Non-ideal Operational
Amplifier
•
•
Various error terms arise in practical operational
amplifiers due to non-ideal behavior.
Some of the non-ideal characteristics include:
1.
2.
3.
4.
5.
6.
7.
Finite open-loop gain that causes gain error
Nonzero output resistance
Finite input resistance
Finite CMRR
Common-mode input resistance
DC error sources
Output voltage and current limits
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1
Finite Open-loop Gain
vo = Av = A(vs − v ) = A(vs − βvo )
1
id
v
A
Av = o =
v s 1 + Aβ
Aβ is called loop gain.
For Aβ >>1,
R
A
= = 1+ 2
ideal β
R
1
v = vs − v = vs − βvo
1
id
v
Aβ
vs = s
= vs −
1 + Aβ
1 + Aβ
1
R
1 v = βv
v =
o
1 R +R o
1 2
is called
R
1
feedback
β=
factor.
R +R
1 2
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No longer zero, vid is small for
large Aβ.
2
Gain Error
Gain Error is given by
GE= (ideal gain)-(actual gain)
For non-inverting amplifier,
GE =
1
1
A
=
β 1+ Aβ β (1+ Aβ )
−
Gain error is also expressed as a fractional or percentage error.
A
1
1
FGE = β 1+ Aβ =
≅
1
1 + Aβ Aβ
1
−
β
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Gain Error: Example
Problem: Find ideal and actual gain and gain error is percent
Given data: Closed-loop gain of 200 (46 dB), open-loop gain of op amp is
10,000 (80 dB).
Approach:Amplifier is designed to give ideal gain and deviations from ideal
case are determined. Hence,
.
R1 and R2 aren’t designed to compensate for finite open-loop gain of
amplifier.
Analysis:
104
A
=
= 196
Av =
4
1 + Aβ
10
1+
200
200 −196
= 0.02
FGE =
200
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Nonzero Output Resistance
Output terminal is driven by test source vx
and current ix is calculated to determine
output resistance (all independent sources
are turned off).The equivalent circuit is same
For both inverting and non-inverting
amplifiers.
v
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Rout = x
ix
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Nonzero Output Resistance
(contd.)
ix = io + i
Analysis:
2
Also, vid= -v1 and
v x - Av
vx
id
i =
io =
2 R +R
Ro
1 2
R
1 v = βv
v =
x
1 R +R x
1 2
i
1 + Aβ
1
1
= x =
+
∴
R
Ro
R +R
vx
out
1 2
Thus, shunt feedback at output reduces Rout.
Ro ⎛
⎞
⎜R + R ⎟
∴Rout =
⎜
⎟
1+ Aβ ⎝ 1 2 ⎠
Since, Ro/(1+Aβ)<<(R1+R2),
Ro
Rout ≅
1 + Aβ
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If A is infinite, Rout=0
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Open-loop Gain Design: Example
• Problem: Design non-inverting amplifier and find
open-loop gain
• Given Data: Av=35 dB, Rout =0.2Ω, Ro = 250Ω
• Analysis:
Av = 1035dB / 20dB = 56.2
β= 1 = 1
Av 56.2
R
Rout = o ≤ 0.2Ω
1 + Aβ
⎛
⎜
⎜
⎜
⎜⎜
⎝
⎞
⎟
⎛ 250
⎞
1 Ro
∴A ≥
−1⎟⎟ = 56.2⎜⎜
−1⎟⎟ = 7.03×104 = 96.9dB
⎜
⎟
β R
⎟⎟
⎝ 0.2
⎠
out ⎠
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Finite Input Resistance: Noninverting Amplifier
Assuming i-<<i2 implies i1 = i2.
Test voltage source vx is applied to
input and current ix is calculated.
vx − v
1
ix =
R
id
v =i R ≅i R
1 1 1 2 1
R
1 v = βv
v ≅
o
1 R +R o
1 2
= β(Av )= Aβ(v x − v )
1
id
Aβ
v
v =
1 1+ Aβ x
Aβ
vx −
vx
vx
1+
A
β
∴ix =
=
(1+ Aβ )R
R
id
id
R = R (1+ Aβ )
in id
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Finite Common-Mode Rejection
Ratio (CMRR)
A real amplifier responds to signal common to both inputs, called
common-mode input voltage. In general,
v + v ⎞⎟
vo = A(v − v ) + Acm 1 2 ⎟
1 2
2 ⎟⎠
= A(v ) + Acm (v )
ic
id
⎛
⎜
⎜
⎜
⎝
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Voltage Follower Gain Error due to
CMRR
Ideal gain for voltage follower is unity,
gain error
A
2CMRR
GE = 1− Av =
⎛
⎞
1
⎟
1+ A⎜⎜1−
⎟⎟
⎜
2
CMRR
⎝
⎠
1−
v = vs − vo
id
v =
ic
vs + v o
2
⎛
(v + v ) ⎞
vo = A⎜⎜ (vs − vo )+ s o ⎟⎟
2CMRR ⎠
⎝
⎛
⎞
1
⎟
A⎜⎜1+
⎟⎟
vo
⎜
2
CMRR
⎝
⎠
=
Av =
⎞
1
vs 1+ A⎛⎜1−
⎟
⎜⎜
⎟
2CMRR ⎟⎠
⎝
Since, both A and CMRR are
normally >>1,
1
1
GE ≅ −
A CMRR
First term is due to finite amplifier
gain, second term shows that
CMRR may introduce an even larger
error.
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DC Error Sources: Input-Offset
Voltage
With inputs being zero, the amplifier
output rests at some dc voltage level
instead of zero. The equivalent dc
input offset voltage is
To include effect of offset voltage
v
⎞
ic
+V ⎟⎟
vo = A v +
id CMRR OS ⎟⎠
⎛
⎜
⎜
⎜
⎝
If vid =0,
V
V = o
OS A
v
⎞
ic
+V ⎟⎟ = A(v )
vo = A
OS
CMRR OS ⎟⎠
v
∴CMRR = v ic μV/V
OS
⎛
⎜
⎜
⎜
⎝
The amplifier is connected as
voltage-follower to give output
voltage equal to offset voltage.
Thus, CMRR is a measure of
how total offset voltage changes
from its dc value when commonmode voltage is applied.
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