2/18/2011
Differential and Common Mode Gain lecture
1/8
Differential and
Common-Mode Gain
Recall that in a previous handout, we analyzed this circuit:
R2
R1
v1
-
v2
ideal
R3
vout
+
R4
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/18/2011
Differential and Common Mode Gain lecture
2/8
Common mode and differential mode
We found that the output is related to the inputs as:
⎛
vout = ⎜ 1 +
⎝
⎛ R2 ⎞
R2 ⎞ ⎛ R4 ⎞
v
−
⎟
⎟⎜
⎜ ⎟v
R1 ⎠ ⎝ R3 + R4 ⎠ 2 ⎝ R1 ⎠ 1
This circuit is a weighted difference amplifier, and typically, it is expressed in
terms of its differential gain Ad and common-mode gain Acm.
To understand what these gains mean, we must first define the difference
signal vd (t ) and common-mode signal vcm (t ) of two inputs v1(t ) and v2(t ) .
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/18/2011
Differential and Common Mode Gain lecture
3/8
Definitions
The difference, as we might expect, is defined as:
vd (t ) v2(t ) − v1(t )
whereas the common-mode signal is simply the average of the two inputs:
vcm (t ) v2(t ) + v1(t )
2
Using these definitions, we can express the two input signals as:
v2(t ) = vcm (t ) +
v1(t ) = vcm (t ) −
Jim Stiles
vd (t )
2
vd (t )
2
The Univ. of Kansas
Dept. of EECS
2/18/2011
Differential and Common Mode Gain lecture
4/8
A new way to express the output
Thus, the differential signal vd (t ) and the common-mode signal vcm (t ) provide
another way to completely specify input signals v1 (t ) and v2 (t ) —if you know
vd (t ) and vcm (t ) , you know v1 (t ) and v2 (t ) .
Moreover, we can express the behavior of our differential amplifier in terms of
vd (t ) and vcm (t ) . Inserting these functions into the expression of the amplifier
output vo (t ) , we find:
⎛ R2 ⎞
R2 ⎞ ⎛ R4 ⎞
v
−
⎟ 2 ⎜ ⎟ v1
⎟⎜
+
R
R
R
⎝
⎝ R1 ⎠
1 ⎠ ⎝ 3
4 ⎠
⎛
vd (t ) ⎞ ⎛ R2 ⎞
R ⎞ ⎛ R4 ⎞ ⎛
v
t
= ⎜1 + 2 ⎟ ⎜
+
(
)
⎟ ⎜ cm
⎟−⎜R ⎟
+
R
R
R
2
⎠ ⎝ 1⎠
⎝
1 ⎠ ⎝ 3
4 ⎠ ⎝
⎛
vout = ⎜ 1 +
vd (t ) ⎞
⎛
v
t
−
(
)
⎜ cm
2 ⎟⎠
⎝
R2 ⎞ ⎛ R4
1 ⎡⎛
1
+
⎢⎜
⎟⎜
2 ⎣⎝
R1 ⎠ ⎝ R3 + R4
⎞ ⎛ R2 ⎞ ⎤
⎟ + ⎜ ⎟ ⎥ vd (t )
⎠ ⎝ R1 ⎠ ⎦
⎡⎛
R ⎞ ⎛ R4 ⎞ ⎛ R2 ⎞ ⎤
+ ⎢⎜ 1 + 2 ⎟ ⎜
⎟ − ⎜ ⎟ ⎥ vcm (t )
+
R
R
R
⎝ R1 ⎠ ⎦
1 ⎠ ⎝ 3
4 ⎠
⎣⎝
R R + R2R3 + 2R2R4
R R − R2R3
= 1 4
vd (t ) + 1 4
v (t )
2R1 (R3 + R4 )
R1 (R3 + R4 ) cm
=
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/18/2011
Differential and Common Mode Gain lecture
5/8
A more “common” form
Thus, we now have an expression for the open-circuit output in the form:
vout (t ) = Ad vd (t ) + Acm vcm (t )
where:
Ad differential gain
Acm common-mode gain
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/18/2011
Differential and Common Mode Gain lecture
6/8
Difference amplifiers should have no
common-mode gain
Note that each of these gains are open-circuit voltage gains.
* An ideal differential amplifier has zero common-mode gain (i.e., Acm =0)!
* In other words, the output of an ideal differential amplifier is
independent of the common-mode (i.e., average) of the two input signals.
* We refer to this characteristic as common-mode suppression.
Typically, real differential amplifiers exhibit small, but non-zero common mode
gain.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/18/2011
Differential and Common Mode Gain lecture
7/8
Common-Mode Rejection ratio
The Common-Mode Rejection Ratio (CMRR) is therefore used to indicate the
quality of a differential amplifier:
CMRR = 10 log10
Ad
2
Acm
2
(dB )
Note the CMRR of a good differential amplifier is very large (e.g., > 40 dB).
For our example circuit, we find that the differential and common-mode gain
are:
Ad =
Acm =
Jim Stiles
R1R4 + R2R3 + 2R2R4
2R1 (R3 + R4 )
R1R4 − R2R3
R1 (R3 + R4 )
The Univ. of Kansas
Dept. of EECS
2/18/2011
Differential and Common Mode Gain lecture
8/8
Common-mode gain depends on design
The ratio of these two gains is thus:
⎛ R R + R2R3 + 2R2R4 ⎞ ⎛ R1R4 − R2R3 ⎞
Ad
=⎜ 1 4
⎟⎟ ⎜⎜
⎟⎟
+
Acm ⎜⎝ 2R1 (R3 + R4 )
R
R
R
(
)
4 ⎠
⎠⎝ 1 3
=
−1
1 R1R4 + R2R3 + 2R2R4
R1R4 − R2R3
2
and therefore CMRR is:
CMRR = 10 log10
1 R1R4 + R2R3 + 2R2R4
2
R1R4 − R2R3
2
(dB )
It is evident that for this example, the common-mode gain Acm is minimized, and
thus the CMRR is maximized, when:
R1R4 = R2R3
so that R1R4 − R2R3 = 0 .
Jim Stiles
The Univ. of Kansas
Dept. of EECS