Operational amplifiers (op-amps) are integrated circuits with two inputs and one output that amplify voltage based on the difference between input signals. They can be used in various applications such as linear and non-linear operations, with configurations including inverting, non-inverting, summing, and difference amplifiers. Ideal op-amps have infinite gain and bandwidth, though practical versions exhibit high gain, high input impedance, and low output impedance.

1. introduction
Operational Amplifiers

2. introduction
Introduction
The operational amplifier or op-amp is a circuit of
components integrated into one chip.
A typical op-amp is powered by two dc voltages
and has an inverting(-) and a non-inverting input
(+) and an output.
An op amp is an electronic device which provides
a voltage output based on the voltage input

3. introduction
Basic op-amp
Op-amp has two inputs that connect to two terminals
and one output

4. introduction
Operational Amplifiers
Five important pins
 2 – The inverting input
 3 – The non-inverting input
 6 – The output
 4 – The negative power supply V-
(-Vcc)
 7 – The positive power supply V+
(+Vcc)

5. introduction
Operational Amplifiers
 The output of the op amp is given by the following
equation:
Vd = E1 – E2
and Vo = AVOL(Vd)
 AVOL is called the open-loop voltage gain because it
is the gain of the op amp without any external feedback
from output to input

6. introduction
Operational Amplifiers
• Positive Saturation –
where the output
voltage exceeds the
positive power input

7. introduction
Operational Amplifiers
• Linear Region –
where the output
voltage is linear
based on A (gain)

8. introduction
Operational Amplifiers
• Negative Saturation
– where the output
voltage would be less
than the negative
power input

9. introduction
Operational Amplifiers

10. introduction
What do they really look like?

11. introduction
IC Circuit

12. introduction
Operational Amplifiers

13. introduction
Operational Amplifiers
An ideal op-amp has infinite gain and bandwidth,
we know this is impossible.
However, op-amps do have:
very high gain
very high input impedance(Zin = ∞)
very low output impedance (Zout = 0)
wide bandwidth.

14. application
Application in op-amp
 There are 2 types of application in op-amp
 Linear application
 Non-linear application
 Linear application is where the op-amp operate
in linear region:
 Assumptions in linear application:
 Input current, Ii = 0
 Input voltage: V+
=V-
 Feedback at the inverting input

15. application
 Non-linear application is where the op-amp
operate in non-linear region
 By comparing these two input voltages:
positive input voltages, V+
and negative input
voltage, V-
where:
VO = VCC if V+
> V-
VO = -VEE if V+
< V-
 Input current, Ii = 0
Application in op-amp

16. application
Applications of op-amp
 Comparator
 Inverter
 Audio amplifier
 Difference Amplifier
 Filter
 Summing Amplifier

17. application
 Inverting Amplifier
 Non-Inverting Amplifier
 Summing Amplifier
 Unity Follower
 Difference Amplifier
 Integrators
 Differentiators
Op-amp Circuit Application

18. application:inverting amplifier
Application: Inverting amplifier
 Provide a constant gain multiplier
 Input signal is connected to the inverting input of the op-
amp. Therefore, the output signal is 180 degree out of
phase from the input signal
 Rf is the feed-back resistor to control the voltage gain of
the op-amp

19. application:inverting amplifier
Summary of op-amp behavior
Vo = A(V+
- V−
)
Vo/A = V+
- V−
Let A infinity
then,
V+
- V−
0

20. application:inverting amplifier
V+
= V−
I+
= I −
= 0
Seems strange, but the input terminals to an
op-amp act as a short and open at the same time
Summary of op-amp behavior

21. application:inverting amplifier
To analyze an op-amp circuit for linear
operation
•Write node equations at + and - terminals
(Ii=I+
= I-
= 0)
•Set V+
= V-
•Solve for Vo

22. application:inverting amplifier
Analysis of inverting amplifier
I1
If
Ii
V+
= 0 Ii = 0
I1 = If + Ii
Vs −V−
R1
=
V−
−Vo
Rf
V−
=V+
= 0
Vo
Vs
= −
Rf
R1
Vo = −
Rf
R1
Vs

23. application:non-inverting amplifier
Application:Non-inverting amplifier

24. Non-inverting configuration






+=
−
=
−
=
−
=
−
=
+=
==
−
−−
−+
1
2
21
21
21
1
0
;
0
:
;0
:
R
R
VV
R
VV
R
V
VVinsert
R
VV
R
V
so
Iwhile
III
KCLuse
VVV
io
oii
i
o
i
i
i
Vi
I1
I2
Ii

25. application:summing amplifier
Application: Summing amplifier
Virtual-ground equivalent circuit.

26. Summing Amplifier
V1
V2
V3
R1
R2
R3
Rf
This circuit is called
a weighted summer






++−=
−
=++=
−
=
−
+
−
+
−
=
+=++
==
−
−−−−
−+
3
3
2
2
1
1
3
3
2
2
1
1
3
3
2
2
1
1
321
;0
:
;0
:
0
R
V
R
V
R
V
RV
R
V
R
V
R
V
R
V
Vinsert
R
VV
R
VV
R
VV
R
VV
so
Iwhile
IIIII
KCLuse
VV
fo
f
o
f
o
i
RfiRRR

27. application:unity-follower
Application: Unity Follower
1VVO =

28. application:difference amplifier
Application: Difference amplifier
)( 21
2
4
VV
R
R
VO −=
43
21
RR
RR
=
=

29. application:instrumentation amplifier
Application: Instrumentation Amplifier
2R
Buffer
2R
1R
1R
AR
BR
AR
Difference amplifier
( )12
1
2 2
1 VV
R
R
R
R
V
B
A
O −





+=
2R

30. application:integrator
Application: Integrator
Feedback component = capacitor : Integrator
I IC
∫−=
+=
+=
dttv
RC
tv
dt
tdv
C
R
tv
III
io
i
Ci
)(
1
)(
)(
0
)( 0
sC
1
Cj
1
X
:impedancecetanCapaci
C =
ω
=

31. application:differentiator
Application: Differentiation
dt
tdv
RCtv
R
tvV
dt
tdv
C
II
i
o
oi
RC
)(
)(
)()(
−=
−
=
=
−

32. exercise
Exercise 1
Find VO?

33. exercise
Exercise 2
Find V2 and V3?

34. exercise
Exercise 3
Find VO?

35. exercise
Exercise 4
Find VO?

36. non-linear application
 Non-linear application is where the op-amp
operate in non-linear region
 By comparing these two input voltages:
positive input voltages, V+
and negative input
voltage, V-
where:
VO = VCC if V+
> V-
VO = -VCC if V+
< V-
 Input current, Ii = 0
Recall: Non-linear application in op-amp

37. non-linear application:comparator
Non-linear application: Comparator