4. Common Mode Rejection Ratio: Part II
4.1
Mismatch in Gain Determining Resistors
Consider again the standard 3 op-amp instrumentation amplifier shown in
Fig. 1.
+
_
R3A
R4A
A1
V1
VO1
R2A
+
R1
_
A3
R2B
_
VO
A2
R3B
+
V2
R4B
VO2
Fig. 1 Schematic Diagram of the Standard Instrumentation Amplifier
If the resistors are mismatched due to manufacturing tolerances, ΔR, such that:
R2 A  R2 1  R  ,
R3 A  R3 1   R  ,
R4 A  R4 1  R  and
R2 B  R2 1  R  ,
R3B  R3 1  ΔR  ,
R4 B  R4 1  R 
When this is allowed for in the analysis of the gain of the two stages we have for
the first stage:
1
 R 
 R 1   R 
R
R 2 1   R 
VO1  1  2A V1  2A V2  1  2
V

V2
 1
R
R
R
R
1 
1
1
1



 R 
 R 1   R 
R
R 2 1   R 
VO 2  1  2B V2  2B V1  1  2
V

V1
 2
R
R
R
R
1 
1
1
1



Once again if we take the input voltages as:
V1  Vic 
Vid
2
V2  Vic 
and
Vid
2
we then get:
 R 1   R 
Vid  R 2 1   R  
Vid 
VO1  1  2
V


V





ic
 ic
R
2
R
2



1
1


 R 1   R 
Vid  R 2 1   R  
Vid 
VO 2  1  2
V


V





ic
 ic
R
2
R
2



1
1


Expanding gives:
 R 2 1   R 
 R 2 1   R  Vid
VO1  1 
 Vic  1 

R
R
1
1



 2

R 2 1   R 
R 1   R  Vid
Vic  2
R1
R1
2
 R 1   R 
 R 2 1   R  Vid
VO 2  1  2
V

 ic 1 

R
R
1
1



 2

R 2 1   R 
R 1   R  Vid
Vic  2
R1
R1
2
2
so that:

R 1   R   Vid

VO1  Vic  1  2 2
R
1

 2

R 1   R   Vid

VO1  Vic  1  2 2
R
1

 2
Note that the error in the resistor values, ΔR, does not change the gain given to
the common-mode component which remains at unity. When expanded further:

R
VO1  Vic  1  2 2
R1

 Vid
R V

 2 R 2 id
R1 2
 2

R V
R V
VO 2  Vic  1  2 2  id  2 R 2 id
R1  2
R1 2

Note that under these circumstances:

R 
VO1  VO 2  1  2 2 Vid
R1 

This shows that the errors in the resistor values tend to cancel each other out
when they are of similar magnitudes keeping the differential gain close to its
design value.
The gain determining resistors in the output stage can also be mismatched.
Previous analysis of the gain of this stage established the output as:
 R  R 4B  R 4A

R 

VO1   4B VO2
VO   3B
 R 3B
 R 3A  R 4A 
 R 3B 
With the mismatches included as outlined above this becomes:
 R 1  Δ R   R 4 1  Δ R  

 R 4 1  Δ R 
R 4 1  Δ R 
VO   3
V


 O1 
 VO 2
R 3 1  Δ R 

  R 3 1  Δ R   R 4 1  Δ R 
 R 3 1  Δ R  
3
Manipulating gives:
 R 1  Δ R   R 3 1  Δ R   R 4 1  Δ R 
 R 4 1  Δ R 
VO   4
V


 O1 
 VO 2
 R 3 1  Δ R    R 3 1  Δ R   R 4 1  Δ R 
 R 3 1  Δ R  
 R 1  Δ R   R 3  Δ R R 3  R 4  Δ R R 4 
 R 4 1  Δ R 
VO   4
V


 O1 
 VO 2




R
1

Δ
R

Δ
R

R

Δ
R
R
1

Δ
R  3
R 3
4
R
4
R 
 3
 3
  ΔRR3  ΔRR 4  

1  
 R 4 1  Δ R    R 3  R 4    R 3  R 4  
 R 4 1  Δ R  
VO  
V

O
1



 VO 2
 R 3 1  Δ R    R 3  R 4  1   Δ R R 4  Δ R R 3  
 R 3 1  Δ R  
 

R

R
 
3
4
 

 R4  R3 
 
1  Δ R 
R 
 R 3  R 4   V   R 4 1  Δ R   V
VO   4  
O1

 O2
 R 3  1  Δ  R 4  R 3  
 R 3 1  Δ R  

R

R

R

4 
 3

Using the Binomial theorem on the denominator factors and neglecting higher
order terms gives:
 R 
 R  R 3  
 R  R 3 
R 
 1  Δ R  4
 VO1   4 1  Δ R 1  Δ R VO 2
VO   4  1  Δ R  4
 R 3 
 R 3  R 4  
 R 3  R 4 
 R3 
The norm with this stage is to make the nominal values R3 = R4 for a balanced
stage so that the coefficients of VO1 involving ΔR become unity. Then neglecting
second order terms gives:
VO 
R4
R
VO1  4 1  2Δ R VO 2
R3
R3
Substituting for the voltages VO1 and VO2 from above gives:
VO 

R4 
R2
Vic  1  2
R3 
R1


 Vid

R V  R
R

 2 R 2 id   4 1  2Δ R Vic  1  2 2
R1 2  R 3
R1
 2


4
 Vid
R V 

 2 R 2 id 
R1 2 
 2
Expanding gives:
VO 
R4
R 
R
Vic  4 1  2 2
R3
R3 
R1
 2 R
 Vid
R R V
R
R 
R

 2 R 4 2 id  4 Vic  4 1  2 2
R 3 R1 2
R3
R3 
R1
 2
R 4 R 2 Vid
R
R 
R
 2 R 4 Vic  2 R 4 1  2 2
R 3 R1 2
R3
R3 
R1
 Vid

 2
 Vid
R R V

 42R 4 2 id
R 3 R1 2
 2
Cancelling and neglecting higher order terms:
VO 
R4
R3

R V
R 
R V
R
R 
R V
1  2 2  id  4 1  2 2  id  2 R 4 Vic  2 R 4 1  2 2  id
R1  2
R3 
R1  2
R3
R3 
R1  2

VO 
R4
R3

R 
R
R 
R V
1  2 2 Vid  2 R 4 Vic  2 R 4 1  2 2  id
R1 
R3
R3 
R1  2

VO 
This is of the form:
R4
R3

R 
R
1  2 2 1   R Vid  2 R 4 Vic
R1 
R3

Vo  A dR Vid  A cR Vic
where:
A dR 
From this:
CMRR R
R4 
R 
1  2 2 1   R 
R3 
R1 
and
AcR  2 R
R4
Vic
R3
AdR 
R 2  1   R  
R2  1





 1  2
 1  2

AcR 
R 1  2 R
R 1  2 R

This can also be expressed in Decibels as:
CMRR R
dB

R2 
1
  20 log 10
 20 log 10 1  2
R1 
2 R

This shows that the CMRR is directly proportional to the gain of the crosscoupled input stage and the mismatch in the resistors in the output stage.
5
4.2 Finite CMRR of Operational Amplifiers
The gain given to a signal applied to the non-inverting input of an op-amp
is always slightly different than the gain given to a signal applied to the inverting
input, i.e. there is a mismatch in the input-to-output gain between the two
channels of the op-amp. This means that if a common-mode signal is applied to
both input terminal of the op-amp the output is not zero but some small finite
value. This implies that the CMRR of the op-amp itself is finite. The small
output voltage which exists can in effect be referred to the input as a differential
signal by taking it as the input common-mode voltage divided by the CMRR of
the op-amp. If this input voltage is applied as a differential signal it gives the
same error voltage at the op-amp output as is generated by the common-mode
input voltage under the influence of the finite CMRR of the op-amp itself. This is
convenient for the analysis of closed loop negative-feedback configurations as
the effect of the limited CMRR of the op-amps can be accounted for in the
analysis, while treating the op-amps as ideal as well as the other properties of the
configuration as appropriate.
±Vic / CMRR
+
~
diff
amp
-
+
opamp
Vic
~
VO
VO
Fig. 2 The Equivalent Effect of Finite CMRR of the Operational Amplifier
The effect of the limited CMRR of the amplifiers in the standard 3 op-amp
instrumentation amplifier can be assessed by using the above principle as shown
in the circuit configuration of Fig. 3 below. In this case the common-mode
voltage, Vic, is still applied at each input since it generates a resulting voltage at
the output of each op-amp even if the op-amp CMRRs were infinite and this
must be accounted for in establishing the common-mode input voltage to the
second stage of the amplifier.
The input voltages to each side of the amplifier can then be described as:
V1  Vic 
Vid
Vic

2 CMRR1
and
V2  Vic 
Vid
Vic

2 CMRR2
The amplifier circuit can then be analysed in the standard manner.
6
Vid/2 ±Vic/CMRR1
~
~
R3
+
A1
_
V1
R4
~
VO1
R2
+
R1
_
R2
-Vid/2 ±Vic/CMRR2
~
~
±Vic/CMRR3
A3
_
VO
A2
R3
+
~ Vic
R4
VO2
V2
For the first stage the output voltages are obtained as:
 R 
V
Vic   R 2 
V
Vic 
  
Vic  id 

VO1  1  2 Vic  id 
R
2
CMRR
R
2
CMRR
1 
1
2 

 1 
 R 
V
Vic   R 2 
V
Vic 
  
Vic  id 

VO 2  1  2 Vic  id 
2 CMRR2   R1 
2 CMRR1 
 R1 
Allowing for a finite CMRR in the second stage gives the output voltage of this
stage as:
 R 
Vid 

VO   4 VO1  VO 2 
CMRR3 
 R 3 
 R 2 

V
Vic   R 2 
V
Vic 
Vic  id 
  
Vic  id 

1 

2 CMRR1   R 1 
2 CMRR 2 
 R 4  R 1 


VO  

Vid
Vic   R 2 
Vid
Vic 
Vic 
 R 3   R 2 


 


 1  R Vic  2  CMRR    R Vic  2  CMRR   CMRR 
1
2
1
1
3









7
Expanding gives:




R 
 R V
 R  Vic
R2 
R V
R  Vic
Vic  1  2  id  1  2 
  2 Vic   2  id   2 
1 

R1 
R1  2 
R1  CMRR1  R1 

 R1  2  R 1  CMRR2 

 R 


R 
 R V
 R  Vic 
R 
R V
R  Vic
 

VO   4  1  2 Vic  1  2  id  1  2 
  2 Vic   2  id   2 

R1 
R1  2 
R 1  CMRR2  R 1 

 R 1  2  R 1  CMRR1 
 R 3  


Vic


CMRR3




 R 


Vic 
R 
R  Vic
R  Vic
VO   4 1  2 2 Vid  1  2 2 
 1  2 2 


R1 
R 1  CMRR1 
R 1  CMRR 2 CMRR3 

 R 3 
The use of the ± symbol in connection with the op-amp CMRRs is simply to
indicate that the channel imbalance can act in either direction and there is some
combination which will result in the worst case. Therefore, the CMRR values
can in effect be treated as positive magnitudes and their degrading effects on the
overall CMRR of the instrumentation amplifier considered as cumulative. Then
the above expression can be taken as of the form:
Vo  AdVid  AcVic
where:
Ad 
R4 
R 
R 
R  1
1 
1 
1  2 2  and Ac  4 1  2 2 
 


R3 
R1 
R 3 
R 1  CMRR1 CMRR 2  CMRR3 
From this the overall CMRR of the amplifier due to finite CMRRs in the opamps can be obtained as:
R4 
R 
1  2 2 
R3 
R1 
A
CMRROP  d 
Ac R 4 
R 2  1
1 
1 




1

2




R 3 
R 1  CMRR1 CMRR 2  CMRR3 
or:
CMRROP 
1
1
1
1


CMRR1 CMRR 2 
R 
1  2 2 CMRR3
R1 

8
The CMRR can also be expressed in dBs as:
CMRROP dB  20 log 10 CMRROP
The expression above is often inverted and written more conveniently as:
1
1
1
1



CMRROP CMRR1 CMRR 2 
R 
1  2 2 CMRR3
R1 

The analysis above shows that the CMRR of the instrumentation amplifier
is limited by the finite CMRRs of the op-amps, with those of the input stage
being the more important, while the effect of the CMRR of the second stage is
reduced by a factor equal to the gain of the first stage, G1. If the CMRRs of the
op amps have the same magnitude then:
1
1
1
1
1 1




 2  
CMRROP CMRR 1 CMRR1 G CMRR1 
G  CMRR1
Then if G >> 1 then the CMRR approximates to:
1
CMRROP  CMRR 1
2
The overall CMRR of the instrumentation amplifier resulting from the
combined effects of the three influencing factors examined can be obtained by
treating the individual factors as influencing the overall CMRR in a parallel
manner. Then:
1
1
1
1



CMRR CMRR Z CMRR R CMRROP
and
CMRR dB  20 log 10 CMRR
If one CMRR factor is low, then this will result in an overall low CMRR for the
amplifier. The overall CMRR can never be higher than that of the lowest
individual factor.
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