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UNIT – II: APPLICATIONS OF OP-AMP PART -A (2 Marks) PDF

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UNIT – II: APPLICATIONS OF OP-AMP PART -A (2 Marks) 1. An input of 3v is fed to the non-inverting terminal of an op-amp. The amplifier has a Ri of 10KΩ and Rf of 10KΩ .Find the output voltage[AUC April 2004] V =[ 0 2. Draw an integrator circuit using op-amp.[ AUC April 2004] 3. Mention two characteristics of instrumentation amplifier.[ AUC May 2005] high gain accuracy high CMRR high gain stability with low temperature coefficient low output impedance 4. Mention two applications of Schmitt trigger.[ AUC May 2005] To convert sine wave to square wave. Over voltage and over current protection circuit. On /off temperature controllers. 5. Draw an adder circuit using op amp to get the output expression as vo=- (0.1+v2+5v3)[ AUC May 2006] EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 1 6. Draw the circuit of a voltage follower using op-amp and prove that its gain is exactly equal to unity.?[ AUC May 2006] Va = Vb = vin Node a is directly connected to output V0= Va V0=Vin For the circuit voltage gain is unity. 7. Draw the block diagram of a multiplier using log and antilog amplifiers.[ AUC May 2006] EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 2 8. State the requirements of an instrumentation amplifier. [AUC Nov 2006] high gain accuracy high CMRR high gain stability with low temperature coefficient low output impedance 9. Name four applications of operational amplifier based comparator.[ AUC Nov 2006,April 2008] ero crossing detector Window detector Marker generator Phase meter 10. What is an antilog amplifier? Draw the circuit diagram of an antilog amplifier.[ AUC Nov 2007] Amplifier that converts logarithmic numbers back to decimal numbers is called antilog amplifiers. RF I 1k D1 A AD7414 2 - -VOS1 1 D1N4001 6 VO I OUT 3 + 5 + VOS2 V4 B U3 VOFF = 7 VAMPL = FREQ = 0 0 11. What is the principle of regenerative comparator.[ AUC Nov2007] If positive feedback is added to the comparator circuit, gain can be increased. Hence ,the transfer curve of comparator becomes more closer to ideal curve. If the loop gain is adjusted to unity then the gain becomes infinity. This results in an abrupt transition between the extreme values of the output voltage. EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 3 12. Draw the circuit diagram of a non-inverting amplifier ?[ AUC May 2008] 13. Draw the circuit of an integrator[AUC Nov 2008] 14. What is a V to I converter?[ AUC Nov 2008] Converter that produces output current which is directly proportional to the input voltage is called as V to I converter. EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 4 15. Draw the Schmitt trigger and give its application?[ AUC May2009] ROM AD7414 2 - -VOS1 1 1k 6 vo OUT V1 VOFF = 3 + 5 VAMPL = + VOS2 FREQ = U1 0 7 R1 R3 1k R2 1k 1k 0 0 Applications o To convert sine wave to square wave. o Over voltage and over current protection circuit. o On /off temperature controllers. 16. Draw the input and output waveforms for an integrator for square wave input.[ AUC May 2010] EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 5 17. Mention the applications of an instrumentation amplifier.[ AUC May 2009] Data acquisition system. Temperature indicator Temperature controller. Light intensity meter. 18. In what way, a precision rectifier using op-amp is superior to a conventional rectifier. [AUC Nov09 ,MAY 2011] Conventional rectifier cannot rectify voltages below 0.6V. Precision rectifiers rectify voltages having amplitude less than 0.7V 19. Draw an op-amp subtractor circuit. . [AUC Nov09] 20. Design an inverter using op-amp [AUC MAY 2010 ,MAY 2011] EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 6 21. Design a peak detector using op-amp [AUC MAY 2010] 22. Distinguish between active and passive filter. [AUC MAY 2011] ACTIVE FILTER PASSIVE FILTER Tuning is easy Difficult to tune Does not cause loading effect Causes loading effect to source or load More economical Cost is high 23. Why active guard drive is necessary for an instrumentation amplifier?[AUC MAY 2012] The common ground is shared by variety of circuits. Due to ground loop interference , additional voltage drop develops and lead to error in low voltage measurement. Due to distributed cable capacitances degradation of CMRR occurs. The active guard drive eliminates all these problems. 24. What is comparator? [AUC MAY 2012] Comparator is a circuit that compares the voltage applied at one of its input to that applied at its other input and to produce an output voltage which is either Vh or vl EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 7 25. Draw a non –inverting amplifier with a voltage gain of 3 [AUC NOV 2013] Ri Rf 1k 2k 45 AD741 2 - 6 Vin 3 + U2 71 26. Give an application for each of the following circuits : voltage follower ,peak detector ,Schmitt trigger , clamper Voltage follower : used in isolator circuit Peak detector : used for amplitude modulation in communication applications. Schmitt Trigger : square wave generators , ON –OFF controllers. Clamper : DC restorer in television receivers. PART-B (16 marks) 1. a)Explain briefly about the working of voltage to current converter[AUC May 2004] Voltage to current converter This circuit converts voltage to current Circuit diagram vf R4 1k I0 RL I0 ib=0 -vee 1k AD7414 2 - -VOS1 1 VD 6 OUT 0 3 + 5 + VOS2 I0 U1 7 V2 vee vin 0 EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 8 Operation Input voltage is applied to the non inverting terminal . Load resistance is connected in place of feedback resistor rf. This circuit is also called as negative feedback amplifier(current series) R1 is directly proportional to Io Applying KVL to the input loop Vin=vd+vf But open loop gain Av of op-amp is very large Vd ~ 0 Vin=Vf But Vin=R1*I0 I0=Vin /r1 Thus vin is converted into proportional output current (I0 = vin / R1) Applications Low voltage dc and ac voltmeters LCD and zener diode testers b) )Explain briefly about the working of triangular wave generator . Triangular wave generators This circuit consist of a square wave generator and an integrator. The square wave generator generate square wave at its output. This square wave is integrated by the integrator to generate a triangular waveform. EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 9 Circuit diagram R8 R9 1k 1k C2 1n AD7414 C1 2AD7-414-VOS1 1 R4 2 - -VOS1 61 V02 1n 6 OUT 3 + +VOOUST2 5 1k 3 U+3 +VOS2 5 U2 7 0 R7 7 R5 1k R6 1k 1k 0 0 Waveform Triangular wave is generated by alternatively charging and discharging a capacitor with a constant current source. Assume V0’ is high at +Vsat . This forces a constant current (+Vsat / R4) through C. When V0’ is low at –Vsat . This forces a constant current (-Vsat / R4) through C. The frequency of the triangular wave is same as the square wave frequency. The amplitude of the triangular wave will decrease as the frequency increases. EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 10

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EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO Phase meter. 10. Conventional rectifier cannot rectify voltages below 0.6V. Comparator is a circuit that compares the voltage applied at one of its input to that The analysis of the circuit can be done by following 2 steps. 1.
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