# Non-ideal Operational Amplifier ```Lecture 11
Non-ideal Operational Amplifiers (op amp)
Objectives
 Study non-ideal op amp behavior.
 Demonstrate circuit analysis techniques for non-ideal op amps.
At the end of this class you should be able to:
* Analyze op amp circuits assuming finite gain, non-ideal input and output resistance
* Understand the effects of common mode gain and input resistance
* Ideal op amp has following assumptions:
 Infinite common-mode rejection, power supply rejection, open-loop gain,
bandwidth, output voltage range, output current capability
 Zero output resistance, input-bias currents and offset current, input-offset
voltage.
*Various error terms arise in practical operational amplifiers due to non-ideal behavior.
 The non-idealities can be classified as follows:
I.
Small signal errors:
1. Finite gain, input and output resistances
2. Common mode gain and input resistance
II.
Frequency Limitation
1. Finite bandwidth
III.
Large signal limitations:
1. Slew rate
2. Output voltage and currents limits
IV.
DC imperfections
1. Offset voltage
2. Biasing and offset currents
3. PSSR
* In this class we will concentrate on the small signal non-idealities. Whereas, other
limitations will be discussed in the following lectures. To simplify non-ideal analysis,
each imperfection can be considered separately.
* Finite Open-loop Gain
Practical op amps have finite gains with typical values in the range of 80 to 100dB.
Finite gain results in error between the ideal and actual gain.
Example 1:
Investigate the effect of finite op amp gain on the non-inverting amplifier performance.
Solution: Replace the op amp with its simple model as shown in Fig. 1. Then, write KCL
equation at v1 as follows:
is
+
vid
vs
vo
Avid
R2
i
Feedback network
v1
R1
Fig. 1: Circuit for analyzing non-inverting amplifier with finite op amp gain
v1 v1 − vo
+
= 0 (1)
R1
R2
Note that for finite A, v − ≠ v +
But
vo = A(v+ − v − ) = A(v s − v1 )
Solving for v1:
v
v1 = − o + v s (2)
A
Substituting (2) in (1) and solving for vo yielding:
vo
1 + R2 / R1
= Anon −ideal =
1
vs
1 + (1 + R2 / R1 )
A
R2
R1
Following this procedure for analyzing the inverting amplifier, it can be shown
that:
− R2
(3)
Anon −ideal =
1
R1 + ( R1 + R2 )
A
You can see that as A → ∞,
Anon −ideal → Aideal = 1 +
Example 2: Consider an inverting amplifier with R1= 1kΩ, R2=100kΩ and vs=+0.1 V.
Calculate the non-ideal gain, error, and v- for different op amp gains of:
A = 10 3 , A = 10 4 and A = 105 .
Solution: 1. The non-ideal gain can be calculated from equation (3)
A − Anon −ideal
2. The percentage error is defended by Ε = ideal
&times; 100
Aideal
vo
as v+=0
A
Using these results, the following table can be developed:
3. Since vo = A(v + − v− ) ⇒ v − = −
A
10 3
10 4
10 5
Anon-ideal
-90.80V/V
-99.00V/V
-99.90V/V
Ε
9.2%
1.0%
0.1%
v-9.1mV
-1.0mV
-0.1mV
It can be seen that as A increases, the non-ideal gain and v- approach their
ideal values of –100V/V and 0V, respectively. Also, it is evidence that as the
gain of the op amp increases, the error is decreased.
* Finite Differential Input Resistance:
For an ideal op amp, the input resistance is infinity forcing i-=i+=0. However, practical
op amp particularly those based on BJT have finite input resistance (in the range of
view mega ohms). This finite differential input resistance will introduce errors in both
the closed loop gain and the input resistance of amplifiers.
Example 3: Consider the inverting amplifier. Calculate its input resistance if the op amp
has finite input resistance Rid.
Solution:
1. Replace the op amp with its model with finite Rid as shown in Fig. 2.
R2
is
vs
R1
v
vid
+
v+
Rid
vo
Avid
Fig. 2: Inverting amplifier model with finite Rid.
2. By definition:
v x i x R1 + v −
v
=
= R1 + −
(4)
ix
ix
ix
v − vo
v
v
v + Av−
= − + −
But i x = i− + i2 = − + −
Rid
R2
Rid
R2
ix
1
1 1+ A
= =
+
v − R Rid
R2
This means
v
R2
(5)
R = − = Rid
ix
1+ A
Substituting equation (5) in (4) yields:
R2
R
Rin = R1 + Rid
≈ R1 + 2
1+ A
1+ A
Rin =
For large A, Rin= R1
Example 4:
Consider the non-inverting amplifier. Calculate its input resistance if the op amp has
finite input resistance Rid.
Solution:
1. Replace the op amp with its model with finite Rid as shown in Fig. 3.
2. By definition:
vx
ix
Test voltage source vx is applied to input and current ix is calculated.
Rin =
ix =
vx − v
1
R
id
v =i R ≅i R
1 1 1 2 1
Assuming i -&lt;&lt;i2 implies i1 = i2
Fig. 3. The non-inverting Amplifier with finite Rid.
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
+
β
A
∴i x =
=
(1+ Aβ )R
R
id
id
R = R (1+ Aβ )
in id
* Nonzero Output Resistance
For an ideal op amp, the output resistance is zero. However, practical op amps have
small output resistance Ro (about view tens of ohms). This output resistance will
introduce errors in both the closed loop gain and the output resistance of amplifiers.
Example 5:
Consider the non-inverting amplifier. Calculate its input resistance if the op amp has nonzero output resistance Ro.
Solution:
1. Replace the op amp with its model with Ro as shown in Fig. 4.
Fig. 4. The non-inverting Amplifier with finite Ro.
2. Output terminal is driven by test source vx and current ix is calculated to determine
output resistance (all independent sources are turned off). By definition:
v
Rout = x
ix
i x = io + i
io =
2
v x - Av
id
Ro
vx
i =
2 R +R
1 2
Also, vid= -v1 and
R
1 v = βv
v =
x
1 R +R x
1 2
i
1+ Aβ
1
1
∴
= x =
+
vx
Ro
R +R
R
out
1 2
∴Rout =
Ro ⎛
⎞
⎜R + R ⎟
⎜
⎟
1+ Aβ ⎝ 1 2 ⎠
Since, Ro/(1+Ab)&lt;&lt;(R1+R2),
R
Rout ≅ o
1+ Aβ
If A is infinite, Rout=0
* Note that the equivalent circuit when calculating the output resistance with vs=0 is the
same for both inverting and non-inverting amplifiers as shown in Fig. 5.
Fig. 5: Circuits for calculating output resistances of inverting and non-inverting
amplifiers
Finite Common-Mode Rejection Ratio (CMRR)
A real amplifier responds to signal common to both inputs, called common-mode input
voltage. In general,
A,Acm
v1
+
vo
v2
Fig. 6: Common mode gain model
v + v ⎞⎟
vo = A(v − v ) + Acm 1 2 ⎟
1 2
2 ⎟⎟⎠
= A(v ) + Acm(v )
ic
id
⎛
⎜
⎜
⎜⎜
⎝
where:
Acm = common-mode gain
vid = differential-mode input voltage
vic = common-mode input voltage
The two inputs can be also expressed as:
v
v = v + id
1 ic 2
v
v = v − id
2 ic 2
Ideal amplifier has Acm = 0, but for a real amplifier,
⎛
A v ⎞⎟ ⎛
v
⎞
⎜
vo = A⎜⎜ v + cm ic ⎟⎟ = A⎜⎜ v + ic ⎟⎟
A ⎟⎠ ⎜⎝ id CMRR ⎟⎠
⎜ id
⎝
CMRR = A
Acm
Example 6: Find voltage follower gain error due to CMRR
v = vs − vo
id
⎛
(v + v ) ⎞
vo = A⎜⎜ (vs − vo )+ s o ⎟⎟
2CMRR ⎠
⎝
⎞
1
⎟
⎟
vo
⎜
⎟
Av = = ⎝ 2CMRR ⎠
⎞
1
vs 1+ A⎛⎜1−
⎟
⎜
⎟
⎜
2CMRR ⎟⎠
⎝
⎛
A⎜⎜1+
Ideal gain for voltage follower is unity. Hence the gain error is given by:
A
1−
2 CMRR
GE = 1 − A v =
⎛
⎞
1
⎟
1 + A ⎜⎜ 1 −
⎟
⎜
⎟
2
CMRR
⎝
⎠
Since, both A and CMRR are normally &gt;&gt;1,
GE ≅
1
1
−
A CMRR
First term is due to finite amplifier gain, second term shows that CMRR may introduce an
even larger error.
Common-mode Input Resistance:
Fig. 7: Model for common mode input resistance
*When a pure common-mode signal vic is applied to amplifier input (vid =0), total
resistance presented to source is Ric or common-mode input resistance. Normally, Ric &gt;&gt;
Rid. Whereas, for a purely differential-mode input signal, input resistance is Rid.
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