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.

ENTC 3320

Op Amp Review

Operational amplifiers (op-amps)

Circuit symbol of an op-amp

Widely usedOften requires 2 power supplies + VResponds to difference between two signals

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)

Model of an ideal op-amp

•Usually used with feedback•Open-loop configuration not used much

V+

V

Vout = A(V+ -V)+-

+

-

I

I+

Summary of op-amp behavior

Vout = A(V+ - V)

Vout/A = V+ - V

Let A infinity

then,

V+ - V 0

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

To analyze an op-amp circuit

•Write node equations at + and - terminals (I+ = I = 0)

•Set V+ = V

•Solve for Vout

Inverting configuration

Very popular circuit

Analysis of inverting configurationI1 = (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 VoGain of circuit determined by external components

I1

I2

Vo/ Vi = -R2/R1

Summing Amplifier

V1

V2

V3

R1

R2

R3

Rf

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,

This circuit is called a weighted summer

Vo = - Rf(V1 / R1 + V2 / R2 + V3 / R3)

To analyze an op-amp circuit

•Write node equations at + and - terminals (I+ = I = 0)

•Set V+ = V

•Solve for Vout

IntegratorI1 = (Vi - V)/R1

I2 =

set I1 = I2,

(Vi - V)/R1 =

but V- = V+ = 0

Vi/R1 =

Solve for VoOutput is the integral of input signal. CR1 is the time constant

I1

I2

dt

VVdC o

dt

VVdC o

dt

dVC o

dtvCR

v io1

1

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

Vi

I

I

Vo = Vi(1+R2/R1)

Buffer amplifier

Vi = V+ = V = Vo

Isolates input from output

Vo = Vi

Analyzing op-amp circuitsWrite 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

Difference amplifier

Use superposition,

• set V1 = 0, solve for Vo

• (noninverting amp)

• set V2 = 0, solve for Vo

• (inverting amp)

V02 = (1 + R2/R1) [R4/(R3+R4)] V2V2 R4/(R3+R4)

Difference amplifier

V01 = -(R2/R1)V1

Difference amplifier

Add the two results

• V0 = V01 + V02

If R1 = R2 = R3 = R4

1

21

43

4

1

22 1

R

RV

RR

R

R

RVVo

1212 12

111 VVVVVo

Design of difference amplifiersFor Vo = V2 - V1

Set R2 = R1 = R, and set R3 = R4 = R

For Vo = 3V2 - 2V1

Set R1 = R, R2 = 2R, then 3[R4/(R3+R4)] = 3Set R3 = 0