# EC_PAPER_1

EC gate paper...

#### Description

:2:

E & TE

PAPER REVIEW Except a few out of scope questions from EDC, remaining questions in the paper can be easily attempted. Particular in this paper selection of questions plays a vital role in securing a good score. For example Section-A is relatively tougher than Section-B, so choosing 3 questions from Section-B will fetch you a big advantage.

SUBJECT WISE REVIEW SUBJECT(S)

LEVEL MARKS

Basic Electronics

Hard

75

Network Theory

Easy

77

Basic Electrical Engineering

Moderate

89

Electronic Measurements & Instrumentation Moderate

74

Analog Electronics & Digital Electronics

Easy

94

Materials Science

Hard

71

Subject Experts, ACE Engineering Academy

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CONVENTIONAL PAPER - 1

SECTION-A 01. (a) In a long semiconductor bar (EG = 2 eV), conduction band electrons come in from the left in the positive x-direction with a kinetic energy of 3 eV. They move from location A to B to C to D. Between A and B, the electric field is zero; between locations B and C, there is a linearly varying voltage increase of 4 V; between C and D, the field is again zero. Assuming no scattering, sketch a simplified band diagram describing the motions of these electrons. Assuming that these electrons can be described as plane waves, with a free-electron mass, write down the wave function of the electrons at location D. Leave your result in terms of an arbitrary normalization constant. Assume the mass of free electron to be 9.11  10–31 kg. (12 M) 01.

(a)

Sol:

Given, band-gap = 2eV Energy of electron, E = 3eV Zero electric field from A to B and C to D Voltage between B and C = 4V m0 = 9.1110–31 kg Band diagram: 2eV

A

B

4eV

D

V0 = +4V W = –V0q = –4eV Time-independent Schrodinger’s wave equation is given by  2  ( x ) 2m  2 E  V( x ) ( x )  0 x 2 h ACE Engineering Academy

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E & TE

V(x) V0 Region-II

Region-I

x

0 Figure: Potential function

The general solution to this can be written as: 1 ( x )  A1e jk1x  B1e  jk1x Where k1 

(x 0)

2mE h2

and  2 ( x )  A 2 e  k 2x  B2 e k 2x Where k 2 

(x 0)

2m(V0  E) h2

01. (b) Calculate the Fermi energy EFO at 0K for copper and estimate the average speed of the conduction electrons in Cu. The density of Cu is 8.96 gm/cm3 and atomic weight is 63.5. Given Avogadro’s number is 6  1023. 01.

(b)

Sol:

Fermi-energy is the maximum energy occupied by electron at 0ok

(12 M)

The conduction electron population for a metal is calculated by multiplying the density of conduction electron states (E) and Fermi function, f(E).

dn  ( E ) . f ( E ) dE  8 2m3 / 2  1  E  ( E  E F ) kT    3 h 1   e ACE Engineering Academy

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 n

CONVENTIONAL PAPER - 1

8 2m 3 / 2  2 3 2   EF  h3 3  23

 h 2  3  2 3   n  E F    8m   

Every atom of Cu contributes one electron  n = Atomic concentration 

A 0d A

; Where A0 is Avogadro’s no.

= 8.46 1022 cm–3

d is density, A is atomic weight

 EF = 7.026 eV 12

 6E  Average speed of conduction electrons in Cu is v e   F 0   5m e  We get Ve = 1.2  106 m/s

01. (c) In the common source amplifier shown, evaluate voltage gain Av, given RD = 2.7 k,  = 50 and rds = 25 k. Derive the expression used.

+

(12 M)

VDD RD +

vi

+ –

RG

RS

CS

v0

:6: 01.

(c)

Sol:

Given circuit is

E & TE

VDD RD +

Vi

+ –

+

V0

RG

CS

RS

– RD = 2.7 k  = 50 rds = 25k We know  = gm rds 50 = gm 25k gm = 2 mA/V Small signal equivalent circuit is

G Vi

+

RG Vgs

+ gmVgs

rds 25k

RD V0

V0 = –gm Vgs (rds ||RD) V0 = –gm Vi (rds||RD)

V0 r  R D  50  R D  A V   g m ds   4.90V / V Vi rds  R D rds  R D ACE Engineering Academy

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CONVENTIONAL PAPER - 1

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E & TE

01. (d) Define lumen and candela. The wavelength of visible light ranges from violet at approximately 380 nm to red at 720 nm. Obtain the bandwidth available of visible light. (12 M) 01.

(d)

Sol:

Lumen:

 It is measure of total quantity of visible light emitted by source”  It is S.1 unit of Luminous flux Candela:

 It is measure of Luminous power per unit solid angle emitted by a point light source in a particular direction  It is S.1 unit of Luminous Intensity 1 lumen = 1 candela Sr Frequency of violet spectrum is given by f1  

3  108 38  10 8 3  1016 38

 0.7895 1015 Hz  789.5 THz Frequency of red spectrum is given by f2  

3  108 72  10 8 3  1016 72

 0.4167 1015 Hz  416.7 THz Band width of visible light = f1 – f2 = 372.8 THz ACE Engineering Academy

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CONVENTIONAL PAPER - 1

01. (e) Implement the following expression using NAND gates only: Y  a  c  a  b  c  01.

(e)

Sol:

Y = (a + c) (a  b  c )

(12 M)

= (a + c) abc

 a (abc)  c(abc)

a b c

 a (abc)  c(abc)

y

 a (abc) . c(abc)

02.

(a) For the MOSFET characteristic shown in the figure, calculate: (i) Linear VT and KN (ii) Saturation VT and KN (iii) The gate oxide thickness and substrate doping.

Assume channel mobility = 500

cm 2 Vs

VFB = 0, Z = 100 m, L = 2 m Where Z is the depth of the channel and L is the length of the channel.

(25 M)

ID(mA) 2.0

VG= 5V

1.0

VG= 4V VG= 3V

0

1

3 2 VD(V)

4

5

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E & TE

02. (b) (i) What is compensated doping? (ii) An n-type semiconductor containing 1016 phosphorus (donor) atoms/cm3 has been doped with 1017 boron (acceptor) atoms/cm3.Calculate the electron and hole concentrations in the semiconductor. 02.

(b) (i)

Sol:

Compensated doping:

(15 M)

Compensated doping is a term used to describe the doping of a semiconductor with both donors and acceptors to control the properties. For example, a p-type semiconductor doped with NA acceptors can be converted to a n-type semiconductor by simply adding donors until the concentration ND exceeds NA. The effect of donors compensates for the effect of acceptors and vice-versa The electron concentration is then given by ND–NA provided the latter is larger than NA. What essentially happens is that electrons from donors recombine with the holes from the acceptors so that the mass action law np  n i2 is obeyed 02.

(b) (ii)

Sol:

ND = 1016 atoms/cm3 NA = 1017 atoms/cm3 Since more acceptors are added into donor atoms it becomes p-type compensated semiconductor So the resultant hole concentration = NA – ND [ valid when NA – ND >> 2ni ] = 1017 – 1016 = 9  1016 atoms/cm3

Now electron concentration =

n i2 hole concentration

n i2 atoms / cm3 9 1016

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CONVENTIONAL PAPER - 1

02. (c) Define Fan-in and Fan-out with an example. Draw the circuit diagram of an NMOS circuit to realize f (a, b, c) = a b  ac.

(20 M)

02.

(c)

Sol:

Fan-In: The maximum number of inputs connected to the logic gate is called fan-in Ex:

k

Here fan-in = k In generally Gates with large fan-in are slower than gates with small fan-in

Fan-Out: fanout is the maximum number of logic gates that can be driven by a logic gate output

by maintaining the output levels without affecting the logic gate performance. (or) Fan-out of a gate specifies no of standard loads that can be connected to the output of the gate without degrading it’s normal operation. Ex:

(o/p= high)

IOH

(o/p= low) IIH n gates

(a) High-level output I OH  n  fanout H I IH ACE Engineering Academy

IOL

IIL

IIL

n gates

(b) low level output I OL  n  fanout L I IL

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E & TE

f (a , b, c)  a b  ac  a  b  ac  a  b  ac

 (a  b). ac VDD

VDD

RD

RD

f

a c VDD RD a

VDD RD

b

03. (a) (i) Consider the ac equivalent circuit of a MOSFET Colpitts oscillator.

R

C2

C1

L Derive the expression for oscillation frequency. Also find the condition on the gain to initiate the oscillations spontaneously. (ii) As per the Barkhausen criterion, the positive feedback exists a particular frequency range and

the resulting feedback signal reinforces the error signal. Explain the

phenomenon which limits the amplitude of the oscillations under steady state. (20 M) ACE Engineering Academy

: 13 :

CONVENTIONAL PAPER - 1

03. (a) (i) Sol:

Given circuit

R

C1

C2

L

Equivalent circuit is sC2 vgs

L G + vgs –

sC2 vgs C2

VD gmvgs

R

C1

1   VD  sL  sC 2 v gs sC 2   VD  s 2 LC2  1v gs ......1

Apply KCL at drain terminal VDsC1 

VD  g m v gs  sc 2 v gs  0 R

1 v gs sC1  s 2 LC2  1  g m v gs  sC 2 v gs  0 R   sC1R  1 s 2 LC  1  g  sC  0 2 m 2  R 

s 3LC1C2R+ s C1R+ s 2LC2+1+gmR+ s C2R= 0 ACE Engineering Academy

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E & TE

Make imaginary part of the above equation is equal to zero –3 LC1C2R + LC1R+ C2 = 0  [RC1+RC2–2LC1C2R] = 0 R C1  C 2  2 LC1C 2   0 

C1  C 2 LC1C 2



CC 1  C eq  1 2 C1  C 2 LCeq

Condition for the gain to initiate the oscillations 1– 2LC2+gmR = 0 1+gmR = 2LC2 1  gmR 

C1  C 2 LC2 LC1C 2

1  gmR  1 

gmR 

03.

C2 C1

C2 C1

(a) (ii)

Sol:

Vi

+ +

A1

V0

2 V0  A1   2  10 Vi Conditions for sustained oscillations as per Backhouses criterion (i) |A| = 1 (ii) A =  1+2 = 0 ACE Engineering Academy

: 15 :

CONVENTIONAL PAPER - 1

Amplitude stabilization:

Given that the circuit is in positive feed back (|A|>1) for the given frequency range of operation. In order to generate and sustain oscillations the gain control mechanism is as follows first to ensure that oscillations will start, initial design is such that A is slightly greater than unity. This corresponds to designing the circuit so that the poles are in the right half of the splane. Thus as the power supply is turned on oscillations will grow in amplitude. When the amplitude reaches the desired level, the non linear network comes into action and causes the loop gain to be reduced to exactly unity. In other words, the poles will be “ pulled back” to j axis. This action will cause the circuit to sustain. oscillations at this desired amplitude. A=1

A>1 X

X 

X

X Generation of oscillations

initially

Sustained oscillations

sustained Generation

A=1

A>1 Amplitude Stabilization:

Vz+ 0.7 V Slope = A2 V0

Slope A1 0

Vf

–(Vz+ 0.7V) Slope A2

: 16 :

E & TE

RB R1 –

RF

+

+

C

Vf

R

C R

C R

+ Vo –

To have amplitude stabilization in the oscillator, it is necessary to limit the output voltage by introducing nonlinearity. Stability can be achieved by adding two zener diodes in series with resistance RB. As long as the magnitude of the voltage V0 across Rf is less than the Vz + 0.7V, the gain of the Amplifier A1 

RF R1

If V0 increases more than Vz + 0.7, then zener diodes are ON A2 

R F || R B R1

V0

A

Zener ON

Zener OFF

A

V0

: 17 :

CONVENTIONAL PAPER - 1

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E & TE

03. (b) A typical 1 MHz quartz crystal has the following properties: fs = 1 MHz, fa = 1.0025 MHz Co = 5 pF,

R = 20

The two frequencies fs and fa are called the series and parallel resonant frequencies. In the equivalent circuit, C0 is parallel with LCR. What are C and L in the equivalent circuit of the crystal? What is the quality factor Q of the crystal? 03.

(b)

Sol:

Quartz crystal equivalent circuit

(20 M)

Co

Rs

Ls

Cs

Given, fs = 1MHz fa = 1.0025MHz C0 = 5pF Rs = 20 fa  fs 1 

Cs C0

 f  C s  C 0  a  f s

2   1.0025  10 6    1  5  10 12  6   1  10 

2     1  

= 25.03125 ×10–15F fs 

1 2 L s C s

 Ls 

1

2fs 

2

 Cs

1

2   110   25 10 6 2

15

 1.013211H ACE Engineering Academy

: 19 :

CONVENTIONAL PAPER - 1

Quality factor Q

2f s L s Rs

2  106  1.013211 20

 318309.8862

03. (c) What is a multiplexer? Write the symbol and truth table of a 4-1 multiplexer. Implement the same using logic gates. 03.

(20 M)

(c)

Sol: (c) Multiplexer: It is a combinational circuit that selects one of several inputs and forwards it to

output. In general a n  1 MUX has n inputs and log2n selection lines. It is also called as Data selector Symbol of 4  1 MUX:

I0 I1 I2 I3

41

output

s1 s0

Truth table:

S1 S0 Output 0

0

I0

0

1

I1

1

0

I2

1

1

I3

: 20 :

E & TE

S0 S1

I0 output

I1

I2 I3

04. (a) Consider a CMOS inverter biased at VDD = 5V with transistor parameters of KN= KP and VTN = –VTP  1 V. Then consider another CMOS inverter biased at VDD = 10 V with the same transistor parameters. Determine the critical voltages on the voltage transfer curve of the CMOS inverter. 04.

(20 M)

(a)

Sol:

V0

VDD

Vin

Slope= –1

Slope = –1

Vout VIL

VIH

VI

Voltage transfer characteristics VIL: By definition, VIL is the smaller of the two input voltage at which the slope of the VTC

becomes equal to (–1), i.e.,

dVout  1 . In this case nMOS transistor is in saturation while the dVin

pMOS transistor is in linear mode. From IDn = IDp we obtain: ACE Engineering Academy

: 21 :

k kn 2 VGSn  VTn 2  p 2VGSn  VTn VDSp  VDSp 2 2

CONVENTIONAL PAPER - 1

Note that VGSn = Vin, VDSn = Vout, VGSp= –(VDD – Vin), and VDSp = –(VDD – Vout) Substituting these values in above equation

k kn Vin  VTn 2  p 2Vin  VDD  VTp Vout  VDD   Vout  VDD 2 …………(1) 2 2

To satisfy

dVout  1 , we differentiate both sides of (1) dVin

  dV k n Vin  VTn   k p Vin  VDD  VTp  out  dVin  Substituting Vin = VIL and

  dV   Vout  VDD   Vout  VDD  out   dVin

  

dVout  1 into, we obtain dVin

k n (VIL  VTn )  k p 2Vout  VIL  v Tp  VDD  Then VIL as a function of Vout can be expressed as: VIL 

2Vout  VTp  VDD  k R VTn 1 kR

…………(2)

where kR 

kn kp

Case-i:For VDD = 5V, VTN = –VTP  1V, k R 

kn 1 kp

Substituting values into (2) produces

VIL 

2V0  1  5  1 11

VIL = V0 – 2.5 …………(3) From (1) and (3)

k kn V0  2.5  12  p 2V0  2.5  5  1V0  5  V0  52 2 2 ACE Engineering Academy

: 22 :

E & TE

V0  3.52  2V0  6.5V0  5  V0  52

V02  (3.5) 2  7V0  2 V02  11.5V0  32.5  V02  25  10V0 

V02  (3.5) 2  7V0  2V02  23V0  65  V02  25  10V0 6V0 = 40 – (3.5)2 = 40 – 12.25

V0 

27.75  4.625V 6

VIL = V0 – 2.5 = 2.125V Case-ii: For VDD = 10V, VTN = –VTP = 1V, k R 

kn 1 kp

Substituting values into (2) produces VIL 

2V0  1  10  1 2

VIL = V0 – 5 …………(3) From (1) and (3)

k kn V0  5  12  p 2V0  5  10  1V0  10  V0  102 2 2

V0  62  2V0  14V0  10  V0  102 

 

V02  36  12V0  2 V02  24V0  140  V02  100  20V0

V02  36  12V0  2V02  48V0  280  V02  100  20V0 16V0 = 180 – 36 V0 = 9V VIL = V0 – 5 = 9 – 5 = 4V

VIH: In this point nMOS is in linear mode and pMOS in saturation. Similarly to VIL we apply

KCL to the output node:

k kn 2 2 2VGSn  VTn  VDSn  VDSn  p VGSp  VTp  2 2

Again, using VGSn = Vin, VDSn = Vout, VGSp= –(VDD – Vin), and VDSp = –(VDD – Vout) ACE Engineering Academy

: 23 :

CONVENTIONAL PAPER - 1

By substituting these values

k kn 2 2 2Vin  VTn Vout  Vout  p Vin  VDD  VTp  …………(4) 2 2

Differentiating both sides with respect to Vin   dV   dV  k n Vin  VTn  out   Vout  Vout  out   k p Vin  VDD  VTp   dVin   dVin   Substituting Vin = VIH and

dVout  1 in above equation, we obtain dVin

k n (VIH  VTn  2Vout )  k p VIH  VDD  VTp  Then VIH as a function of Vout can be expressed as: VIH 

VDD  v Tp  k R (2Vout  VTn ) 1  kR

…………(5)

Case-i:For VDD = 5V, VTN = –VTP = 1V, k R 

kn 1 kp

Substituting values into equation(5) produces

VIL 

5  1  2V0  1 = V0 + 2.5 ………(6) 2

From (4) and (6) 2V0  2.5  1(V0 )  V02  V0  2.5  5  1

2

2(V0  1.5)(V0 )  V02  V0  1.5

2

2V02  3V0  V02  V02  3V0  2.25 6V0 = 2.25 V0 = 0.375 VIH = V0 + 2.5 = 2.875 V Case-ii: For VDD = 10V, VTN = –VTP = 1V, k R 

VIH  ACE Engineering Academy

kp kn

1

10  1  2V0  1 = 5 + V0 ………(7) 2

: 24 :

E & TE

From (4) and (7) 2V0  5  1(V0 )  V02  V0  5  10  1

2

2(V02  4V0 )  V02  V0  4 

2

2V02  8V0  V02  V02  16  8V0 16V0 = 16 V0 = 1V VIH = V0 + 5 = 6V

Critical voltages: For VDD = 5V

For VDD = 10V

04. (b) If

VIL = 2.125V, VIH = 2.875V VIL = 4V,

VIH = 6V

1 is defined as the mean probability per unit time that an electron scattered, show that τ

the mean time between collisions is .

(15 M)

04.

(b)

Sol:

Consider an infinitesimally small interval dt at time t. Let N be the number of unscattered 1 electrons at time ‘t’. The probability of scattering during dt is   dt, and the no. of scattered  1 1 electrons during dt is N  dt . The change dN in N is thus dN   N  dt   The negative sign indicates a reduction in N because, as electrons becomes scattered, N decreases. Integrating this equation, we can find N at any time t, given that at time t = 0, N0 is the total number of unscattered electrons.  N  N 0 e t   This equation represents the no. of unscattered electrons at time t. It reflects an exponential decay law for the no. of unscattered electrons the mean free time t can be calculated from the mathematical definition of t

: 25 :

CONVENTIONAL PAPER - 1

t

 t N dt 0 



 N dt 0

Where we have used N  N 0 e  t   Clearly, (1/) is the mean probability of scattering per unit time

04. (c) (i) The transistor circuit shown has  = 100, VBE(ACTIVE)= 0.7 V. Find the operating point (VCE, IC) and the mode of operation when S1, S2, are closed and S1, S2, are open.

S1 IC 1k

470k

+ – 10V

IB

S2

1k

(10 M) (ii) Find the diode current ID in the circuit shown below when the diode has cut-in voltage, V = 0.7 and forward resistance, Rf = 25

100 + 10V

(15 M)

ID 100

100

: 26 :

04.

(c) (i)

Sol:

Given data is  =100

E & TE

VBE (active) = 0.7V S1 IC 1k

470k

+ – 10V

IB

S2

1k

When S1,S2 are open 10V IC 1k 470k 1k

Transistor is in cutoff mode so IC = 0 10V IC 1k

VCE = 10V Operating point (VCE,IC) = (10,0) When S1&S2 are closed

470k IB 0.7V

1k

: 27 :

CONVENTIONAL PAPER - 1

Let us consider transistor is in active region 10V IC 1k

470k IB + 0.7V – IB 

10  0.7  19.7A 470k

IC = IB = 1.977mA VE = 0V VC =10 –1k(IC) = 8.03V VCE > 0.2  Transistor is in active region Operation point (VCE,IC) = (8.03V,1.97mA) 04.

(c) (ii)

Sol:

Given circuit is

100 + 10V

ID 100

100

After applying thevenin’s equivalent at 100 50 5V

ID 100

: 28 :

E & TE

Given Rf = 25 V =0.7V 50 + 5V –

ID 

ID

+ 0.7–

25 100

5  0.7  24.57mA 175

: 29 :

CONVENTIONAL PAPER - 1

: 30 :

E & TE

SECTION-B 05.

(a) Find the Thevenin resistance for the circuit shown below by zeroing the sources. Then, find the short-circuit current and the Thevenin equivalent circuit.

5 20V +–

20

2A (12M)

05.

(a)

Sol:

Finding Rth:

5 20

Rth = 5//20 = 4 Finding Isc:

5 20V

+ –

0V

20

2A

Isc

0  20 0  0   2  I sc  0 5 20 – 4 – 2 + Isc = 0 Isc = 6A ACE Engineering Academy

: 31 :

CONVENTIONAL PAPER - 1

Finding VOC:

5

VOC +

20V

+ –

20

2A

VOC –

VOC  20 VOC  20 5 20 4VOC – 80 + VOC – 40 = 0 5VOC = 120 VOC = 24V OR

VOC = ISCRth = 64 = 24V 4 24V

+ –

05. (b) A voltmeter and an ammeter are to be used to determine the power dissipated in a resistor. Both the instruments are guaranteed to be accurate within  1% at full-scale deflection. If the voltmeter reads 80 V on its 150 V range and the ammeter reads 70 mA on its 100 mA range, Determine the limiting error for the power calculation. 05.

(b)

Sol:

Given data:

(12 M)

Both voltmeter and ammeter accurate within  1% of fsd. Voltmeter reads 80 V on 150 V range ACE Engineering Academy

: 32 :

E & TE

Ammeter reads 70 mA on 100 mA range For voltmeter error =

1  150  1.5V 100

Relative error for voltmeter = For ammeter error =

1.5  100  1.875 % 80

1  100 mA  1mA 100

Relative error for ammeter =

1  100  1.43% 70

% limiting error in power measurement % P = %V + %I = 1.875 % + 1.43 % = 3.305 % Power = 8070 mA = 5.6 W Limiting Error = 5.6 

3.305  0.18W 100

Power = 5.6  0.18 W 05. (c) The following measurements pertain to a two-port circuit operating in the sinusoidal steady state. With port 2 open, a voltage equal to 150 cos 4000t V is applied to port 1. The current into port 1 is 25 cos (4000t-45) A, and the port 2 voltage is 100 cos (4000t+15) V. With port 2 short-circuited, a voltage equal to 30 cos 4000t V is applied to port1. The current into port 1 is 1.5 cos(4000t+30) A, and the current into port 2 is 0.25cos(4000t+150)A. Find the parameters that can describe the sinusoidal steady-state behaviour of the circuit. (12M) 05.

(c)

Sol:

25–450 15000

I2=0 N/W

+ –

+ 100150 –

V1 = AV2 – BI2 I1= CV2 – DI2 ACE Engineering Academy

: 33 :

I2 = 0  A  C

CONVENTIONAL PAPER - 1

V1 1500 0   1.5  150 V2 100150

I1 25  450 1     600 0 V2 10015 4

Now

1.5300

3000

V2 = 0  B  D

0.251500

N/W

+ –

 V1  300 0   12030 0 0 I2 0.25150

 I1  1.530 0   660 0 0 I2 0.25150

The parameters that can describe the sinusoidal steady state behaviour are A B   1.5  150 12030 0    C D   0 660 0    0.25  60

05. (d) For the circuit shown in the figure, find the branch currents I1,I2 and I3 using Mesh analysis.

I1

I2 5

(12 M)

6

I3

15V +–

10

4

+– 10V 05.

(d)

5

Sol:

+ 15V

+ –

6

I1 –

I3

+ – 10 – + + –

+

I2 + –

4

10V

: 34 :

E & TE

I3 = I1 – I2 15 – 5I1 – 10(I1–I2) – 10 = 0 15I1 – 10I2 = 5 3I1 – 2I2 = 1 --------(1) 6I2 + 4I2 – 10 +10(I2–I1) = 0 2I2 – I1= 1 -------(2) from (1) & (2) I1 = 1A I2 = 1A I3 = I1 – I2 = 1–1 I3 = 0A 05. (e) (i) Briefly discuss only the basic principles of a Thermistor and Thermocouple. (ii) Explain why a semiconductor has a negative resistance coefficient. (12 M) 05.

(e) (i)

Sol:

Thermistors are essential semiconductor devices that behave as resistors with high negative temperature coefficient and are atleast 10 times as sensitive as the platinum resistance thermometer. Thermisters can detect very small changes in temperature which could not be observed with an RTD or a thermocouple. Sensitivity of thermister is higher than both RTD and thermocouple. Sensitivity of thermistor is higher than both RTD and thermocouple. Thermistors exhibits a highly nonlinear characteristic of resistance versus temperature. The approximate relationship between resistance and temperature applying to most thermistors is R1 = R 0 e

 1 1     T1 T0

  

 = thermistor constant in Kelvin R1, R0  Resistance at absolute temperature T1 and T0 . Thermistors widely involve in temperature measurement in range of –60oC to 15oC. ACE Engineering Academy

: 35 :

CONVENTIONAL PAPER - 1

The response time of thermistors can vary from a fraction of second to minutes depending on the size of the detecting mass and thermal capacity of the thermistor. Thermocouples consists of two dissimilar metal wires A and B insulated from each other but welded or brazed together at their ends forming two junctions. Metal A Hot T1

Cold T2

Metal B Thermo junctions If two wires of different metals are joined together at each form a complete circuit in which current flows if two junctions are kept at two different temperature called seeback effect. The relation between the thermo emf setup and temperature difference of hot and cold junction is given by E =  (T – T0) + ( T 2  T02 ) E = Thermoelectric emf (Volts) T = Absolute temperature of hot junction T0 = Absolute temperature of cold junction ,  - Constants depends upon the metals used.

05.

(e)(ii)

Sol:

Conductivity in a semiconductor is given by i  n i  n   p   n i

[Let  = n + p]

Let ni is intrinsic carrier concentration,  be the mobility of intrinsic semiconductor as temperature increases  ni increases   decreases But here dominant factor is ni [∵ ni > ] ACE Engineering Academy

: 36 :

E & TE

 Conductivity of intrinsic semiconductor increases and hence its resistivity decreases there by resistance decreases from this we can say that intrinsic semiconductor has negative resistance coefficient 

ni

T

Temperature coefficient 

i

R

R  Ve T

06. (a) (i) Use source transformations to aid in solving for the currents i1 and i2 shown in the circuit below.

i1

5

(15 M)

i2 20V

+ –

1A

10

06.

(a) (i)

Sol:

Applying source transformation to current source and resistance in parallel i1

20V

i1 

5

10

+ –

+ –

10V

20  10 2  A 15 3

Applying source transformation to voltage source and resistance in series i2 4A

i2 

5

10

1A

5 5  4  1  A 5  10 3

: 37 :

CONVENTIONAL PAPER - 1

06. (a) (ii) Find the phasor voltage and the phasor current through each element in the circuit shown in the figure.

(10 M)

iC 5 cos (200t)

06.

iL

iR

1H

100

+

v –

100F

(a) (ii)

Sol:

iC

iL L

C

5cos200t

iR 100

Y(Admittance) 1 j  jC  R L

Y 

1 j j   100 50 200

1 3j  100 200

2  3j 200

I = Y.V

V

I 5 cos 200t 1000   200  cos 200 t  56.30 Y 2  3j 13

iR 

V 10  cos200t  56.30  R 13

iL 

V 5  cos 200t  146.30 jL 13

: 38 :

i C  V( jC) 

1000  j  200  10 4 cos 200t  56.30 13

20 j cos200t  56.30  13

20 cos 200t  33.7 0 13

E & TE

06. (b) Find the equivalent resistance and capacitance that causes a Wien bridge to null with the following component values:

(15 M)

R1 = 3.1 k , R3 = 25k, R4 = 100k C1 = 5.2 F and f = 2.5 kHz 06.

(b)

Sol:

I1

C2

b

R2

R4 I1

a

G

I2

c

R1

R3 C1

d

I2

 E

Wien’s bridge is used to determine the frequency at a AC source and also used in many applications like capacitance having series resistance at balance.    R2 j   R 3    R 1  R 4    1 j C R C 2 2  1   

i.e.,

 R 3 R1 C2 1      j C1R 2  C 2 R 1  R 4 R 2 C1 

: 39 :

CONVENTIONAL PAPER - 1

Equating the real and imaginary parts, R 3 R1 C2   R 4 R 2 C1 and 

1 R 1 R 2 C1 C 2

f

1 2 R 1R 2C1C 2

6.25 106  R 2C2 

1 4  (3.110 )  (5.2 10 6 )  R 2C 2 2

3

1 4  (3.1  10 )  (5.2  10 6 )  6.25  106 2

3

= 0.25110–6 ------- (1) C2 R 3 R1   --------- (2) C1 R 4 R 2 Using equation (1) & (2) C2 is 20 pF R2 = 12.5 k

06. (c) (i) The voltage source Vg drives the circuit shown in the figure. The response signal is the voltage across the capacitor, V0. Calculate the numerical expression for the transfer function.

(10 M)

1000

+ 250

+ Vg –

50mH

1F V0 –

: 40 :

06.

(c) (i)

Sol:

Voltage gain transfer function is given by R1

E & TE

+ R2

Vg

+ –

sL

1/cs V0 –

  1    R 2  SL  // CS     V0 (s)  Vg (s)   1   R  SL  //  R1     2  CS    SL  R 2  V0 (s)  Vg (s)   2  R 1LCS  (R 1R 2 C  L)S  R 1  R 2  

Now substitute the given values where, R1 = 1000 R2= 250 L = 50mH C = 1F then the voltage gain T.F is V0 (s) 250  (0.05  s)  Vg (s) 5  10 5 s 2  0.3s  1250

: 41 :

CONVENTIONAL PAPER - 1

: 42 :

E & TE

06. (c) (ii) Write down the incidence matrix and cut-set matrices for the network shown below. (10 M)

5

5

4

10V 4

4

5

06.

(c) (ii)

Sol:

Writing graph for given network

C

6

1

2

D

4 A Incident matrix given by

5 B

3

1 2 3 4 5 6 A 1 0 0  1 0 1 B 0 1 1 0 1 0   C 1 1 0 0 0  1  D0 0 0 1 1 1   Now writing tree for given graph and identifying basic cut sets C3

C

6

1

2 5

4 A C1 ACE Engineering Academy

B

3 C2

: 43 :

CONVENTIONAL PAPER - 1

Now writing down cutest matrix 1 2 C1 C2 C3

3

4 5 6

  1 0 1 1 0 0  0 1  1 0 1 0    1 1 0 0 0 1

07. (a) A certain 5-hp three-phase induction motor operates from a 440-V-rms (line-to-line) threephase source and draws a line current of 6.8 A rms at a power factor of 78 percent lagging under rated full load conditions. The full load speed is 1150 rpm. Under no-load conditions, the speed is 1195 rpm, and the line current is 1.2 A rms at power factor of 30 percent lagging. Find the power loss and efficiency with full load, the input power with no load, and the speed regulation. 07.

(a)

Sol:

Given

(20 M)

VLL = 440V If1 = 6.8A cosf1= 0.78lag Nf1 = 1150 rpm I0 = 1.2A cos0= 0.3lag N0 = 1195 rpm Power rating of machine = 5hp Pf10 = 5735 = 3675W Power output at full load (P0) = 3675W Power input at full load Pin = Pin =

3VL I f1 cos f1

3  440  6.8  0.78

Pin = 4042.19W

: 44 :

E & TE

Power loss Ploss = Pin – P0 = 4042.19 – 3675 PLoss = 367.19W efficiency  = 

Power output Power input

3675 4042.19

= 90.91%

Input power at no load is Pin O  3VL I O cos  0

=

3  440  1.2  0.3

Pin O = 274.36W speed regulation  

No load speed  full load speed full load speed

1195  1150 1150

speed regulation = 3.9%

07. (b) (i) Explain the operation of a Voltage-to Frequency Converter. Given the primary advantages and limitations of voltage-to-frequency converters. (ii) The relationship between the input voltage vi and the output frequency f for the VCO is given as vi = f/50. If 530 pulses are passed by the AND gate during 0.1 sec gating pulse, what is the amplitude of vi?

(20 M)

: 45 :

CONVENTIONAL PAPER - 1

07.

(b) (i)

Sol:

In voltage to frequency converter the analog input is applied to an integrator. C R

Comparator

Counter

R Digital Output Pulse Generator

Pulse trigger output

Output of integrator Zero Level Trigger Level Time between two threshold levels The integrator produces a ramp signal whose slope is proportional to the input voltage signal level. When this ramp signal reaches a present threshold level, a trigger pulse is produced. Also a current pulse is produced which discharges the capacitor of the integrator, after which a new ramp is initiated. The time between successive threshold level crossings is inversely proportional to the slope of the ramp. Since the slope of the ramp is proportional to the input analog voltage, hence the frequency pf output pulses from the comparator is directly proportional to the input voltage. Advantages:

(1) V to F converters integrate noise and so are useful under circumstances similar to dual slope units (2) V to F are precise, accurate, simple, inexpensive, low powered and mostly run for a wide range of supply voltages ACE Engineering Academy

: 46 :

E & TE

Limitations:

(1) Due to changeover switch there is integrator transient problem. (2) The switch isn’t closed for long time as there will be residual potential on the integrator and the time to integrate to threshold will be reduced. (3) It relatively slow. 07.

(b) (ii)

Sol:

Vi =

f 50

530 pulses  0.1 sec 5300 pulses  1 sec  f = 5300 Vi =

5300 530 f   106V = 50 5 50

Vi = 106 V

07. (c) For the circuit shown in the figure below, let vc (0) = 15V (i) Find vc,vx and ix for t > 0.

5 0.1F

07.

(10 M)

8

ix + 12 vx –

+ –vc

(c) (i)

Sol:

Vc(s) 5

+ –

15/s 10/s

8 Ix(s) + 12 Vx(s) –

: 47 :

Vc (s)  5

CONVENTIONAL PAPER - 1

15 s  Vc (s)  0 10 / s 20

Vc (s) 

1 3 1 s Vc (s)       5 10 20  2  4  2s  1 3  Vc (s)   20  2 Vc (s) 

30 15  2s  5 s  5 2

v c ( t )  15e

5  t 2

; t>0

Vx(s) = Ix(s).12  

Vc (s)  12 20 15 12  5 20 s 2 9 s

5 2

 v x ( t )  9e I x (s)  

5  t 2

; t>0

Vc (s) 20 15 1  20 s  5 2 5

3  t i x ( t )  e 2 ; t >0 4

: 48 :

E & TE

: 49 :

CONVENTIONAL PAPER - 1

07. (c) (ii) Compute V1 and V2 in the circuit shown below. (10 M)

10450V +–

V1

300A

4

–j3

07.

(c) (ii)

Sol:

Using super node analysis

j6

V2 12

V1 V2 V2    30 0 -------(1)  3 j 6 j 12 V1 –V2 = 10450 V1 V V j  2 ( j)  2  30 0 3 6 12 (4 j)V1  V2 (2 j)  V2  360 0 V1 (4 j)  V2 (1  2 j)  360 0 4 j(V1 )  4 j(V2 )  401350    0 V2 (2 j  1)  360  401350 360 0  401350 V2  1 2 j

= 31.4–87.180 V1 = V2+10450 = 25.78–70.470

: 50 :

E & TE

08. (a) (i) Two coils are wound on a toroidal core as illustrated in the figure below. The reluctance of the core is 107 (ampere-turns)/Wb. Determine the self-inductances and mutual inductance of the coils. Assume that the flux is confined to the core so that all of the flux links both the coils.

(10 M)

i2

i1 +

+ e1

e2 –

– N2 = 200

N1 = 100

08. (a) (i) Sol:

Given reluctance of core c = 107 AT/wb N1 = 100 N2 = 200 Self inductance of coil 1 L1   L1 L1 

N12 

2  100 

10 7

10 4 107

L1 = 10–3H Self conductance of coil 2 L 2  L2

N 22 

2  200 

L2 

2002 

L2 = 410–3H Since the flux is confined only to core coefficient of coupling K = 1 ACE Engineering Academy

: 51 :

CONVENTIONAL PAPER - 1

Mutual inductance M = K L1L 2 M = 1 10 3  4  10 3 M = 210–3H

08. (a) (ii) Consider the source, transformer and load shown in the figure below. Determine the rms values of the currents and voltages, (case-1) with the switch open and (case-2) with (10 M)

the switch closed. I1 +

110 Vrms 50 Hz –

I2

N1:N2

+

+

V1

V2

– N1 5 N2

08. (a) (ii) Sol:

10

Case I: switch open I1 +

110 Vrms 50 Hz –

I2

N1:N2

+

+

V1

V2

10

N1 5 N2 Assuming zero voltage drop V1= 110V When secondary is open I2 = 0A From current transformation formula I1 N 2 1   I 2 N1 5 I1 = 0A ACE Engineering Academy

: 52 :

but

E & TE

V1 N1  (Voltage transformation) V2 N 2

110 5 V2 V2 = 22V V1= 110V

I1 = 0A

V2 = 22V

I2= 0A

Case II: When switch is closed: I1 + 110 Vrms 50 Hz –

I2

N1:N2

+

+

V1

V2

10

– N1 5 N2

Assuming negligible voltage drop V1= 110V From voltage transformation V1 N1  V2 N 2 V2 

N2 1  110V  110 N1 5

V2= 22V From secondary side we can say that V2 = I2RL = I2(10) 22 = I2(10) I2= 2.2A ACE Engineering Academy

: 53 :

CONVENTIONAL PAPER - 1

From current transformation I1 N 2 1   I 2 N1 5 I1  I 2  I1 

1 5

2.2 5

I1= 0.44A

08.

V1= 110V

I1= 0.44A

V2= 22V

I2= 2.2A

(b) (i) Describe sampling oscilloscope and storage oscilloscope in brief. (ii) If, in the figure given below, the distance Y1 is 1.8 cm and Y2 is 2.3 cm, what is the

phase angle using the X-Y mode of oscilloscope? (15 M)

Y1

Y2

08.

(b) (i)

Sol:

A sampling oscilloscope generally used to test fast varying signals. The advantage of a sampling oscilloscope is that it can measure very high speed events, which require sweep speeds of the order of 10 psec per division and amplifier bandwidths of 15 GHz. While the disadvantage of a sampling oscilloscope is that it can only make measurements on repetitive wave form signals, continuous display for frequencies in the range of 50 – 300 MHz. The sampling oscilloscope is able to respond and store rapid bits of information and present them in a continuous display. However it should be understood that sampling techniques cannot be used for the display of transient waveforms.

: 54 :

E & TE

The horizontal deflection of the electron beam is obtained by application of a staircase waveform to X deflection plates Input Waveform Sampling Plates

Stair case time base waveform

Figure: Sampling Oscilloscope A digital oscilloscope digitises the input signal, so that all subsequent signals are digital. A conventional CRT is used and storage occurs in electronic digital memory. Input Signal

Amplifier

Digitiser

Memory

Analyser Circuitry

Vertical Plates

Waveform Reconstruction

CRT Trigger Clock

Time Base

Horizontal Amplifier

Horizontal Plates

Figure: Block diagram of basic DSO The input signal is digitised and stored in memory in digital form as shown in block diagram. In this state it is capable of being analysed to produce a variety of different information. To view the display on the CRT the data from memory is reconstructed in analog form. Digitising occurs by taking a sample of the input waveform at periodic intervals. In order to ensure that no information is lost, sampling theory states that the sampling rate must be atleast twice as fast as the highest frequencies in the input signal. To full fill this requirement we require digitiser which is flash A/D converter which is fastest in conversion rate many different input channels are used with DSO. However if all these channels share a common store, through a multiplexer then the memory available to each channel is reduced. ACE Engineering Academy

: 55 :

08.

(b) (ii)

Sol:

Y Phase angle  = sin 1  int  Ymax

CONVENTIONAL PAPER - 1

 Y  1.8  o  = sin 1  1  = sin 1   = 51.5  2.3   Y2  

But as direction of movement of Lissagious figure not given phase angle may be two possible cases  1.8  o 1 = sin 1   = 51.5  2.3 

or

 1.8  o 1  1.8  2 = – sin 1   = 360 – sin   = 308.5  2.3   2.3 

08. (c) For the S-domain circuit shown in the figure, find: (i) the transfer function H(s) = V0/V1, (ii) the impulse response, (iii) the response when vi(t) = u(t) V, and (iv) the response when vi(t) = 8 cos 2t V. 1

Vi

a

S + 1 V0 –

1

+ –

(25 M)

b

08.

(c)

Sol: (i)

1

Vi

+ –

Va a 1

S + 1 V0 –

Va  Vi Va V   a  0 -------(1) 1 1 s  1 V0 

1 Va ------(2) s 1

: 56 :

(1)  2Va – Vi+

E & TE

Va 0 s 1

1   Va 2   Vi s  1  Va s 1  Vi 2s  3 (2) 

V0 1  Va s  1

V0 V0 Va 1 s 1     Vi Va V1 s  1 2s  3 V0 1  Vi 2s  3

(ii)

V0 1 / 2  Vi s  3 2 3

h(t) 

1 2t e ;t0 2

(iii) Vi(t) = u(t)  Vi (s) 

1 s

1     1/ 2 1 1 1  2  V0 (s)  Vi (s)      3  3  3 s s  3  s s s   2 2   2 V0 ( t ) 

1 1  e 3 t / 2 u ( t ) 3

(iv) When Vi(t) = 8cos2t H( j) 

1 3  2 j

: 57 :

CONVENTIONAL PAPER - 1

When input is sine, output is also sine for LTI system   1 Magnitude of output = 8    at  =2 2   9 4  

8 5

 2  Angle of output =  Tan 1    3  2 4  Tan    3 = –53.130 So, output = 1.6cos(2t–53.130)

: 58 :