 2 is TRUE c. Both the stateme s tatements nts are TRUE d. Both the Statements Statements are FALSE.

GATE-2003 One Mark Questions

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 discriminatory

d. voltage doubler

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Analog Electronics Old GATE ECE

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 comparator has a duty cycle of  Input

Output

Vref =2V

a. 1/2

b. 1/3

c. 1/6

d. 1/12

38. If the differential voltage gain and the common mode v oltage gain of a differential amplifier 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 GATE-2003 Two Marks Questions

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 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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Analog Electronics Old GATE ECE

41. In the amplifier circuit shown in the figure, the values of R 1 and R2 are such that the transistor is operating at V CE = 3V and I C = 1.5mA when its β is 150.

for a

transistor with β of 200, the operating point (V CE, 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

R

C

R

C

R

a.

b.

c.

d.

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

17

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Analog Electronics Old GATE ECE

+ 1K 15V DC Unregulated Power source

Vz=3V

20K

Regulated DC Output

40K

-

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 V out will be equal to

18

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Analog Electronics Old GATE ECE

5K

1K

2V 3V

1K 8K

a. 1V

b. 6V

c. 14 V

d. 17V

46. Three identical amplifiers with each one having a volta ge 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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Analog Electronics Old GATE ECE

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 i mpedance and decrease in output impedance impedance d. decrease in input impedance impedance and increase in output impedance

50. The circuit in the figure is a 5K

R

R

a. low-pass filter

b. high-pass high-pass filter

c. band-pass filter

d. band-reject filter

51. Assuming VCEsat= 0.2V and β = 50, the minimum base current (I B) required to drive the transistor in the figure to saturation is 3V

↓I

c

1K

IB



a. 56 µ A

b.140 mA

c. 60 mA

d. 3 mA

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GATE-2004 Two Marks Questions

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 (V T) is

25 mV, the transconductance (g m) and the input resistance (r π) of the transistor in the common emitter configuration are a. gm = 25mA/V and r π = 15.625kΩ b. gm = 40/V and r π = 4.0kΩ c. gm = 25mA/V and r π = 2.5kΩ d. gm = 40 mA/V and r π = 2.5kΩ

53. The value of C required for sinusoidal oscillations of frequency 1 kHz in the circuit of the figure is 2.1K

1K

C 1K 1K

C

a.

 b. 2πµ F

c.

d. 2 π √6 µ F

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

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Analog Electronics Old GATE ECE

↓ iL

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 R

+ 12V

5V -

Variable Variable Load 100 to 500 mA

a. 7Ω

 b. 70 Ω

c. 70/3 Ω

d. 14 Ω

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Analog Electronics Old GATE ECE

56. In a full-wave rectifier rectifier using two ideal diodes V dc and Vm are the dc and peak  values of the voltage respectively across a resistive load. If P IV is the peak inverse voltage of the diode, then the appropriate relationships for the rectifier are a.

b.

c.

d.

GATE-2005 One Mark Questions

57. The effect of current shunt feedback in an amplifier 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 R i 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

23

b. CE-CB nd

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c. CB-CC

Analog Electronics Old GATE ECE

d. CE-CC

GATE- 2005 Two Marks Questions

60. For an npn transistor connected as shown in the figure V BE = 0.7 volts. Given 0

that reverse saturation current of the junction at room temperature 300 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 e 0 indicated in the figure has been measured by an ideal voltmeter. Which of the following can be calculated? 1M

e0

1M

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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Analog Electronics Old GATE ECE

62. The OP-amp circuit shown in the figure is a filter. The type of filter and its cut-off frequency are respectively respectively 10 K

10 K

1 F

1K

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 (R E)

-VEE

a. increases both the differential and common mode gains. b. increase the common-mode common-mode gain only. c. decrease the differential-mode differential-mode gain only d. decrease the common-mode common-mode gain only

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Analog Electronics Old GATE ECE

64. For an n-channel MOSFET and its transfer curve shown in the figure, the threshold voltage is VD= 5V

ID ↑

D Transfer characteristics

VG=3V G |





|

V GS

1V

S Vs= 1V

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 V BE = 0.7V is shown in the figure.

The base current I B and collector voltage V c are respectively 20 V 2k

430K

10 F 1K

40 F

a. 43 µ A and 11.4 Volts

b. 40µA 40µ A and 16 Volts

c. 45µA 45µ A and 11 Volts

d. 50µA 50µ A and 10 Volts

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Analog Electronics Old GATE ECE

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 1k

↑ Load

=5.8V

20-30

a. 23.7 mA

b. 14.2 mA

c. 13.7 mA

d. 24.2 mA

67. Given the ideal operational amplifier circuit circui t shown in the figure indicate the correct transfer characteristics assuming ideal diodes with zero cut-in voltage. +10V

2K

-10V

0.5K

2K

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Analog Electronics Old GATE ECE



+10V (a) ←→ -8V

-5V

→ ←→

-10V



+10V (b) ←→ -5V

+8V

→ ←→

-10V



+5V (c)

←→

-5V

+5V

→ ←→

-10V



+10V (d) ←→ -5V

+5V

→ ←→

-5V

Common Data Questions 68, 69, 70 Given rd = 20 kΩ, I DSS = 10 mA, VP = - 8V

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Analog Electronics Old GATE ECE

68. Zi and Z0 of the circuit are respectively a. 2M Ω and 2k Ω 2k  Ω c. infinity and 2 M

b. 2M Ω and 20/11 k Ω k Ω

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 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 (Z i) and the output impedance (Z 0) of an ideal transconductance transconductance (voltage controlled current source) s ource) amplifier are a. Zi = 0, Z0 = 0

b. Zi = 0, Z0 = ∞

c. Zi =∞, Z0 = 0

d. Zi =∞, Z0 = ∞

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Analog Electronics Old GATE ECE

72. An n-channel depletion MOSFET has following two points on its I D – VGS curve. (i) VGS = 0 at I D = 12 mA and (ii) VGS = - 6 Volts at Z 0 = ∞

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

GATE-2006 Two Marks Questions

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 V C across the capacitor at t = 1 millisecond is C=1 F

S -

+

1K 10V

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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Analog Electronics Old GATE ECE

6V + 12sin t

~

VR

R

-

(a) 6V

(b) -12V 12V (c) -6V

(d) -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 C c can be assumed to be infinite. 12v

1K 53K +

5.3K Cc

Vc -

~

In the figure above, the ground has been shown by the symbol ▼ 31

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Analog Electronics Old GATE ECE

75. Under the DC conditions, the collect to-emitter voltage voltage drop is a. 4.8 Volts

b. 5.3 volts

c. 6.0 volts

d. 6.6 volts

76. If β DC 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 v c /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 generates a regulated re gulated output V out. Use the component values shown in the figure

15V (UR)

Q1

+ 12 1K

12

-

6V

24

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

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Analog Electronics Old GATE ECE

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%

GATE-2007 One Mark Questions

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

Analog Electronics Old GATE ECE

83. For the Op-Amp circuit shown in the figure, V 0 is 2K

1K 1V 1K 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

10K

+ 10V 2V

+ -

1K

a. cut-off

b. saturation

c. normal active

d. reverse active

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Analog Electronics Old GATE ECE

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 V i = 4 V, V 0 = V02. the relationship between V 01 and V02 is

2K

a. V02 = √2 V01

b. V02 = e V01

c. V02 = V01 In 2

d. V01 – V02 = VT In 2

2

86. In the CMOS inverter circuit shown, if the if  the transconductance parameters parameters of the NMOS and PMOS transistors are k n = k p = µ n Cox Wn /Ln = µ n Cox Wp /Lp = 40 µ  2

A/V and their threshold voltages are V THn = |VTHp| = 1V, the current is 5V PMOS ↓

2.5 V

I

NMOS

a. 0A

b. 25µA

c. 45 µ A

d. 90 µ A

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 (V i) range is from 10 to 16 V, the output voltage voltage (V 0) 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 volta ge (C-V) characteristics of  a Metal/SiO2 /silicon (MOS) capacitor having an area of 1 x 10 the permitivities

-4

2

cm . assume that

(ε0 εr) of silicon and SiO 2 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 1µ m

89. The maximum depletion layer width in silicon is a. 0.143 µ m

b. 0.857 µ m

c. 1 µ m

d. 1.143 µ m

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

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Analog Electronics Old GATE ECE

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.

R C

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

b.

c.

d.

92. If V i = V1 sin (ωt) and V 0 = V2 sin (ωt + φ), then the minimum and maximum

values of φ (in radians) are respectively a. – π/2  – π/2 and π/2

 b. 0 and π/2

 – π and 0 c. – π

 – π/2 and 0 d. – π/2

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GATE-2008 One Mark Questions

93. In the following limiter circuit, an input voltage V i = 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. 1K

D1 D2 Z

6.8V

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

GATE-2008 Two Marks Questions

94. Consider the following circuit using an ideal OPAMP. The I-V characteristics characteristics of the diode is described by the relation Where VT = 25m V, I0 = 1µ A and V is the voltage across the diode (taken as positive for forward bias). D

4K

=-1V 100 K

For an input voltage V i = -1V, the output voltage voltage V 0 is a. 0 V

b. 0.1 V

c. 0.7 V

d. 1.1 V

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Analog Electronics Old GATE ECE

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 g m of the pair is defined to be

at constant V out.

The equivalent g m of the pair is a. the sum of individual g m’s of the transistors b. the product of individual g m’s of the transistors c. nearly equal to the g m of M1 d. nearly equal to g m /g0 of M2

97. Consider the Schmidt trigger circuit shown below

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Analog Electronics Old GATE ECE

I out

M2

V

bias

M

1

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, V BE = 0.7 V, r e = 25 mV/IE and β and all =9V 3K

20K

Cc2 C c1 10K

↓I

E

2.3K

CE

3K

98. The value of DC current I E is a. 1mA

40

b. 2 mA nd

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c. 5 mA

Analog Electronics Old GATE ECE

d. 10 mA

GATE-2009 Two marks Questions

100. In the circuit below, the diode is ideal. The voltage V is given by + V1

1 + -

↓ 1A

a. min (Vi 1)

b. max (Vi 1)

c. min (-V i 1)

d. max (-Vi 1) 6

101. A small signal source v i (t) = A cos 20t + B sin 10 t is applied to a transistor amplifier as shown below. The transistor has

β = 150 and h ie = 3KΩ. Which

expression best approximates v 0 (t)? 12V 3K

100 K

(t)

(t)

100 nF

20 K

900K

10

F

6

a. v0 (t) = -1500 (A cos 20t + B sin s in 10 t) 6

b. v0 (t) = -150 (A cos 20t +B sin 10 t) 6

c. v0 (t) = -1500 B sin 10 t 6

d. v0 (t) = -150 B sin 10 t Common Data for Questions 102 and 103.

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Analog Electronics Old GATE ECE

Consider a silicon p-n junction at room temperature having the following parameters: Doping on the n-side = 1 x 10

17

-3

cm

Depletion width on the n-side = 0.1 µm Depletion width one the p-side = 1.0 µm Intrinsic carrier concentration concentration = 1.4 1. 4 x 10

10

-3

cm

Thermal voltage =26 mV Permittivity of free space = 8.85 x 10

-14

-1

F.cm

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 -1

a. 0.15 MV.cm , directed from p-region to n-region n-r egion -1

b. 0.15MV. cm , directed from n-region to p-region p -region -1

c. 1.80 MV.cm directed from p-region to n-region -1

d. 1.80 MV. cm directed from n-region to p-region

Statement for Linked Answer Question 104 and 105.

Consider the CMOS circuit shown, where the gate voltage V G 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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Analog Electronics Old GATE ECE

and the product of the transconductance parameter parameter and the (W/L) ratio, i.e. the -2

quantity µCox(W/L). is 1mA. V . 5V

← 3V

VG



104. For small increase in V G 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 V 0 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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Analog Electronics Old GATE ECE

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 R i ↑ & R 0 ↓ by 1 + AB -- Ri = 100 x 1K = 100 K

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Analog Electronics Old GATE ECE

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., V c becomes zero. ie1 = ie2 = 0 ib2 < < ic2 So ic2 ~ ie2

10. (d) In positive feedback op-amp act in its satuation region ± V sat. Here applied voltage is positive. V0 = + Vsat = + 15 V

11. (c)

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Analog Electronics Old GATE ECE

12. (a)

At low frequency, A i = - 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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Analog Electronics Old GATE ECE

C Control Gate

17. (d) 15V Rc R th 10/3 Vth

5V

Since β is large

430

IB ≈ 0, R thth = 5 || 10

18. (c) V00 = Vi0.A = 5 mV x 10,000 = 50 V But V00 = ± 15V, V00 can never be greater than ± V sat

19. (c)

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Analog Electronics Old GATE ECE

hfe = gm .rπ

22. (b)

(β0 = hfe) 23. (C) I1 20 →

↓ Iz

↓ IC

IB



20-30V

RB

+

=10V

VBE - ↓IE

(i.e. when Iz = 0) IE = IC +Iz IB = Iz (as no current flows in R B)

From (i) IE = βIz + Iz = (99 +1) Iz IE = 100 Iz

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Analog Electronics Old GATE ECE

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)

50

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Analog Electronics Old GATE ECE 10

10mH

F

10 1

=10cos =10cos (100t) ( 100t)

100

+ 2

3

KCL at node 1,

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

27. (c) Ri increases by factor of 1 + Aβ and R 0 decreases by 1+ Aβ.

28. (b)

Gain X B. ω = 1x10 6

20 log x = 20 dB X = 10

29. (a) f L = 20 Hz

51

f H = 1 KHz for single stage. nd

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Analog Electronics Old GATE ECE

For cascaded stage

30. (c)

Slew rate = A. 2π fV m V = A.Vm sin ωt

20 log X = 40 X = 100 = A

Vm = 79. 5 mV

31. (a)

1

KCL at node 1

52

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Analog Electronics Old GATE ECE

32. (a) +

I1



↓I



z

↓I

L

RL ZL ↓

-

(Iz + IL = I1) When Vin = 30 V

When Vin = 50 V

53

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Analog Electronics Old GATE ECE

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 V i > Vth When V0 = 0, CMOS inverter has I/P = 1 i.e. 5 V So T 1 is in saturation and conducts.

35. (b) 36. (d) 37. (b)

8Vp-p So, Vi = 4 sin ω t

54

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Analog Electronics Old GATE ECE

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 R 2 1.5 mA x R2 = 3 R2 = 2 KΩ

55

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Analog Electronics Old GATE ECE

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 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)

56

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Institute of Engineering Studies (IES,Bangalore) 250

250

1K

+ -

+ 50

250

1K +-

1K

Analog Electronics Old GATE ECE

V

4

50

50

st

Volt. Across 1 K after 1 stage =

Similarly

Therefore A V = 40 x 40 x 50 = 8 x 10

4

4

AV in dB = 20 log (8 x 10 ) = 98 dB 47. (d)

I = 3mA

50. (a)

At ω = ∞,

& at ω = 0

51. (a)

57

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Analog Electronics Old GATE ECE

VCE = VCC -ICRC 0.2 = 3-IC x 1 K IC = 2.8 mA

52. (d)

hfe = gm. rπ, hfe = β

53. (a) 2.1K 1K C 1K ←

1K

↓ ↓

C

R= 1K

58

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Analog Electronics Old GATE ECE

For oscillation imaginary part is zero.

i.e.

ω2 C2 R2 – 1 = 0

54. (a)

1K V



iL

59

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Analog Electronics Old GATE ECE

…..(i) ….. (ii) Putting V0 from (1)

55. (d)

When IL = 100 mA,

When IL = 500mA, 500mA,

Therefore R= 14 Ω (choosing minimum one)

56. (b)

57. (d)

60

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Analog Electronics Old GATE ECE

58. (b)

59 (b)

60. (c) When two terminals of a transistor are shorted it acts as diode.

61. (c) 1M

-

+

IB2 ↓ e0 ↑

IB1

1M

V1 = - IB1 x 1M, V2 = V1 = - IB1 x 1M (due to virtual ground) Drop in feedback resistor 1M = I B2 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,

61

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Analog Electronics Old GATE ECE

63. (d) Only common mode gain depends on R E and differential mode gain is independent of RE.

64. (c) From the graph its clear that V th = 1V VGs = 3-1 = 2V VDS = 5-1= 4V Since V Ds ≥ VGS – VT S. MOSFET is in saturation region. 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) I B RE

IB = 40 µA IC = β = 50 x 40

mA µA = 2000 µ A = 2 mA

VC = VCC- ICRC = 20-2 mA X 2K VC = 16 V

66. (a) I1



1K

IL

↓I

→ z

20-30V 5.8V

62

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Analog Electronics Old GATE ECE

VL= ‘5.8 V

Maximum load current will be when V 1 = 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)

63

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Analog Electronics Old GATE ECE

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. ↑ ID

12mA →

-6V (Z 0 =

V

GS

)

Transfer character of n-channel D-MOSFET D -MOSFET

73. (d)

64

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Analog Electronics Old GATE ECE

1 F

1K

+

1 10V 10V

Applying KCL at node 1

74. (b) Zener diode works as normal diode in FB. So, when V in < 0, VR = Vin When 0 < Vin < 6, Diode is OFF and VR = 0. When Vin > 6, Diode conducts and V R = 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 65

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Analog Electronics Old GATE ECE

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

77. (a)

78. (d) 15V

Q1

+ (3v)

1K

12K

6v

6V

(6V)

12K

-

24K

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= V CE . IC = 6 V x 1A P= 6 Watts

79. (b)

66

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Analog Electronics Old GATE ECE

New unregulated voltage = 18V Therefore V CE = 18-9 = 9 V IC = 1A Therefore P = 9 x 1 = 9 watts

80. (c)

81. (a)

82. (b) IPC1

IPE

→ P →

n

P

InE

+

+

-

VCB

VEB

Transport factor

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

67

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Analog Electronics Old GATE ECE

83. (c) 2K

1k

1V X

1k

1V

y

1K

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

84. (b)

Given β is large so I B = 0 & IE = IC Assuming BJT is in active Applying KVL in Base. Emitter loop

 – 0.7 = 1 K Ω x I E 2 – 0.7 IE = 1.3 mA

Now ICsat =

As ICactive > ICsat So BJT is in saturation

85. (d) Applying KCL

68

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Analog Electronics Old GATE ECE

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/v So ID = K (VGS – VT)

2

2

= 40 ID = 90 µA

87. (c) VZ = 7 volt IK = 0, R z = 10 Ω Range of V i = 10 to 16V

Range of voltage across 200 Ω = V i- Vz = 3 to 9 volt Range of current through 200 Ω = 15 to

45 mA

Range of variation in output voltage

69

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Analog Electronics Old GATE ECE

= (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

70

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Analog Electronics Old GATE ECE

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 V i D1 is forward biased and Zener diode is in breakdown stage V0 = 0.7 + 6.8 = 7.5 V For the negative half of V i D2 is forward biased. V0 = - 0.7 V

94. (b)

71

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Analog Electronics Old GATE ECE

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 V 1 Applying KCL at non-inverting input end

15-V1 + V0 – V1 = V1 + 15

Since, V0 swings from – 15V to + 15 V, Therefore V 1 switches between  – 5V & + 5V.

98. (a) The given circuit can be redesigned as shown below

72

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Institute of Engineering Studies (IES,Bangalore)

Analog Electronics Old GATE ECE

= 9V 3K RTh

IE

VTh

↓ 2.3K

= 6.67 k 

Since β is large I B can be ignored

99. (d) Mid-band voltage gain,

= - 60

100. (a)

101. (b) The best approx answer for output voltage v 0 is V0 = Ar. vi

73

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≈ - 150 (A cos 20t + B sin 106

Analog Electronics Old GATE ECE

)

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. (*?)

74

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