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E

k

E

k

Detailed solution of IES 2014 (ECE) Conventional PaperI

Sol. 1 (a)

(i) Conductivity is a degree to which a specified material conducts electricity and gives idea how much

smooth flow is of electricity by a carrier. Mobility is degree to which specified material can move freely

and easily and it gives idea about ability of movement of a carrier.

(ii) Zener break down occurs in highly doped Zener diode and it is due to tunneling phenomenon whileAvalanche breakdown occurs in lightly doped Zener diode and it is avalanche multiplication due to

successive collisions of electrons in depletion region of Zener diode. Zener occurs at smaller value of break

down voltage while avalanche occurs at higher value of break down voltage.

(iii) Piezo-electric materials which are insulators become electrically polarized in presence of mechanical

stress and produce voltage which is reversible process. Ceramic materials are in organic materials and are

combination of metal and non metals which are generally formed by action of heat and subsequent cooling.

(iv) Direct band gap:

Minima of C.B coincides with Maxima of VB and here energy is emitted in form of light by photons

Indirect band gap:

Minima of C.B does not coincides with Maxima of VB and here energy is emitted in form of heat.

(v) Polarisability is the ability of molecules to be polarized and express as dipole moment per unit electric

Field. Permittivity is the measure of resistance which is encountered when forming and electric field in

a medium. Unit of polarisability is F-m2 while permittivity has no unit.

Sol: 1(b)

(i) Given condition that NMOS is in Saturation region

2

D GS TI K V V GS TV 2.1V & V 1V

Then 23

DI 0.8 10 2.1 1 =0.968 mA which is nearly equal to 1 mA

(ii) dm GS Tgs

dIg 2K V V

dV

3mg 2 0.8 10 1.1 =1.76mA/ V

(iii) If VI=10 mV then new VGS=2.11 Volt so transistor will still remains in saturation region

2

D GS TI K V V =

230.8 10 2.11 1 0.9856mA

Output voltage 0 DD DV V RI 9 0.9856mA 2 7.0288Volt

Sol.1(c):j

A D

2V 1 1W

q N N

Since A DN N soA D

1 1

N N

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2

+

CLK

CHold

Vout

Vin

1

So herej

A

2VW

qN

j p

p A p

p

2VBut qN so W

Sol.1(d) :To synthesize a driving point immittance function z(s) the first step is to decompose it into a sum of simpler

positive real functions z1(s), z2(s), z3(s), z4(s),.zn(s), and then to synthesize these individual z(s) as

elements of the overall network whose driving point impedance is z(s).

A function is said to be positive real function if it satisfies the following conditions:

1. F(s) is real for real s i.e F() is real

2. F(s) may have only simple poles on the jw axis with real and positive residues

3. Re F(jw) 0 for all w

6 s 3 s 9 27 9Z s 6s s 6 s (s 6)

Sol.1(e): 2t th t e u t , x t e u t

Y s X s .H s

1 1

Y ss 1 s 2

t 2ty t e u t e u t

Sol.1(f): when clock is high complete circuit responds similarly to an OPMAP in unity gain feedback

configuration when clock is low input voltage at that time is stored on capacitor. By use of OPAMP in

feedback loop input impedance of sample and hold circuit is greatly increased.

Figure:

Sol.1 (g):Propagation constant (P) R j L G j C

As frequency is not mentioned so problem cant be solved

Sol.1 (h):It is a transducer which uses change in the electrical resistance to measure strain. Here electrical

resistance is proportional to instantaneous spatial average strain over its surface.

Applications :

1.

Vibration measurement

2. Compression and tension measurement

3. Contractions in muscles in medical science

4.

Blood pressure measurement5. Used in volumetric differential low pressure

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3

5V

D1

V1

D2

V2+ +

Force

R

R

Active gauge

Dummy or

compensatinggauge

Temperature compensation in strain gauge:

The active strain gauge is installed on the test specimen while the dummy or compensating gauge is installed

on a like piece of material and is not specified to strain.

Sol. 3(a) As 2inp n

A Dn N p N

butA

N 0

Dn p N

2D i2 2

D D i

n n N n

N N 4nn

2

But n cannot be negative so

2

DD i

i

NN n 4

nn

2

AsD i

N n

2

D

i

N

n

can be neglected.

D iN 2n

n2

2 22

D D i D

i

N N 4n Np neglecting

2 n

D iN 2n

p2

Sol. 3(b)

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n type S.C

light

x < 0 x > 0x = 0

10cm

D1 is R.B and Non conducting

D2is F.B but Non conducting

D1is not in breakdown as VB= 50 V but 1 25 V V

Both diodes are in series, in D1 reverse current will flow from N to P while in D2current will flow

from P to N so here1 2

I I

2

T

V

V

0 0 2 TI I e 1 V 2V n2

= 34.66 mV=0.03466 Volt

So here 1 2V 5 V 4.9653V

Sol. 3(c)

0

p p

p pdp p

dt

since it is N-type SC so major change will be in concentration of holes only

and not electrons.

Here given 21 310 / / secdp

EHP cmdt

So 21 6 1510 10 10 /p cc n

here, dx = 34.6 m

here; Initially in N type s.c, holes are minority and only contribution which is dominant is

after the following of light.

So dx = 34.6 m

3 2

pJ 5.536 10 A/m

And p pI J area

pI 5.536mA

Now, Initially 15 3DN 4.5 10 /cm

Now 21

n

n10

Then 15 3n 10 / cm

Change in eis 15 15n ' 4.5 10 10 15 33.5 10 / cm

Thus Applying continuity equation.

15

n

6

dJ1 3.5 10

q dx 10

Now, 3 2nJ 19.376 10 amp / m

n nI J A

nI 19.376mA

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5

VCC

RCR2

R1 RE

R ||R1 2

VT Re

Rc

Vcc

Sol. 3. (d) CC CE E CV 20V, V 8V, V 6V, S 10, 200, I 5mA

Now

E

E B

1s

R1

R R

E E EV R I

ER 1.2k

As E CI I ; ' ' is very high

Then CC CE C C E

V V I R R C

R 1.2k

E

E TH

1s

R1

R R

E

E B

R 11

R R S

BR 11.365k

Now, 1 2

1 2

R R11.365k

R R

ET BE 1 2 E EI

V V R || R I R

Then; 1

1 2

R0.3492

R R

1 2

1 2

R R11.365

R R

2R 32.545k

1R 17.462k

Sol. 3.(e) Initially T is switch off then capacitor get charged to 10V in steady state. So capacitor is

charged to 10 Volt . at t=0 capacitor voltage will remain at 10 volt Now T becomes ON by 4

volt Here VDS=10 Volt and VGS=4 Volt so MOSFET will be replaced by resistance then

ds onm 0 m

1 1 1r

g r g where 3m GS Tg 2K V V 2 5 10 2 20mA / V

ds

1r 50

20mA / V Now rdsand C will be in parallel and capacitor will discharge through rds

t / t / 12

0

12

V(t) V e 5 10e where 100 10 50 s

t ln2 100 10 50 0.6932s 3466psec 3.47nsec

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BE V

N

Sol. 4(a) Z transform offers significant advantage relative to time domain procedures. By this we can

model discrete time physical systems with linear differential equations with constant

coefficients one example is linear time variant digital filter. The Z transform of a difference

equation gives us a good description of the characteristics of the equation and hence of

physical system. In addition transformed difference equations are algebraic and therefore

easier to manipulate.

The Z transform of a sampled signal or sequence is defined as : kk 0

Z f (kT) f (kT)z

Sol.4(b) 2

sY s

s s 3s 2

( There is misprint in this question)

2

1Y s

s 3s 2

1 1Y s

s 1 s 2

t 2ty t e u t e u t

Sol.4(d)

E only Educated

B both

V only voters

N None

Probability of educated = 4/10

Probability of both =2

10

Probability of voters =5

10

E B V N 100%

V B 40%B 20%

B E 50%

E = 30%

V = 20%

N= 30%

(i) 2 /10 1VoterP Educated 4 /10 2

(ii)

Not educated 2 /10 2

P Voter 5 /10 5

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100V

s s

I1 I2

[Y]1

[Y]2

(iii) 30 3

P N100 10

Sol.4(e) By duality theorem

f t F

F t 2 f

2sgn t

j

22 sgn

jt

1

sgnj t

Sol.5(a)

Applying KVL in loop 1

1 1 1 2100 / s 10I sI 10 I I (1)

1 2I ,I are in 's ' domain

Applying KVL in loop 2

2 2 2 1sI 10I 10 I I 0

1 2

s 20I I

10

.(2)

Put (2) in (1)

2

10 / 3 5 / 3 5I

s s 30 s 10

30t 10ti t 3.333 1.67e 5e u t

Sol.5.(b) It is a parallelparallel combination hence total Yparameter will be added.

1 2

Y Y Y

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+R1 R2

R3

I1 I2

dR I2 1

+

V2

+

V1

dxx

1

2 / 3 1/ 3Y

1/ 3 2 /3

Similarly 2

2 / 3 1/ 3Y

1/ 3 2 / 3

So 1 24 / 3 2 / 3

Y Y Y 2 / 3 4 / 3

So over all Y parameter of above network is : U

Y + L

Y =

4 2

3 3

2 4

3 3

sol.5(c)

1 1 3 1 3 2V R R I R I ----(1)

2 2 3 1 2 3 2V dR R I R R I ----(2)

We gate

11 1 3 12 3

21 2 3 22 2 3

Z R R , Z R

Z dR R , Z R R

As12 21

Z Z so network is not reciprocal.

1 3 32 3 2 3

R R RZ

dR R R R

Sol. 6 (a) Charge density 2500

20 c / m25

dq 20 2 xdx

2

0

1 dqdE

4 x

6

x 52x 0

0

1 20 10 2 xE dx4 x

61.82 10 N / C 6 6F Q.E 50 10 1.82 10

F 91N

Sol. 6(b) 2 2 20

Q QV a h h where

2 .2 4

9 92 2 2 2

0

Q 40 10 9 10V a h h 2 5 5

8 2

180 0.385 69.3Volt

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9

9Q 40nC 40 10 C where r 5m

Potential0

1 QV

4 r

99 40 10V 9 10 V 72V

5

Sol.6(c)1 0115m, Z 300

2 023m z 150

f 50MHz

8

6

c 3 106m

f 50 10

Length of quarter wave line =4

=

61.5m

4

Characteristic impedance

0 01 02

0

Z Z .Z 300 150

Z 212.13

Reflection coefficient 02 01

02 01

Z Z

Z Z

150 300 150 1

150 300 450 3

11

1 | | 43VSWR

11 | | 213

VSWR 2

Sol.6(d) a b = 2.3 cm 1.0 cm

Operating frequency f = 1.5 fc

Guide wave length g= ?, Phase velocity vp= ?

Asa

2b

So first five modes will be

10 20 01 11 02TE , TE , TE ,TE and TE

Cut of frequency

2 2

c

r r

c m nf

a b2

For TE10mode

2

c 10

c 1f TE 0

2 1 2.3

= 6.52 GHz

cf 1.5f 1.5 6.52 9.78GHz

10

9

c 3 103.067cm

f 9.78 10

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g2 2

c c

c

3.067

f f1 1

f 1.5f

g

3.0674.12cm

0.7453

p g

9 2

p

8

p

v f

v 9.78 10 4.12 10 m /s

v 4.024 10 m /s

Similarly we calculate guided wavelength (g) & phase velocity (vp) for next modes

g p8

10

8

20

801

8

11

8

02

Mode Guided wavelength Phase velocity v

TE 4.12cm 4.02 10 m / s

TE 2.06cm 4.02 10 m / s

TE 1.79cm 4.02 10 m / s

TE 1.64cm 4.02 10 m / s

TE 0.89cm 4.02 10 m / s

Sol.7(a)

(i) Absolute error= Actual value-measured value =20-18=2 mA

(ii)

Percentage error =Absoluteerror 2

100% 100% 11.11%Measured value 18

(iii)Relative Accuracy =Actual value Absoluteerror

Actual value

=20 2

0.920

(iv)Percentage accuracy =90%

(v) Precision=16 19 20 17 21 18 15 16 18 17

17.710

Sol. 7(b)

Total deflection in beam is given by d2

eV L Ld D

m s v 2

where 2 a

2eVv

m

Here s is separation between deflecting plates and s=5mmHere D is distance of screen from centre of plates and D=30 cm

Here L is length of each deflecting plate and L=2 cm

Here d=3 cm and Vd should be calculated

192 15a

31

d

2

2eV 2 1.6 10 2000 6.4v 10

m 9 10 9

eV L Ld D

m s v 2

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d

a

2 22 2d

3

V L Ld D

s 2V 2

V 2 10 2 103 10 30 10

5 10 2 2000 2

Solve Vdfrom here and that will be answer :

Sol. 7(c) Given 0 40R 4k , R 800

1 2T T

1 2

1 1R R exp

T T

So0 40

1 1R R exp

273 313

3438.96per K

Now0 50

1 1R R exp

273 323

50R 569.1

And for R100 0 1001 1

R R exp 3438.96273 373

100R 136.57

So range of resistance 569.1 to 136.57

Sol.7(d) Given,

2 4 2A 2.5cm 2.5 10 m

Separation d 33mm 3 10 m 4 2P 10 N / m , deflection = 0.3 mm = 43 10 m

C = 300 PF

Capacitance after a pressure of 104N/m2

AC

d

3A 300PF 3 10

3

3

A 300PF 3 10

C' d x 3 0.3 10

C' 333.33PF