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

  • Analog Electronic Circuits - AEC
  • 2014
  • PYQ Solution
  • Biju Patnaik University of Technology Rourkela Odisha - BPUT
  • Electronics and Communication Engineering
  • B.Tech
  • 162 Offline Downloads
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Subject: Analog Electronics Circuit Subject Code: PCEC 4201 Semester: Third Semester Applicable Branch (S): AEIE, BIOMED, CSE, EC, EEE, ELECTRICAL, ETC, IEE, IT Question Code: H 397 Year of Examination: 2014 Exam Type: (Special/Regular): Regular Solution Prepared by Sl.No. Faculty Name Email ID 1. SANTOSH KUMAR ACHARYA santosh.acharya1988@gmail.com 2. MITU BARAL mitubaral@gmail.com 3. MANOJ KUMAR SENAPATI manojsenapati@nist.edu 4. MUKESH KUMAR SUKLA mukesh_ele02@yahoo.co.in 5. MALABIKA PATTNAIK mala2007_patt@yahoo.co.in 6. DEBABANDANA APTA apta.debabandana@gmail.com 7. RAKESH ROSAN yasrakesh@gmail.com 8. G.GIRISH girish@nist.edu 9. RAJESH KUMAR DASH r_kdash@yahoo.com NATIONAL INSTITUTE OF SCIENCE &TECHNOLOGY

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Analog Electronics Circuits (Regular Examination 2014) PCEC 4201 PALUR HILLS, BERHAMPUR, ORISSA – 761008, INDIA Q.1 Answer the following questions: a). Write Shockley’s equation. How it is used to design DC biasing of JFET? Ans: Shockley’s equation:  V  I D  I DSS 1  GS   VP  2 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. b) The voltage gain of an amplifier with negative feedback is 100. If the feedback factor is 40%. Find the voltage without feedback. Ans: With negative feedback amplifier gain, Af  A 1  A Where Af = Gain with feedback, A = Open loop Gain, β = Feedback factor Given Af = 100, β = 40 % Now, Af  Remark: The answer violating the principle of negative feedback. For a negative feedback closed loop A 1  A  100  A  100  40 A  A 1  0.4 A 39 A  100  A  2.56 gain Af is always less than the open loop gain A but as per the data given it’s coming to be low. c) Define CMRR and Slew rate of ideal op-amp. Ans: CMRR:  The relative sensitivity of an Op-amp to a differential signal as compared to a common mode signal is called CMRR. Slew Rate:  Slew rate is defined as the maximum rate of change of output voltage per unit time. National Institute of Science and Technology, Palur Hill, Berhampur, Odisha-761008. 2 For an ideal Op-amp common mode rejection ratio is infinite (CMRR=∞). Page 

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Analog Electronics Circuits (Regular Examination 2014)  PCEC 4201 For an ideal Op-amp, infinite slew rate. d) Write the difference BJT and FET. Ans: BJT(Bipolar Junction Transistor) FET(Field Effective Transistor) 1. BJT is a bipolar semiconductor device, 1. FET is a unipolar Device (Current (Current conduction due to both type of conduction due to only majority charge charge carriers i.e. due to both majority carriers i.e. due to electrons or holes) and minority electrons and holes) 2. BJT is current control device as the 2. FET is voltage control device as the output current is control by input current. output current is control by input voltage. 3. Less thermal stability 3. Better thermal stability 4. Low input impendence (Due to forward 4. High input impendence (Due to reverse bias) bias) e) Show that the dynamic resistance of a diode varies inversely with current. Ans: Dynamic resistance of a diode is rd  VD 26mv or rd  I D ID The equation says that the dynamic resistance of diode is is varying inversely with current. f) Design an RC phase shifter that introduces a phase shift of π/4 radians. Ans:  XC  R  1   .If X C  R , then   tan 1  4  Page Let C=0.1µF and f=1 KHz. 3 1 Here   tan  National Institute of Science and Technology, Palur Hill, Berhampur, Odisha-761008.

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Analog Electronics Circuits (Regular Examination 2014) R  XC  1 2 fc PCEC 4201  1591  1.6 K  g) What is the linear amplification factor of a transistor if its gain 100? Ans: Here given β=100 Now linear amplification factor,   100 100   0.99   1 100  1 101  h) 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 i) Write two advantages of push-pull power amplifier. Ans: Advantages of push-pull power amplifier:  The efficiency of a class-B push-pull amplifier is higher than class-A amplifier. (The reason for this is that no power is drawn from the D.C. power supply VCC under no signal condition in class-B push-pull amplifier, as the Q-point is at cut off region in class-B power amplifier)  It helps in impedance matching between stages. j) Why a fixed bias is called so? Validate. Ans: For the given circuit, Applying KVL to the input loop, VCC  VBE RB Page 4 VCC  I B RB  VBE  0  I B  National Institute of Science and Technology, Palur Hill, Berhampur, Odisha-761008.

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