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Schedule…. Give to Receive. Alma 34:28

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  1. Schedule… Discussion #18 – Operational Amplifiers

  2. Give to Receive Alma 34:28   28 And now behold, my beloved brethren, I say unto you, do not suppose that this is all; for after ye have done all these things, if ye turn away the needy, and the naked, and visit not the sick and afflicted, and impart of your substance, if ye have, to those who stand in need—I say unto you, if ye do not any of these things, behold, your prayer is vain, and availeth you nothing, and ye are as hypocrites who do deny the faith. Discussion #18 – Operational Amplifiers

  3. Lecture 18 – Operational Amplifiers Answer questions from last lecture Continue with Different OpAmp configurations Discussion #18 – Operational Amplifiers

  4. i1 – – vin + io + v– – + + vo – + v+ – i2 + – Op-Amps – Open-Loop Model • How can v– ≈ v+ when vo is amplifying (v+ - v-) ? • How can an opAmp form a closed circuit when (i1 = i2 = 0) ? i1 v– – – vin + Rin Rout + vo – AOLvin + v+ i2 NB: op-amps have near-infinite input resistance (Rin) and very small output resistance (Rout) AOL – open-loop voltage gain Discussion #17 – Operational Amplifiers

  5. + – Op-Amps – Open-Loop Model • How can v– ≈ v+ when vo is amplifying (v+ - v-) ? • How can an opAmp form a closed circuit when (i1 = i2 = 0) ? i1 v– – – vin + Ideally i1 = i2 = 0 (since Rin → ∞) Rin Rout + vo – AOLvin + v+ i2 What happens as AOL → ∞ ? → v– ≈ v+ Discussion #17 – Operational Amplifiers

  6. RF – iF RS v– + i1 iS + – vS(t) + vo – v+ Op-Amps – Closed-Loop Mode • How can v– ≈ v+ when vo is amplifying (v+ - v-) ? • How can an opAmp form a closed circuit when (i1 = i2 = 0) ? Discussion #17 – Operational Amplifiers

  7. RF – iF RS v– + i1 iS + – vS(t) + vo – v+ Op-Amps – Closed-Loop Mode • How can v– ≈ v+ when vo is amplifying (v+ - v-) ? • How can an opAmp form a closed circuit when (i1 = i2 = 0) ? NB: if AOL is very large these terms → 0 NB: if AOL is NOT the same thing as ACL Discussion #17 – Operational Amplifiers

  8. Op-Amps – Closed-Loop Mode How can v– ≈ v+when vo is amplifying(v+ - v-)? How can an opAmp form a closed circuit when (i1 = i2 = 0) ? R R vb va R R R – – R1 – vo + + + R2 NB: Current flows through R1 and R2 Discussion #17 – Operational Amplifiers

  9. Op-Amps – Closed-Loop Mode How can v– ≈ v+when vo is amplifying(v+ - v-)? How can an opAmp form a closed circuit when (i1 = i2 = 0) ? R R vb va R R R – – R1 – vo + + + R2 NB: Inverting amplifiers and (RS = RF) → vo = -vi Discussion #17 – Operational Amplifiers

  10. Op-Amps – Closed-Loop Mode How can v– ≈ v+when vo is amplifying(v+ - v-)? How can an opAmp form a closed circuit when (i1 = i2 = 0) ? iF R R1 -va – vo + i1 R2 -vb i2 Discussion #17 – Operational Amplifiers

  11. More OpAmp Configurations Discussion #18 – Operational Amplifiers

  12. RF – iF v– RS + i1 + – + – iS vS1(t) vS2(t) v+ + vo – i1 RS RF Op-Amps – Closed-Loop Mode The Differential Amplifier: the signal to be amplified is the difference of two signals Discussion #18 – Operational Amplifiers

  13. RF – iF v– RS + i1 iS vS1(t) + – vS2(t) + – v+ + vo – i2 RS RF Op-Amps – Closed-Loop Mode The Differential Amplifier: the signal to be amplified is the difference of two signals NB: an ideal op-amp with negative feedback has the properties Discussion #18 – Operational Amplifiers

  14. RF – iF v– RS + i1 iS vS1(t) + – vS2(t) + – v+ + vo – i2 RS RF Op-Amps – Closed-Loop Mode The Differential Amplifier: the signal to be amplified is the difference of two signals Discussion #18 – Operational Amplifiers

  15. RF – iF v– RS + i1 iS vS1(t) + – vS2(t) + – v+ + vo – i2 RS RF Op-Amps – Closed-Loop Mode The Differential Amplifier: the signal to be amplified is the difference of two signals Discussion #18 – Operational Amplifiers

  16. RF – Vsensor v– RS + + – Vref v+ + vo – Op-Amps – Level Shifter Level Shifter: can add or subtract a DC offset from a signal based on the values of RS and/or Vref AC voltage with DC offset DC voltage Discussion #18 – Operational Amplifiers

  17. RF – Vsensor v– RS + + – Vref v+ + vo – Op-Amps – Level Shifter Example1: design a level shifter such that it can remove a 1.8V DC offset from the sensor signal (Find Vref) RS = 10kΩ, RF = 220kΩ, vs(t) = 1.8+0.1cos(ωt) Discussion #18 – Operational Amplifiers

  18. RF – Vsensor v– RS + + – Vref v+ + vo – Op-Amps – Level Shifter Example1: design a level shifter such that it can remove a 1.8V DC offset from the sensor signal (Find Vref) RS = 10kΩ, RF = 220kΩ, vs(t) = 1.8+0.1cos(ωt) Find the Closed-Loop voltage gain by using the principle of superposition on each of the DC voltages Discussion #18 – Operational Amplifiers

  19. – + + – Vsensor Op-Amps – Level Shifter Example1: design a level shifter such that it can remove a 1.8V DC offset from the sensor signal (Find Vref) RS = 10kΩ, RF = 220kΩ, vs(t) = 1.8+0.1cos(ωt) RF DC from sensor: Inverting amplifier v– RS v+ + vo – Discussion #18 – Operational Amplifiers

  20. – + + – Vref Op-Amps – Level Shifter Example1: design a level shifter such that it can remove a 1.8V DC offset from the sensor signal (Find Vref) RS = 10kΩ, RF = 220kΩ, vs(t) = 1.8+0.1cos(ωt) RF DC from reference: Noninverting amplifier v– RS v+ + vo – Discussion #18 – Operational Amplifiers

  21. RF – Vsensor v– RS + + – Vref v+ + vo – Op-Amps – Level Shifter Example1: design a level shifter such that it can remove a 1.8V DC offset from the sensor signal (Find Vref) RS = 10kΩ, RF = 220kΩ, vs(t) = 1.8+0.1cos(ωt) Discussion #18 – Operational Amplifiers

  22. – + + – Vref Op-Amps – Level Shifter Example1: design a level shifter such that it can remove a 1.8V DC offset from the sensor signal (Find Vref) RS = 10kΩ, RF = 220kΩ, vs(t) = 1.8+0.1cos(ωt) RF Since the desire is to remove all DC from the output we require: Vsensor v– RS v+ + vo – Discussion #18 – Operational Amplifiers

  23. CF – iF(t) RS v– + i1 + – vS(t) iS(t) + vo(t) – v+ Op-Amps – Ideal Integrator The Ideal Integrator: the output signal is the integral of the input signal (over a period of time) The input signal is AC, but not necessarily sinusoidal NB: Inverting amplifier setup with RF replaced with a capacitor Discussion #18 – Operational Amplifiers

  24. CF – iF(t) RS v– + i1 + – vS(t) iS(t) + vo(t) – v+ Op-Amps – Ideal Integrator The Ideal Integrator: the output signal is the integral of the input signal (over a period of time) Discussion #18 – Operational Amplifiers

  25. CF – A iF(t) RS v– + T/2 T i1 + – vS(t) iS(t) + vo(t) – v+ -A Op-Amps – Ideal Integrator Example2: find the output voltage if the input is a square wave of amplitude +/–A with period T T = 10ms, CF = 1uF, RS = 10kΩ Discussion #18 – Operational Amplifiers

  26. CF – iF(t) RS v– + i1 + – vS(t) iS(t) + vo(t) – v+ Op-Amps – Ideal Integrator Example2: find the output voltage if the input is a square wave of amplitude +/–A with period T T = 10ms, CF = 1uF, RS = 10kΩ Discussion #18 – Operational Amplifiers

  27. CF – iF(t) RS v– + i1 + – vS(t) iS(t) + vo(t) – v+ Op-Amps – Ideal Integrator Example2: find the output voltage if the input is a square wave of amplitude +/–A with period T T = 10ms, CF = 1uF, RS = 10kΩ NB: since the vs(t) is periodic, we can find vo(t) over a single period – and repeat Discussion #18 – Operational Amplifiers

  28. CF – iF(t) RS v– + i1 + – vS(t) iS(t) + vo(t) – v+ Op-Amps – Ideal Integrator Example2: find the output voltage if the input is a square wave of amplitude +/–A with period T T = 10ms, CF = 1uF, RS = 10kΩ NB: since the vs(t) is periodic, we can find vo(t) over a single period – and repeat Discussion #18 – Operational Amplifiers

  29. CF T/2 T – iF(t) RS v– + i1 + – vS(t) iS(t) + vo(t) – v+ -50AT Op-Amps – Ideal Integrator Example2: find the output voltage if the input is a square wave of amplitude +/–A with period T T = 10ms, CF = 1uF, RS = 10kΩ Discussion #18 – Operational Amplifiers

  30. RF – iF(t) + CS v– + – vS(t) i1 iS(t) + vo(t) – v+ Op-Amps – Ideal Differentiator The Ideal Differentiator: the output signal is the derivative of the input signal (over a period of time) The input signal is AC, but not necessarily sinusoidal NB: Inverting amplifier setup with RS replaced with a capacitor Discussion #18 – Operational Amplifiers

  31. RF – iF(t) + CS v– + – vS(t) i1 iS(t) + vo(t) – v+ Op-Amps – Ideal Differentiator The Ideal Differentiator: the output signal is the derivative of the input signal (over a period of time) NB: this type of differentiator is rarely used in practice since it amplifies noise Discussion #18 – Operational Amplifiers

  32. RF RS1 – vS1 + – + + vo – RS2 vS2 + – RSn vSn + – RF RS – + vo – + vS + – Op-Amps – Closed-Loop Mode Discussion #18 – Operational Amplifiers

  33. RF RS – – + vo – + + + vo – R + – + – vS Op-Amps – Closed-Loop Mode Discussion #18 – Operational Amplifiers

  34. RF RS – RS v1 + vo – + + – + – v2 RF Op-Amps – Closed-Loop Mode Discussion #18 – Operational Amplifiers

  35. RF CF CS RS – – + + vo(t) – + + vo(t) – vS vS + – + – Op-Amps – Closed-Loop Mode Discussion #18 – Operational Amplifiers

  36. CF iF(t) vs(t) R2 R1 v+ + vo(t) i1 i1(t) i2(t) CS – v– iS(t) Op-Amps Example3: find an expression for the gain if vs(t) is sinusoidal CF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F Discussion #18 – Operational Amplifiers

  37. Op-Amps Example3: find an expression for the gain CF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F ZF=1/jωCF • Transfer to frequency domain • Apply KCL at nodes a and b Node b IF(jω) Vs(jω) Z1 Z2 v+ Vo(jω) + Iin I1(jω) I2(jω) ZS – v– IS(jω) Node a NB: v+ = v– = vo and Iin = 0 Discussion #18 – Operational Amplifiers

  38. Op-Amps Example3: find an expression for the gain CF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F ZF=1/jωCF • Transfer to frequency domain • Apply KCL at nodes a and b Node b IF(jω) Vs(jω) Z1 Z2 v+ Vo(jω) + Iin I1(jω) I2(jω) ZS – v– IS(jω) Node a Discussion #18 – Operational Amplifiers

  39. Op-Amps Example3: find an expression for the gain CF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F ZF=1/jωCF • Transfer to frequency domain • Apply KCL at nodes a and b • Express Vo in terms of Vs IF(jω) Vs(jω) Z1 Z2 v+ Vo(jω) + Iin I1(jω) I2(jω) ZS – v– IS(jω) Discussion #18 – Operational Amplifiers

  40. Op-Amps Example3: find an expression for the gain CF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F ZF=1/jωCF • Transfer to frequency domain • Apply KCL at nodes a and b • Express Vo in terms of VS • Find the gain (Vo/VS) IF(jω) Vs(jω) Z1 Z2 v+ Vo(jω) + Iin I1(jω) I2(jω) ZS – v– IS(jω) Discussion #18 – Operational Amplifiers

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