Analog circuits
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This document provides a test paper on analog electronics with multiple choice questions and solutions. It begins with an introduction stating it is Test Paper-1 on Analog Electronics from the GATE Multiple Choice Questions ECE source book. It then provides 20 multiple choice questions related to topics in analog electronics like diodes, transistors, op-amps, and filters. Each question is followed by a short solution explaining the reasoning. The questions cover concepts such as diode and transistor biasing, amplifier configurations, filter design, and more.
![Page 9 Analog Electronics Test Paper-1
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gm 2 ( )K V Vn GS TN= −
2(1 )m= (1.51 0.8) 1.42− = mS
Av (1.42 )(7 )m k=− 9.9=−
Hence (B) is correct option.
MCQ 1.16 Consider the common-source circuit shown below The transistor parameters are
.V 0 8TN = V, K 1n = mA/V2 and 0λ = . The small-signal voltage gain is
(A) .10 83− (B) .8 96−
(C) .5 76− (D) .3 28−
SOL 1.16 Form the DC analysis :
VGSQ 1.5 V= ,
IDQ 0.5 mA=
gm 2 ( )K V Vn GS TN= −
2(1 )(1.5 0.8) 1.4 /mA Vm= − =
ro [ ]IDQ
1
3λ= =−
The resulting small-signal equivalent circuit is shown below
vo ,g u R v v g v Rm gs D i gs m gs S=− = +
&
v
v
i
o
g R
g R
1 m S
m D
=
+
−](https://image.slidesharecdn.com/analogcircuits-130731120148-phpapp02/85/Analog-circuits-9-320.jpg)
Analog circuits
- 1. Test Paper-1 Analog Electronics Source Book: GATE Multiple Choice Questions ECE Author: RK Kanodia Edition: 6th ISBN: 9788192276205 Publisher : Nodia and Company Visit us at: www.nodia.co.in Q. No. 1 - 10 Carry One Mark Each MCQ 1.1 Consider the given a circuit and a waveform for the input voltage. The diode in circuit has cutin voltage 0V =γ . The waveform of output voltage vo is SOL 1.1 Hence (C) is correct option. Diode will be off if 2 0v >i + . Thus 0vo = For 2 0v <i + V, 2, 2 3v v v<i o i− = + =− V MCQ 1.2 In the shunt regulator shown below, the 8.2VZ = V and 0.7VBE = V. The regulated output voltage Vo is
- 2. Page 2 Analog Electronics Test Paper-1 www.gatehelp.com (A) 11.8 V (B) 7.5 V (C) 12.5 V (D) 8.9 V SOL 1.2 Hence (D) is correct option. Vo 8.2 0.7 8.9V VZ BE= + = + = V MCQ 1.3 The Early voltage of a BJT is 75VA = V. The minimum required collector current, such that the output resistance is at least 200ro = kΩ, is (A) 1.67 mA (B) 5 mA (C) 0.375 mA (D) 0.75 mA SOL 1.3 Hence (C) is correct option. ro I V CQ A = & ICQ 200 . kr V 75 0 375 o A = = = mA MCQ 1.4 In the given circuit of figure if 0.4VTH = V, the transistor M1 is operating in (A) Linear region (B) Saturation region (C) M1 is off (D) Cannot be determined SOL 1.4 Hence (B) is correct option. For P-channel MOSFET V ( )SD sat (1 0) 0.4 0.6V VSG TH= + = − − = VSD 1 0.3 0.7V VS D= − = − = Here, V V> ( )SD SD sat So, M1 is in saturation region.
- 3. Page 3 Analog Electronics Test Paper-1 www.gatehelp.com MCQ 1.5 In the circuit shown below the PMOS transistor has parameter 1.5 VVTP =− , 25 /A Vk' p 2 μ= , 4 mL μ= and 0λ = . If 0.1 mAID = and 2.5 VVSD = , then value of W will be (A) 15 mμ (B) 1.6 mμ (C) 32 mμ (D) 3.2 mμ SOL 1.5 Hence (C) is correct option. VSD VSG= , ID ( ) k L W V V 2 ' p GS TP 2 = + 10 4− (2.5 1.5) 32 mW W 2 25 4 2 & μ= − =b bl l MCQ 1.6 Which of the following amplifier has high input impedance, low output impedance and low voltage gain (A) Common-gate (B) Common-Drain (C) Common-Source (D) None of these SOL 1.6 Hence (B) is correct option. For common drain amplifier output impedance, R g 1 o m = (low) Input impedance Rin 3= (high) Voltage gain ( )Av 1 g R g R 1 m L m L .= + (low) MCQ 1.7 What is the input resistance to the common gate circuit shown in figure if I 1D = mA, I 5DSS = mA and V 2TH =− V (A) 2.24 kΩ (B) 447 Ω (C) 27 kΩ (D) 224 Ω SOL 1.7 Hence (B) is correct option. Input resistance is given by
- 4. Page 4 Analog Electronics Test Paper-1 www.gatehelp.com Rin I v g 1 D i m = = gm V I I2 TH D DSS#= ( )( ) 2.24 /mA V 2 2 1 5= = Rin ( . ) 447 2 24 1 mA/V Ω= = MCQ 1.8 For the circuit shown below the value of A v v v i o = is (A) 10− (B) 10 (C) 13.46 (D) 13.46− SOL 1.8 Hence (A) is correct option The noninverting terminal is at ground level. Thus inverting terminal is also at virtual ground. There will not be any current in 60 kΩ. Av 10 40 400=− =− MCQ 1.9 For the circuit shown below the true relation is (A) v vo o1 2= (B) v vo o1 2=− (C) 2v vo o2= (D) 2v vo o1 2= SOL 1.9 At second stage input to both op-amp circuit is same. The upper op-amp circuit is buffer having gain 1Av = . Lower op-amp circuit is inverting amplifier having gain 1A R R v =− =− . Therefore v vo o1 2=− .
- 5. www.gatehelp.com Page 5 Analog Electronics Chapter 1 Hence (B) is correct option. MCQ 1.10 The following op-amp circuit is (A) Band-pass filter (B) Second order low pass filter (C) Second order high pass filter (D) Band Reject filter SOL 1.10 Hence (A) is correct option. Transfer function of above circuit is v v i o .R R R sC R R C C s R C C sC 1 11 2 1 1 1 2 1 2 2 2 1 2 1 = − + + +b l> H So, this is a band pass filter Q. No. 11- 21 Carry Two Mark Each MCQ 1.11 In the following circuit minimum required value of R R 1 2 to sustain oscillation is (A) 1.5 (B) 1 (C) 2 (D) 4 SOL 1.11 This is a wein-bridge oscillator circuit, loop gain is given as ( )T s ( / )R R sRC sRC 1 3 1 1 1 2 = + + +b cl m Condition for oscillation ( )T j oω 1 ( / )R R j RC j RC 1 3 1 1 o o1 2 ω ω = = + + +b cl m
- 6. Page 6 Analog Electronics Test Paper-1 www.gatehelp.com oω RC 1= So, 1 , 2 R R R R1 3 1 1 2 1 2 &= + =b bl l So, to sustain the oscillations, we must have ( / ) 2R R >2 1 Hence (C) is correct option. MCQ 1.12 The analog multiplier X shown below has the characteristics v v vp 1 2= . The output of this circuit is (A) v vs ss (B) v vs ss− (C) v v ss s − (D) v v ss s SOL 1.12 Hence (C) is correct option. v+ 0 v= = −, Let output of analog multiplier be vp . R vs , R v v v v v vp s p p ss o&=− =− = vs v vss o=− , vo v v ss s =− MCQ 1.13 For the circuit shown below the value of io is (A) 12 mA (B) 8.5 mA (C) 6 mA (D) 7.5 mA SOL 1.13 The circuit is as shown below
- 7. Page 7 Analog Electronics Test Paper-1 www.gatehelp.com v+ 0, 3 mAv i 4 12 k1= = = =− i2 3 2 5 mA= + = , vo (5)(3) 15 V=− =− i2 i io L= + , 5 i 6 15 o= + − , io .7 5= mA Hence (D) is correct option. MCQ 1.14 For the op-amp circuit shown below the voltage gain /A v vv o i= is (A) 8− (B) 8 (C) 10− (D) 10 SOL 1.14 The circuit is as shown below R v R v0 0i 1− + − 0= , v1 vi=− R v R v v R v01 1 2 1− + − + 0= ,
- 8. Page 8 Analog Electronics Test Paper-1 www.gatehelp.com 3 , 3v v v vi1 2 2= =− r v v R v R v vo2 1 2 2− + + − 0= 3 3 3v v v vi i i i− + − − vo= & v v i o 8=− Hence (A) is correct option. MCQ 1.15 Consider the common-source circuit with source bypass capacitor. The signal frequency is sufficiently large. The transistor parameters are .V 0 8TN = V, K 1n = mA/V2 and 0λ = . The voltage gain is (A) .15 6− (B) .9 9− (C) .6 8− (D) .3 2− SOL 1.15 Since the DC gate current is zero, v Vs GSQ=− IDQ ( )I K V VQ n GSQ TN 2 = = − & 0.5 1( 0.8)VGSQ 2 = − VGSQ 1.51 V vs= =− VDSQ 5 (0.5 )(7 ) ( 1.51) 3.01m k= − − − = V The transistor is therefore biased in the saturation region. The small-signal equivalent circuit is shown below vo (7 )g v km gs=− vgs , (7 )v v v A g ki i o v m= = =−
- 9. Page 9 Analog Electronics Test Paper-1 www.gatehelp.com gm 2 ( )K V Vn GS TN= − 2(1 )m= (1.51 0.8) 1.42− = mS Av (1.42 )(7 )m k=− 9.9=− Hence (B) is correct option. MCQ 1.16 Consider the common-source circuit shown below The transistor parameters are .V 0 8TN = V, K 1n = mA/V2 and 0λ = . The small-signal voltage gain is (A) .10 83− (B) .8 96− (C) .5 76− (D) .3 28− SOL 1.16 Form the DC analysis : VGSQ 1.5 V= , IDQ 0.5 mA= gm 2 ( )K V Vn GS TN= − 2(1 )(1.5 0.8) 1.4 /mA Vm= − = ro [ ]IDQ 1 3λ= =− The resulting small-signal equivalent circuit is shown below vo ,g u R v v g v Rm gs D i gs m gs S=− = + & v v i o g R g R 1 m S m D = + −
- 10. Page 10 Analog Electronics Test Paper-1 www.gatehelp.com (1.4 ) ( . )( . ) ( ) 5.76 1 1 4 0 5 7 m m k k =− + =− Hence (C) is correct option. MCQ 1.17 Inthecircuitshownbelowthetransistorparametersare 1 VVTN = and 36 /A Vk' n 2 μ= If 0.5 mAID = , 5 VV1 = and 2 VV2 = then the width to-length ratio required in each transistor is L W 1 b l L W 2 b l L W 3 b l (A) 1.75 6.94 27.8 (B) 4.93 10.56 50.43 (C) 35.5 22.4 5.53 (D) 56.4 38.21 12.56 SOL 1.17 Each transistor is biased in saturation because V VDS GS= and V V V>DS GS TN− For , 2M V3 2 = V VGS3= ID 0.5 (2 1) L W 2 36 10 3 3 2#= = − − b bl l & L W 3 b l 27.8= For ,M VGS2 2 5 2 3V V1 2= − = − = V ID 0.5 (2 1) L W 2 36 10 3 3 2#= = − − b bl l & L W 3 b l 27.8= For M2, VGS2 5 2 3V V1 2= − = − = V ID 0.5 (3 1) L W 2 36 10 3 2 2#= = − − b bl l
- 11. Page 11 Analog Electronics Test Paper-1 www.gatehelp.com & L W 2 b l 6.94= For M1, VGS1 10 10 5 5V1= − = − = V ID 0.5 (5 1) L W 2 36 10 5 1 2 #= = −− b bl l & L W 1 b l 1.74= Hence (A) is correct option. MCQ 1.18 A p-channel JFET biased in the saturation region with 5 VVSD = has a drain current of 2.8 mAID = , and 0.3 mAID = at 3 VVGS = . The value of IDSS is (A) 10 mA (B) 5 mA (C) 7 mA (D) 2 mA SOL 1.18 Hence (B) is correct option. ID I V V1DSS P GS 2 = −b l 2.8 m ,I V 1 1 DSS P 2 = −b l 0.3m I V 1 3 DSS p 2 = −c m & . . 0 3 2 8 3.97 V V V 1 3 1 1 P P P&= − − = b b l l V 2.8 . 5I I1 3 97 1 DSS DSS&= − =b l mA MCQ 1.19 The transistor in the circuit shown below has parameters 8IDSS = mA and 4VP =− V. The value of VDS is (A) 2.7 V (B) 2.85 V (C) 1.30− V (D) 1.30 V SOL 1.19 Hence (B) is correct option. VG (20) 6 60 140 60= + = V
- 12. Page 12 Analog Electronics Test Paper-1 www.gatehelp.com Assume the transistor in saturation, ID I V V1DSS P GS 2 = −b l ID R V R V V V 2 6 kS S S G GS GS = = − = − 6 VGS− (2 )(8 )k m= V1 4 GS 2 − −b l & VGS 1.3=− V ID (8 )m= . 3651 4 1 3 2 − − − =b l mA VDS 20 (2.7 2 )I k kD= − + 20 (3.65)(2.7 2) 2.85= − + = V V ( )DS sat 1.30 ( 4) 2.7V VGS p= − =− − − = V V V> ( )DS DS sat Assumption is correct. MCQ 1.20 In the following circuit, impedance Rb seen through base of the transistor is (A) 100 Ω (B) 9.90 Ω (C) 502 Ω (D) 900 Ω SOL 1.20 Input impedance through base is given as Rb r gm β = =π gm V I T C = So, Rb r I V C Tβ = =π
- 13. Page 13 Analog Electronics Test Paper-1 www.gatehelp.com By DC analysis IE 10= mA, I IC Eα= α 0.955 1β β = + = IC (0.995)= (10 mA) = 9.95 mA So, Rb ( . ) ( )( ) 502r 9 95 200 25 mA mV Ω= = =π Hence (C) is correct option. MCQ 1.21 For the circuit shown below each diode has 0.6V =γ V and 0rf = . Both diode will be ON if (A) 3.9v >s V (B) 4.9v >s V (C) 6.3v >s V (D) 5.3v >s V SOL 1.21 The circuit is as follows For vs small, both diode are OFF. For 0.6v >s V, D1 is ON. For 0.6v >1 V, both diode will be ON. .v v v v 5 0 6 5 s s1 1= − + − . . .v v 0 5 0 5 0 61 1 = + − v1 . 0.6u 22 2 5 4 >s = + V 3.9v >s& V Hence (A) is correct option.
- 14. Page 14 Analog Electronics Test Paper-1 www.gatehelp.com Common Data for Q. 22-23 : For the circuit given 3 10IC 17 #= − and VA 3= and 1IC = mA. MCQ 1.22 The value of VB is (A) 26 mV (B) 809.6 mV (C) 726 mV (D) 0 mV SOL 1.22 Hence (B) is correct option. IC I e V V1/ S V V A CEB T = +: D As, VA 3= , VB lV I InT S C = b l 26 10 ln 0.8096 3 10 1 103 17 3 # ##= =− − − c m V MCQ 1.23 If 5VA = V and 1IC = mA for 15VCE = V, the value of VB is (A) 802.8 mV (B) 796 mV (C) 809.6 mV (D) 26 mV SOL 1.23 The collector current is IC I e V V1/ S V V A CEB T = +b l 10 3− 3 10 .e 0 5 1 5/V V17 B T #= +− b l e /V VB T .3 9 1014 = VB lnV 39 10 T 14 = b l 26 10 ln . 802.8 3 9 103 14 .#= − b l mV Hence (A) is correct option. Statements for Linked Answer : 24 & 25
- 15. Page 15 Analog Electronics Test Paper-1 www.gatehelp.com Consider following an ideal op-amp circuit MCQ 1.24 Load current iL is (A) 5 mA (B) 0.5− mA (C) 5− mA (D) 0.5 mA SOL 1.24 This is a voltage to current converter circuit with 10 kR1 Ω= , 1 kR2 Ω= , 1 kR3 Ω= , 10 kRF Ω= vs 5 V=− a Here R R R RF 2 1 3= So, load current is given as iL ( ) 5 mA R v 1 5 k s 2 Ω =− =− − = Hence (A) is correct option. MCQ 1.25 Output voltage vo is (A) 5 V (B) 4 V (C) 6 V (D) 4.5 V SOL 1.25 By solving the circuit vL 100 (5 10 )(100) 0.5 ViL 3 # #= = =− i4 ( ) . 0.5 mAv 1 1 0 5 k k L Ω = = =
- 16. Page 16 Analog Electronics Test Paper-1 www.gatehelp.com i3 0.5 5 5.5 mAi iL4= + = + = Output voltage is vo (1 )i vk L3 Ω#= + vo (5.5 10 )(1 10 ) 0.53 3 # #= +− − vo 6 V= Hence (C) is correct option. Answer Sheet 1. (C) 6. (B) 11. (C) 16. (C) 21. (A) 2. (D) 7. (B) 12. (C) 17. (A) 22. (B) 3. (C) 8. (A) 13. (D) 18. (B) 23. (A) 4. (B) 9. (B) 14. (A) 19. (B) 24. (A) 5. (C) 10. (A) 15. (B) 20. (C) 25. (C)