Analog Elec
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ESE Online Test Series : 2015Conventional Paper
E & T : ELECTRONICS & TELECOMMUNICATION ENGG.
TEST - 8 | Analog Electronic CircuitsDate: 15-03-2015 | Subjectwise Test
Duration: 1 hrs. Maximum Marks: 100
Read the following instructions carefullyRead the following instructions carefullyRead the following instructions carefullyRead the following instructions carefullyRead the following instructions carefully
1. Candidate should attempt any FOURFOURFOURFOURFOUR questions out of five. Each question carries 25 marks.
2. Marks carried by each subdivision of a question is indicated at the end of subdivision.
3. Answers must be written only in ENGLISH.
4. Assume suitable data, if necessary, and indicate the same clearly.
5. Neat sketches may be drawn, wherever required.
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Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata
2 E & T Engineering Analog Electronic Circuits
IOTET15
Q.1Q.1Q.1Q.1Q.1 (a)(a)(a)(a)(a) Design a self bias circuit for a CE amplifier using an NPN transistor with = 100, the other details are:VCC = 12 V, VCEQ = 6 V, ICQ = 4 mA.
[15 marks][15 marks][15 marks][15 marks][15 marks]
(b)(b)(b)(b)(b) Consider the following circuit
RD
+VCC
VS
RS
Rf
RS
V0
gm = 5 mSRD = 5 kRS = 0.75 kRF = 20 k
Calculate the ratio of voltage gain with feedback to voltage gain without feedback is [10 marks] [10 marks] [10 marks] [10 marks] [10 marks]
Q.2Q.2Q.2Q.2Q.2 (a)(a)(a)(a)(a) In the basic differential amplifier of figure , Given : RC = 2 k; RE = 4.3 k, VCC = VEE = 5 V; 0 = 200,VBE = 0.7 V.
VBE1
B1
V1 V2
B2
VBE2
VCC
RCRC iC1 iC2
VO2VO1
Q1 Q2
iE1 iE2
REIQ
VEE
Determine :(i) For V1 = V2 = 0, that is for both the inputs grounded, the values of quiescent currents and voltages,
IBQ, ICQ, V01, V02, VCEQ.(ii) ADM, ACM and CMRR.
[15 Marks][15 Marks][15 Marks][15 Marks][15 Marks]
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3ESE Online Test Series 2015 : Conventional Paper
Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata IOTET15
(b)(b)(b)(b)(b) Assuming forward voltage drop across diodes to be 0.7 V, draw the transfer characteristic of theclipper circuit shown in figure.
1 k
Vi V = 10 V
1 k
D
5 V
V0
+
[10 marks][10 marks][10 marks][10 marks][10 marks]
Q.3Q.3Q.3Q.3Q.3 (a)(a)(a)(a)(a) For the circuit shown below,
R2RD CC
VL
+
+
Vi
CCriii
iL
RL
CSRS
+VDD
R1
Let gm = 2 mS, rds = 30 k, Rs = 3 k, RD = RL = 2 k, R1 = 200 k, R2 = 800 k and ri = 5 k. IfCC and CS are large and the amplifier is biased in the pinch off region find(a) Zin (b) AV (c) Ai
[15 Marks][15 Marks][15 Marks][15 Marks][15 Marks]
(b)(b)(b)(b)(b) A single phase full-wave rectifier with silicon diodes has a load of 2 kW. The transformer primaryvoltage is 230 rms with frequency of 50 Hz. If the step down transformer has a turns ratio of 10 : 1, whatis(i) The load voltage(ii) The load current(iii) The output power(iv) The frequency of the load voltage.
[10 Marks][10 Marks][10 Marks][10 Marks][10 Marks]
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Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata
4 E & T Engineering Analog Electronic Circuits
IOTET15
Q.4Q.4Q.4Q.4Q.4 (a)(a)(a)(a)(a) Derive an expression for the closed loop gain V0\Vi of the circuit shown below. Assume ideal OP-AMP.
R1Vi
V0
R2 R4R3
[10 marks][10 marks][10 marks][10 marks][10 marks]
(b)(b)(b)(b)(b) For the circuit shown, neglecting base currents find I0 and I1, I2, I3.
I3I2I1+6 V
10 k
1.4 k Ro R1 300 R2 400 R3 500
I0
6 V
[15 marks][15 marks][15 marks][15 marks][15 marks]
Q.5Q.5Q.5Q.5Q.5 (a)(a)(a)(a)(a) For the circuit shown below, if the input is a constant V, show that the output V0(t) is given by adifferential equation
+
R2
R3
C
V0
ZF
ViR1
[15 Marks][15 Marks][15 Marks][15 Marks][15 Marks]
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5ESE Online Test Series 2015 : Conventional Paper
Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata IOTET15
(b)(b)(b)(b)(b) Shown below is a positive feedback arrangement to generate sinusoidal oscillation, 21
5RR
= .
+
R1
R2
N/W( ) f
+
V f0( )+( )
V ff
The feedback factor, 0
( )( )
( )fV f f
V f= is
[5 marks][5 marks][5 marks][5 marks][5 marks]
(c)(c)(c)(c)(c) Given below is the shunt regulator circuit.
VL
IC IL
RL= 100 + _VBE
8.3 V+
Iz
Is
125 Vi
(+21 V)
The collector current IC is[5 marks][5 marks][5 marks][5 marks][5 marks]