lecture 9 1

Loop Analysis (3.2)Circuits with Op-Amps (3.3)

Prof. Phillips

February 19, 2003

lecture 9 2

Op Amps

• Op Amp is short for operational amplifier.

• An operational amplifier is modeled as a voltage controlled voltage source.

• An operational amplifier has a very high input impedance and a very high gain.

lecture 9 3

Use of Op Amps

• Op amps can be configured in many different ways using resistors and other components.

• Most configurations use feedback.

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Applications of Op Amps

• Amplifiers provide gains in voltage or current.

• Op amps can convert current to voltage.

• Op amps can provide a buffer between two circuits.

• Op amps can be used to implement integrators and differentiators.

• Lowpass and bandpass filters.

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The Op Amp Symbol

+

-

Non-inverting input

Inverting input

Ground

High Supply

Low Supply

Output

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The Op Amp Model

+

–Inverting input

Non-inverting input

Rin

v+

v-

+–

A(v+ -v- )

vo

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Typical Op Amp

• The input resistance Rin is very large (practically infinite).

• The voltage gain A is very large (practically infinite).

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“Ideal” Op Amp

• The input resistance is infinite.

• The gain is infinite.

• The op amp is in a negative feedback configuration.

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The Basic Inverting Amplifier

–

+Vin+

–Vout

R1

R2

+–

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Consequences of the Ideal

• Infinite input resistance means the current into the inverting input is zero:

i- = 0

• Infinite gain means the difference between v+ and v- is zero:

v+ - v- = 0

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Solving the Amplifier Circuit

Apply KCL at the inverting input:

i1 + i2 + i-=0

–R1

R2

i1 i-

i2

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KCL

0i

111 R

v

R

vvi inin

222 R

v

R

vvi outout

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Solve for vout

Amplifier gain:

21 R

v

R

v outin

1

2

R

R

v

v

in

out

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Recap

• The ideal op-amp model leads to the following conditions:

i- = 0 = i+

v+ = v-

• These conditions are used, along with KCL and other analysis techniques, to solve for the output voltage in terms of the input(s).

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Where is the Feedback?

–

+Vin+

–Vout

R1

R2

+–

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Review

• To solve an op-amp circuit, we usually apply KCL at one or both of the inputs.

• We then invoke the consequences of the ideal model.

– The op amp will provide whatever output voltage is necessary to make both input voltages equal.

• We solve for the op-amp output voltage.

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The Non-Inverting Amplifier

+

–

vin

+

–

vout

R1

R2

+–

lecture 9 18

KCL at the Inverting Input

+

–

vin

+

–

vout

R1

R2

i-

i1 i2

+–

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KCL

0i

111 R

v

R

vi in

222 R

vv

R

vvi inoutout

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Solve for Vout

021

R

vv

R

v inoutin

1

21R

Rvv inout

lecture 9 21

A Mixer Circuit

–

+v2+

–vout

R2

RfR1

v1

+–

+–

lecture 9 22

KCL at the Inverting Input

–

+v2+

–vout

R2

RfR1

v1

i1

i2

if

i-

+–

+–

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KCL

1

1

1

11 R

v

R

vvi

2

2

2

22 R

v

R

vvi

lecture 9 24

KCL

0i

f

out

f

outf R

v

R

vvi

lecture 9 25

Solve for Vout

02

2

1

1 f

out

R

v

R

v

R

v

22

11

vR

Rv

R

Rv ff

out

lecture 9 26

Class Example