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- 1. Applications of Op-Amps Dr. C.SARITHA Lecturer in ElectronicsS.S.B.N. DEGREE & PG.COLLEGE ANANTAPUR
- 2. OVERVIEW Introduction Definitions Circuit Diagrams Derivations Applications Conclusion
- 3. Integrator The circuit in which the output wave form is the integral of input wave form is known as an integrator Such type of circuit is obtained by using basic inverting amplifier configuration where we use a capacitor in feed back
- 4. Circuit diagramIR Ic Vs
- 5. Explanation Input is applied to inverting terminal of the op-amp. Non inverting terminal is grounded. If sin wave is applied terminal then the output will be cosine wave.
- 6. For an ideal op-amp Ri=infinite R0=0For an ideal op-am input current=output current IR=IC IR=Vi-Vs/RiThe capacitor current Ic=c(d/dt(VS-V0) Ic=-c(d/dtV0-VS)
- 7. Input current =Output current(Vi-VS)/Ri=-C(d/dtV0-VS)Vi/Ri-VS/Ri =-C(d/dtV0-VS)V0=-AVSVS=-V0/ASubstitute this Vs in the above equationVi/Ri+V0/ARi=-C(d/dt(V0+VS/A) =-C(d/dtV0)-C(d/dtVS/A)
- 8. V0/A and V0/ARiAre very less compare to V0 and hence thisterms are neglected.Vi/Ri=-C(d/dtV0)Integrating on both sides b/w 0 to t t tVi/Ridt=- o Cd/dtV0o
- 9. 1 tRi 0 Vidt= -CV0V0= -1 t RiC 0 VidtThe output of the integrator is the integral ofinput voltage with time constant that is V0 isdirectly proportional to integral of Vidt andinversely proportional to the time constant.
- 10. The input is sine wave the output become cosine waveInput=Output=
- 11. Similarly the input is square wave theoutput become Triangle waveInput=Output=
- 12. Applications Analog computer A to D converters Many linear circuits Wave Shapping circuit
- 13. Differentiator The differentiator is the circuit whose output wave form is the differential input wave form. The differentiator may be constructed form the basic inverting amplifier Here we replace the input resistor by a capacitor.
- 14. Circuit Diagram
- 15. The capacitor current Ic=CI(d/dtVi-VS)Current through the feed back resistor IR=(VS-V0)/RF Input current=Output Current IC=IR C(d/dt(Vi-VS)=(VS-V0)/RF CiRF(d/dt(Vi-VS)=VS-V0 V0=-CiRF(d/dt(Vi-VS)+VS
- 16. Gain A=-V0/VS VS=-V0/ASubstitute VS in the above equation -CiRFd/dt(Vi-VS)+VS=V0-CiRFd/dt(Vi-VS)-VO/A=V0V0/A is very small and hence neglected VO=-CiRFd/dt(Vi)
- 17. The input is cosine wave the output becomesine waveInput=Output=
- 18. Similarly the input is Triangle wave theoutput become Square waveInput=Output=
- 19. Active Filters Filter is a circuit which gives the DC from the given input AC. The filter which constructed by using an op- amp is known as the active filter. (Because the op-amp is an active component)
- 20. Types of filters High pass filter: It allows the high frequency signals and filter them (convert them into DC). Low pass filter: It allows the low frequency signals and filter them (convert them into DC).
- 21. Low pass filtersR C
- 22. Working A first order filter consists of a single RC network connected non inverting terminal of the op-amp At low frequency the capacitor appears open and the circuit acts like an inverting amplifier with a voltage a gain of (1+R2/R1)
- 23. …Continues As the frequency increases the capacitive reactance the capacitive reactance decreases causing a decrease in the voltage at the non inverting input and hence at the output.
- 24. Frequency Response Here fcutoff the value of input signal frequency at which the output decrease to 0.707 times its low frequency value
- 25. High pass filters
- 26. Working A first order active high pass RC filter also consist of a RC network connected to the non inverting terminal of the op-amp in case low pass. Here R and C are inter changed.
- 27. …Continues In this case at low frequencies the reactance of the capacitor is infinite and it blocks the input signal. Hence the output is zero. As we increase the frequency, capacitive reactance decreases, and out put increase
- 28. Frequency Response Here fcutoff the value of input signal frequency at which the output increase to 0.707 times its low frequency value

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