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ECE

IES MASTER

Ans. (d)

Sol. If f(x) = x

1 2 1 2x x x x ; for x 1

if f(x) = e–x

1 2 1 2(x x ) x xx e e ; for x 1

if f(x) = 1x

1 2

1x x

1 2

1 1 ; for x 1x x

if f(x) = ex

1 2x xe 1 2x xe e ; for x 1

Hence, option (d) is correct.

3. The Canadian constitution requires that equalimportance be given to English and French.Last year Air Canada lost a lawsuit, and had topay a six-figure fine to a French-speaking coupleafter they filed complaints about formal in-flightannouncements in English lasting 15 seconds,as opposed to informal 5 second messages inFrench.The French-speaking couple were upset at______.(a) the in-flight announcements being made in

English(b) the English announcments being clearer

than the French ones.(c) equal importance being given to English and

French.(d) the English announcments being longer

than the French ones.Ans. (d)

4. The following figure shows the data of studentsenrolled in 5 years (2014 to 2018) for two schoolsP and Q. During this period, the ratio of theaverage number of the students enrolled inschool P to the average of the difference of thenumber of students enrolled in schools P and Qis _____.

GENERAL APTITUDE1. A circle with centre O is shown in the figure.

A rectangle PQRS of maximum possible area isinscribed in the circle. If the radius of the circleis a, then the area of the shaded portion is ____

P Q

RS

O

(a) 2 2a 2a (b) 2 2a 2a

(c) 2 2a 3a (d) 2 2a a

Ans. (a)Sol. radius of circle is a.

area of circle = 2a

area of shaded region = area of circle –area of square

P Q

RS

2aO

side of square = 2a

area of shaded portion = 2 2a ( 2a)

= 2 2a 2a Hence, (a) is correct.

2. A superaddivitive function f ( ) satisfies thefollowing property

1 2 1 2f(x x ) f(x ) f (x )

Which of the following function is asuperadditive function for x > 1?

(a) x (b) e–x

(c) 1/x (d) ex




ECE

IES MASTER

Num

ber o

f stu

dent

s (in

thou

sand

s) 9876543210

2014 2015 2016 2017 2018Year

School PSchool Q

(a) 31 : 23 (b) 8 : 23(c) 23 : 8 (d) 23 : 31

Ans. (c)Sol. average number of students enrolled in

school P = 235

average of difference of the number ofstudents enrolled is schools P and Q

= 1 31 235

= 85

=

average number of students enrolled in school P 23

average of difference of 8the number enrolled in

school P and Q

5. He was not only accused of theft ___ ofconspiracy.(a) rather than (b) but also(c) rather (d) but even

Ans. (b)

6. The untimely loss of life is a cause of seriousglobal concern as thousands of people get killed_____ accidents every year while many otherdie ___ diseases like cardio vascular disease,cancer, etc.(a) during, from (b) from, from(c) from, of (d) in, of

Ans. (d)

7. Select the word that fits the analogy:Explicit: Implicit : : Express: ______(a) Impress (b) Repress(c) Compress (d) Suppress

Ans. (d)

8. It is quarter past three in your watch. Theangle between the hour hand and the minutehand is _______.(a) 22.5° (b) 0°(c) 15° (d) 7.5°

Ans. (d)

Sol. = 11m 30h

2

quarter past three = 3 : 15

=11 15 30 3 7.5

2

9. a, b, c are real numbers. The quadratic equationax2 – bx + c = 0 has equal roots, which is ,then

(a) 3 2bc / (2a ) (b) 2 ac

(c) b2 4ac (d) = b/a

Ans. (a)Sol. ax2 – bx + c = 0

roots of the equation = 2b b 4ac

2a

Let , be the root of the equation.

2 = 2b c,a a

32 = 2bca

3

2bc

2a

10. The global financial crisis in 2008 is consideredto be the most serious worldwide financial crisis,which started with the sub-prime lending crisisin USA in 2007. The sub-prime lending crisisled to the banking crisis in 2008 with the collapse




ECE

IES MASTER

of Lehman Brothers in 2008. The sub-primelending refers to the provision of loans to thoseborrowers who may have difficulties in repayingloans, and it arises because of excess liquidityfollowing the East Asian crisis.Which one of the following sequences shows thecorrect precedence as per the given passage?

(a) Global financial crisis East Asian crisis banking crisis subprime lendingcrisis.

(b) Subprime lending crisis global financialcrisis banking crisis East Asian crisis.

(c) East Asian crisis subprime lendingcrisis banking crisis global financialcrisis.

(d) Banking crisis subprime lendingcrisis global financial crisis East Asiancrisis

Ans. (c)

ECE TECHNICAL1. In an 8085 microprocessor, the number of

address lines required to access a 16 k bytememory bank is _____

Ans. (14_14)

Sol. Size of memory bank = 16kB = 24 × 210B

= 214B

If N is the length of address line then the sizeof memory it can address is 2N B.

So, 2NB = 214B

Hence, N = 14

2. The random variable

Y W(t) (t)dt,

where

1; 5 t 7(t)

0; Otherwise

and W(t) is a real white Gaussian noise processwith two-sided power spectral density SW(f) = 3W/Hz, for all f. The variance of Y is ______.Given __

Ans. (6_6)

Sol. Given Y = W(t) (t)dt

where 1; 5 t 7

(t)0; otherwise

So,

Y = 7

5W(t) dt

Variance of Y is

7 721 1 2 25 5

E(Y ) E W(t )dt W(t )dt

E(Y2) = 7 7

1 2 1 25 5E(W(t )W(t ))dt dt …(i)

E(Y2) = 7 7

W 2 1 1 25 5R (t t ) dt dt

Given SW(f) = 3W/Hz

W W 2 1 2 1R ( ) R (t t ) 3 (t t )

So from (i)

E(Y2) =7 7

2 1 1 25 53 (t t ) dt dt

= 71 1 15

3 4(7 t ) u(5 t ) dt = 3 × 2 = 6

3. In the circuit shown below, all the componentsare ideal and the input voltage is sinusoidal.The magnitude of the steady-state output Vo(rounded off to two decimal places) is ______V.

C = 0.1 µF1

D1

D2

C = 0.1 µF

2230 V (rms) V0

Ans. (644_657)Sol–3:

Given circuit is voltage doubler,

230(rms)

C =0.1 f1

+–Vm

D1

C =0.1 f2 +– 2Vm

+

–

+

V=V

=2V

0C2

m

–

D2




ECE

IES MASTER

Thus, voltage drop across C1, is Vm and voltagedrop across C2 is 2Vm.

Vm = Vrms 2

= 230 2

= 325.269

Vout = VC2 = 2Vm = 650.5 V

4. The output y(n) of a discrete-time system for aninput x[n] is

k n

y n max x k

The unit impulse response of the system is(a) unit step signal u[n](b) 0 for all n.

(c) unit impulse signal [n]

(d) 1 for all n.Ans. (a)

Sol. Given y[n] = k nmax x[k]

In question x[k] = (k)

so, y[n] = k nmax [k]

when n < 0

y[n] =k n

max [k]

= 0

when n = 0

y[n] = (0) 1

when n > 0

y[n] =0 kmax (n) 1

so, y[n] =0 n 01 0 n

So, y[n] = u[n]

5. A single crystal intrinsic semiconductor is at atemperature of 300 K with effective density ofstates for holes twice that of electrons. Thethermal voltage is 26 mV. The intrinsic Fermilevel is shifted from mid-bandgap energy levelby

(a) 13.45 meV (b) 18.02 meV(c) 9.01 meV (d) 26.90 meV

Ans. (c)Sol–5:

Given, T = 300 K, Nv = 2NC, VT = 26 mVwhere NV = Effective density of states of holesNC = Effective density of states of electrons

Gfi

E E2 = C

V

NkT n2 N

Given KT = VT = 26 mVNV = 2 NC

Gfi

E E2 =

3

C

C

N26 10 n2 2N

= 313 10 n 0.5

= –9.01 × 10–3 eV= –9.01 meV

Gfi

EE2 = 9.01 meV

Thus intrinsic fermi level is shifted from mid-bandgap energy level by 9.01 meV

Therefore option (c) is correct.

6. If v1, v2, … v6 are six vector in 4 , which oneof the following statements is FALSE?(a) It is not necessary that these vectors span

4

(b) These vectors are not linearly independent.

(c) If {v1, v3, v5, v6} spans 4 , then it forms a

basis for 4 .(d) Any four of these vectors form a basis for

4

Ans. (d)

7. A digital communication system transmits ablock of N bits. The probability of error indecoding a bit is . The error event of each bitis independent of the error events of the otherbits. The received block is declared erroneous if

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ECE

IES MASTER

at least one of its bits is decoded wrongly. Theprobability that the received block is errorneousis

(a) N1 (1 ) (b) N1

(c) N(1 ) (d) N

Ans. (a)Sol. Given

size of block = N bitsAll bits are independentGiven probability (Pe) of error in a bit = As block is decleared erroneous if at least oneof bits is decoded wrongly.So, probability Pr of right block =

NN 0 N0C 1 (1 )

So, Ne rP 1 P 1 1

8. A 10-bit D/A converter is calibrated over thefull range from 0 to 10V. If the input to the D/A converter is 13A (in hex), the output (roundedoff to three decimal places) is _____ V.

Ans. (3.050_3.080)

Sol. Given 10 bit D/A converter have VP–P = 10V

Resolution aD/A =

P PN

V 1010232 1

Input to the D/A converter (13A)16

(13A)16 = (162 + 3 × 16 + 10) = (314)10

So output = Resolution × Input

=

10 314 3.0691023

9. The components in the circuit shown below areideal. If the op-amp is in positive feedback andthe input voltage Vi is a sine wave of amplitude1V, the output voltage vo is

+–

Vi

1k

–5V

+5V

Vo

1k

1V0

–1V

(a) a square wave of 5V amplitude.(b) a non-inverted sine wave of 2V amplitude.(c) an inverted sine wave of 1 V amplitude(d) a constant of either +5V or –5V.

Ans. (d)Sol.

+1O

–1V0

1k

+5VVbVa

+–

–5V

Vi

1k

The given circuit is of schmitt trigger, thusoutput is either +Vsat or –Vsat i.e +5V or –5V.

Applying KCL at Vb

We get,

b i b oV V V V

1k 1k = 0

Vb – Vi + Vb – V0 = 0

o ib

V VV2

Case (i) When V0 = + 5V

Then, Vb = i5 V2

Vb = iV2.52

Given input (Vi) varies from +1V to –1V thus,1. WhenVi = +1V

Vb = 12.52 = 3V

Vd = Vb – Va = 3V – 0V = 3VAs Vd is positive hence V0 = +5V2. WhenVi = –1V

Vb =

12.52

= 2.5 – 0.5

Vb = 2V




ECE

IES MASTER

Vd = Vb – Va = 2V – 0V = 2VAs Vd is positive hence V0 = +5VTherefore when V0 = +5V irrespective of input

variation output voltage remains +5V.Case (ii): When V0 = –5V

Then Vb = i5 V2

= iV2.52

Vb = iV2.52

Given input (Vi) varies from +1V to –1V thus,1. WhenVi = +1V

Vb = 12.52

= –2.5 + 0.5Vb = –2V

Vd = Vb – Va = –2V – 0V= –2V

As Vd is negative hence V0 = –5V2. WhenVi = –1V

Vb =

12.52

= 2.5 – 0.5= –3V

Vd = Vb – Va

= –3V – 0VVd = –3V

As Vd is negative hence V0 = –5VTherefore when V0 = –5V irrespective of input

variation output voltage remain –5V.Thus, option (d) is correct.

10. The general solution of 2

2d y dy6 9y 0

dxdx is

(a) y = C1e3x

(b) y = (C1 + C2x) e–3x

(c) y = C1 e3x + C2e–3x

(d) y = (C1 + C2x) e3x

Ans. (d)

Sol.2

2d y dy6 9y 0

dxdx

(D2 – 6D + 9)y = 0Characteristics equation = D2 – 6D + 9 = 0 (D–3)2 = 0

D = 3,3

y 3,3

For equal roots, y = (C1 + C2 x) e3x

11. The loop transfer function of a negative feedbacksystem is

K(s 11)G(s)H(s)s(s 2)(s 8)

The value of K, for which the system is mar-ginally stable, is ____.

Ans. (160_160)Sol. Given:

K s 11G s H s

s s 2 s 8

CalculatingCharacteristics equation:s(s2+10s+16) + Ks + 11 K = 0s3 + 10s2 + 16s + Ks + 11 K = 0s3 + 10s2 + (16+K)s + 11 K = 0

3

2

1 16 KS10 11KS

S 016 K 10 11K10

For marginally stable system:10 (16 + K) = 11 K16 + K = 1.1 K16 = 0.1 K

K 160




ECE

IES MASTER

12. Consider the recombination process via bulktraps in a forward biased pn homojunction diode.The maximum recombination rate is Umax. Ifthe electron and the hole capture cross-sections are equal, which one of the following isFALSE?(a) Umax occurs at the edges of the depletion

region in the device(b) With all other parameters unchanged, Umax

increases if the thermal velocity of thecarriers increases.

(c) With all other parameters unchanged, Umaxdecreases if the intrinsic carrier density isreduced.

(d) Umax depends exponentially on the appliedbias.

Ans. (a)

13. In the circuit shown below, the Thevenin voltageVTH is

2V

1A

2 4

22A1 VTH

+

–

– +

(a) 3.6 V (b) 2.4 V(c) 2.8 V (d) 4.5 V

Ans. (a)Sol. Case (i): Only current source, 1A is there

and voltage source, 2V and current source,2A are short circuited and open circuitedrespectively.

1A

2 4

21 VTH1

+

–

I2

I1I=0

from current division rule,

I1 = 1 11 A4 1 5

VTH1 = 2 × I1 = 2 × 1 2 A5 5

Case II: Only voltage source 2V is thereand both current source 1A and 2A areopen circuited.

2 4

2 VTH2

+

–

I2– +2V

1

From ciruit, I3 = 2 2 A2 2 1 5

2TH2 4V 2 V5 5

Case III: Only current source, 2A is thereand voltage source 2V and current source1A are short circuited and open circuitedrespectively

2 4

2 VTH3

+

–

I4

1 2A

I5

From current division rule,

I5 = (1 2) 62 A

(1 2 2) 5

3TH 5

12V 2 I A5

Superposition thereom,

VTH = 1 2 3TH TH THV V V

=2 4 12 18 3.6V5 5 5 5

14. For a vector field A

, which one of the followingis FALSE?




ECE

IES MASTER

(a) A

is another vector field.

(b) 2( A) ( A) A

(c) A

is irrotational if 2A 0

.

(d) A

is solenoidal if A 0.

Ans. (c)

Sol. Divergence and curl operator is performed on

a vector field A

.

• Curl operation provides a vector

orthogonal to the given vector field A

.

• 2( A) ( A) A

• If a vector field is irrortational thenA 0

• If a vector field is solenoidal thenA 0

• If a field is scalar A, then 2A 0 , is alaplacian equation.Hence option (c) is incorrect.

15. A binary random variable X takes the value +2or –2. The probability P(X = +2) = . The valueof (rounded off to one decimal place), for whichthe entropy of X is maximum, is ____.

Ans. (0.5_0.5)Sol. Binary raddom variable x takes –2 and +2.

The probability P(X = +2) =

So, P(x = –2) = 1 As we knew entropy is maximum whenP(X = +2) = P(X = –2)

1

0.5

16. The two sides of a fair coin are labelled as -0and 1. The coin is tossed two timesindependently. Let M and N denote the labelscorresponding to the outcomes of those tosses.For a random variable X, defined as X = min(M, N), the expected value E(X) (rounded off totwo decimal places) is ________.

Ans. (0.25_0.25)Sol. X = min (M, N)

When coin is tossed two times then possiblevalues of ‘x’.X1 = 0,0 min(0,0) 0X2 = 0,1 min(0,1) 0X3 = 1,0 min(1,0) 0X4 = 1,1 min(1,1) 1

P(0) = 3 ,4 P(1) = 1

4

E(X) = i iiX P X

= 3 10 14 4

= 1 0.254

17. The pole-zero map of a rational function G(s) isshown below. When the closed countor ismapped into the G(s)-plane, then the mappingencircles.

Xs-plane

Re

Im

X

X

XX

oo

oo

(a) the origin of the G(s)-plane once in thecounter-clockwise direction

(b) the point –1 + j0 of the G(s)-plane once inthe clockwise direction

(c) the origin of the G(s)-plane once in theclockwise direction

(d) the point –1 + j0 of the G(s)-plane once inthe counter-clockwise direction.

Ans. (c)

Sol. According to principles of the Argument :

The corresponding locus mapped in the G(s)-plane will encircle the origin as many times asthe difference between the number of zero andpoles of G(s) that are encircled by the S-planeLocus . So

N = Z – P = 3 – 2 = 1

So, it encircle the origin once in clockwisedirection.

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18. The figure below shows a multiplexer where S1and S0 are the select lines, I0 to I3 are the inputdata lines, EN is the enable line, and F(P, Q,R) is the output, F is

ENI0

I1I1

I2

I3

0R0R1

MUX

S1 S0

P Q

F

(a) PQ QR (b) Q PR

(c) PQR PQ (d) P QRAns. (a)

Sol. Given multiplexer is

I0

I1

I2

I3

R

0

R

1

4 : 1 F

0 EN

P Q

F = PQR 0PQ RPQ PQ

F = PQR RPQ PQ

F = RQ P P PQ P P 1

F = RQ PQ

19. The current in the RL-circuit shown below isi(t) = 10 cos (5t /4)A . The value of theinductor (rounded off to two decimal places) is________H.

+

–200 cos (5t) V

R i(t)

L

Ans. (2.80_2.85)Sol. Given

i(t)R

L200 cos 5t V+

–

V(t) = 200 cos 5t V

=200 0 V

2

i(t) = 10 cos 5t A4

=10 A

42

Here, = 5 rad/secImpedance of the circuit,

z = V(t) 200 / 2 0i(t) 10 / 2

4

= 204

= 20 cos jsin4 4

=20 20j

2 2

= R + jXL

XL =20

2

L =20

2

L =20 4 2.83H2 5 2

20. The partial derivatives of the function

21 x cos y 1/(1 y )f (x,y,z) e xze

With respect to x at the point (1, 0, e) is(a) 1 (b) 0

(c) –1 (d) 1e

Ans. (b)




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IES MASTER

Sol. f(x,y,z) = 21 1 y1 x cos ye xze

21 1 y1 x cos ydf cos ye xz edx

at (1,0,e)

21

1 x cos y 1 ydf cosy e z edx

= –1 + e0

= –1 + 1= 0

21. In the circuit shown below, all the componentsare ideal. If Vi is +2 V, the current Io sourcedby the op-amp is ____ mA

Vi

1k

–5V

5V

1k

1k

IO

Ans. : (6_6)Sol.

1k

1k

5V

bVb

a2V

–

+

I0

1k

–5VGiven Vi = 2V,Thus by virtual ground concept Vb = 2VApplying KCL at node b:We get,

b b oV 0 V V

1k 1k = 0

Vb = 2V2 + 2 – V0 = 0

oV 4V

Applying KCL at V0 node,

We get,

o b o

oV V V I

1k 1k = 0

o4 2 4 I1k 1k = 0

2 + 4 – I0(1k) = 0

I0 (1k) = 6

oI 6mA

22. The impedances Z = jX, for all X in the range( , ) , map to the Smith chart as

(a) a circle of radius 0.5 with centre at (0.5, 0).(b) a line passing through the centre of the

chart(c) a circle of radius 1 with centre at (0, 0)(d) a point at the centre of the chart.

Ans. (c)

Sol. Smith chat is constructed within a circle ofunit radius 1 .

22 11

x =

21x

For Z = jX, R = 0

R = 0 circle is a unit cycle with centre at (0,0).

x

y

(1, 0)(0, 0)

Hence, option (c) is correct.

23. A transmission line of length 3 /4 and havinga characteristic impedance of 50 is terminatedwith a load of 400 . The impedance (roundedoff to two decimal places) seen at the input endof the transmission line is ______ .




ECE

IES MASTER

Ans. (6.25_6.25)Sol. Given,

3 /4 l

Z0, characteristics impedance = 50

ZL = 400We know that,

Zin = L 00

0 L

Z jZ tanZZ jZ tan

ll

for 3 2 3,4 2

ll l =

tan 32

Zin =2 20

L

Z (50) 6.25Z 400

24. Which one of the following pole-zero plotscorresponds to the transfer function of an LTIsystem characterized by the input-outputdifference equation given below?

3k

k 0y[n] ( 1) x[n k]

(a)

3 order polerd1

–1

–1

1 Re

Im

(b)

4 order poleth1

–1

–1

1 Re

Im

(c)

3 order polerd1

–1

–1

1 Re

Im

(d)

3 order polerd1

–1

–1

1 Re

Im

Ans. (c)

Sol.

Given,

y[n] =

3

k

K 01 x n k

y[n] =

2 31 x n 1 x n 11 n 2 1 x n 3

= x n x n 1 x n 2 x n 3Taking Z-transform, we get

Y(z) = 1 2 3X z Z X z Z X z Z X z

Y zX z = 1 2 31 z z z

= 3 2

3z z z 1

zs

= 2

3z 1 z 1

zFrom above derived transfer function,

Number of poles at origin

= 3[i.e., 3rd order pole at origin]

Number of Zeros = 3

Two of the zeros are at imaginary axis andanother is at real axis (z = 1).

Hence, option (c) is correct.

25. In the given circuit, the two-port network has

the impedance matrix [z] = 40 6060 120

. The

value of ZL for which maximum power istransferred to the load is ________ .

[Z]+

–V1

120V ZLV2

+

–

I1 I210




ECE

IES MASTER

Ans. (48_48)Sol. Given,

[Z] = 40 6060 120

V1 = 40 I1 + 60 I2 …(i)V2 = 60 I1 + 120 I2 …(ii)

For maximum power transfer,ZL = RTH

To find the RTH, the given circuit isredrawn as:

[Z]+

–V1 RTHV2

+

–

I1 I2

From the above circuit,

V1 = –10 I1 and RTH = 2

2

VI

Putting V1= 10 I1 in eq. (i), we get– 10I1 = 40 I1 + 60 I2

–5I1 = 6 I2 I1 = 26 I

5

Putting I1 = 26 I

5 in eq. (ii), we get

V2 = 2 2660 I 120I

5

V2 = –12 I2 + 120 I2 = 45 I2

2

2

VI = RTH = 48

26. The transfer function of a stable discrete-time

LTI system is K(z )H(z)z 0.5

, where K and

are real numbers. The value of (rounded offto one decimal place) with 1, for which themagnitude response of the system is constantover all frequencies, is ________.

Ans. (-2_-2)

Sol.

Given

H(z) = k z

z 0.5

We know that, for an all pass filter,

Zero = *1

pole

Pole at Z = 12

Zero =

1 212

From given transfer function, = –2

27. Using the incremental low frequency small-signal model of the MOS device, the Nortonequivalent resistance of the following circuit is

R

VDD

g , rm ds

(a) dsm

1r Rg

(b) rds + R

(c) ds

m ds

r R1 g r

(d) ds m dsr R g r R

Ans. (c)Sol.

R

VDD

g , rm ds

AC equivalent,

G+ g Vm gs rds

RD

+–IxVx

–

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ECE

IES MASTER

Above circuit can be replaced by,

g Vm gs+–

+–

Ix

Vx

–

+

Vgs

rds

Vgs = –Vx

By applying voltage transformation,

+–

rds

–g V rm x ds RIx

Vx

Applying kVL we get,

x x ds m x dsV I r R g V r = 0

x

x

VI =

ds

m ds

r R1 g r

Thus, option (c) is correct.

28. For an infinitesimally small dipole in free space,the electric field E in the far field isproportional to jkr(e /r)sin , where k 2 / . A vertical infinitesimally small electric dipole( l ) is placed at a distance h (h > 0) abovean infinite ideal conducting plane, as shown inthe figure. The minimum value of h, for whichone of the maxima in the far field radiationpattern occurs at 60 , is

Il

h

z

y0

Infinite conducting plane(a) 0.25 (b)

(c) 0.5 (d) 0.75

Ans. (b)Sol. As plane is conducting i.e. from image theory

the image of small electric dipole will be format the same distance under the plane.

h

h

I dl

I dl

as we know that

|A.F| =

sin N 2sin 2

=

sin2 2sin 2

sin2a = 2 sin (a) cos (a)

|A.F| =

2sin 2 cos 2

sin 2

|A.F| = 2cos 2

|A.FN| =

max

A F 2cos 2A F 2

|A.FN| = cos 2

2d cos ,&

2 2h cos

60 given

2 12h

2

N 602 hA F cos

2

N 60hA F cos

if h n ,

where n = 0, 1, 2....

|A.FN| will be maximum




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IES MASTER

eh n

For hmin, n = 1

minh 1

minh

29. The state diagram of a sequence detector isshown below. State S0 is the initial state of thesequence detector. If the output is 1, then

1/0 0/00/0

0/0 0/01/0

1/0

1/0

1/0

0/1

s3s2s1s0

s4

(a) the sequence 01001 is detected(b) the sequence 01110 is detected(c) the sequence 01010 is detected(d) the sequence 01011 is detected

Ans. (c)

Sol. By following sequence S0 S1 S2 S3 S4 S3, output is 1.

Hence, sequence [0 10 10]2 can be design withnearly circuit have state diagram given below

S0 S1 S2 S3

1/0 0/0 0/0

1/00/0

1/0

S4

1/0

0/11/01

0 0/0

So, option (c) is correct.

30. For the components in the sequential circuitshown below, tpd is the propagation delay, tsetupis the setup time, and thold is the hold time. The

maximum clock frequency (rounded off to thenearest integer), at which the given circuit canoperate reliably, is _______ MHz.

Flip Flop 1t = 2nspd t = 2nspd

t =3nspdt =5nst =1ns

setup

hold

t =8nspdt =4nst =3ns

setup

hold

Flip Flop 2

IN

Clk

Ans. (76_78)Sol.

Total maximum propagation delay= (Tpd + Tsetup)max= 8ns + 5ns= 13 ns

Frequency of operation

=1000 MHz

13

= 76.92 MHz

31. The characteristic equation of a system iss3 + 3s2+ (K + 2)s + 3K = 0In the root locus plot for given system, as Kvaries from 0 to , the break-away or break-in point(s) lie within(a) (–3, –2) (b) (–2, –1)

(c) (–1, 0) (d) , 3

Ans. (c)Sol.

s3 + 3s2 + (K+2)s + 3K = 0 s3 + 3s2 + 2s + K(s+3) = 0

2

K s 31s s 3s 2

= 0

K s 31

s s 2 s 1

= 0

1 + G(s) H(s) = 0




ECE

IES MASTER

Since:

0–1–2–3

G(s) =

K s 3

s s 2 s 1

According to unity feedback forward transferfunction and pole zero plot; there is onebreakaway point between (–1,0)

32. In the voltage regulator shown below, V1 is theunregulated input at 15V. Assume VBE = 0.7Vand the base current is negligible for both theBJTs. If regulated output V0 is 9V, the value ofR2 is ________ .

V = 15V1 V = 9V0

R = 1k3 R = 1k1

R2V = 3.3VZ

Ans. (800_800)Sol.

loop (1)V = 3.3Vz

VBE

+– R2

R =1k3 R =1k1

V =9V0

Given, VBE = 0.7V, BI 0A

Thus, resistance R1, and R2 are in series

V0 will divide between R1 and R2 usingvoltage division

2RV =

o 2

1 2

V RR R

2RV =

2

2

9 R1k R ...(i)

Applying KCL in loop (1):

2R3.3 0.7 V = 0

2RV 4V

Putting value of VR2 in equation (i)we get,

4 = 2

2

9R1k R

4k + 4R2 = 9R2

5R2 = 4k

R2 = 40005

2R 800

33. P, Q and R are the decimal integerscorresponding to the 4-bit binary number 1100considered in signal magnitude, 1’s complement,and 2’s complement representations, respectively.The 6-bit 2’s complement representation of(P+Q+R) is(a) 111101 (b) 111001(c) 110010 (d) 110101

Ans. (d)

Sol. Given 4 bit binary number = 1100

If it is represented in signed magnitude then1100

P = –4

If it is represented in 1’s complement then

Q = – 3




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IES MASTER

If it is represented in 2’s complement then

R = –4

So,

(P + Q + R) = –11

(–11)10 is represented in 2’s complement as(110101)2

34. X is the Fourier transform of x(t) shown

below. The value of 2X d

(rounded off to

two decimal places) is _______.

t43210–1–2–3

1

2

3

x(t)

Ans. (58.5_58.8)

Sol.

x(t)FT X

2x t dt FT 21 X d2

From above property,

2x d = 2 Energy [x(t)]

Energy [x(t)] =

2x t dt

–1 < t < 0; x(t) = t + 1

0 < t < 1; x(t) – 1 = 2(t – 0)

x(t) = 2t + 1

1 < t < 2; x(t) = –2t + 5

2 < t < 3; x(t) = –t + 3

E x t =

0 12 2

1 02 3

2 2

1 2

t 1 dt 2t 1 dt

2t 5 dt t 3 dt

=

0 12 2

1 01 3

2 2

1 2

t 1 2t dt 4t 1 4t dt

4t 25 20t dt t 9 6t dt

=

0 13 2 3 2

1 02 33 2 3 2

1 2

t 2t 4t 4tt t3 2 3 34t 20t t 6t25t 9t3 2 3 2

=

1 41 1 1 1 23 3

4 8 1 25 10 4 131 27 8 9 3 9 43

= 1 13 13 13 3 3 3

= 2 26 28 9.333 3 3

2X d

= 2 9.33 58.64

35. SPM(t) and SFM(t) as defined below, are the phasemodulated and the frequency modulatedwaveforms, respectively, corresponding to themessage signal m(t) shown in the figure,

PM pS t cos 1000 t K m t

and t

FM fS t cos 1000 t K m d

where Kp is the phase deviation constant inradians/volt and Kf is the frequency deviationconstant in radians/second/volt. If the highestinstantaneous frequencies of SPM(t) and SFM(t)

are same, then the value of the ratio p

f

KK is

________ seconds.




ECE

IES MASTER

V m(t)

10

01 3 6 7 t(sec)

Ans. (2_2)Sol. Given:

PM p 1S t cos 1000 t K m t cos t

t

FM fS t cos 1000 t K m d

2cos t

Highest instantaneous frequencies ofSPM(t) = fPM i

=

pK dm t1 1000

2 d t 2

Highest instantaneous frequencies of FM FM iS t f

f1 1000K m t2 2

Given

m(t) =

m(t)

10

0 1 3 6 7t

dm tdt =

5

0 t

–10

dm tdt

From questionfPM i = fFM i

p f1 dm t 1K 500 K m t 5002 dt 2

We will take dm t 5

dt [because highest

frequency]

p fdm tK K m t

dt

p

f

K m t 10 2K 5dm t

dt

36. The current I in the given network is

+–

+– Z

Z

Z = (80 – j35)

120 –90°V

120 –30°V

(a) 2.38 96.37 A

(b) 2.38 143.63 A(c) 2.38 23.63 A (d) 0A

Ans. (b)

https://iesmaster.org/results/ese




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IES MASTER

Sol.

Z

Z

Z=(80–j35)

I2

I1

–+120 –30°V

120 –90°V

IA

From the circuit,

I1 =120 90 120 90

Z 80 j35

= 1.374 –66.37°A

I2 =120 30 120 30

Z 80 j35

= 1.374 –6.37°AApplying KCL at node A, we get:I + I1 + I2 = 0 I = – (I1 + I2)= (1.374 66.37 1.374 6.37 )

= 2.38 143.63 A

37. A pn junction solar cell of area 1.0 cm2

illuminated uniformly with 100 mW cm–2, hasthe following parameters: Efficiency = 15%, opencircuit voltage = 0.7V, fill factor = 0.8, andthickness = 200µm. The charge of an electronis 1.6×10–19C. The average optical generationrate (in cm3 s–1) is(a) 1.04 × 1019 (b) 83.60 × 1019

(c) 0.84 × 1019 (d) 5.57 × 1019

Ans. (c)Sol.

Given, Area = 1cm2 , efficiency = 15%, Voc =0.7V, Fill factor (FF) = 0.8, thickness =

200 m

Pin = 2 2100mW cm 1cm

Pin = 100 mW

Fill factor (FF) = out

oc s c

PV I

Pout = Fill factor × Voc × Is/c ...(i)

Efficiency = out

in

PP

Substituting equation (i) in above equationWe get,

= oc s c

in

FF V IP

0.15 =

s c

30.8 0.7 I

100 10

Is/c = 0.026785 AOptical generation rate (GL)

=

s cIq area thickness

= 19 40.026785

1.6 10 1 200 10

= 8.37 × 1018

= 0.837 × 1019 /cm3/secThus, option (c) is correct.

38. In a digital communication system, a symbol Srandomly chosen from the set {s1, s2, s3, s4} istransmitted. It is given that s1 = –3, s2 = –1, s3= +1 and s4 = +2. The received symbol is Y =S + W. W is a zero-mean unit-variance Gaussianrandom variable and is independent of S.Pt isthe conditional probability of symbol error forthe maximum likelihood (ML) decoding whenthe transmitted symbol S = si. The index i forwhich the conditional symbol error probabilityPi is the highest is ________.

Ans. (3_3)Sol. as ML detector is used, the decision boundry

between two adjacent signal points will betheir arthmetic mean. for s1 = –3, the probability of error (p1):

–3 –2 –1 +1 2




ECE

IES MASTER

P1 = 1 – (shaded area)for s2 : The probability of error (P2)

–3 –2 –1 +1 20

P2 = 1 – (shaded area)for s3: the probability of error P3.

–3 –2 –1 1.5 20 +1P3 = 1 – (shaded area)

for s4: The probability of error (P4)

–3 –2 –1 +1.5 +20P4 = 1 – (shaded area)By concluding above graph P3 i.e.probability of error when s3 is transmittedis larger among the four.

i 3

39. The magnetic field of a uniform plane wave invacuum is given by

x y zˆ ˆ ˆH x,y,z,t a 2a ba cos t 3x y z

The value of b is _______.Ans. (1_1)Sol. Given

x y zˆ ˆ ˆH (a 2a ba )cos( t 3x y z)

Wave propagation direction,

p x y zˆ ˆ ˆa 3a a a

H x y zˆ ˆ ˆa a 2a ba

As we know that E, H, P

are orthogonal toeach other.

Hence, H pa a 0

x y z x y zˆ ˆ ˆ ˆ ˆ ˆ(a 2a ba ) ( 3a a a ) 0

–3 + 2 + b = 0

b 1

40. For the BJT in amplifier shown below, VBE =0.7 V, kT/q = 26 mV. Assume that BJT outputresistance (r0) is very high and the base currentis negligible. The capacitors are also assumedto be short circuited at signal frequencies. Theinput i is direct coupled. The low frequency

voltage gain 0 i of the amplifier is

RE CE

RL

RC

C1

100µF

10k

20k

1µF

10k

V = –10VEE

V = 10VCC

i

0

(a) –89.42 (b) –128.21(c) –178.85 (d) –256.42

Ans. (a)Sol.

200k E

C =1 f1 R = 10kC

V = 10VCC

R=1

0kL

C =100 F Vi

V =–10VEE

This is common emitter with by passcapacitor amplifier.




ECE

IES MASTER

Given, VBE = 0.7, kTq = 26 mV

For DC analysis,All capacitors are open circuitedThus

10k I C

V = 10VCC

IE

20k

–10V

1

c EI I

Applying KCL in loop (1)We get,0 – 0.7 – IC (20k) + 10 = 0

IC = 10 0.720k

IC = 0.465 mA

gm = C

T

IV = 0.465mA

26mV

= 0.01788 A/V

Low frequency voltage gain (AV) = m Lg R

AV = m C Lg R R

= 0.01788 10k 10k

= –0.08942 k

VA 89.42

Thus, option (a) is correct.

41. A system with transfer function

1G s ,s 1 s a

a > 0 is subjected to aninput 5cos3t. The steady state output of the

system is 1 cos 3t 1.89210

. The value of a

is ______.Ans. (4_4)Sol.

G(s)R(s) C(s)

Given:

C(t) = 1 cos 3t 1.89210

r(t) = 5cos(3t)

1G jj 1 j a

Amplitude of c(t) = Amplitude of 3r t G j

2 2 23

1 1510 1 a

25 10 10 9 a

25 = 9 + a2

a2 = 16

a 4

42. For a 2-port network consisting of an ideallossless transformer, the parameter S21 (roundedoff to two decimal places) for a referenceimpedance of 10 , is _______.

2 Port Network

2:1Port 2Port 1

Ans. (0.8_0.8)Sol. For ideal lossless transformer, scattering

matrix,




ECE

IES MASTER

x : 1

2 2

11 12 2

21 22 2 2

x 1 2x x 1S S2x 1 xS S

x 1 1 x

21 22xS

x 1

According to questionx = 2

212 2 4S4 1 5

21S 0.8

43. For the modulated signal cx t m t cos 2 f t ,the message signal m t 4cos 1000 t andthe carr ier frequency fc is 1 MHz. The signalx(t) is passed through a demodulator, as shownin the figure below. The output y(t) of thedemodulator is

x(t)

cos(2 (f +40)t) c

–510 510 f(Hz)

1Ideal LPF with cut-off 510Hz y(t)

(a) cos 460 t (b) cos 1000 t(c) cos 920 t (d) cos 540 t

Ans. (c)Sol.

x(t)

cos( + 2 × 40t) c

–510 510 f(Hz)

1y(t)z(t)

cx t m t cos 2 f t

m c4cos t cos t

m c c mx t 2 cos t cos t

cZ t x t cos t 2 40t

m c c2 cos t cos t 2 40t

c m c2 cos t cos t 2 40t

m c mcos 2 2 40t cos t 2 40t

c m mcos 2 t t 2 40t cos t 2 40t

After passing through low pass filter higherfrequency will be attenuated.

So, my t cos 2 40 t

y t cos 1000 80 t

y t cos 920 t

44. An enhancement MOSFET of threshold voltage3V is being used in the sample and hold circuitgiven below. Assume that the substrate of theMOS device is connected to 10V. If the inputvoltage V1 lies between 10V s, the minimumand the maximum values of GV required forproper sampling and holding respectively, are

VG

V1 V0

(a) 10V and –10V (b) 3V and –3V(c) 13V and –7V (d) 10V and –13V

Ans. (c)Sol. For proper sampling and holding

VDS VGS – Vth

V1 – V0 (VG – V0) – Vth

V1 VG – Vth

VG V1 + Vth

when V1 = +10V VGmin = 10 + 3 = 13V

when V1 = –10V VGmax= –10+3 = –7V

45. Which one of the following options contains twosolutions of the differential equation




ECE

IES MASTER

dy y 1 x?dx

(a) ln|y–1| = 0.5x2 + C and y = 1(b) ln|y–1| = 2x2 + C and y = 1(c) ln|y–1| = 0.5x2 + C and y = –1(d) ln|y–1| = 2x2 + C and y = –1

Ans. (a)

Sol. dy y 1 xdx

dy x dxy 1

Integrating both sides, we get

1 dy xdx

y 1

2xn y 1 C2

l Eqn. (1)

Now,

dydx = 0 when y = 1

It means y = constant C. y = 1 Eqn. (2)So, Eqn. (1) and Eqn. (2) are two differentsolution of given differential equation

46. The base of an npn BJT T1 has a lineardoping profile NB(x) as shown below. The baseof another npn BJT T2 has a uniform dopingNB of 1017 cm–3. All other parameters areidentical for both the devices. Assuming thatthe hole density profile is the same as that ofdoping, the common-emitter current gain of T2is

Emitter Base Collector

10 (cm )14 –3

10 (cm )17 –3

N (x)B

0 W

(a) approximately 0.3 times that of T1(b) approximately 0.7 times that of T1(c) approximately 2.0 times that of T1(d) approximately 2.5 times that of T1

Ans. (*)

47. A one-sided abrupt pn junction diode has adepletion capacitance CD of 50 pF at a reversebias of 0.2V. The plot of 2

D1 C versus the appliedvoltage V for this diode is a straight line asshown in the figure below. The slope of the plotis _______ × 1020 F–2 V–1.

V0

2D1 C

(a) –0.4 (b) –3.8(c) –1.2 (d) –5.7

Ans. (Data Insufficient)Sol.

Depletion Capacitance, CD = AW

W = Width of depletion region

As,

w = 1/2

dA D

2 1 1 V Vq N N

[Vd = Cut in voltage]After puytting the value of w in CD.

CD = 1/2

dA D

A2 1 1 V Vq N N

CD = 1/2d

D A

E2 1 1 V Vq N N

https://iesmaster.org/results/gate




ECE

IES MASTER

CD = 1/2d

KV V

2D

1C = d2

1 V VK

[1/K2 = m = slope of the plot]

2D

1C = dm V V

Given, CD = 50 PF at V = 0.2 V

212

1

50 10= dm V 0.2

m = 212

d

1

50 10 V 0.2

m = 24

d

102500 V 0.2

We need Vd for finding the slope of plot.Vd is not given. So data is insufficient.

48. X is a random variable with uniform probabilitydensity function in the interval [–2, 10]. For Y= 2X – 6, the conditional probability P Y 7|X 5 (rounded off to three decimal

places) is _______.Ans. (0.3_0.3)Sol. Given:

x

1 12 2 x 10f x0 otherwise

=

–2 10

f (x)x

112

x

As Y = 2x – 6

So, y

1 10 x 14f y 240 otherwise

=

124

–10 14

f (y)y

y

If x 5 then y 4

So, P y 7|x 5 = P y 7 / y 4

=

P 4 y 7P 4 y 14

=

1/24

–10 140 4 7y

4 y 7

4 y 14

So, P y 7|x 5 =

P 4 y 7P 4 y 14

= 310

= 0.3

49. For the solid S shown below, the value of

S

xdxdydz (rounded off to two decimal places)is ________.

y0 1

1

z

x

3

Ans. (2.25_2.25)Sol. x : 0 to 3

y : 0 to 1




ECE

IES MASTER

z : 0 to 1 – y

I = x dx dy dz

= 1 y1 3

y 0 z 0 x 0

x dx dy dz

= 31 2

1 y0

0y 0

x z dy2

= 1

0

9 1 y dy2

= 12

0

9 yy2 2

= 9 2.254

50. Consider the following closed loop control system.

Y(s)C(s) G(s)R(s)+

–

where

1G ss s 1

and s 1C s K

s 3

. If the

steady state error for a unit ramp input is 0.1,then the value of K is _________.

Ans. (30_30)Sol. Calculating closed loop transfer function:

C s G s H s1 C s G s

K s 1 H ss s 1 s 3 K s 1

Obtaining open loop transfer function from

closed loop function = G

1 GH G

OLTF =

K s 1

s s 1 s 3 K s 1 K s 1

OLTF =

K s 1 K

s s 1 s 3 s s 3

Type-1:So, for ramp input,

ssV

AeK

[A = 1 unit ramp given]

ssV

1e 0.1K

(given) VK 10

V s 0K Lim s OLTF

s 0

K KLim ss 3 s 3

K 10 K 303

51. A finite duration discrete-time signal x[n] isobtained by sampling the continuous-time signal x t cos 200 t at sampling instants t = n/

400, n = 0,1, ...,7. The 8-point discrete Fouriertransform (DFT) of x[n] is defined as

kn7 j4

n 0X k x n e ,k 0,1,...,7

Which one of the following statements is TRUE?(a) Only X[3] and X[5] are non-zero(b) Only X[2] and X[6] are non-zero(c) All X[k] are non-zero(d) Only X[4] is non-zero

Ans. (b)

Sol. Given, kn7 j4

n 0X[k] x[n] e ,k 0,1, ,7

x(t) = cos (200 t)

for t =n

400

X[n] = cos 200 n

400

=ncos2

x(k) =kn7 j4

n 0

ncos e2




ECE

IES MASTER

x(n) =2 3cos(0),cos ,cos ,cos ,

2 2 2

5 6 7cos(2 ),cos ,cos , cos2 2 2

= {1, 0, –1, 0, 1, 0, –1, 0}

X(0)

X(1)

X(2)

X(3)

X(4)

X(5)

X(6)

X(7)

x(0)

x(4)

x(2)

x(6)

x(1)

x(5)

x(3)

x(7)

1 2

01

0

0–1

–2–1

–1 0

00

00

00

00–1

4–1

0–1

0–1

0

0

–1

–10

–1

–1

–1

–1

04W 1

24W 1

24W 1

04W 1

04W 114W j

24W 1

34W j

2 2j j jN 4 2

4W e e e j

04W 1 2

4W 1

14W j 3

4W j

X(0) = 0 X(4) = 0

X(1) = 0 X(5) = 0

X(2) = 4 X(6) = 4

X(3) = 0 X(7) = 0

X(K) = {0,0,4,0,0,0,4,0}Hence, option (b) is correct.

52. The components in the circuit given below areideal. If R 2k and C = 1µf, the –3dB cut-offfrequency of the circuit in Hz is

–+2R V (j )0

C

R

R

2C

V (j )i

(a) 79.58 (b) 14.92(c) 34.46 (d) 59.68

Ans. (a)Sol.

+–

Vb

Va

R

2R

C

2C

RVi Vo

Given R = 2k , C 1 F

The given circuit is op-amp active filter ofinverting type

3dB cutoff frequency = 12 RC

= 3 61

2 2 10 10 = 79.577 Hz

Thus, option (a) is correct.

53. For the given circuit, which one of the followingis the correct state equation?

0.5 H

i+

–V

0.25Fi2i1 2 1

(a)1

2

iV 4 4 V 0 4didt i 2 4 i 4 4

(b)1

2

iV 4 4 V 4 0didt i 2 4 i 0 4

(c)1

2

iV 4 4 V 0 4didt i 2 4 i 4 0

(d)1

2

iV 4 4 V 4 4didt i 2 4 i 4 0

Ans. (c)Sol. given figure




ECE

IES MASTER

(i – i)i

i1 2

0.5H V A

L1

0.25F+– 1 i2

i

applying KVL in loop L1

–2 (i1 – i) + 0.5 di V 0dt

1di2i 2i 0.5 V 0dt

1 1di 12i 2i V 4i 4i 2Vdt 0.5

1di 2V 4i 4idt

…(i)

Applying KVL at node A

2(0 V) VI i (0.25)

(1) t

4 [I2 – V + i] = Vt

Vt

= 4I2 – 4V + 4i …(ii)

Arranging eq. (i) and eq. (ii)

Vt

= –4 V + 4i + 4I2

idt

= –2V – 4i + 4i1

Vt i

= 1

2

i4 4 V 0 4i2 4 i 4 0

So, option c is correct.

54. Consider the following system of linearequations.x1 + 2x2 = b1 ; 2x1 + 4x2 = b2; 3x1 + 7x2 = b3; 3x1+ 9x2 = b4

Which one of the following conditions ensuresthat a solution exists for the above system?(a) b2 = 2b1 and 6b1 – 3b3 + b4 = 0(b) b2 = 2b1 and 3b1 – 6b3 + b4 = 0

(c) b3 = 2b1 and 3b1 – 6b3 + b4 = 0(d) b3 = 2b1 and 6b1 – 3b3 + b4 = 0

Ans. (a)Sol. By observing first two equation, b2 = 2b1

from equation (i), (iii) & (iv)

1

3

4

1 2 b3 7 b3 9 b

2 2 1 3 3 1R R 3R & R 3R 3R

1

3 1

4 1

1 2 b0 1 b 3b0 3 b 3b

3 3 2R R 3R

1

3 1

4 3 1

1 2 b0 1 b 3b0 0 b 3b 6b

For solution to exist:b4 – 3b3 + 6b1 = 0 & b2 = 2b1

55. The band diagram of a p-type semiconductorwith a band-gap of 1 eV is shown. Using thissemiconductor, a MOS capacitor having VTH of–0.16V, oxC of 100 nF/cm2 and a metal workfunction of 3.87 eV is fabricated. There is nocharge within the oxide. If the voltage acrossthe capacitor is VTH, the magnitude of depletioncharge per unit area (in C/cm2) is

Vacuum level

4 eV

EC0.5 eV

EiEFs

EV0.2 eV

(a) 1.70 × 10–8 (b) 1.41 × 10–8

(c) 0.93 × 10–8 (d) 0.52 × 10–8

Ans. (a)Sol. EG = 1 eV, VTh = –0.16 V = VT




ECE

IES MASTER

Thus, GE 0.5 eV2

Ei – EV = 0.5 eVThus, Ei – EFs = (0.5 – 0.2)eV = 0.3 eV

Vacuum level(E )Vac

4 eV

EC0.5 eV

Ei

EFs0.2 eV0.3 eV

As,

VT = ox dms fp

ox ox

Q Q 2C C

...(i)

ms m s

Given, m 3.87eV

s vac FSE E

= 4 eV + 0.5 eV + 0.3 eV= 4.8 eV

ms 3.87eV 4.8eV

ms 0.93eV

fp i FSE E 0.3 eV

Substituting value of ms ox fp,C , and VT inequation (i)We get,

ox dT ms fp

ox ox

Q QV 2C C

d

9Q0.16 0.93eV 0 2 0.3eV

100 10

d9

Q 0.17100 10

8 2dQ 1.7 10 C cm

Thus, the magnitude of depletion charge perunit area is 1.7×10–8

Hence, option (a) is correct.

solution - ies master · which started with the sub-prime lending crisis in usa in 2007. the...

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