1

Operational Amplifiers(Op Amps)

Dr. Mustafa Kemal Uyguroğlu

2

Introduction

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.

3

Use of Op Amps

Op amps can be configured in many different ways using resistors and other components.Most configurations use feedback.

4

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.

5

More Applications

Lowpass and bandpass filters.

6

The Op Amp Symbol

7

Schematic diagram of op amp

8

The Op Amp Model

+

-Inverting input

Non-inverting

input

Rin

v+

v-+-

A(v+ -v- )

voRo

+

-v-

+

+

--

v+ vo

9

Typical Op Amp

5 to 24Supply voltage, vcc

0 Ω10 to 100Output resistance, Ro

∞106 to 1013Input resistance, Rin

∞105 to 108Open-loop gain, A

Ideal Values

Typical Range

Parameter

10

Example

A 741 op amp has an open-loop voltage gain of 2 x 105, input resistance of 2 MΩ, and output resistance of 50 Ω.The op amp is used in the circuit shown below. Find the closed- loop gain v0/vs. Find i0when vs = 1 V.

+

-5 kΩ

vs 40 kΩ

io

20 kΩ vo

+-

+

-

+

-

vs

2 MΩvin 2 ×105 vin

50 Ω io

40 kΩ

5 kΩ 20 kΩ vo

Equivalent circuit

11

Example cont.

io+-

+-

vs 2 MΩ

vin

5 kΩ

v1

40 kΩ

2 ×105 vin50 Ω vo

20 kΩ

1 1 016 3 3 02 10 5 10 40 10sv v v vv− −+ + =

× × ×

•Redrawn for clarityKCL at v1

KCL at v05

0 0 1 0 13 3

2 10 ( ) 020 10 40 10 50

sv v v v v v− − × −+ + =× ×

v0 = 9.0004vsi0 = 0.675 mA

12

“Ideal” Op Amp

The input resistance is infinite, Rin=∞The gain is infinite, A = ∞Zero output resistance, Ro= 0The op amp is in a negative feedback configuration.

13

Consequences of the Ideal

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

i- = 0 = i+Infinite gain means the difference between v+ and v- is zero:

v+ - v- = 0

14

Example

vo

io

+

-

i- = 0

i+ = 0

vs

vs

vs40 kΩ

5 kΩ 20 kΩ

vo

0

0

040 5

9

s s

s

v v v

v v

−+ =

=

KCL at noninvertingterminal:

If vs = 1 V then i0 = 0.65 mA

0 00 20 40

sv v vik k

− = +

KCL at v0:

15

Inverting Amplifier

1 2 0v v= =

01 2

1

0 0if

v vi iR R− −

= ⇒ =

Since the noninverting terminal is grounded

KCL at v1: 01

fi

Rv v

R= −

16

Where is the Feedback?

-

+Vin+

-+

-Vout

R1

R2

17

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.

i- = 0 = i+v+ - v- = 0

We solve for the op amp output voltage.

18

The Non-Inverting Amplifier

+

-

vin+

-

+

-

vout

R1R2

19

KCL at the Inverting Input

+

-

vin+

-

+

-

vout

R1R2

i-

i1 i2

20

KCL

0=−i

111 R

vRvi in

−=

−= −

222 R

vvR

vvi inoutout

−=

−= −

Since v- = v+ = vin

21

Solve for Vout

021

=−

+−

Rvv

Rv inoutin

+=

1

21RR

vv inout

22

The Voltage Follower

23

Inverting Summer

_ = 0

24

KCL at the Inverting Input

-

+v2+

-+

-vout

R2

RfR1

v1+

-

i1

i3

if

i-

v3+

-

R3

i2

25

KCL

since 0v− =1

1

1

11 R

vR

vvi =−= −

2

2

2

22 R

vR

vvi =−= −

3 33

3 3

v v viR R

−−= =

26

KCL

0=−i

f

out

f

outf R

vR

vvi =

−= −

27

Solve for Vout

31 2

1 2 3

0outf

v vv vR R R R

+ + + =

1 2 31 2 3

f f fout

R R Rv v v v

R R R= − − −

28

Noninverting Summer

+

-v2+

-+

-

vout

R2

Ra

R1

v1+

-

i1

i3

ia

i-

v3+

-

R3

i2

Rf

if

29

1 2 3

11

1

0i i iv vi

R+

+ + =−

=

22

2

v viR

+−=

33

3

v viR

+−=

KCL at noninverting input: KCL at inverting input:

0f aout

ff

i iv vi

R−

+ =−

=

aa

aout

a f

viR

Rv vR R

v v

−

−

− +

=

=+

=

30

1 2 3

31 2

1 2 3 1 2 3

1 2 3

31 2

1 2 3

0

1 1 1

1 1 1 1

1T

aout

T a f

i i i

vv v vR R R R R R

R R R Rv Rv v v

R R R R R R

+

+ + =

+ + = + +

= + +

+ + =+

1 2 31 2 3

1 f T T Touta

R R R Rv v v vR R R R

= + + +

31

The difference amplifier

32

2

3 4

42

3 4

KCL at node :bb b

b a

vv v v

R RRv v v

R R

−=

= =+

33

1

1 2

12 1 2 1

2 2 2 4 21 2 1

1 1 1 3 4 1

1

22 2 2 20 2 1 2 1

3 31 1 1 1

4 4

KCL at

0

1 1 1 1

1 1

111

1 1

a

a a o

o a

o a

vv v v v

R R

v v vR R R R

R R R R Rv v v v vR R R R R R

RRR R R Rv v v v vR RR R R R

R R

− −+ =

= + −

= + − = + − +

+ = + − = − + +

34

Since a difference must reject a signal common to the two inputs, the amplifier must have the property that v0 = 0 when v1 = v2. This implies that

4

3

2

1

RR

RR

=

When R1 = R2 and R3 = R4 it acts like a subtractor

12 vvvo −=

35

Interconnecting of Op Amps

36

Example

+

+ +

- - -

v1 v2 v3

10 kΩ

10 kΩ

40 kΩ

20 kΩ

20 kΩ

50 kΩ

25 kΩ

25 kΩ

12 V

Find the voltage transfer equation of the following circuit