1. ECEN 301Discussion #18 – Operational Amplifiers1 DateDayClass No. TitleChaptersHW Due date Lab Due date Exam 3 NovMon18Operational Amplifiers 8.4 LAB 6 4 NovTue 5 NovWed19Binary Numbers13.1 – 13.2 6 NovThu 7 NovFri Recitation HW 8 8 NovSat 9 NovSun 10 NovMon20Exam Review LAB 7EXAM 2 12 NovTue Schedule…

2. ECEN 301Discussion #18 – Operational Amplifiers2 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.

3. ECEN 301Discussion #18 – Operational Amplifiers3 Lecture 18 – Operational Amplifiers Answer questions from last lecture Continue with Different OpAmp configurations

4. ECEN 301Discussion #17 – Operational Amplifiers4 Op-Amps – Open-Loop Model 1.How can v – ≈ v + when v o is amplifying (v + - v - ) ? 2.How can an opAmp form a closed circuit when (i 1 = i 2 = 0) ? – + +v+–+v+– +v––+v–– +vo–+vo– ioio i2i2 i1i1 – v in + – + +–+– R out R in i1i1 A OL v in +vo–+vo– – v in + v–v– v+v+ NB: op-amps have near-infinite input resistance (R in ) and very small output resistance (R out ) A OL – open-loop voltage gain i2i2

5. ECEN 301Discussion #17 – Operational Amplifiers5 Op-Amps – Open-Loop Model 1.How can v – ≈ v + when v o is amplifying (v + - v - ) ? 2.How can an opAmp form a closed circuit when (i 1 = i 2 = 0) ? – + +–+– R out R in i1i1 A OL v in +vo–+vo– – v in + v–v– v+v+ i2i2 Ideally i 1 = i 2 = 0 (since R in → ∞) What happens as A OL → ∞ ? → v – ≈ v +

6. ECEN 301Discussion #17 – Operational Amplifiers6 Op-Amps – Closed-Loop Mode RFRF – + +vo–+vo– i1i1 RSRS v S (t) +–+– v+v+ v–v– iFiF iSiS 1.How can v – ≈ v + when v o is amplifying (v + - v - ) ? 2.How can an opAmp form a closed circuit when (i 1 = i 2 = 0) ?

7. ECEN 301Discussion #17 – Operational Amplifiers7 Op-Amps – Closed-Loop Mode RFRF – + +vo–+vo– i1i1 RSRS v S (t) +–+– v+v+ v–v– iFiF iSiS NB: if A OL is very large these terms → 0 NB: if A OL is NOT the same thing as A CL 1.How can v – ≈ v + when v o is amplifying (v + - v - ) ? 2.How can an opAmp form a closed circuit when (i 1 = i 2 = 0) ?

8. ECEN 301Discussion #17 – Operational Amplifiers8 Op-Amps – Closed-Loop Mode – + vava R R – + vbvb R R – + R1R1 R R2R2 vovo 1.How can v – ≈ v + when v o is amplifying (v + - v - ) ? 2.How can an opAmp form a closed circuit when (i 1 = i 2 = 0) ? NB: Current flows through R 1 and R 2

9. ECEN 301Discussion #17 – Operational Amplifiers9 Op-Amps – Closed-Loop Mode – + vava R R – + vbvb R R – + R1R1 R R2R2 vovo 1.How can v – ≈ v + when v o is amplifying (v + - v - ) ? 2.How can an opAmp form a closed circuit when (i 1 = i 2 = 0) ? NB: Inverting amplifiers and (R S = R F ) → v o = -v i

10. ECEN 301Discussion #17 – Operational Amplifiers10 Op-Amps – Closed-Loop Mode 1.How can v – ≈ v + when v o is amplifying (v + - v - ) ? 2.How can an opAmp form a closed circuit when (i 1 = i 2 = 0) ? – + R1R1 R R2R2 vovo -v a -v b iFiF i1i1 i2i2

11. ECEN 301Discussion #18 – Operational Amplifiers11 More OpAmp Configurations

12. ECEN 301Discussion #18 – Operational Amplifiers12 Op-Amps – Closed-Loop Mode The Differential Amplifier: the signal to be amplified is the difference of two signals – + +vo–+vo– i1i1 RFRF v S2 (t) +–+– v+v+ v–v– iFiF RSRS iSiS i1i1 v S1 (t) +–+– RFRF RSRS

13. ECEN 301Discussion #18 – Operational Amplifiers13 Op-Amps – Closed-Loop Mode The Differential Amplifier: the signal to be amplified is the difference of two signals RFRF – + +vo–+vo– i1i1 v S2 (t) +–+– v+v+ v–v– iFiF RSRS iSiS i2i2 v S1 (t) +–+– RFRF RSRS NB: an ideal op-amp with negative feedback has the properties

14. ECEN 301Discussion #18 – Operational Amplifiers14 Op-Amps – Closed-Loop Mode The Differential Amplifier: the signal to be amplified is the difference of two signals RFRF – + +vo–+vo– i1i1 v S2 (t) +–+– v+v+ v–v– iFiF RSRS iSiS i2i2 v S1 (t) +–+– RFRF RSRS

15. ECEN 301Discussion #18 – Operational Amplifiers15 Op-Amps – Closed-Loop Mode The Differential Amplifier: the signal to be amplified is the difference of two signals RFRF – + +vo–+vo– i1i1 v S2 (t) +–+– v+v+ v–v– iFiF RSRS iSiS i2i2 v S1 (t) +–+– RFRF RSRS

16. ECEN 301Discussion #18 – Operational Amplifiers16 Op-Amps – Level Shifter Level Shifter: can add or subtract a DC offset from a signal based on the values of R S and/or V ref – + +vo–+vo– RFRF V ref +–+– v+v+ v–v– RSRS V sensor AC voltage with DC offset DC voltage

17. ECEN 301Discussion #18 – Operational Amplifiers17 Op-Amps – Level Shifter Example1: design a level shifter such that it can remove a 1.8V DC offset from the sensor signal (Find V ref ) R S = 10kΩ, R F = 220kΩ, v s (t) = 1.8+0.1cos(ωt) – + +vo–+vo– RFRF V ref +–+– v+v+ v–v– RSRS V sensor

18. ECEN 301Discussion #18 – Operational Amplifiers18 Op-Amps – Level Shifter Example1: design a level shifter such that it can remove a 1.8V DC offset from the sensor signal (Find V ref ) R S = 10kΩ, R F = 220kΩ, v s (t) = 1.8+0.1cos(ωt) – + +vo–+vo– RFRF V ref +–+– v+v+ v–v– RSRS V sensor Find the Closed-Loop voltage gain by using the principle of superposition on each of the DC voltages

19. ECEN 301Discussion #18 – Operational Amplifiers19 Op-Amps – Level Shifter Example1: design a level shifter such that it can remove a 1.8V DC offset from the sensor signal (Find V ref ) R S = 10kΩ, R F = 220kΩ, v s (t) = 1.8+0.1cos(ωt) – + +vo–+vo– RFRF v+v+ v–v– RSRS V sensor +–+– DC from sensor: Inverting amplifier

20. ECEN 301Discussion #18 – Operational Amplifiers20 Op-Amps – Level Shifter Example1: design a level shifter such that it can remove a 1.8V DC offset from the sensor signal (Find V ref ) R S = 10kΩ, R F = 220kΩ, v s (t) = 1.8+0.1cos(ωt) – + +vo–+vo– RFRF v+v+ v–v– RSRS V ref +–+– DC from reference: Noninverting amplifier

21. ECEN 301Discussion #18 – Operational Amplifiers21 Op-Amps – Level Shifter Example1: design a level shifter such that it can remove a 1.8V DC offset from the sensor signal (Find V ref ) R S = 10kΩ, R F = 220kΩ, v s (t) = 1.8+0.1cos(ωt) – + +vo–+vo– RFRF V ref +–+– v+v+ v–v– RSRS V sensor

22. ECEN 301Discussion #18 – Operational Amplifiers22 Op-Amps – Level Shifter Example1: design a level shifter such that it can remove a 1.8V DC offset from the sensor signal (Find V ref ) R S = 10kΩ, R F = 220kΩ, v s (t) = 1.8+0.1cos(ωt) – + +vo–+vo– RFRF V ref +–+– v+v+ v–v– RSRS V sensor Since the desire is to remove all DC from the output we require:

23. ECEN 301Discussion #18 – Operational Amplifiers23 Op-Amps – Ideal Integrator The Ideal Integrator: the output signal is the integral of the input signal (over a period of time) – + + v o (t) – i1i1 CFCF RSRS v S (t) +–+– v+v+ v–v– i F (t) i S (t) The input signal is AC, but not necessarily sinusoidal NB: Inverting amplifier setup with R F replaced with a capacitor

24. ECEN 301Discussion #18 – Operational Amplifiers24 Op-Amps – Ideal Integrator The Ideal Integrator: the output signal is the integral of the input signal (over a period of time) – + + v o (t) – i1i1 CFCF RSRS v S (t) +–+– v+v+ v–v– i F (t) i S (t)

25. ECEN 301Discussion #18 – Operational Amplifiers25 Op-Amps – Ideal Integrator Example2: find the output voltage if the input is a square wave of amplitude +/–A with period T T = 10ms, C F = 1uF, R S = 10kΩ – + + v o (t) – i1i1 CFCF RSRS v S (t) +–+– v+v+ v–v– i F (t) i S (t) A -A T/2T

26. ECEN 301Discussion #18 – Operational Amplifiers26 Op-Amps – Ideal Integrator Example2: find the output voltage if the input is a square wave of amplitude +/–A with period T T = 10ms, C F = 1uF, R S = 10kΩ – + + v o (t) – i1i1 CFCF RSRS v S (t) +–+– v+v+ v–v– i F (t) i S (t)

27. ECEN 301Discussion #18 – Operational Amplifiers27 Op-Amps – Ideal Integrator Example2: find the output voltage if the input is a square wave of amplitude +/–A with period T T = 10ms, C F = 1uF, R S = 10kΩ – + + v o (t) – i1i1 CFCF RSRS v S (t) +–+– v+v+ v–v– i F (t) i S (t) NB: since the v s (t) is periodic, we can find v o (t) over a single period – and repeat

28. ECEN 301Discussion #18 – Operational Amplifiers28 Op-Amps – Ideal Integrator Example2: find the output voltage if the input is a square wave of amplitude +/–A with period T T = 10ms, C F = 1uF, R S = 10kΩ – + + v o (t) – i1i1 CFCF RSRS v S (t) +–+– v+v+ v–v– i F (t) i S (t) NB: since the v s (t) is periodic, we can find v o (t) over a single period – and repeat

29. ECEN 301Discussion #18 – Operational Amplifiers29 Op-Amps – Ideal Integrator Example2: find the output voltage if the input is a square wave of amplitude +/–A with period T T = 10ms, C F = 1uF, R S = 10kΩ – + + v o (t) – i1i1 CFCF RSRS v S (t) +–+– v+v+ v–v– i F (t) i S (t) T/2T -50AT

30. ECEN 301Discussion #18 – Operational Amplifiers30 Op-Amps – Ideal Differentiator The Ideal Differentiator: the output signal is the derivative of the input signal (over a period of time) – + + v o (t) – i1i1 CSCS RFRF v S (t) +–+– v+v+ v–v– i F (t) i S (t) The input signal is AC, but not necessarily sinusoidal NB: Inverting amplifier setup with R S replaced with a capacitor

31. ECEN 301Discussion #18 – Operational Amplifiers31 Op-Amps – Ideal Differentiator NB: this type of differentiator is rarely used in practice since it amplifies noise The Ideal Differentiator: the output signal is the derivative of the input signal (over a period of time) – + + v o (t) – i1i1 CSCS RFRF v S (t) +–+– v+v+ v–v– i F (t) i S (t)

32. ECEN 301Discussion #18 – Operational Amplifiers32 Op-Amps – Closed-Loop Mode Circuit DiagramA CL Inverting Amplifier Summing Amplifier – + +vo–+vo– +–+– vSvS RSRS RFRF – + +vo–+vo– +–+– +–+– +–+– R Sn R S2 R S1 v Sn v S2 v S1 RFRF

33. ECEN 301Discussion #18 – Operational Amplifiers33 Op-Amps – Closed-Loop Mode Circuit DiagramA CL Noninverting Amplifier Voltage Follower – + +vo–+vo– +–+– R RSRS RFRF vSvS – + +vo–+vo– +–+–

34. ECEN 301Discussion #18 – Operational Amplifiers34 Op-Amps – Closed-Loop Mode Circuit DiagramA CL Differential Amplifier – + +vo–+vo– +–+– +–+– RSRS RSRS RFRF RFRF v1v1 v2v2

35. ECEN 301Discussion #18 – Operational Amplifiers35 Op-Amps – Closed-Loop Mode Circuit DiagramA CL Ideal Integrator Ideal Differentiator – + + v o (t) – +–+– vSvS CSCS RFRF – + + v o (t) – +–+– vSvS RSRS CFCF

36. ECEN 301Discussion #18 – Operational Amplifiers36 Op-Amps Example3: find an expression for the gain if v s (t) is sinusoidal C F = 1/6 F, R 1 = 3Ω, R 2 = 2Ω, C S = 1/6 F + – v o (t) i1i1 CFCF R2R2 v+v+ v–v– i F (t) i 2 (t) R1R1 i 1 (t) CSCS i S (t) v s (t)

37. ECEN 301Discussion #18 – Operational Amplifiers37 Op-Amps + – Vo(jω)Vo(jω) I in Z F =1/jωC F Z2Z2 v+v+ v–v– IF(jω)IF(jω) I2(jω)I2(jω) Z1Z1 I 1 (jω) ZSZS IS(jω)IS(jω) Vs(jω)Vs(jω) Node a Node b 1.Transfer to frequency domain 2.Apply KCL at nodes a and b NB: v + = v – = v o and I in = 0 Example3: find an expression for the gain C F = 1/6 F, R 1 = 3Ω, R 2 = 2Ω, C S = 1/6 F

38. ECEN 301Discussion #18 – Operational Amplifiers38 Op-Amps + – Vo(jω)Vo(jω) I in Z F =1/jωC F Z2Z2 v+v+ v–v– IF(jω)IF(jω) I2(jω)I2(jω) Z1Z1 I 1 (jω) ZSZS IS(jω)IS(jω) Vs(jω)Vs(jω) Node a Node b 1.Transfer to frequency domain 2.Apply KCL at nodes a and b Example3: find an expression for the gain C F = 1/6 F, R 1 = 3Ω, R 2 = 2Ω, C S = 1/6 F

39. ECEN 301Discussion #18 – Operational Amplifiers39 Op-Amps + – Vo(jω)Vo(jω) I in Z F =1/jωC F Z2Z2 v+v+ v–v– IF(jω)IF(jω) I2(jω)I2(jω) Z1Z1 I 1 (jω) ZSZS IS(jω)IS(jω) Vs(jω)Vs(jω) 1.Transfer to frequency domain 2.Apply KCL at nodes a and b 3.Express V o in terms of V s Example3: find an expression for the gain C F = 1/6 F, R 1 = 3Ω, R 2 = 2Ω, C S = 1/6 F

40. ECEN 301Discussion #18 – Operational Amplifiers40 Op-Amps + – Vo(jω)Vo(jω) I in Z F =1/jωC F Z2Z2 v+v+ v–v– IF(jω)IF(jω) I2(jω)I2(jω) Z1Z1 I 1 (jω) ZSZS IS(jω)IS(jω) Vs(jω)Vs(jω) 1.Transfer to frequency domain 2.Apply KCL at nodes a and b 3.Express V o in terms of V S 4.Find the gain (V o /V S ) Example3: find an expression for the gain C F = 1/6 F, R 1 = 3Ω, R 2 = 2Ω, C S = 1/6 F