Operational amplifiers (op amps) are voltage amplifiers that can be configured in various ways using feedback. Common op amp configurations include inverting amplifiers, non-inverting amplifiers, summing amplifiers, and difference amplifiers. These configurations allow op amps to perform functions like amplification, filtering, integration, differentiation, and mathematical operations. Op amp analysis involves applying Kirchhoff's laws and the assumptions of the ideal op amp model, namely infinite gain, infinite input impedance, and zero output impedance.

1. 1
Operational Amplifiers
(Op Amps)
Dr. Mustafa Kemal Uyguroğlu

2. 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. 3
Use of Op Amps
{ Op amps can be configured in many
different ways using resistors and
other components.
{ Most configurations use feedback.

4. 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. 5
More Applications
{ Lowpass and bandpass filters.

6. 6
The Op Amp Symbol

7. 7
Schematic diagram of op amp

8. 8
The Op Amp Model
+
-
Inverting
input
Non-
inverting
input
Rin
v+
v-
+
-
A(v+ -v- )
vo
Ro
+
-
v-
+
+
-
-
v+ vo

9. 9
Typical Op Amp
5 to 24
Supply voltage, vcc
0 Ω
10 to 100
Output resistance, Ro
∞
106 to 1013
Input resistance, Rin
∞
105 to 108
Open-loop gain, A
Ideal
Values
Typical
Range
Parameter

10. 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 i0
when vs = 1 V.
+
-
5 kΩ
vs
40 kΩ
io
20 kΩ vo
+
-
+
-
+
-
vs
2 MΩvin 2 ×10
5
vin
50 Ω io
40 kΩ
5 kΩ 20 kΩ vo
Equivalent circuit

11. 11
Example cont.
io
+
-
+
-
vs
2 MΩ
vin
5 kΩ
v1
40 kΩ
2 ×10
5
vin
50 Ω
vo
20 kΩ
1 1 0
1
6 3 3
0
2 10 5 10 40 10
s
v v v v
v
− −
+ + =
× × ×
•Redrawn for clarity
KCL at v1
KCL at v0
5
0 0 1 0 1
3 3
2 10 ( )
0
20 10 40 10 50
s
v v v v v v
− − × −
+ + =
× ×
v0 = 9.0004vs
i0 = 0.675 mA

12. 12
“Ideal” Op Amp
{ The input resistance is infinite,
Rin=∞
{ The gain is infinite, A = ∞
{ Zero output resistance, Ro= 0
{ The op amp is in a negative
feedback configuration.

13. 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. 14
Example
vo
io
+
-
i- = 0
i+ = 0
vs
vs
vs
40 kΩ
5 kΩ 20 kΩ
vo
0
0
0
40 5
9
s s
s
v v v
v v
−
+ =
=
KCL at noninverting
terminal:
If vs = 1 V then i0 = 0.65 mA
0 0
0
20 40
s
v v v
i
k k
−
 
= +
 
 
KCL at v0:

15. 15
Inverting Amplifier
1 2 0
v v
= =
0
1 2
1
0 0
i
f
v v
i i
R R
− −
= ⇒ =
Since the noninverting terminal is grounded
KCL at v1:
0
1
f
i
R
v v
R
= −

16. 16
Where is the Feedback?
-
+
Vin
+
-
+
-
Vout
R1
R2

17. 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. 18
The Non-Inverting Amplifier
+
-
vin
+
-
+
-
vout
R1
R2

19. 19
KCL at the Inverting Input
+
-
vin
+
-
+
-
vout
R1
R2
i-
i1 i2

20. 20
KCL
0
=
−
i
1
1
1
R
v
R
v
i in
−
=
−
= −
2
2
2
R
v
v
R
v
v
i in
out
out −
=
−
= −
Since v- = v+ = vin

21. 21
Solve for Vout
0
2
1
=
−
+
−
R
v
v
R
v in
out
in








+
=
1
2
1
R
R
v
v in
out

22. 22
The Voltage Follower

23. 23
Inverting Summer
_ = 0

24. 24
KCL at the Inverting Input
-
+
v2
+
-
+
-
vout
R2
Rf
R1
v1
+
-
i1
i3
if
i-
v3
+
-
R3
i2

25. 25
KCL
since 0
v− =
1
1
1
1
1
R
v
R
v
v
i =
−
= −
2
2
2
2
2
R
v
R
v
v
i =
−
= −
3 3
3
3 3
v v v
i
R R
−
−
= =

26. 26
KCL
0
=
−
i
f
out
f
out
f
R
v
R
v
v
i =
−
= −

27. 27
Solve for Vout
3
1 2
1 2 3
0
out
f
v v
v v
R R R R
+ + + =
1 2 3
1 2 3
f f f
out
R R R
v v v v
R R R
= − − −

28. 28
Noninverting Summer
+
-
v2
+
-
+
-
vout
R2
Ra
R1
v1
+
-
i1
i3
ia
i-
v3
+
-
R3
i2
Rf
if

29. 29
1 2 3
1
1
1
0
i i i
v v
i
R
+
+ + =
−
=
2
2
2
v v
i
R
+
−
=
3
3
3
v v
i
R
+
−
=
KCL at noninverting input: KCL at inverting input:
0
f a
out
f
f
i i
v v
i
R
−
+ =
−
=
a
a
a
out
a f
v
i
R
R
v v
R R
v v
−
−
− +
=
=
+
=

30. 30
1 2 3
3
1 2
1 2 3 1 2 3
1 2 3
3
1 2
1 2 3
0
1 1 1
1 1 1 1
1
T
a
out
T a f
i i i
v
v v
v
R R R R R R
R R R R
v R
v v
v
R R R R R R
+
+ + =
 
+ + = + +
 
 
 
= + +
 
 
+ + =
+
1 2 3
1 2 3
1
f T T T
out
a
R R R R
v v v v
R R R R
  
= + + +
  
  

31. 31
The difference amplifier

32. 32
2
3 4
4
2
3 4
KCL at node :
b
b b
b a
v
v v v
R R
R
v v v
R R
−
=
= =
+

33. 33
1
1 2
1
2 1 2 1
2 2 2 4 2
1 2 1
1 1 1 3 4 1
1
2
2 2 2 2
0 2 1 2 1
3 3
1 1 1 1
4 4
KCL at
0
1 1 1 1
1 1
1
1
1
1 1
a
a a o
o a
o a
v
v v v v
R R
v v v
R R R R
R R R R R
v v v v v
R R R R R R
R
R
R R R R
v v v v v
R R
R R R R
R R
− −
+ =
 
= + −
 
 
   
= + − = + −
   
+
   
 
+
 
   
= + − = −
 
  + +

34. 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
R
R
R
R
=
When R1 = R2 and R3 = R4 it acts like a subtractor
1
2 v
v
vo −
=

35. 35
Interconnecting of Op Amps

36. 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