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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

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ENTC 3320

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  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

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