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Solution to Previous Year Exam Questions for Analog Electronic Circuits - AEC of CEC by Mitu Baral

  • Analog Electronic Circuits - AEC
  • 2015
  • PYQ Solution
  • Biju Patnaik University of Technology Rourkela Odisha - BPUT
  • Electronics and Communication Engineering
  • B.Tech
  • 377 Offline Downloads
  • Uploaded 1 year ago
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Analog Electronics Circuit (Regular/Back Exam-2015) PCEC 4201 Q.1 Answer the following Question: a). Why a fixed bias is called so? Justify. Ans: For the given circuit, Applying KVL to the input loop, VCC  IB R B  VBE  0  IB  VCC  VBE RB Here VCC is the supply voltage (fixed), RB is the base resister (fixed) and VBE (0.7V) does not vary significantly during use. Hence base current (IB) is fixed. So this type of circuit is called as fixed bias circuit. b). An amplifier bursts into oscillation when the loop gain Aβ=1, but for sustained oscillation Aβ>1. Why so? Ans: In reality, no input signal is needed to start the oscillator, only the condition βA=1 must be satisfied for self-sustained oscillation. But in practice, βA>1 and the system is started oscillating by amplifying the noise voltage, which is always present in an amplifier. Saturation factors in the practical circuit provide an „average‟ value of βA of 1. The resulting wave due to noise are never exactly sinusoidal, but when βA is closer to one, more nearly closer the sinusoidal waveform. [Noise signal results in build-up steady state oscillation condition] National Institute of Science and Technology, Berhampur-761008 Page 2

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Analog Electronics Circuit (Regular/Back Exam-2015) PCEC 4201 c). Which h-parameters one can determine from input characteristics and the output characteristics of a BJT? Ans: Input impedance (hie) and reverse voltage gain (hre) can be calculated from the input characteristics of BJT. hie  h re  VBE IB VBE VCE  VCE Const  IB Const VBE2  VBE1 IB2  IB1 VBE2  VBE1 VCE2  VCE1 VCE  VCEQ IB  IBQ Output admittance (h0e) and forward current gain (hfe) can be calculated from the input characteristics of BJT. h fe  h oe  IC IB  VCE Const IC VCE IC2  IC1 IB2  IB1  IB Const VCE  VCEQ IC2  IC1 VCE2  VCE1 I B  I BQ d). Give the load line of BJT amplifier of VCC =+9V and RC=1.8KΩ. Ans: e). Explain the origin of crossover distortion? How can this be minimized? Ans: Crossover distortion is a type of distortion which is caused by switching between devices driving a load, most often when the devices (such as a transistor) are matched. It is most commonly seen in complementary, or "push-pull", Class-B amplifier stages. National Institute of Science and Technology, Berhampur-761008 Page 3

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Analog Electronics Circuit (Regular/Back Exam-2015) PCEC 4201 To eliminate crossover distortion, both transistors in the push-pull arrangement must be biased slightly above cut-off when there is no signal. f). What are the minimum values of gain in inverting and non inverting amplifiers? Ans: Inverting amplifier: Av  RF  0 When RF <<< R1 R1 Non-inverting amplifier: Av  1  RF  1 When RF <<< R1 R1 g). Write Shockley’s equation. How it is used to design d.c. biasing of JFET? Ans: Shockley’s equation:  V  I D  I DSS 1  GS  VP   2 Where, ID is the drain current, IDSS is the drain to source saturation current, VGS is the gate to source voltage and VP is the pinch off voltage of the JFET. Shockley‟s equation gives the information about to set the JFET output current (ID) by the controlling the input voltage (VGS), with a constant value of IDSS and VP. National Institute of Science and Technology, Berhampur-761008 Page 4

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Analog Electronics Circuit (Regular/Back Exam-2015) PCEC 4201 h). Which is better buffer, a BJT or FET, Justify? Ans: A FET is a better buffer than BJT. As the input impedance of a FET is very high than BJT, so there will be very low loading effect will be occurs. i). Design an RC phase shifter that introduces a phase shift of 450 degree. Ans: X  Here   tan 1  C  .  R  If XC  R , then   tan 1 1   4 Let C=0.1µF and f=1 KHz. R  XC  1  1591  1.6K 2fc j). What is the linear amplification factor of a transistor if its gain is 100? Ans: Here given β=100 Now linear amplification factor,    100 100    0.99   1 100  1 101 Q.2 (a) Consider a general feedback system with parameters A=106 and Af=100. If the magnitude of A decreases by 20percentage what is the corresponding percentage in Af. Ans: Given that: open loop gain A=106 and gain with feedback Af=100 Now feedback factor can be calculated as: Af  Again  A 1  1  A   1  0.01 Af dAf 1 dA 1 dA   Af 1  A A A A dAf 1   20%  0.002% Af 0.01106 National Institute of Science and Technology, Berhampur-761008 Page 5

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