INSTRUMENTATION

ENGINEERING

GATE | PSUs

ANALOG
ELECTRONICS

Volume - I : Study Material with Classroom Practice Questions

 Page No. 01

Classroom Practice solutions

To
Analog Electronics

CONTENTS

Name of the Chapter Page No.

Analog Electronics 02 – 21
 Analog Electronics
Class Room Practice Solutions 5V

5K
01.
Sol: 1V V0 = 1V
(1–2)
D1
1V 2V
3V
2V D2 (1–3)
Vo
3V D3  D2 & D3 are reverse biased and ‘D1’ is
forward biased.
i.e., D1 only conduct
5k
Io 5 1
 I0 = = 0.8mA
5K
(1–3)
03.
1V Sol: Let diodes D1 & D2 are forward biased.
2V V0 = 3V
(2–3)  V0 = 0 volt
10  0
3V I2 = = 2mA
5K
5K 0  (10)
I0 I3 = = 1mA
10K
Apply KVL at nodes ‘V0’:
 D1,D2 are reverse biased and D3 is –I1 + I3 – I2 = 0
I1 = –(I2–I3) = –1mA
forward biased.
+10V
i.e., D3 only conducts.
5k
 I0 = 3/5K = 0.6mA
Vo

I1 I2
02.
Sol:
+5V
I3 10k
Io 5k

D1 –10V
1V Vo So, D1 is reverse biased & D2 is forward
2V D2 biased

3V D3  ‘D1’ act as an open circuit & D2 is act as

short circuit.
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:3: Postal Coaching Solutions

Then circuit becomes Transfer characteristics:

+10V V0

3V
5K
20 V0
I A
15k
10K –5V 2V 5V Vin

–10V
05.
 20  Sol:
 V0  10k     10
 15k  R
 V0 = 3.33V +
+
Vin Vo
04. – 2V –
Sol:
2V
+ –
+ For Vi < – 2Volt, Diode ON
+ +
Vin Vx Vo
RL  V0 = – 2Volt
– – –
For Vi > – 2Volt, Diode OFF
Apply KVL to the loop:  V0 = Vi
Vin – 2 – Vx = 0
 Vx = Vin–2 ----- (1) V0
Given, Vin range = –5V to 5V
+5V
 Vx range = –7V to 3V [∵ from eq (1)]
Diode ON for Vx > 0V t
 V0 = Vx –2V
Diode OFF for Vx < 0V –5V
 V0 = 0 V
V0 range = 0 to 3V 06.
Sol: Consider a half wave peak detector the
Output wave form: calculate average value for triangular
waveform
V0 charges discharges
3V
Vx 0
–2 t
+
Vin R  100
–7V – C= 100C

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:4: Analog Electronics

VP 9.3
(d) I R  =
R 100
10V Total ID(max) = IC+IR
Vx Vy
9.3V 9.3
8.77 V  4
 100
0.7V = 4.093 A

–10V 07.
Sol: For positive half cycle diode Forward
T=1/f msec biased and Capacitor start charging towards
peak value.
When diode is OFF, the capacitor
discharges through the resistor  VC = Vm= 5V
Vy  Vx e  t RC t  T  V0 = Vin – VC = Vin – 5
 1m 
Vy  9.31   = 8.37 V Vin range = – 5V to +5V
 100  100 
Ripple amplitude, Vr = Vx–Vy  V0 range = – 10V to 0V
= 9.3 – 8.77 = 0.93 V
V0
0 t
0.93 (ripple)
–5V
9.3V
Area 8.37V –10V

1msec
08.
1 Sol: For +ve cycle, diode ‘ON’, then capacitor
9.3(1m)  (0.93)(1m)
Area 2 starts charging
(a) VAvg  
Base 1m  VC = Vm – 7 = 10 – 7 = 3V
0.93 Now diode OFF for rest of cycle
 9.3  = 8.84V  V0 = –VC +Vin
2
10  Vr  = Vin –3
(b) Tan  Vin range : –10V to +10V
T 4  t 
 V0 range: –13V to 7V
10 0.93
  V0
0.25m t
t = 0.023 msec
7Volts
V
(c) I C ( avg )  C
t
–3Volts t
0.93
 100
0.023m
–13Volts
IC ( avg )  4A
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09. 11.
Sol: For +ve cycle, diode ON, then capacitor Sol:
+10V
starts charging  VC = VP + 9
= 12 + 9 = 21V 4.7K
Now diode OFF for rest of cycle.
 V0 = VC + Vin 4V
= 21 + Vin  = 100
Vin range: –12 to +12V
V0 range: 9V to 33V
3.3K
V0
Given,
33V VB = 4V
VBE = 0.7
21V VB–VE = 0.7
t
VE = VB –7= 3.3V
9V 3.3
IE = = 1mA
3.3K
Let transisotr in active region
10.  IC = /(+1) . IE = 0.99mA
Sol: During positive cycle, IB = IC/ = 9.9A
D1 forward biased& D2 Reverse biased.
VC = 10 – 4.71030.9910-3 = 5.347V
V
+ C–  VC >VB
+  Transistor in the active region.
Vin
– 12.
Sol:
+10V
VC1 = Vin = 6volt
4.7K
During negative cycle,
D1 reverse biased & D2 forward biased. 6V
+
+6V – 0.7 6–0.7=5.3V
– –
+ 3.3K
6V VC2
+ –

VC2 = – 6 – 6 = –12V
VE = VB – VBE = 6 – 0.7 = 5.3V
Capacitor C2 will charge to negative voltage 5.3
IE = = 1.6mA
of magnitude 12V 3.3K
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:6: Analog Electronics

Let transistor is active region VC = 10V

 VB = 0V
IC = I
1   E
IC = 1.59mA 14.
VC = 2.55V Sol:
+10V
VC < VB
 Transistor in saturation region 5K
 VCE(sat) = 0.2V
+
VC–VE = 0.2 0.7
VC = 5.3+0.2 –
 VC = 5.5V
VC
VC= 5.5 V 1K

+
VB= 6V –10V
0.2 V = VCE
–
VE=5.3 V VE = 0.7V [∵ VB = 0V]

10  5.5 10  0.7
 IC = = 0.957mA  IE = = 1.86mA
4.7K 5K
IB = 1.6 – 0.957 = 0.643mA Let transistor in active region.
I C 0.957 mA 
=  =1.483  IC = I = 1.84mA
I B 0.643 mA   1 E
forced < active  VC = –10 + 1K  1.84m
VC = –8.16V
13. VEC = VE – VC = 8.86V
Sol: +10V VEC > VEB
 Transistor in active region
4.7K
15.
Sol: +10V

1K
VE= –0.7V
+
3.3K 0.7
–  = 100

VE = –0.7V
Transistor in cut off region 5K
IC = IB = IE = 0A
VCE = 10V –10V
VE = 0V
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:7: Postal Coaching Solutions

Let transistor in active region 43  5.5

VB = = 3.125V
VE = 0.7V [∵ VB = 0V] 12
10  0.7 3.125
IE   9.3mA IB = = 0.3125mA
1k 10K

IC  .I E  9.2mA
 1 VC = VB+0.5 = 3.625V
 VC = –10+5K  9.2m
VE = 3.825V
VC = 36V
VEC < VEB  IE = 1.175mA
Transistor in saturation region
 VEC = 0.2  IC = 0.862mA
VE–VC = 0.2  VC = 0.5V
0.5  10 17.
 IC = = 2.1mA
5K
Sol: Here the lower transistor (PNP) is in cut off
IB = IE – IC = 7.2mA
I  region.
forced = c sat
IB +5V
2.1
= IC
7.2
= 0.29 10K
forced < active i.e., saturation region 5V
+
IB 0.7 V0
16. –
Sol: 1K
+5V IE

1K IE
IB Apply KVL to the base emitter loop:
0.7 + + VE = (VB+0.7)
10K – 5 – 10K. IB – 0.7 – 1K. (1+)IB = 0
0.2
4.3
– V = (V +0.5)  IB =
c B (101)K  10K
10K
IC = 38.73A
IC = 3.87mA
–5V IE = 3.91mA
 VE = V0 = IE(1k) = 3.91 V
IE = IC+IB
5  (VB  0.7) (V  0.5)  5 VB VC = 5V
 = B 
1k 10k 10k
VB = 5 – 10 k (IB) = 4.61 V
10(5–VB–0.7) = VB+0.5+5+VB
43–10VB = 2VB+5.5
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:8: Analog Electronics

18. +12V 1
Sol: Zeq 
1
sC 
r  sL
r  sL
9K 4K 3.3K 
srC  s 2 LC  1
Vc1 r  jL

3V Q1 Q2 (1  2 LC)  jrC
Vc2 Q3
Zeq 

(r  jL) 1  2 LC  jrC 
3K 2.3K
4K
Q4
V0 1   LC  rC
2 2 2

2 rLC  r  2 rLC  jL[1  2 LC]  jr 2 C


I C1  I 1 
2.3V
 1m Amp
1   LC  rC
2 2 2

2.3k
VC1  12V  4  103  1  10 3  8V Equate Imaginary terms:

V2  8  0.7 V  8.7V L – 3L2C – r2C = 0

12V  V 2 12V  8.7 L – 2L2C – r2C = 0
I 2    1m Amp
3.3k 3.3k
VC 2  4k  1mA  4V 2L2C = L – r2C
V 3  4V  0.7  3.3V 1 r2C
 
V4  3.3  0.7  2.6V L C L2 C
V0=2.6 V 2
1 r
  
19. LC  L 
Sol:
Equate real:
r r  2 LCr  2 LCr
C Zeq 
R1 L (1  2 LC) 2  2 r 2 C 2
r
 2
r C  1
2
r2 
     2  r 2C2
 L   LC L 
r
R2  4 2
r C r C2 r 4C 2 2
  2
L2 LC L
L
Z
rC
r
1/sC (ii) AV = –gmZ
sL L
 g m  
 rC 

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20. Apply small signal analysis:

Sol:
+10V

4.7k 4.7 k
Vo
Vo

4V
Vi  25 
+ –
Vin 3.3k CE
– 3.3 k

For D.C Analysis:

V0  Rc  4700
VB = 4 V  
VB – VE = 0.7  VE = 4 – 0.7 = 3.3 V Vi re  R E 25  3300
3.3 V
IE =  1mA  A V  0  1.413
3.3k Vi
V 25 mV
re = T   25  22.
IE 1mA Sol: To calculate re value apply D.C analysis
To apply small signal analysis set D.C Vth  VBE
source equal to zero. IE 
R
R E  th
 1
Rc = 4.7k
3  0.7
Vo   0.991mA
2k
2.3k 
101
VT 25
+ re    25.22
Vin re= 25 I E 0.991
–
Now apply small signal analysis.:
V0 = –icRc
Vi = icre
V 2k
 AV  0
Vi Vo
i R  R c  4.7 k 2k
 c c  
i c re re 25
= –188
Vi  RB 2k re
–
21.
RE
Sol: D.C calculation is same as previous
question
v0  R C  2k || 2k 
IE = 1 mA AV     39.65
re = 25  vi re 25.22
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: 10 : Analog Electronics

Ri = RB || re 24.
Ri = 1.116k Sol: V0 = –icRC
 Vi ic
i v R R ie  ic = V0
AI  0  0  i  AV  i re
ii R L vi RL
10k 10k
 39.5  1.116  103 V 
 = –22.322 V0   i R C
2  103  re 
Ro = RC = 2k V0 R C

Vi re ie
23.
Sol: Given IE = 1mA re = 25
+12V
25mV
 re   25
1mA
20k R ~ Vi
AV = C
re
10k // 10k 5000
CB AV    200
25 25
CE R 0  R C  10k
+ Vo Ri = re = 25
Vi 20k 8k
i v R
– AI  0  0  i
ii R L vi
R 200  25
 AV  i   0.5
Apply KVL at input Loop: RL 10 4
6 – 10k (IB) – 0.7 – 8 k(1+)IB = 0
25.
6  0.7 Sol: For the given differential amplifier,
IB =  6.47 A
10 k  8 k  101 IE = 1mA
IE = 0.65 mA V
re  T  25
V 25 IE
re = T   38.5  V  R c  3000
I E 0.65 Ad  0   (or) –gmRc
Vi re 25
Apply small signal analysis Ad = –120
V RE 26.
Av = 0  = 0.995
Vi re  R E Sol: Io

Ri = RB ||  R E Total 3k

R E Total = (RE + re) VB +

0.7 –
Ri = 10 k || 803.85k = 9.87 k 9k 8.3k
R0 = RE || re = 38.3 
–12V
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0   12
I1   1mA 1mH
12k
0  VB –1000
I1 
3K Leq
VB = –3V
VB – VE = 0.7 1mH
L eq   1H
VE = VB – 0.7 1  1000
VE = –3.7 Volt
1F
 3.7  12
I0  –106
8.3k
I o  1mA Ceq
I E  0.5mA

C eq  1F 1  10 6  1F 
25mV
re   50
0.5mA 30.
 R C  2000 Sol: Maximum power across the Transistor will
Ad  
re 50 be at the middle of active region
V
A d  40 Ideally at VCE  CC
2
27. VCE = 12V
Sol: Voltage shunt feedback amplifier and 24  12
IC   1.5mA
8k
V0  R f  10k
   10 PTmax = VCE  IC
Vin RS 1k = 12  1.5
= 18mW

28. 31.
Sol: Current – series feedback amplifier and  R 
 R C  4.7k Sol: V0  1  f Vi
AV    1.4242  R1 
RE 3.3k
 2k 
V0  1  2
29.  3k 
Sol: 10
V0  volt
1k 3
2k
–100 3k
 I2
Req Vo
+ Io I1
using millers effect, 1k
2V
1k
R eq   9.9
1  100
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: 12 : Analog Electronics

V0 10 V01
I1   mA &  I 2  V02
1k 3 1
10 V01  I 2  V02
2
V 2 3 2
I2  0   mA  I1  I 2  V02
2k 2k 3
 I 0  I1  I 2  4mA V02  (I1  I 2 ) volt
I01  I1  I x
32.  V01 
 R2 I01  I1  V01  I x  1 
Sol: V0  Vin  
R1
I01  V01  I1

33. 1k
I01  2I1  V01  I1 
Sol: I 02  I 2  I x 

I02  I 2  V01 
+ Io
Iin V0
I02  I1  I 2 A
– + 0.5k
IL 2k

V0 = –Iin 1K 35.
I  1K Iin Sol: 0.5k
IL  i  I1 2k
2K 2

I0 + Iin+ IL = 0 Va Vb
3k V
b
+ Vo
I 1mA
I0  Iin  in  0 I2
2 1k
2I0  2Iin  Iin = 0
Apply KCL at Va :
2I0 = –3Iin
Va  Vb Va  Vb
I0  3 1m  
 = –1.5 2k 3K
Iin 2
3Va  3Vb  2Va  2Vb
1m 
6k
34.
Sol: 6 = 5Va – 5Vb
1 6
 Va – Vb =
+ 1
I1  1 5
– + I01 V01 +
Ix I2 Va  Vb  1.2Volt
– + I02 V02
Va  Vb 1.2
I1    0.6mA
V01 = –I1 2k 2k
Apply KCL: 1.2
I2   0.4mA
0  V02 3k
Ix  I2 
1 Vb = 0.4m  1k = 0.4 Volt

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: 13 : Postal Coaching Solutions

Vb  V0 39.
I1 
0.5k Sol: SR= 2fmax V0max
0.4  V0 SR
0.6m  V0 max 
0.5k 2f max
0.3 = 0.4 – V0 106
  7.95Volt
V0 = 0.1 Volt 2  20 103
V
V0  A  Vi  Vi  0  79.5mV
36. A
I  10  10 3
Sol: VC  .t =  0.5  10 3 40.
C 10 6 Sol:
C
VC = –5Volt R2
R1 L
37. 
+ Vo
Sol: Given open loop gain = 10 Vin 
– +
 Rf 
1  
V0  R 1 
 1 R2
Vi  R  1 z2 = R2|| =
1  1  f   sC sCR 2  1
 R1  A0L
z1 = R1 + sL
V0 1  3 R2

Vi 1  4 V0 sCR 2  1
10 
Vi R1  sL
4
V0  Vi  V0 R2
4 
1 Vi (sCR 2  1)(R1  sL)
10
2 4 It represent low pass filter with
V0   5.715Volt
4 R
1 D.C gain = 2
10 R1

38. 41.
 Rf
V0 R1 Sol: (i)
Sol: 
Vi
1
1  R f / R 1 
A OL 
V0 9 Ii Vi
 Vi +
Vi 10 I1 20k
1
10 Vx
V0  9 Ii 10k
 I2 5k
Vi 2 Ri
.V0 = –4.5Volt

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: 14 : Analog Electronics

Apply KCL at Vx : R 2  R1  R 2 
Ix =  
Vx
 Ii  I1 R 2  R s  R1R 2 
5k
Vx Vi  Vx Vi  Vx 1  Rs  R2 
R0   R1
5k

10k

20k I x  R1  R 2 
Vx 3Vi  3Vx

5 20 42.
3 Sol: VE = Vin
Vx = Vi
7 VCE = VC – VE
V  Vx VCE = 15 – Vin
Ii = i given Vin 0 to 5 Volt
10k
3 Transistor is in active region
Vi  Vi Vin  15 17
Ii = 7 IE = I0 =   1.7 A [Vin = 2V ]
10k 10 10
Vi I0 1.7
 17.5 k IB = = A
Ii 1   100
VB = Vin + 0.7 = 2.7V
(ii) V  VB
Vp
 IB = op
R1 Ix 100
Vp
+ Vp Vop  2.7 1.7

I1 100 100
R2
Vop = 4.4 Volt
RS + 1V
–

43
Sol:
1 (a) n-stage
R0 =
Ix
Rs K K K
Vp =
R2  Rs
K
LPF single stage gain =
1  Vp 1  Vp f
Ix =  1 j
R2 R1 fc
n
 1 1   
Ix = (1–Vp)     K 
 R 2 R1  For n stages gain =  f 
1  j 
 R s  R1  R 2   fc 
Ix = 1    
 R 2  R s  R1R 2  3dB cut-off frequency (f3dB) is given by

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: 15 : Postal Coaching Solutions

 
n
1M
f 3dB  100kHz , f3dB = 100kHz(for
  10
n  K 
K single stage)
 2 
2  1   f 3dB  
  f   9k
 c  9k
 
1k
 f3dB = fc 21/ n  1 –
– V0
1k +
K +
K
2
1

f
f3dB ft = Kf3dB Overall BW = f3dB 21/ 2  1
= 100k (0.04)
(b) Gain = 40dB = 100, ft = 1MHz = Gain BW = 64 kHz
f 106
BW  f 3dB  t   10kHz 44
Gain 100 Sol:
(a)
R2=1M 
100
R1
100 –
2
+ V0
+
10kHz 1MHz Vin
f –
99k

V0 1M
– Gain= 1  100  R 1  10.1k
1k Vin R1
+
IB
1M
10.1k
–
IB oV 100 V0
I=0 +
(c)
Gain=10 Gain=10
ft=1MHz ft=1MHz

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: 16 : Analog Electronics

V0=IB(1M) 1
 100nA1M 
=100nA(1M) 10
= 0.1V = 0.01V
(b) (d)
 op-amp draws current R2
 op-amp CKT the curve doesn’t pass
through ‘0’ (transfer characteristics) R V0
1
–

1M  +

10.1k R2//R
–
100nA +
+ Vo
–s
+
V0s=1mV
– V0  V0 Offset Voltage  V0 Bios current
= 0.1 + 0.01
V0  V0 Bios current  V0 Offset Voltage = 0.11

 R  45.
 1MI B   1  2 Vos Sol: Given
 R1 
R1 = R3 = 10k
= 1M (100nA) +100(1mV)
R2 = R4 = 1M
= 0.2V
(c) R2 R2
Vin1 R1
–
R1 Vin2 V0
– +
R3
+
R4

Rcomp = R1//R2 R2
+ V0  Vin 2  Vin1 
R1
Vin 1M
–  Vin 2  Vin1 
10k
 Rcomp = R1//R2, then V0 = (IB1 – IB2) R2
Given Vos = 4mV
= Ios R2
IB = 0.3 A
V0 = (IB1 –IB2) R2
Ios = 50 nA
= Ios R2
= 1/10 (IBR2)
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: 17 : Postal Coaching Solutions

R2 1
=
1
[3  SCR  ]
R1 SCR
–
1
= (S = j)
+  1 
3  j RC  
+  RC 
Vos
– Rx
R

R3 R4
Vo
+

 R  Vf
V0  1  2  Vos  I os R 2
 R1 
 1M  V0 R
 1  4mV  50nA1M  A= = 1+ x
 10k  Vf R

= 454mV Loop gain =1  A = 1/

46. A = 1
Sol:
Vx R Vf Rx  1 
1+ = 3 + j  RC  
1/sC + R  RC 
+
V0 R 1/sC Vf
Equate img. parts
–
– 1
0 = RC –
RC
KCL
1
2 =
Vx  V0 Vx Vx  Vf R C2 2
  = 0 -------(1)
(1 / SC) R R 1
f= frequency of oscillation
2RC
Vf  Vx Vf
 = 0 -------(2)
R (1 / SC)
Equate A=3

Rx
From (1) and (2) eliminate Vx 1+ 3
R
Vf SCR = 1/3
= = 2 2 2 Rx = 2R
V0 [S C R  3SCR  1]

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: 18 : Analog Electronics

47.  96
Vth – VC = 2103I  I  
Sol: 0 =
1
 3 k 
LC
Vth – VC = 2 V
VF 0.5 k VC = Vth – 2 = 4 V

V0 R x  0.5 1
Vtrigger = VCC = 3 V
9k 3
A = 1  10
1k VC = 3 V to 4 V
A = 1 for sustained oscillations
0.5 k 50.
 10  1 Sol:
R x  0.5 k + 4V –
 Rx = 4.5 k +
+
Vi  RL Vo
48. –
1 –
Sol: Given  =
6 Vi = 8 sint V
R During –Ve cycle, Zener is Forward
A = 1 2
R1 biased and act as short circuit.
A = 1 for sustained oscillations V0 = Vi
 R2  1 During + Ve cycle,
1  .  1
 R 1  6 For 0 < Vi < 4, Zener OFF Since
R2 Zener is not in break down
=5
R1 V0 = 0
R2 = 5 R1 For Vi > 4, Zener is in break down.
V0 = Vi – 4
Vi
49.
Sol: 8
+9V

4
R1=3k
VCC
Threshold
R2 =2k
Output
trigger
–8
Discharge
V0
+ 3k
VC 4
–

2 2
Vth = VCC =  9  6 V –8
3 3
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51. 30  10
Sol: I z max   0.22A I L min  0A 
90.9
I 300
+ Pz = Vz Izmax
+ + Pz = 10  0.22
VS  10V 1K Vo Pz = 2.2W
– –
–
53.
+18V
Iz = 1mA to 60mA Sol:

Vs  Vz I1
I 300
300
Vs min  10 +
V0
I min  _______(I) Iz 10
300 – RL 1k
Vsmax  10
I max  _____(II)
300
VB = 10volt
 V  VE = 10 – 0.7 = 9.3volt
Imin = Izmin + IL  I L  z  10mA 
 1k 
IE = 9.3mA
Imin = 1mA + 10mA = 11mA IE 9.3mA
IB    92.07A
Imax = 60mA + 10mA = 70mA 1  101

From equation (1) and (2) required range of 18  10

I1   26.67mA
VS is 13.3 to 31 volt. 300
I z  I1  I B  26.57 mA
52.
54.
Sol: RS Sol:
+ IL 0 –100mA 30Volt IE
Vo
Vs:20 to 30volt  RL
10V – IL
300

I1 1k 100
+
The current in the diode is minimum when + –
the load current is maximum and vs is 10V
–
minimum. VP
V  Vz I1 5k
R s  s min Vp = 10volt
I z min  I L max
10
20  10 I1   2mA
Rs  5k
10  100mA V0 = (6k) I1=12V = VE
Rs = 90.9 VC = 30volt
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: 20 : Analog Electronics

 VCE = VC – VE =18 volt. V0 = –gmVgs (rd||RD)

IE = I1 + I L
 AV = –gm(rd||RD)
12
IE = 2 m   122mA  V 
2
100 I D  I DSS 1  GS 
 VP 

 IC  IE VG = – 2V
1 
VGS = –2 – 0 = – 2V (VS = 0V)
 IC = 0.120Amp  - 2
2
I D  10m 1  
 PT = IC  VCE  -5
 PT = 2.17W ID = 3.6 mA
2 ID IDSS
55. gm 
VP
Sol:
20Volt 2 (3.6m)(10m) 2  6m
   2.4ms
5 5
+ IC 1k  AV = – (2.4  10–3)[30 k || 6.8 k]
5V VP V0
– I = AV = – 13.3
VP I=0 – Vgs
+ R2
10k 10 6
I Vgs = Vs
RL 10 6  10 4
= 0.99 Vs
Io
V0
20  5 15 = AVs
I  mA Vs
10k 10 = – 13.30.99
VP = 10k I = 15volt = –13.16
VG = – 2V
20  VP 20  15
IC    5mA VGS = –2 – 0 = – 2V (VS = 0V)
1k 1k 2
 - 2
I D  10m 1  
 large  IB  0A  -5
 IC = I0 = 5mA ID = 3.6 mA
2 ID IDSS
56. gm 
VP
Sol: Given IDSS = 10 mA, VP = – 5V,
VGG = – 2V and rd = 30 k 2 (3.6m)(10m) 2  6m
   2.4ms
10k
5 5
G D
 AV = – (2.4  10–3)[30 k || 6.8 k]
RS + + +
V0
RG
Vin
RD
= AV = – 13.3
Vs Vgs rd=30k V Vgs
=1M gmVgs =6.8k 0
10 6
– – – Vgs = Vs
S 10 6  10 4

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: 21 : Postal Coaching Solutions

57. I DS  V  1 
Sol: DC Equivalent (ii) gm =  2I DSS 1  GSQ   
VGS  VP   VP 
VDD
 0.607   1 
 2(1.65 mA) 1 
Rd  2   2 
IG = 0 = 1.149 ms
+
VGS ID (iii) VG = VGS + IDSRS = 0
RG – R  – 0.607 + 0.8 mA(RS) = 0
S
 0.607
RS =  758.75 
 0.8 mA
AC Equivalent
(iv) Voltage gain (AV) = – gmRd
+
V0 Gain (db) = 20 log AV
Rd
20 = 20 log AV
+ –  AV = 10
Vin RG
–  10 = gmRd
 10 = (1.149 m)Rd
Device equation  Rd = 8.7 k
2
 
(i) IDS = IDSS 1  VGS 
 VP 
2
 0.8 mA = 1.65 mA 1  VGS 
  2 
 VGSQ = – 0.607 V

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