1

3. Common Mode Rejection Ratio: Part I

3.1 Introduction In general, an instrumentation amplifier is required to amplify the difference

between two input signals or voltages, V1 and V2 as shown in Fig.1. However, as

discussed, in the case of many transducers there is often a potential at both

inputs present when the input parameter to be measured is zero. This signal is

the same at both inputs and is present regardless of the value of the input

parameter. This signal is defined as the Common-Mode input signal, Vic and

should make no contribution to the output voltage of the amplifier. On the other

hand, when the value of the input parameter is non-zero, the potential at one

input terminal will increase, while that at the other input terminal will decrease

proportionately, giving a difference in the potentials at the two input terminals.

This difference in the two input potentials is defined as the Differential-Input

signal, Vid. Fig. 1 below shows the common-mode voltage as applied centrally to

both input terminals of the amplifier while the differential voltage is considered

split between the two inputs as half the differential potential one each side, Vid

/2. It is applied as a positive sense signal to the non-inverting input of the

amplifier and as a negative sense signal to the inverting input of the amplifier.

Fig. 1 Differential and Common Mode Inputs to a Differential Amplifier

This distribution of common-mode and differential mode signal is

illustrated in Fig, 2. The potentials shown represent steady-state or dc levels, but

could equally be time-varying or dynamic signals. For example if the signal of

interest to be measured were the Electrocardiogram obtained from two

electrodes placed on the surface of the body, then this would form the

differential input signal. The common-mode signal in this scenario is often

composed of mains interference from the electricity supply. This gives rise to an

unwanted signal at the input of the amplifier which must be rejected in favour of

the wanted differential signal which should be amplified. The ability of the

amplifier to discriminate against the common- mode signal and prevent it from

making any contribution to the output voltage of the amplifier is termed

+

diff

amp

-

~

~

-Vid / 2

Vid / 2

V2 V1 V

Vic ~

2

Common Mode Rejection Ideally, the common-mode input signal should produce

no response at the output but in practice it does make some contribution. A

figure of merit used to quantify the extent of an amplifier’s ability to reject or

supress the common-mode input signal is the Common-Mode-Rejection-Ratio or

CMRR of the amplifier.

Fig. 2 The Distribution of Differential and Common Mode Input Voltages

3.2 Common-Mode-Rejection-Ratio From the above definitions and from Fig. 1 and Fig. 2 we have:

2

id

ic

VVV +=1

and 2

id

ic

VVV −=2

so that:

21 VVVid

−= and 2

21 VVV

ic

+=

If the differential amplifier were ideal, it would suppress the common-mode

component of the input signal so that the output has no contribution from the

common mode input and the output voltage would be simply be given by:

idddoVAVVAV =−= )( 21

where Ad is the differential gain of the amplifier as given, for example by Eq. 6

of the previous lecture. However, in practice the amplifier does not fully reject

the common mode component of the input signal and this consequently makes

some contribution to the output. A common-mode gain, Ac can therefore also be

specified so that the output voltage of the amplifier is given in practice as:

icciddoVAVAV +=

Ideally Ac →0 but in practice Ac << Ad but is finite.

Vid /2

Vid /2

V2 Vic

V1

Vid

3

The measure of the ability of the amplifier to reject the common mode input

component, Vic , in favour of the differential component is the Common-Mode-

Rejection Ratio or CMRR of the amplifier. This is defined as:

∞→→= CMRRand0 AideallyA

A CMRR c

c

d ;

Then expressing the common-mode gain in terms of the differential gain we

have:

CMRR

A A d

c =

In this case:

ic

d

iddOV

CMRR

AVA V +=

The left hand term is the wanted output signal while the right hand term is

essentially an error component. The error in the output is then given as the ratio

of these components:

100%xV

V

CMRR

1

V

/CMRRVε

id

ic

id

ic ==

In order to obtain a desired fractional error ε the required CMRR is:

id

ic

V

V

ε

1CMRR =

For example, if Vic=1V and Vid =1mV, then for a 1% error we need:

dBxCMRR 1001010

1

01.0

1 5

3≡==

−

This is substantial but typical of the CMRR required in bio-amplifiers.

4

3.3 Determining Factors of CMRR Consider again the standard 3 op-amp instrumentation amplifier as

shown below:

Fig. 3 The standard 3 op-amp Instrumentation amplifier

From previous work:

211 1 VVVO

1

2

1

2

R

R

R

R−

+=

and

122 1 VVVO

1

2

1

2

R

R

R

R−

+=

for the input stage.

If

2

id

ic

VVV +=1

and 2

id

ic

VVV −=2

then this gives:

VO

R2B

V1

VO2 V2

R1

R2A

A2

+

_

A1

+

_

VO1

A3 _

+

R3A R4A

R4B R3B

5

−−

+

+=

2R

R

2R

R

1

2

1

2 id

ic

id

icO

VV

VVV 11

2R

R

R

R

2R

R

R

R

1

2

1

2

1

2

1

2 id

ic

id

icO

VV

VVV +−

++

+= 111

so that

2R

R

1

2 id

icO

VVV

++= 211

Similarly:

+−

−

+=

2R

R

2R

R

1

2

1

2 id

ic

id

icO

VV

VVV 12

2R

R

R

R

2R

R

R

R

1

2

1

2

1

2

1

2 id

ic

id

icO

VV

VVV −−

+−

+= 112

2R

R

1

2 id

icO

VVV

+−= 211

From this it can be seen that the common-mode signal gets amplified by a factor

of only unity on both sides of the first stage of this amplifier structure. On the

other hand, the differential input component on each side of the amplifier

receives a gain of (1+2R2/R1). This improves the overall common mode rejection

ratio of the amplifier because it boosts the wanted differential signal compared

to the unwanted common-mode signal before being passed to the output stage

which performs the differential input-to-single-ended output. In the final stage

we have:

( )21 OOOVVV −

=

3

4

R

R

6

Substituting gives:

+−−

++

=

2R

R

2R

R

R

R

1

2

1

2

3

4 id

ic

id

icO

VV

VVV 2121

so that finally:

idOVV

+

=

1

2

3

4

R

R

R

R21

This shows that in the final stage, under perfect conditions, the common-

mode signal is rejected completely by the amplifier, while the differential signal

gets a further gain of R4 / R3. This implies that under these ideal conditions this

amplifier structure has infinite common-mode rejection ratio.

In practice, of course, conditions are not ideal and the CMRR of the amplifier is

finite. There are three main factors which act to limit the CMRR attainable:

• mismatch in the source and input impedances of the amplifier

• manufacturing tolerances, in the gain-determining resistors

• finite CMRR of the individual operational amplifiers.

3.4 Source and Input Impedance Mismatch The input stage of any instrumentation amplifier can be modelled as

shown in Fig. 4. Each input has an impedance associated with the corresponding

source. For, example the impedance of the electrodes used to measure an ECG

signal is significant and will also show a variation from one electrode to another,

so that there is a mismatch between the two source impedances. The amplifier

has an impedance between each input terminal and ground, which is referred to

as the common-mode input impedance. This can also be slightly different on

each side. Finally, there is a finite impedance between the two input terminals of

the amplifier, referred to as the differential input impedance. If a common-mode

signal alone is applied to the amplifier inputs, via different source impedances to

different input common-mode impedance on each side, then the signal appearing

at the two input terminals of the amplifier will be slightly different. This means

that the mismatch in the impedances leads to the applied common-mode input

signal being effectively converted into a signal having a differential component

at the input to the amplifier. This differential component will subsequently be

given the full differential gain in passing through the amplifier and will give rise

to an error in the overall output signal which is due to the unwanted common-

mode input signal which has not been completely rejected.

7

Fig. 4 The Equivalent Circuit of an Instrumentation Amplifier Input

The principle of superposition can be used to establish the potential at the

inverting and non-inverting inputs of the amplifier. If the input impedance of

the amplifier itself is taken as ideal, then no current flows into it but is confined

to the network of impedances. Consider the case where the input potential V1 is

applied with V2 = 0V so that the latter input can then be considered grounded.

In this case the potential at the non-inverting input, V+ and that at the inverting

input due to the input, V1 can be found as:

( )

( ) 1

CACBSBDSA

CACBSBD1 V

Z//Z//ZZZ

Z//Z//ZZV

++

+=

+

and

( )

( )

( )[ ] 1

CACBSBDSA

CACBSBD

CBSBD

CBSB1 V

Z//Z//ZZZ

Z//Z//ZZ

Z//ZZ

Z//ZV

++

+

+=

−

Similarly, expressions for the potentials at the inverting and non-inverting

inputs due to the input V2 can be obtained by symmetry as:

( )

( ) 2

CBCASADSB

CBCASAD2 V

Z//Z//ZZZ

Z//Z//ZZV

++

+=

−

and

( )( )

( )[ ] 22////

////

//

//V

ZZZZZ

ZZZZ

ZZZ

ZZV

CBCASADSB

CBCASAD

CASAD

CASA

++

+

+=

+

V1

V2

ZD

ZSA

Amplifier

_

+

ZCA

ZCB

ZSB

ZSA

V1

ZD

V+

ZCA

ZCB ZSB

V-

8

If the common-mode input impedances have a tolerance of ±∆C but are intended

to be purely resistive where possible and the source impedances have a

mismatch of ±∆S then the worst case mismatch arises when:

( )SSSA 1ZZ ∆+=

( )SSSB 1ZZ ∆−=

( )CCCA 1RZ ∆−= ( )CCCB 1RZ ∆+=

Then with:

2

id

ic

VVV +=1

2

id

ic

VVV −=2

When these substitutions are made, the potential at the input of the amplifier

from the network can be obtained in the form:

iddZiCZc VAVAVV +=−−+

The CMRR due to input impedance mismatch can then be defined in the usual

manner and when the appropriate analysis is carried out it can be shown that:

( )SCS

2

CC

SACBSBCA

CBCA

Zc

dZZ

Z2

)1(R

ZRZR

RR

A

ACMRR

∆+∆

∆−=

−==∆

With a small value of ∆C this can be expressed in Decibels as:

( )SC

10

S

C10dBZ

2

1log20

Z

Rlog20CMRR

∆+∆+=

∆

If all of the other factors influencing the overall CMRR of the amplifier can be

taken as ideal then the mismatch in impedance values will determine the CMRR

which can be attained by the instrumentation amplifier in rejecting unwanted

common mode signals.

9

3.5 Design Example: An ECG recording amplifier is used with high impedance electrodes in a

portable application. The smallest component of interest in the ECG signal has

an amplitude of 1mV at the point of pick-up. It is desired to maintain a

minimum signal-to-interference ratio of 20dB when observing this component.

Due to interference in the recording environment mains supply hum at a

frequency of 50Hz is present on the patient’s body at an amplitude which can be

as high as 2Vrms. If the recording electrodes have an impedance 0.5MΩ ± 20% at

50Hz, determine the value of common-mode input impedance required in the

instrumentation amplifier used.

Solution:

The differential input signal level must be taken as Vid = 1mV peak

The common-mode interfering signal is the mains hum at an rms level of 2V

The peak value of the common-mode signal is then Vic = 2 x √2 = 2.83V

The signal-to-interference ratio required at the output of the amplifier is 20dB

which corresponds to an absolute value of 10. This implies that the error in the

signal observed should be less than 1 part in 10 or 1/10 = 0.1. Then:

4

310832

10

832

10

11×=×==

−.

.

.V

V

εCMRR

id

ic

If the CMRR is taken as being limited only by impedance mismatch, then this

implies that:

( )4

2

1083.22

)1(×=

∆+∆

∆−=∆

SCS

CC

Z

Z

RCMRR

The tolerance in the electrode impedance is ±20% so that ∆S = 0.2. If a 5%

tolerance in the resistors is assumed then ∆C = 0.05. This gives:

4

61083.2

25.0105.02×=

×××

CR

So that:

Ω×≈×××=946 1071083.21025.0

CR