Applications of op amps

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Applications of op amps

  1. 1. Applications of Op-Amps Dr. C.SARITHA Lecturer in ElectronicsS.S.B.N. DEGREE & PG.COLLEGE ANANTAPUR
  2. 2. OVERVIEW Introduction Definitions Circuit Diagrams Derivations Applications Conclusion
  3. 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. 4. Circuit diagramIR Ic Vs
  5. 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. 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. 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. 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. 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. 10.  The input is sine wave the output become cosine waveInput=Output=
  11. 11. Similarly the input is square wave theoutput become Triangle waveInput=Output=
  12. 12. Applications Analog computer A to D converters Many linear circuits Wave Shapping circuit
  13. 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. 14. Circuit Diagram
  15. 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. 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. 17. The input is cosine wave the output becomesine waveInput=Output=
  18. 18. Similarly the input is Triangle wave theoutput become Square waveInput=Output=
  19. 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. 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. 21. Low pass filtersR C
  22. 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. 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. 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. 25. High pass filters
  26. 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. 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. 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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