1 / 41

Give to Receive

Give to Receive. Alma 34:28

arlo
Télécharger la présentation

Give to Receive

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 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

  2. Lecture 18 – Operational Amplifiers Continue with Different OpAmp configurations Discussion #18 – Operational Amplifiers

  3. 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

  4. + – 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

  5. 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

  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) ? NB: if AOL is very large these terms → 0 NB: if AOL is NOT the same thing as ACL Discussion #17 – Operational Amplifiers

  7. 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

  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: Inverting amplifiers and (RS = RF) → vo = -vi 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) ? iF R R1 -va – vo + i1 R2 -vb i2 Discussion #17 – Operational Amplifiers

  10. More OpAmp Configurations Discussion #18 – Operational Amplifiers

  11. 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

  12. 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

  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 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 – 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

  16. 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

  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) Find the Closed-Loop voltage gain by using the principle of superposition on each of the DC voltages Discussion #18 – Operational Amplifiers

  18. – + + – 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

  19. – + + – 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

  20. 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

  21. – + + – 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

  22. 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

  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) Recall Discussion #18 – Operational Amplifiers

  24. 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

  25. 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

  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 NB: since the vs(t) is periodic, we can find vo(t) over a single period – and repeat 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 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

  29. 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

  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) NB: this type of differentiator is rarely used in practice since it amplifies noise Recall Discussion #18 – Operational Amplifiers

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

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

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

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

  35. CF iF(t) vs(t) R2 R1 v+ + vo(t) i1 i1(t) i2(t) CS – v– iS(t) Op-Amps Example 3: 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

  36. Op-Amps Example 3: 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

  37. Op-Amps Example 3: 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

  38. Op-Amps Example 3: 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

  39. Op-Amps Example 3: 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

  40. Example 3: find an expression for the gain CF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F Op-Amps Frequency Response ZF=1/jωCF IF(jω) Vs(jω) Z1 Z2 v+ Vo(jω) + Iin I1(jω) I2(jω) ZS – v– IS(jω) 2nd order Lowpass filter ECEN 301 Discussion #18 – Operational Amplifiers • 40

  41. Instrumentation Op-Amp • Special configuration used for common-mode voltage rejection Connect sensor with twist pair in differential configuration to minimize external noise pickup. Using grounded shield also helps. Advantages: - Common-mode voltage rejection - Very high input impedance for V1 and V2 - Set gain with one resistor

More Related