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Op Amps

Op Amps. Introduction. Op Amp is short for operational amplifier. An operational amplifier is modeled as a voltage controlled voltage source. An operational amplifier has a very high input impedance and a very high gain. Use of Op Amps.

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Op Amps

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  1. Op Amps Lecture 30

  2. Introduction • Op Amp is short for operational amplifier. • An operational amplifier is modeled as a voltage controlled voltage source. • An operational amplifier has a very high input impedance and a very high gain. Lecture 30

  3. Use of Op Amps • Op amps can be configured in many different ways using resistors and other components. • Most configurations use feedback. Lecture 30

  4. Applications of Op Amps • Amplifiers provide gains in voltage or current. • Op amps can convert current to voltage. • Op amps can provide a buffer between two circuits. • Op amps can be used to implement integrators and differentiators. Lecture 30

  5. More Applications • Lowpass and bandpass filters. Lecture 30

  6. + - The Op Amp Symbol High Supply Non-inverting input Output Inverting input Ground Low Supply Lecture 30

  7. + - The Op Amp Model v+ Non-inverting input + vo Rin v- Inverting input - A(v+ -v- ) Lecture 30

  8. Typical Op Amp • The input resistance Rin is very large (practically infinite). • The voltage gain A is very large (practically infinite). Lecture 30

  9. “Ideal” Op Amp • The input resistance is infinite. • The gain is infinite. • The op amp is in a negative feedback configuration. Lecture 30

  10. R2 R1 - + + + Vin Vout - - The Basic Inverting Amplifier Lecture 30

  11. Consequences of the Ideal • Infinite input resistance means the current into the inverting input is zero: i- = 0 • Infinite gain means the difference between v+ and v- is zero: v+ - v- = 0 Lecture 30

  12. Solving the Amplifier Circuit Apply KCL at the inverting input: i1 + i2 + i-=0 R2 i2 R1 - i1 i- Lecture 30

  13. KCL Lecture 30

  14. Solve for vout Amplifier gain: Lecture 30

  15. Recap • The ideal op amp model leads to the following conditions: i- = 0 v+ = v- • These conditions are used, along with KCL and other analysis techniques, to solve for the output voltage in terms of the input(s). Lecture 30

  16. R2 R1 - + + + Vin Vout - - Where is the Feedback? Lecture 30

  17. Review • To solve an op amp circuit, we usually apply KCL at one or both of the inputs. • We then invoke the consequences of the ideal model. • The op amp will provide whatever output voltage is necessary to make both input voltages equal. • We solve for the op amp output voltage. Lecture 30

  18. + + - + vin - vout R2 R1 - The Non-Inverting Amplifier Lecture 30

  19. KCL at the Inverting Input + + - + i- vin - vout i1 i2 R2 R1 - Lecture 30

  20. KCL Lecture 30

  21. Solve for Vout Lecture 30

  22. R1 Rf R2 - + + + + v2 v1 vout - - - A Mixer Circuit Lecture 30

  23. + v2 - KCL at the Inverting Input R1 Rf i1 if + R2 v1 i2 - - i- + + vout - Lecture 30

  24. KCL Lecture 30

  25. KCL Lecture 30

  26. Solve for Vout Lecture 30

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