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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.

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Presentation on theme: "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."— Presentation transcript:

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2 ENTC 3320 Op Amp Review

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

4 Ideal op-amp Characteristics of an ideal op-amp R in = infinity R out = 0 A vo = infinity (A vo is the open-loop gain, sometimes A or A v of the op-amp) Bandwidth = infinity (amplifies all frequencies equally)

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

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

7 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

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

9 Inverting configuration Very popular circuit

10 Analysis of inverting configuration I 1 = (V i - V  )/R 1 I 2 = (V  - V o )/R 2 set I 1 = I 2, (V i - V  )/R 1 = (V  - V o )/R 2 but V  = V + = 0 V i /R 1 = -V o /R 2 Solve for V o Gain of circuit determined by external components I1I1 I2I2 V o / V i = -R 2 /R 1

11 Summing Amplifier V1V2V3V1V2V3 R1R1 R2R2 R3R3 RfRf Current in R 1, R 2, and R 3 add to current in R f (V 1 -V  )/ R 1 + (V 2 -V  )/ R 2 + (V 3 -V  )/R 3 = (V  - V o )/ R f Set V  = V + = 0, V 1 / R 1 + V 2 / R 2 + V 3 / R 3 =  V o / R f solve for V o, This circuit is called a weighted summer V o = - R f (V 1 / R 1 + V 2 / R 2 + V 3 / R 3 )

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

13 Integrator I 1 = (V i - V  )/R 1 I 2 = set I 1 = I 2, (V i - V  )/R 1 = but V- = V+ = 0 V i /R 1 = Solve for Vo Output is the integral of input signal. CR1 is the time constant I1I1 I2I2

14 Noninverting configuration (0 - V  )/R 1 = (V  - V o )/R 2 But, V i = V + = V , (- V i )/ R 1 = (V i - V o )/R 2 Solve for V o, (- V i )/R 1 - (-V i )/R 2 = (-V o )/R 2 V i (1/R 1 + 1/R 2 ) = (V o )/R 2 V o = V i (R 2 /R 1 + R 2 /R Vi I I Vo = Vi(1+R2/R1)

15 Buffer amplifier V i = V + = V  = V o Isolates input from output V o = V i

16 Analyzing op-amp circuits 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. OR

17 Difference amplifier Use superposition, set V 1 = 0, solve for Vo (noninverting amp) set V 2 = 0, solve for Vo (inverting amp)

18 V 02 = (1 + R 2 /R 1 ) [R 4 /(R 3 +R 4 )] V 2 V 2 R 4 /(R 3 +R 4 ) Difference amplifier

19 V 01 = -(R 2 /R 1 )V 1 Difference amplifier

20 Add the two results V 0 = V 01 + V 02 If R 1 = R 2 = R 3 = R 4

21 Design of difference amplifiers For V o = V 2 - V 1 Set R 2 = R 1 = R, and set R 3 = R 4 = R For V o = 3V 2 - 2V 1 Set R 1 = R, R 2 = 2R, then 3[R 4 /(R 3 +R 4 )] = 3 Set R 3 = 0


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