Operational Amplifiers
(Op Amps)
Dr. Mustafa Kemal Uyguroğlu
1
Introduction
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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.
2
Use of Op Amps
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Op amps can be configured in many
different ways using resistors and
other components.
Most configurations use feedback.
3
Applications of Op Amps
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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.
4
More Applications
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Lowpass and bandpass filters.
5
The Op Amp Symbol
6
Schematic diagram of op amp
7
The Op Amp Model
Noninverting
input
v+
+
Inverting
input
v+
-
+
Rin
v+
v-
-
Ro
+
-
vo
+
vo
A(v+ -v- )
-
8
Typical Op Amp
Parameter
Typical
Range
Ideal
Values
Open-loop gain, A
105 to 108
∞
Input resistance, Rin
106 to 1013
∞
Output resistance, Ro
10 to 100
0Ω
Supply voltage, vcc
5 to 24
9
Example
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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.
50 Ω
io
-
2 MΩ vin
+
vs
40 kΩ
5 kΩ 20 kΩ
+
vo
-
+
-
io
5
2 ×10 vin
40 kΩ
vs
5 kΩ
20 kΩ
+
vo
-
Equivalent circuit
10
Example cont.
•Redrawn for clarity
KCL at v1
v1 − vs
v1 − v0
v1
+
+
=0
2 ×106 5 ×103 40 ×103
40 kΩ
vs
2 MΩ
-
v1
5
2 ×10 vin50 Ω
io
vin +
5 kΩ
+
-
20 kΩ
vo
KCL at v0
v0
v0 − v1 v0 − 2 × 105 (v1 − vs )
+
+
=0
20 ×103 40 × 103
50
v0 = 9.0004vs
i0 = 0.675 mA
11
“Ideal” Op Amp
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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.
12
Consequences of the Ideal
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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
13
Example
i- = 0
io
vs
vs
vs
KCL at noninverting
terminal:
vs − v0 vs
+ =0
40
5
9vs = v0
i+ = 0
40 kΩ
5 kΩ 20 kΩ
vo
+
vo
-
KCL at v0:
v −v 
 v
i0 =  0 + 0 s 
40k 
 20k
If vs = 1 V then i0 = 0.65 mA
14
Inverting Amplifier
Since the noninverting terminal is grounded
v1 = v2 = 0
KCL at v1:
v − 0 0 − v0
i1 = i2 ⇒ i
=
R1
Rf
v0 = −
Rf
R1
vi
15
Where is the Feedback?
R2
R1
Vin
+
-
+
+
Vout
-
16
Review
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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.
17
The Non-Inverting Amplifier
+
+
vin
+
R1
vout
R2
-
18
KCL at the Inverting Input
+
+
vin
+
-
ii1
i2
R1
vout
R2
-
19
KCL
i− = 0
− v − − vin
=
i1 =
R1
R1
Since v- = v+ = vin
v out − v − v out − vin
i2 =
=
R2
R2
20
Solve for Vout
− vin vout − vin
+
=0
R1
R2
v out

R2 

= vin 1 +
R1 

21
The Voltage Follower
22
Inverting Summer
_=0
23
KCL at the Inverting Input
R1
v1
+
i1
R2
v2
-
i2
i3
+
v3
+
-
Rf
if
R3
i- +
+
vout
-
24
KCL
v1 − v − v1
i1 =
=
R1
R1
since v− = 0
v2 − v− v2
i2 =
=
R2
R2
v3 − v− v3
=
i3 =
R3
R3
25
KCL
i− = 0
vout − v − vout
if =
=
Rf
Rf
26
Solve for Vout
v1 v2 v3 vout
+
+ +
=0
R1 R2 R3 R f
vout = −
Rf
R1
v1 −
Rf
R2
v2 −
Rf
R3
v3
27
Noninverting Summer
R1
v1
i1
R2
+
v2
i3
+
-
i2
v3
+
R3
i- +
+
if
-
vout
Rf
Ra
ia
-
28
KCL at noninverting input:
i1 + i2 + i3 = 0
v1 − v+
i1 =
R1
v2 − v+
i2 =
R2
v3 − v+
i3 =
R3
KCL at inverting input:
if + i a = 0
v− − vout
if =
Rf
v−
ia =
Ra
Ra
v− =
vout
Ra + R f
v− = v+
29
i1 + i2 + i3 = 0
v1 v2 v3  1
1
1 
+
+
= +
+  v+
R1 R2 R3  R1 R2 R3 
1  1
1
1 
= +
+ 
RT  R1 R2 R3 
Ra
v1 v2 v3
1
+
+
=
vout
R1 R2 R3 RT Ra + R f
vout
 R f   RT
RT
RT 
= 1 +
v2 +
v3 
  v1 +
Ra   R1
R2
R3 

30
The difference amplifier
31
KCL at node vb :
vb − v2 vb
=
R3
R4
R4
vb =
v2 = va
R3 + R4
32
KCL at va
va − v1 va − vo
+
=0
R1
R2
1
1
1 
1
vo =  +  va − v1
R2
R1
 R1 R2 
 R2 
 R2  R4
R2
R
vo =  1 +  va − v1 =  1 + 
v2 − 2 v1
R1 
R1
R1  R3 + R4
R1


 R1 
 + 1
 R2  1
R2
R2  R2 
R
v0 = 1 + 
v2 − v1 =
v2 − 2 v1
R1  R3 + 1
R1
R1 R3 + 1
R1

R4
R4
33
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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
R1 R3
=
R2 R4
When R1 = R2 and R3 = R4 it acts like a subtractor
vo = v2 − v1
34
Interconnecting of Op Amps
35
Example
Find the voltage transfer equation of the following circuit
50 kΩ
40 kΩ
20 kΩ
+
25 kΩ
10 kΩ
v1
10 kΩ
+
v2
20 kΩ
25 kΩ
+
v3
12 V
-
-
-
36