Lic lab manual 1

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Lic lab manual 1

  1. 1. SAVEETHA ENGINEERING COLLEGE SAVEETHA NAGAR, THANDALAM, CHENNAI - 602105DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING EC2258-LINEAR INTEGRATED CIRCUITS LAB LABORATORY MANUAL PREPARED BY A.HEMA MALINI, ASST. PROF. Department of ECE
  2. 2. LIST OF EXPERIMENTSDesign and Testing of 1. Inverting, Non – inverting and Differential amplifiers. 2. Integrator and Differentiator. 3. Instrumentation amplifier. 4. Active low pass and band pass filter. 5. Astable, monostablemultivibrator and Schmitt trigger using Op – amp. 6. Wein bridge and RC Phase shift oscillator using Op – amp. 7. Astable and monostablemultivibrator using NE 555 timer. 8. Frequency multiplier using PLL. 9. DC power supply using LM 317. 10.Study of SMPS control IC SG 3524 / SG3525. 11.Simulation of Experiments 3, 4, 5, 6 and 7 using PSpicenetlists
  3. 3. Expt. No.1 ( INVERTING AND NON – INVERTING AMPLIFIER) 1. a. INVERTING AMPLIFIERAIM: To design an Inverting Amplifier for the given specifications using Op-Amp IC 741.APPARATUS REQUIRED:S.No Name of the Apparatus Range Quantity 1. Function Generator 3 MHz 1 2. CRO 30 MHz 1 3. Dual RPS 0 – 30 V 1 4. Op-Amp IC 741 1 5. Bread Board 1 6. Resistors As required 7. Connecting wires and probes As requiredTHEORY:The input signal Vi is applied to the inverting input terminal through R1 and the non-invertinginput terminal of the op-amp is grounded. The output voltage Vo is fed back to the invertinginput terminal through the Rf - R1 network, where Rf is the feedback resistor. The outputvoltage is given as, Vo = - ACL Vi 0Here the negative sign indicates that the output voltage is 180 out of phase with the inputsignal.
  4. 4. PROCEDURE: 1. Connections are given as per the circuit diagram. 2. + Vcc and - Vcc supply is given to the power supply terminal of the Op-Amp IC. 3. By adjusting the amplitude and frequency knobs of the function generator, appropriate input voltage is applied to the inverting input terminal of the Op-Amp. 4. The output voltage is obtained in the CRO and the input and output voltage waveforms are plotted in a graph sheet.RESULT:The design and testing of the inverting amplifier is done and the input and output waveformswere drawn.
  5. 5. PIN DIAGRAM:CIRCUIT DIAGRAM OF INVERTING AMPLIFIER:DESIGN:We know for an inverting Amplifier ACL = RF / R1Assume R1 ( approx. 10 KΩ ) and find RfHence Vo = - ACL Vi
  6. 6. OBSERVATIONS: Input Time period (seconds) Output S.No Amplitude(volts) Practical Theoretical 1. 2. 3.MODEL GRAPH:
  7. 7. 1. b. NON - INVERTING AMPLIFIERAIM:To design a Non-Inverting Amplifier for the given specifications using Op-Amp IC 741.APPARATUS REQUIRED:S.No Name of the Apparatus Range Quantity 1. Function Generator 3 MHz 1 2. CRO 30 MHz 1 3. Dual RPS 0 – 30 V 1 4. Op-Amp IC 741 1 5. Bread Board 1 6. Resistors As required 7. Connecting wires and probes As requiredTHEORY: The input signal Vi is applied to the non - inverting input terminal of the op-amp. Thiscircuit amplifies the signal without inverting the input signal. It is also called negative feedbacksystem since the output is feedback to the inverting input terminals. The differential voltage Vdat the inverting input terminal of the op-amp is zero ideally and the output voltage is given as, Vo = ACL ViHere the output voltage is in phase with the input signal.PROCEDURE: 1. Connections are given as per the circuit diagram. 2. + Vcc and - Vcc supply is given to the power supply terminal of the Op-Amp IC. 3. By adjusting the amplitude and frequency knobs of the function generator, appropriate input voltage is applied to the non - inverting input terminal of the Op-Amp. 4. The output voltage is obtained in the CRO and the input and output voltage waveforms are plotted in a graph sheet.
  8. 8. PIN DIAGRAM:CIRCUIT DIAGRAM OF NON INVERITNG AMPLIFIER:RESULT:The design and testing of the Non-inverting amplifier is done and the input and outputwaveforms were drawn.
  9. 9. DESIGN:We know for a Non-inverting Amplifier ACL = 1 + ( RF / R1)Assume R1 ( approx. 10 KΩ ) and find RfHence Vo = ACL ViOBSERVATIONS: Input Time period (seconds) Output S.No Amplitude(volts) Practical Theoretical 1. 2. 3.MODEL GRAPH:
  10. 10. Expt. No.2 (DIFFERENTIATOR AND INTEGRATOR)2. a. DIFFERENTIATORAIM:To design a Differentiator circuit for the given specifications using Op-Amp IC 741.APPARATUS REQUIRED:S.No Name of the Apparatus Range Quantity 1. Function Generator 3 MHz 1 2. CRO 30 MHz 1 3. Dual RPS 0 – 30 V 1 4. Op-Amp IC 741 1 5. Bread Board 1 6. Resistors 7. Capacitors 8. Connecting wires and probes As requiredTHEORY: The differentiator circuit performs the mathematical operation of differentiation; that is,the output waveform is the derivative of the input waveform. The differentiator may beconstructed from a basic inverting amplifier if an input resistor R1 is replaced by a capacitor C1 .The expression for the output voltage is given as, Vo = - Rf C1 (dVi/dt)Here the negative sign indicates that the output voltage is 1800 out of phase with the inputsignal. A resistor Rcomp= Rf is normally connected to the non-inverting input terminal of the op-
  11. 11. amp to compensate for the input bias current. A workable differentiator can be designed byimplementing the following steps: 1. Select fa equal to the highest frequency of the input signal to be differentiated. Then, assuming a value of C1< 1 µF, calculate the value of Rf. 2. Choose fb= 20 fa and calculate the values of R1 and Cf so that R1C1 = Rf Cf.The differentiator is most commonly used in waveshaping circuits to detect high frequencycomponents in an input signal and also as a rate–of–change detector in FM modulators.PIN DIAGRAM:CIRCUIT DIAGRAM OF DIFFERENTIATOR:
  12. 12. DESIGN :To design a differentiator circuit to differentiate an input signal that varies in frequency from 10Hz to about 1 KHz. If a sine wave of 1 V peak at 1000Hz is applied to the differentiator , drawits output waveformGiven fa= 1 KHzWe know the frequency at which the gain is 0 dB, fa= 1 / (2π Rf C1)Let us assume C1 = 0.1 µF ; thenRf = _________Since fb = 20 fa,fb = 20 KHzWe know that the gain limiting frequency fb = 1 / (2π R1 C1)Hence R1 = _________Also since R1C1 = RfCf ; Cf = _________Given Vp = 1 V and f = 1000 Hz, the input voltage is Vi = Vp sin ωtWe know ω = 2πfHence Vo= - Rf C1 (dVi/dt) = - 0.94 cosωtPROCEDURE: 1. Connections are given as per the circuit diagram. 2. + Vcc and - Vcc supply is given to the power supply terminal of the Op-Amp IC. 3. By adjusting the amplitude and frequency knobs of the function generator, appropriate input voltage is applied to the inverting input terminal of the Op-Amp. 4. The output voltage is obtained in the CRO and the input and output voltage waveforms are plotted in a graph sheet.
  13. 13. OBSERVATIONS: Time period(msec) Output Time Input Amplitude(volts)S.No Amplitude(volts) period(msec) .MODEL GRAPH: DIFFERENTIATORRESULT:-Thus the Differentiator Circuit is designed and constructed using op-amp and its performance isstudied.
  14. 14. 2. b. INTEGRATORAIM: To design an Integrator circuit for the given specifications using Op-Amp IC 741.APPARATUS REQUIRED:S.No Name of the Apparatus Range Quantity 1. Function Generator 3 MHz 1 2. CRO 30 MHz 1 3. Dual RPS 0 – 30 V 1 4. Op-Amp IC 741 1 5. Bread Board 1 6. Resistors 7. Capacitors 8. Connecting wires and probes As requiredTHEORY: A circuit in which the output voltage waveform is the integral of the input voltagewaveform is the integrator. Such a circuit is obtained by using a basic inverting amplifierconfiguration if the feedback resistor Rf is replaced by a capacitor Cf . The expression for theoutput voltage is given as, Vo = - (1/RfC1 ) ∫Vi dtHere the negative sign indicates that the output voltage is 1800 out of phase with the inputsignal. Normally between faandfb the circuit acts as an integrator. Generally, the value of fa<fb
  15. 15. . The input signal will be integrated properly if the Time period T of the signal is larger than orequal to RfCf . That is, T ≥ RfCfThe integrator is most commonly used in analog computers and ADC and signal-wave shapingcircuits.PIN DIAGRAM:CIRCUIT DIAGRAM OF INTEGRATOR:
  16. 16. DESIGN: To obtain the output of an Integrator circuit with component values R1Cf = 0.1ms , Rf = 10 R1and Cf = 0.01 µF and also if 1 V peak square wave at 1000Hz is applied as input.We know the frequency at which the gain is 0 dB, fb= 1 / (2π R1Cf)Therefore fb = _____Since fb = 10 fa, and also the gain limiting frequency fa = 1 / (2π RfCf)We get , R1 = _______ and hence Rf = __________PROCEDURE: 1. Connections are given as per the circuit diagram. 2. + Vcc and - Vcc supply is given to the power supply terminal of the Op-Amp IC. 3. By adjusting the amplitude and frequency knobs of the function generator, appropriate input voltage is applied to the inverting input terminal of the Op-Amp. 4. The output voltage is obtained in the CRO and the input and output voltage waveforms are plotted in a graph sheet.OBSERVATIONS: Time Output Input Amplitude(volts) Time period(msec)S.No period(msec) Amplitude(volts)
  17. 17. MODEL GRAPH: INTEGRATORRESULT:-Thus the integrator Circuit is designed and constructed using op-amp and its performance isstudied.
  18. 18. Expt. No.4SECOND ORDER ACTIVE LOW PASS FILTERAIM: Design a second order active Butterworth low pass filter, High pass filter and Bandpassfilterand also determine its frequency response using IC 741.APPARATUS REQUIRED :S.NO ITEM RANGE Q.TY 1 OP-AMP IC741 1 2 RESISTOR 10K , 1 1.5K 1 5.6 K 1 3 Capacitor 0.1 F 1 4 CRO - 1 5 RPS DUAL(0-30) V 1 DESIGN: Given: fH = 1 KHz = 1/ (2 RC) Let C = 0.1 F, R = 1.6 K For n = 2, (damping factor) = 1.414, Passband gain = Ao = 3 - =3 – 1.414 = 1.586. Transfer function of second order butterworth LPF as: 1.586 H(s) = --------------------------- S2 + 1.414 s + 1 Now Ao = 1 + (Rf / R1) = 1.586 = 1 + 0.586 Let Ri = 10 K , then Rf = 5.86 K
  19. 19. CIRCUIT DIAGRAM: LOW PASS FILTER MODEL GRAPH:Frequency Response Characteristics:Gain - 3 dB IndB fc = 1KHz Frequency (Hz)
  20. 20. TABULAR COLUMN: Vi = 1 Volt O/P voltage Gain=Vo/Vi Av=20 log Vo/ViS.No. FREQUNCY VO(Volts) dB (Hz)CIRCUIT DIAGRAM: HIGHPASS FILTER
  21. 21. MODEL GRAPH:Frequency Response Characteristics: TABULAR COLUMN: Vi = 1 Volt O/P voltage Gain=Vo/Vi Av=20 log Vo/Vi S.No. FREQUNCY VO(Volts) dB (Hz)
  22. 22. THEORY:An improved filter response can be obtained by using a second order active filter. Asecond order filter consists of two RC pairs and has a roll-off rate of -40 dB/decade. Ageneral second order filter (Sallen Kay filter) is used to analyze different LP, HP, BP andBSF.PROCEDURE : The connections are made as shown in the circuit diagram. The signal which hasto be made sine is applied to the RC filter pair circuit with the non-inverting terminal.The supply voltage is switched ON and the o/p voltages are recorded through CRO byvarying different frequencies from 10 Hz to 100 KHz and tabulate the readings.Calculating Gain through the formula and plotting the frequency response characteristicsusing Semi-log graph sheet and finding out the 3 dB line for fc.OBSERVATION: Vi = 1 Volt O/P voltage Gain=Vo/Vi Av=20 log Vo/ViS.No. FREQUNCY VO(Volts) dB (Hz)RESULT: Thus the second order Active Low Pass filter is designed and its frequencyresponse characteristic curves are drawn.

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