Op_amp Review_2

8/11/2019 Op_amp Review_2

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ENTC 3320

Op Amp Review


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Operational amplifiers (op-amps)Circuit symbol of an op-amp

Widely used

Often requires 2 power supplies + V

Responds to difference between two signals


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Ideal op-ampCharacteristics of an ideal op-amp

Rin= infinity

Rout= 0

Avo= infinity (Avois the open-loop gain,

sometimes A or Avof the op-amp)

Bandwidth = infinity (amplifies all frequencies

equally)


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Model of an ideal op-amp

Usually used with feedback

Open-loop configuration not used much

V+

V

Vout = A(V+-V)+

-

+

-

I

I+


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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


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To analyze an op-amp circuit

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

Set V+= V

Solve for Vout


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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/ V

i= -R

2/R

1


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Summing Amplifier

V1

V2

V3

R1

R2

R3

Rf

Current in R1, R2, and R3add 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 weightedsummer

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


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To analyze an op-amp circuit

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

Set V+= V

Solve for Vout


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Integrator I1= (Vi- V)/R1

I2=

set I1= I2,

(Vi- V)/R1 =

but V- = V+ = 0

Vi/R1=Solve for Vo

Output is the integral of input

signal. CR1 is the time constant

I1

I2

dt

VVdC

o

dt

VVdC

o

dt

dV

C

o

dtvCRv io1

1


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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= V

i(R

2/R

1+ R

2/R

Vi

I

I

Vo = Vi(1+R2/R1)


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Buffer amplifier

Vi= V+= V= Vo

Isolates input from

output

Vo= Vi


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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


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Difference amplifier

Use superposition,

set V1= 0, solve for

Vo

(noninverting

amp)

set V2= 0, solve for

Vo

(inverting amp)


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

Difference amplifier


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V01 = -(R2/R1)V1

ifference amplifier


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Add the two results

V0 = V01 +V02

If R1 =R2 =R3 =R4

1

2

1

43

4

1

2

2 1

R

RV

RR

R

R

RVV

o

1212

12

111 VVVVV

o


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Design of difference amplifiersFor Vo= V2- V1SetR2= R1= R, and set R3= R4= R

For Vo= 3V

2- 2V

1

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

Set R3= 0

Op_amp Review_2

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