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GATE-2014Exam Solutions (15 Feb)Electronics Engineering(Morning Session)

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Section - I (Technical)

Q.1 In the voltage regulator shown in figure op-amp is ideal. The BJT has VBE

=0.7V

and = 100 and Zener voltage Vzis 4.7 V for a regulated output of 9 V the valueof R in is

+

V = 4.7 Vz

1 k

V = 12 Vi V = 9 Vo

1 K = R1

R = R2

Solution: (1093.0232)

Given circuit is a op-amp series regulator

Vo is given by

Vo = 1

z

2

R1 V

R

+

9 V =2

1k1 4.7

R

+

R2 = 1093.0232

Q.2 A depletion type N-channel MOS is biased in its linear region to use as a voltage

controlled resistor. Assume Vth = 0.5 V, VGS = 20 V, VDS = 5 V, =W

100,L

COX = 108 F/m2, n= 800 cm

2/V-s. Find the resistance of voltage control resistor

in ().

Solution: (641025.641)

Voltage controlled resistor rDSis given by

rDS = n OX GS t

1

W( C ) (V V )

L

rDS = 4 81

800 10 10 100 (20 0.5)

rDS = 641.02 k


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Q.3 Capacity of binary symmetric channel with cross-over probability 0.5 ____.

Solution: (0)

Channel capacity of BSC is

C = Plog2P + (1 P) log2(1 P) + 1

C = 0.5log20.5 + 0.5log20.5 + 1

C = 0

It is the case of channel with independent input and output, hence C = 0.

Q.4 In BJT transistor VBE = 0.7 V and VT = 25 mV and reverse saturation current is

1013A. Find the transconductance in mA .V

Solution: (5785.0282)

We know that

gm =c

T

I

V

where Ic =

BE

T

V

VsI e

So, Ic =

0.713 0.02510 e

Ic = 144.6257 mA

Hence, gm =144 mA

5785.02820.025 V

=

Q.5 Find the RMS value of the given pulse

0

1

T/2 T t3T/2


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Solution: (0.4082)

RMS value = T

2

0

1 f (t)dtT

where T is time period

= +

2T/ 2 T2

0 T /2

1 2t dt (0) dt

T T

= T/2

2

20

1 4t dt

T T

So, RMS value =1

or 0.4086

Q.6 Let x(n) =

n n1 1

u(n) u( n 1)9 3

ROC of z-transform is

(a) 1 1Z3 9

(d) does not exist

Solution: (c)

x(n) =

> >1 1

Z3 9

Q.7 The amplifier shown in figure. The BJT parameters are, VBE = 0.7 V, = 200 V,

VT = 250 mV. Find the gain =o

i

V_______.

V


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1 F

Vi

R1 RC = 5 k

R2 11 k

10 kRs

Re 11 k

33 k

1 mF

V = +12 Vcc

Solution: (0.4889)

RB

re

ib

+

Vi= R R1 2

ib

R = 10 ks

R = 5 kC

+

V0

Vi =ibre+ (ib+ ib) RsVi =ibre+ ib(1 + ) RsVi = ib[re+ (1 + ) Rs] ...(i)

V0 = Rcib ...(ii)

0

i

V

V=

c b

b e s

R i

i [ r (1 )R ]

+ +

AV =c

e s

R

r (1 )R

+ + ...(iii)

where re is given by

re =T

E

V

I


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from dc analysis

V = 3 Vth

R = 8.25 kth

10 k

11 k

5 k

12 V

Rth = 33k 11k = 8.25k

Vth =11 12

3 V44

=

3 = 8.25k IB+ 21k IE

3 =E

E

I8.25 21 k I

1+

+

3 = E8.25

I 21 k201

+

IE = 0.142 mA

re =25 mV

0.142 mA

re = 176.0563

So, A v =5 k 200

200 176.0563 201 10 k

+

= 0.4889

Q.8 A transmission line has characteristic impedance is 50and length l=/8. Ifload ZL = (R + j30), then what is the value of R, if input impedance of transmissionline is real is ______ .

Solution: (40)

Zin = + +

l

l

L oo

o L

Z jZ tanZ

Z jZ tan


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Here =

=

l2

and8

tan l =

=tan 14

Thus, Zin = + + +

= + +

L oo

o L

Z jZ R j30 j50Z 50

Z jZ 50 jR 30

=++

50(R j80)

(20 jR)

= +

+50(R j80)

(20 jR)

For Zinto be real

Zin = + +

=+ +2

50(R j80) (20 jR) 50(R j80) (20 jR)

(20 jR) (20 jR) (R 400)

= +

+50(R j80)

(20 jR)

For Zinto be real

Zi = + +

=

+ +2

50(R j80) (20 jR) 50(R j80) (20 jR)

(20 jR) (20 jR) (R 400)

jR2+ j1600 = 0

or R = = 1600 40

Q.9 Which of the following equation is correct?

(a) [ ]>22E[x ] E(x) (b) [ ]

22E[x ] E(x)

(c) [ ]

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