operational amplifiers (op amps)2 introduction {op amp is short for operational amplifier. {an...
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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.
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More Applications
Lowpass and bandpass filters.
6
The Op Amp Symbol
7
Schematic diagram of op amp
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The Op Amp Model
+
-Inverting input
Non-inverting
input
Rin
v+
v-
+-
A(v+ -v- )
voRo
+
-v-
+
+
--
v+ vo
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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
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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
vs40 kΩ
io
20 kΩ vo
+-
+
-
+
-
vs
2 MΩvin 2 ×105 vin
50 Ω io
40 kΩ
5 kΩ 20 kΩ vo
Equivalent circuit
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Example cont.
io
+-
+-
vs2 MΩ
vin
5 kΩ
v1
40 kΩ
2 ×105 vin
50 Ωvo
20 kΩ
1 1 016 3 3 0
2 10 5 10 40 10sv v v vv− −
+ + =× × ×
•Redrawn for clarityKCL at v1
KCL at v0
50 0 1 0 1
3 3
2 10 ( ) 020 10 40 10 50
sv v v v v v− − × −+ + =
× ×
v0 = 9.0004vs
i0 = 0.675 mA
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“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.
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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
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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:
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Inverting Amplifier
1 2 0v v= =
01 2
1
0 0i
f
v vi iR R− −
= ⇒ =
Since the noninverting terminal is grounded
KCL at v1: 01
fi
Rv v
R= −
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Where is the Feedback?
-
+Vin+
-+
-Vout
R1
R2
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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.
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The Non-Inverting Amplifier
+
-
vin+
-
+
-
vout
R1
R2
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KCL at the Inverting Input
+
-
vin+
-
+
-
vout
R1
R2
i-
i1 i2
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KCL
0=−i
111 R
vRvi in−
=−
= −
222 R
vvR
vvi inoutout −
=−
= −
Since v- = v+ = vin
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Solve for Vout
021
=−
+−
Rvv
Rv inoutin
+=
1
21RR
vv inout
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The Voltage Follower
23
Inverting Summer
_ = 0
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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
−−= =
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KCL
0=−i
f
out
f
outf R
vR
vvi =
−= −
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Solve for Vout
31 2
1 2 3
0out
f
v vv vR R R R
+ + + =
1 2 31 2 3
f f fout
R R Rv v v v
R R R= − − −
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Noninverting Summer
+
-v2+
-+
-
vout
R2
Ra
R1
v1+
-
i1
i3
ia
i-
v3+
-
R3
i2
Rf
if
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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 a
outf
f
i iv vi
R−
+ =−
=
aa
aout
a f
viR
Rv vR R
v v
−
−
− +
=
=+
=
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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 Tout
a
R R R Rv v v vR R R R
= + + +
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The difference amplifier
32
2
3 4
42
3 4
KCL at node :b
b b
b a
vv v v
R RRv v v
R R
−=
= =+
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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
− −+ =
= + −
= + − = + − +
+ = + − = − + +
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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
4
3
2
1
RR
RR
=
When R1 = R2 and R3 = R4 it acts like a subtractor
12 vvvo −=
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Interconnecting of Op Amps
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Example
+
+ +
- - -
v1 v2v3
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