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Module 12Operational Amplifiers – Part II

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Review from Operational Amplifiers I:

Negative input

Positive inputOutput

VPOS

–VNEG

Power SupplyVoltages

Anatomy of an “Op-Amp”

3

Dependent Source Model

• rin is on the order of several Megohms:

• Av is on the order of 105 to 106

These features motivate the Ideal Op-Amp approximation

v+

v–

Av(v+ – v–)

Equivalent model for the circuit inside an op-amp

rin

vOUT

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Dependent Source Model

• vOUT must lie between Vpos and –Vneg

VPOS

–VNEG

vOUT

Vpos

–Vneg

Upper Limit

Lower Limit

Range

• Otherwise, the op-amp becomes saturated.

• Saturated op-amp vOUT = Vpos or –Vneg limit

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The Ideal Op-Amp ApproximationVPOS

–VNEG

rin =

v+

v–

rin = Av = Very Large

vOUT

This model greatly simplifies op-amp analysis

–Vneg < VOUT < Vpos

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A Consequence of Infinite rin

VPOS

VNEG

rin = i+ = 0

i = 0

Currents i+ and i to (or from) input terminals are zero

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A Consequence of Large Av

VPOS

–VNEG

If vOUT lies between Vpos and –Vneg …

(v+ v–) 0

Defines the Linear Region of operation

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Example: The Non-Inverting Amplifier Revisited

Use the Ideal Op-Amp approximation:

v– = vOUT R1 + R2

R1 Via voltage division (works because i = 0)

vIN

R2

R1

vOUT R1 + R2

R1

vOUT vIN

=i = 0

v vIN When vOUT in linear region:

–vneg< vOUT < vPos

vIN= vOUT R1 + R2

R1 R1 + R2

R1

vOUT vIN

= Done!

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+ –

vIN

vOUT R2

R1

vOUT vIN

=

Example: The Inverting Amplifier Revisited

R2

R1 i = 0

Use the Ideal Op-Amp approximation:

v 0v+ = 0

i1

i1 = vIN v

R1

vIN

R1

=

i2

i1 = i2 Via KCL (with i = 0)

Ohm’s Law

vOUT = i2 R2 = i1 R2vOUT =

R2

R1

vIN Done!

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+ –

The Summation Amplifier

vOUT

+

–+_

R2

R1

+_v1

v2

RF

Another Example:

Use the Ideal Op-Amp Approximation…

i2

i1 iF

KCL: i1 + i2 = iF

i1 = v1

R1

i2 = v2

R2

vOUT = iFRF

iF = + v1

R1

v2

R2

vOUT = RF

R1

RF

R2

v2v1 +

Output is weighted, inverted sum of inputs

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Can extend result to arbitrary number of input resistors:

Output is weighted, inverted sum of inputs:

vOUT

+

–

R2

R1

+_v2

RF

i2

i1 iF

...

+_v1

+_v3 +_vn

R3

Rn

vOUT = RF

R1

RF

R2

v2v1 +RF

R3

RF

Rn

v3+ vn+ …+

iF = i1 + i2 + i3+ … + in

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Difference AmplifierAnother Example:

vOUT

+

–

+_

R1

R2+_v2

v1

R2

R1

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vOUT

+

–

+_

R1

R2+_v2

v1

R2

R1

Use Superposition:

Set v2 to zero

i+ = 0 v+ = 0 v = 0

We have an inverting amplifier

vOUT =R2

R1

v1 1st Partial result for vOUT

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vOUT

+

–

+_

R1

R2+_

v1

R2

R1

Use Superposition, con’t:

Set v1 to zero

We have an non-inverting amplifier2nd Partial result for vOUT

v+ = v2 R1 + R2

R2 Via voltage division

R1 + R2

R1

vOUT v+= = v2 R1 + R2

R1 R1 + R2

R2 = v2

R2

R1

v2

i+

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vOUT

+

–

+_

R1

R2+_v2

v1 R1

Add together the 2nd and 1st partial results:

R2

= v2 R2

R1

vOUT R2

R1

v1

= (v2 v1) R2

R1

vOUT Amplifies difference between v2 and v1

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Summary

• Ideal Op-Amp Approximation simplifies circuit analysis

• “Ideal” implies rin = and v+ = v in the linear region

• Summation Amplifier

vOUT = RF

R1

RF

R2

v2v1 +RF

R3

RF

Rn

v3+ vn+ …+

= (v2 v1) R2

R1

vOUT

• Difference Amplifier

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End of This Module

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