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
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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 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%
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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
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
3
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+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
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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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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
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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 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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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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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
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=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,
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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
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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
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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
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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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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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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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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
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+ 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
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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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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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↓ 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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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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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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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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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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↑
+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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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
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32. (a) +
I1
→
↓I
↑
z
↓I
L
RL ZL ↓
-
(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 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
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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 R 2 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 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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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)
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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) 2.1K 1K C 1K ←
1K
↓ ↓
C
R= 1K
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For oscillation imaginary part is zero.
i.e.
ω2 C2 R2 – 1 = 0
54. (a)
1K V
↓
iL
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…..(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)
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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,
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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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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)
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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. ↑ ID
12mA →
-6V (Z 0 =
V
GS
)
Transfer character of n-channel D-MOSFET D -MOSFET
73. (d)
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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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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)
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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
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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
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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/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
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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 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)
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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 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
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= 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
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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. (*?)
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