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OPERATIONAL AMPLIFIERSPart 1

Mikroelektronik Devreler-1Yrd.Do.Dr. Mutlu BOZTEPE

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

Rin = infinity

Rout = 0

Avo = infinity (Avo is the open-loop gain)

Bandwidth = infinity (amplifies all frequencies equally)

2.2 Ideal op-amp

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Model of an ideal op-amp

V+

V-

Vout = A(V+ - V-)+

-

+

-

Usually used with feedback

Open-loop configuration not used much

I-

I+

Summary of op-amp behavior

Vout = A(V+ - V-)

Vout/A = V+ - V-

Let A infinity

then,

V+ -V- 0

V+ = V-

I+ = I- = 0

Seems strange, but theinput terminals to an

op-amp act as a short and

open at the same time

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Gain of circuit determined by

external components; R2 and R1

2.3 Inverting configuration

Analysis of inverting configuration;

open-loop gain is infinite

Vo/ Vi = -R2/ R1

open-loop gain is finite

Vo/ Vi -R2/ R1

I1

I2

Virtual

ground

Input resistance; Ri = vI/i1 = R1

Outpur resistance is zero

2.4.2 Integrator

I1 = (Vi - V-)/R1

I2 = C d(V- - Vo)/dt

set I1 = I2,

(Vi - V-)/R1 = C d(V- - Vo)/dt

but V- = V+ = 0

Vi/R1 = -C d(Vo)/dtSolve for Vo

Vo = -(1/CR1)( Vi dt)

Output is the integral of

input signal.

I1

I2

CR1 is the time constant

1/CR1 is integrator frequency Negative

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Problem 2.28

t

mS

10.5 1.5

1V

-1V

0

Vo = -(t/1mS), for 0 < t < 0.5mS

Vo = (t/1mS), for 0.5 < t < 1.0mS

When t = 0.5mS, Vo = 0.5V

CR1 = 1mS

CR1 = 0.5mS

Exercise 2.6

Design an integrator (find RC) such that the output has a 20Vp-p

amplitude given the input below.

Vo = -(1/CR)( Vi dt)

Vi = 10V, 0 < t < 1.0mS

Vo = -(1/RC)

Vo = -10t/RC

let t = 1.0mS, and Vo = -20

20 = 0.01/RC

RC = 0.5mS

t2mS

10V

-10V

Vi

t

dt

0

10

Vout

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2.5 Noninverting configuration

(0 - V-)/R1 = (V- - Vo)/R2

But, Vi = V+ = V-,

( - Vi)/R1 = (Vi - Vo)/R2

Solve for Vo,

Vo = Vi(1+R2/R1)

Vi

I

I

Positive

VirtualVi

Buffer amplifier

Vi = V+ = V- = Vo

Vo = Vi

Isolates input

from output

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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:

model, let A infinity

Solve for Vout

Works for every op-amp circuit.

OR

Input resistance of noninverting amplifier

Vout =

A(V+ -V-)

V-

V+

Rin = Vin/ I, from definition

Rin = Vin/ 0

Rin = infinity

I

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Input resistance of inverting amplifier

Vout =

A(V+ - V-)

V-

V+

Rin = Vin/ I, from definition

I = (Vin - Vout)/R

I = [Vin - A (V+ - V-)] / R

But V+ = 0

I = [Vin - A( -Vin)] / R

Rin = VinR / [Vin (1+A)]

As A approaches infinity,

Rin = 0

I

Summary of op-amp behavior

Vi

Inverting configuration Noninverting configuration

Vo/Vi = 1+R2/R1

Rin = infinity

Vo/Vi = - R2/R1

Rin = R1

Rin = 0 at

this point

Vi

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Fig. 2.21 A difference amplifier.

Difference amplifier

Use superposition,

set V1 = 0, solve for Vo

(noninverting amp)

set V2 = 0, solve for Vo

(inverting amp)

Add the two results

Vo = -(R2/R1)V1 + (1 + R2/R1) [R4/(R3+R4)] V2

Difference amplifier

Vo1 = -(R2/R1)V1 Vo2 = (1 + R2/R1) [R4/(R3+R4)] V2

V2 R4/(R3+R4)

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Improving the input resistance of amplifiers

Add buffer amplifiers to the inputsRin = infinity at both V1 and V2

Instrumentation Amplifier

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Prob. 2.55

(a) Find Rin at V1.Ground V2, Rin = R.

(b) Find Rin at V2.Ground V1, Rin = 2R.

Rin = V/I from definition.

Write voltage loop,

V = IR + IR, I = V/2R

Rin = V/I = V2R/V

Rin = 2R

+

V

-

I

I

(c) Find Rin between V1 and V2.

Rin = V/I from definition

I1 = V/2R from Ohms law

I2 = (V - V/2)R = V/2R because V+ = V-

Rin = V/ (I1 + I2) = V2R/2V

Rin = R

I

+

V

-

I1 = V/2R

I2

V/2

(d) Find Rin between V1 and V2

connected together and ground.

Prob. 2.55