1 / 20

ENTC 3320

ENTC 3320. Op Amp Review. Operational amplifiers (op-amps). Circuit symbol of an op-amp Widely used Often requires 2 power supplies + V Responds to difference between two signals. Ideal op-amp. Characteristics of an ideal op-amp R in = infinity R out = 0

kasimir-harvey
Download Presentation

ENTC 3320

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. ENTC 3320 Op Amp Review

  2. Operational amplifiers (op-amps) Circuit symbol of an op-amp Widely used Often requires 2 power supplies + V Responds to difference between two signals

  3. Ideal op-amp Characteristics of an ideal op-amp Rin = infinity Rout = 0 Avo = infinity (Avo is the open-loop gain, sometimes A or Av of the op-amp) Bandwidth = infinity (amplifies all frequencies equally)

  4. I+ V+ + + Vout = A(V+ -V) I - V - Model of an ideal op-amp • Usually used with feedback • Open-loop configuration not used much

  5. Summary of op-amp behavior Vout = A(V+ - V) Vout/A = V+ - V Let A infinity then, V+ - V 0

  6. Summary of op-amp behavior V+ = V I+ = I  = 0 Seems strange, but the input terminals to an op-amp act as a short and open at the same time

  7. To analyze an op-amp circuit • Write node equations at + and - terminals (I+ = I = 0) • Set V+ = V • Solve for Vout

  8. Inverting configuration Very popular circuit

  9. I2 Analysis of inverting configuration I1 = (Vi - V)/R1 I2 = (V - Vo)/R2 set I1 = I2, (Vi - V )/R1 = (V - Vo)/R2 but V = V+ = 0 Vi /R1 = -Vo/R2 Solve for Vo I1 Gain of circuit determined by external components Vo/ Vi = -R2/R1

  10. Rf R1 V1 V2 V3 R2 R3 Summing Amplifier Current in R1, R2, and R3 add to current in Rf (V1-V)/ R1 + (V2-V)/ R2 + (V3-V)/R3 = (V - Vo)/ Rf Set V  = V+ = 0,V1/ R1 + V2/ R2 + V3/ R3 = Vo / Rf solve for Vo, Vo = - Rf(V1 / R1 + V2 / R2 + V3 / R3) This circuit is called a weighted summer

  11. To analyze an op-amp circuit • Write node equations at + and - terminals (I+ = I = 0) • Set V+ = V • Solve for Vout

  12. Integrator I1 = (Vi - V)/R1 I2 = set I1 = I2, (Vi - V)/R1 = but V- = V+ = 0 Vi/R1 = Solve for Vo I2 I1 Output is the integral of input signal. CR1 is the time constant

  13. I I Vi Noninverting configuration (0 - V)/R1 = (V - Vo)/R2 But, Vi = V+ = V , (- Vi)/ R1 = (Vi - Vo)/R2 Solve for Vo, (- Vi)/R1 - (-Vi)/R2 = (-Vo)/R2 Vi (1/R1+ 1/R2) = (Vo)/R2 Vo = Vi (R2/R1+ R2/R Vo = Vi(1+R2/R1)

  14. Buffer amplifier Isolates input from output Vi = V+ = V = Vo Vo = Vi

  15. Write node equations using: V+ = V I + = I = 0 Solve for Vout Usually easier, can solve most problems this way. Write node equations using: op amp model. Let A infinity Solve for Vout Works for every op-amp circuit. Analyzing op-amp circuits OR

  16. Difference amplifier • Use superposition, • set V1 = 0, solve for Vo • (noninverting amp) • set V2 = 0, solve for Vo • (inverting amp)

  17. V2 R4/(R3+R4) Difference amplifier V02 = (1 + R2/R1) [R4/(R3+R4)] V2

  18. Difference amplifier V01 = -(R2/R1)V1

  19. Add the two results • V0 = V01 +V02 • If R1 =R2 =R3 =R4

  20. Design of difference amplifiers For Vo = V2 - V1 SetR2 = R1 = R, and set R3 = R4 = R For Vo = 3V2 - 2V1 Set R1 = R, R2 = 2R, then 3[R4/(R3+R4)] = 3 Set R3 = 0

More Related