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Loop Analysis (3.2) Circuits with Op-Amps (3.3)

Loop Analysis (3.2) Circuits with Op-Amps (3.3). Prof. Phillips February 19, 2003. Op Amps. 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.

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Loop Analysis (3.2) Circuits with Op-Amps (3.3)

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  1. Loop Analysis (3.2)Circuits with Op-Amps (3.3) Prof. Phillips February 19, 2003 lecture 9

  2. Op Amps • 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 9

  3. Use of Op Amps • Op amps can be configured in many different ways using resistors and other components. • Most configurations use feedback. lecture 9

  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. • Lowpass and bandpass filters. lecture 9

  5. + - The Op Amp Symbol High Supply Non-inverting input Output Inverting input Ground Low Supply lecture 9

  6. The Op Amp Model v+ Non-inverting input + vo Rin + – Inverting input – A(v+ -v- ) v- lecture 9

  7. Typical Op Amp • The input resistance Rin is very large (practically infinite). • The voltage gain A is very large (practically infinite). lecture 9

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

  9. The Basic Inverting Amplifier R2 R1 – + – + + Vin Vout – lecture 9

  10. 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 9

  11. Solving the Amplifier Circuit Apply KCL at the inverting input: i1 + i2 + i-=0 R2 i2 R1 – i1 i- lecture 9

  12. KCL lecture 9

  13. Solve for vout Amplifier gain: lecture 9

  14. Recap • The ideal op-amp model leads to the following conditions: i- = 0 = i+ 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 9

  15. Where is the Feedback? R2 R1 – + – + + Vin Vout – lecture 9

  16. 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 9

  17. The Non-Inverting Amplifier + + – + – vin vout R2 R1 – lecture 9

  18. KCL at the Inverting Input + + – + – i- vin vout i1 i2 R2 R1 – lecture 9

  19. KCL lecture 9

  20. Solve for Vout lecture 9

  21. A Mixer Circuit R1 Rf + – R2 v1 – + – + + v2 vout – lecture 9

  22. KCL at the Inverting Input R1 Rf i1 if + – R2 v1 i2 – i- + – + + v2 vout – lecture 9

  23. KCL lecture 9

  24. KCL lecture 9

  25. Solve for Vout lecture 9

  26. Class Example lecture 9

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