Operational Amplifiers

Supplemental lecture

Rick Matthews

The inverting amplifier

• R2 provides negative feedback.

The inverting amplifier

• R2 provides negative feedback.

• This means V- is adjusted to V+.

The inverting amplifier

• R2 provides negative feedback.

• This means V- is adjusted to V+.

• V+ is zero, so V- must be zero, too.

The inverting amplifier

• R2 provides negative feedback.

• This means V- is adjusted to V+.

• V+ is zero, so V- is zero.

1

22

1

/in

out in

I V R

RV IR V

R

I

The inverting amplifier

• R2 provides negative feedback.

• This means V- is adjusted to V+.

• V+ is zero, so V- is zero.

1

22

1

/in

out in

I V R

RV IR V

R

I

The inverting amplifier

• R2 provides negative feedback.

• This means V- is adjusted to V+.

• V+ is zero, so V- is zero.

1

22

1

/in

out in

I V R

RV IR V

R

I

More generally,…

More generally,…

• Whatever sits in the place of R1 serves to create a current I that is a function of Vin.

I=f(Vin)

More generally,…

• Whatever sits in the place of R1 serves to create a current I that is a function of Vin.

• And whatever sits in place of R2 serves to create a voltage Vout that is a second function of I.

I=f(Vin)

Vout= -g(I)

More generally,…

• Whatever sits in the place of R1 serves to create a current I that is a function of Vin.

• And whatever sits in place of R2 serves to create a voltage Vout that is a second function of I.

I=f(Vin)

( )out inV g f V

Vout= -g(I)

Example

2

11

2 2

2

1

/ , so ( ) .

, so ( ) .

( ) .

inin in

R

out in in

VI V R f V

R

V IR g I IR

RV g f V V

R

Example

2

11

2 2

2

1

/ , so ( ) .

, so ( ) .

( ) .

inin in

R

out in in

VI V R f V

R

V IR g I IR

RV g f V V

R

Example

2

11

2 2

2

1

/ , so ( ) .

, so ( ) .

( ) .

inin in

R

out in in

VI V R f V

R

V IR g I IR

RV g f V V

R

Example: Exponentiating amp

exp 1 ,

so ( ) exp 1 .

, so ( ) .

( ) exp 1

exp .

o

in o

R

inout in o

ino

eVI I

kT

eVf V I

kT

V IR g I IR

eVV g f V I R

kT

eVI R

kT

Example: Exponentiating amp

exp 1 ,

so ( ) exp 1 .

, so ( ) .

( ) exp 1

exp .

o

in o

R

inout in o

ino

eVI I

kT

eVf V I

kT

V IR g I IR

eVV g f V I R

kT

eVI R

kT

Example: Exponentiating amp

exp 1 ,

so ( ) exp 1 .

, so ( ) .

( ) exp 1

exp .

o

in o

R

inout in o

ino

eVI I

kT

eVf V I

kT

V IR g I IR

eVV g f V I R

kT

eVI R

kT

Example: Exponentiating amp

2

exp 1 ,

so ( ) exp 1 .

, so ( ) .

( ) exp 1

exp .

o

in o

R

inout in o

ino

eVI I

kT

eVf V I

kT

V IR g I IR

eVV g f V I R

kT

eVI R

kT

Example: Exponentiating amp

exp 1 ,

so ( ) exp 1 .

, so ( ) .

( ) exp 1

exp .

o

in o

R

inout in o

ino

eVI I

kT

eVf V I

kT

V IR g I IR

eVV g f V I R

kT

eVI R

kT

Example: log amp

Box 1 is the resistor.

, so ( ) .

Box 2 is the diode.

exp 1

log 1

log .

( ) log .

Therefore,

( ) log .

in inin

o

o

o

o

inout in

o

V VI f V

R R

eVI I

kT

kT IV

e I

kT I

e I

kT Ig I

e I

VkTV g f V

e I R

Example: log amp

Box 1 is the resistor.

, so ( ) .

Box 2 is the diode.

exp 1

log 1

log .

( ) log .

Therefore,

( ) log .

in inin

o

o

o

o

inout in

o

V VI f V

R R

eVI I

kT

kT IV

e I

kT I

e I

kT Ig I

e I

VkTV g f V

e I R

Example: log amp

Box 1 is the resistor.

, so ( ) .

Box 2 is the diode.

exp 1

log 1

log .

( ) log .

Therefore,

( ) log .

in inin

o

o

o

o

inout in

o

V VI f V

R R

eVI I

kT

kT IV

e I

kT I

e I

kT Ig I

e I

VkTV g f V

e I R

Example: log amp

Box 1 is the resistor.

, so ( ) .

Box 2 is the diode.

exp 1

log 1

log .

( ) log .

Therefore,

( ) log .

in inin

o

o

o

o

inout in

o

V VI f V

R R

eVI I

kT

kT IV

e I

kT I

e I

kT Ig I

e I

VkTV g f V

e I R

Example: log amp

Box 1 is the resistor.

, so ( ) .

Box 2 is the diode.

exp 1

log 1

log .

( ) log .

Therefore,

( ) log .

in inin

o

o

o

o

inout in

o

V VI f V

R R

eVI I

kT

kT IV

e I

kT I

e I

kT Ig I

e I

VkTV g f V

e I R

Example: log amp

Box 1 is the resistor.

, so ( ) .

Box 2 is the diode.

exp 1

log 1

log .

( ) log .

Therefore,

( ) log .

in inin

o

o

o

o

inout in

o

V VI f V

R R

eVI I

kT

kT IV

e I

kT I

e I

kT Ig I

e I

VkTV g f V

e I R

Example: log amp

Box 1 is the resistor.

, so ( ) .

Box 2 is the diode.

exp 1

log 1

log .

( ) log .

Therefore,

( ) log .

in inin

o

o

o

o

inout in

o

V VI f V

R R

eVI I

kT

kT IV

e I

kT I

e I

kT Ig I

e I

VkTV g f V

e I R

A Multiplier

Recall

log( ) log( ) log( )ab a b

Log Amp

Log Amp

SummingAmp

ExponentialAmp

Vin1

Vin2

Vout

log(a)

log(b)

log(a)+log(b) =log(ab)

ab

A Divider

Recall

log( / ) log( ) log( )a b a b

Log Amp

Log Amp

DifferentialAmp

ExponentialAmp

Vin1

Vin2

Vout

log(a)

log(b)

log(a)-log(b) =log(a/b)

a/b

Calculus

R11k

VoutVinC11uF

Differentiator

Calculus

R11k

VoutVinC11uF

VoutVin

C11uF

R11k

Differentiator Integrator

Etc.

• Can you think of a circuit to take cube roots?

• We can fashion sophisticated analog computers this way.