Download 3. Common Mode Rejection Ratio: Part I 3.1 Introduction

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.
Vid / 2
+
diff
amp
-
~
~
Vic
~
-Vid / 2
V2
V1
V
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
1
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.
Vid /2
Vid
Vid /2
V1
Vic
V2
Fig. 2
3.2
The Distribution of Differential and Common Mode Input Voltages
Common-Mode-Rejection-Ratio
From the above definitions and from Fig. 1 and Fig. 2 we have:
V1 = Vic +
Vid
2
V2 = Vic −
and
Vid
2
so that:
Vid = V1 − V2
and
Vic =
V1 + V2
2
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:
Vo = Ad (V1 − V2 ) = AdVid
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:
Vo = AdVid + AcVic
Ideally Ac →0 but in practice Ac << Ad but is finite.
2
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-ModeRejection Ratio or CMRR of the amplifier. This is defined as:
CMRR =
Ad
Ac
; ideally A c → 0 and CMRR → ∞
Then expressing the common-mode gain in terms of the differential gain we
have:
Ac =
Ad
CMRR
In this case:
VO = A d Vid +
Ad
Vic
CMRR
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:
ε=
Vic /CMRR
1
Vic
=
x 100%
Vid
CMRR Vid
In order to obtain a desired fractional error ε the required CMRR is:
CMRR =
1 Vic
ε Vid
For example, if Vic=1V and Vid =1mV, then for a 1% error we need:
CMRR =
1
1
x −3 = 105 ≡ 100dB
0.01 10
This is substantial but typical of the CMRR required in bio-amplifiers.
3
3.3
Determining Factors of CMRR
Consider again the standard 3 op-amp instrumentation amplifier as
shown below:
+
_
R3A
R4A
A1
V1
VO1
R2A
+
R1
_
A3
R2B
_
VO
A2
R4B
R3B
+
VO2
V2
Fig. 3 The standard 3 op-amp Instrumentation amplifier
From previous work:

R 
R
VO1 = 1 + 2 V1 − 2 V2
R1 
R1

and
 R 
R
VO 2 = 1 + 2 V2 − 2 V1
R1 
R1

for the input stage.
If
V1 = Vic +
Vid
2
V2 = Vic −
and
4
Vid
2
then this gives:

R 
V
VO1 = 1 + 2  Vic + id
R 1 
2

VO1
Vid 
 R2 
−
V
−

 ic

2 
 R1 


R2 
R 2  Vid R 2
R V
Vic + 1 +

= 1 +
−
Vic + 2 id
R1 
R1  2 R1
R1 2


so that

R
VO1 = Vic + 1 + 2 2
R1

 Vid

 2
Similarly:

V  R 
V 
R 
VO 2 = 1 + 2  Vic − id  − 2  Vic + id 
R 1 
2  R1 
2 



R 
R V
R
R V
VO 2 = 1 + 2 Vic − 1 + 2  id − 2 Vic − 2 id
R1 
R1  2 R1
R1 2



R V
VO1 = Vic − 1 + 2 2  id
R1  2

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:
R 
VO =  4 (VO1 − VO 2 )
 R3 
5
Substituting gives:
 R 4   


R 2  Vid  
R 2  Vid  





  
VO = 
 Vic + 1 + 2
 − Vic − 1 + 2


R1  2  
R 1  2  


 R 3   
so that finally:
 R 
R 
VO =  4 1 + 2 2 Vid
R1 
 R 3 
This shows that in the final stage, under perfect conditions, the commonmode 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 commonmode input signal which has not been completely rejected.
6
ZCA
ZSA
V+
+
ZSA
ZD Amplifier
_
V1
ZD
V1
ZSB
ZCA
ZCB
ZSB
ZCB
V2
Fig. 4
V-
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:
V1+ =
(Z D + ZSB // ZCB ) // ZCA V
1
ZSA + (Z D + ZSB // Z CB ) // Z CA
and
ࢂ
૚
(Z D + ZSB // ZCB ) // ZCA V
ZSB // Z CB
V1− =
(Z D + ZSB // ZCB ) [ZSA + (Z D + ZSB // ZCB ) // ZCA ] 1
Similarly, expressions for the potentials at the inverting and non-inverting
inputs due to the input V2 can be obtained by symmetry as:
V2− =
(Z D + ZSA // ZCA ) // ZCB V
2
ZSB + (Z D + ZSA // ZCA ) // ZCB
and
V2+ =
Z SA // Z CA
(Z D + Z SA // Z CA ) // Z CB V
(Z D + Z SA // Z CA ) [Z SB + (Z D + Z SA // Z CA ) // Z CB ] 2
7
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:
ZSA = ZS (1 + ∆ S )
ZSB = ZS (1 − ∆ S )
Z CA = R C (1 − ∆ C )
Z CB = R C (1 + ∆ C )
Then with:
V1 = Vic +
Vid
2
V2 = Vic −
Vid
2
When these substitutions are made, the potential at the input of the amplifier
from the network can be obtained in the form:
V + − V − = A c Z ViC + A dZ Vid
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:
CMRR ∆Z
A dZ
R CA R CB
R C (1 − ∆2C )
=
=
=
A c Z R CA Z SB − R CB Z SA 2 Z S (∆ C + ∆ S )
With a small value of ∆C this can be expressed in Decibels as:
CMRR ∆Z
dB
= 20 log 10
RC
1
+ 20 log 10
ZS
2(∆ C + ∆ S )
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.
8
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:
CMRR =
1 Vic
1 2.83
= × −3 = 2.83 ×104
ε Vid 0.1 10
If the CMRR is taken as being limited only by impedance mismatch, then this
implies that:
CMRR ∆Z
RC (1 − ∆2C )
=
= 2.83 × 10 4
2 Z S (∆ C + ∆ S )
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:
RC
4
=
2
.
83
×
10
2 × 0.5 × 10 6 × 0.25
So that:
RC = 0.25 × 10 6 × 2.83 × 10 4 ≈ 7 × 10 9 Ω
9