Differential Amplifier, Differential Mode and Common Mode

Gain of an amplifier is defined as VOUT/VIN. For the special case of a differential amplifier, the input VIN is the difference between its two input terminals, which is equal to (V1-V2) as shown in the following diagram.

 So the gain of this differential amplifier is

Gain = VOUT/(V1-V2). -------------- (1) We can find the expression of VOUT in term of V1 and V2 by using superposition theorem: VOUT = [R3/(R1+R3)] [(R4 + R2)/R2] V1 - [R4/R2] V2 -------------- (2) However, we will not be able to re-arrange this expression in the form of eqn (1) to find the gain of the amplifier (except in the special case of R1 = R2 and R3 = R4).
 
 

Instead of applying superposition theorem with V1 and V2 separately, a better way is to first combined V1 and V2 in a different format, viz. (V1-V2). This is known as the differential mode input - Vd. Associated with this differential mode component will be the common mode input - Vcm., which is equal to the average value of V1 and V2.

Differential mode component : Vd = (V1-V2)

Common mode component : Vcm = (V1+V2)/2

By using these alternate representation of the input components (Vd and Vcm) instead of the original components (V1 and V2), we can re-express eqn (2) in terms of Vd and Vcm as follows.

Vcm = (V1+V2)/2     Þ      2Vcm = V1 + V2 ---- (3)

Since                                     Vd = V1 - V2         ---- (4)

Therefore (3) + (4)                    Þ        V1 = Vcm + Vd/2 ---- (5) and               (3) - (4)                    Þ         V2 = Vcm - Vd/2 ---- (6)
 
 

Substitute eqns (5) & (6) into eqn (2) :

                      VOUT   =   1/2[R3/(R1+R3)] [(R4 + R2)/R2 + R4/R2] Vd +

                    [R3/(R1+R3)] [(R4 + R2)/R2 - R4/R2] Vcm -------------- (7)
 
 
From this expression, we can find the gain of the differential amplifier Gain = VOUT/(V1-V2)

        = VOUT/Vd

        = 1/2[R3/(R1+R3)] [(R4 + R2)/R2 + R4/R2]

This gain is known as the Differential Gain (Ad) as it is based on the differential input alone, i.e.

                     Ad = 1/2[R3/(R1+R3)] [(R4 + R2)/R2 + R4/R2]

As there is another component in VOUT due to the common-mode component Vcm of the input, we define another gain for the differential amplifier, the Common Mode Gain (Acm=VOUT/ Vcm). From eqn (7), this is

Acm = [R3/(R1+R3)] [(R4 + R2)/R2 - R4/R2]
 
 
So although a differential amplifier is supposed to amplify the differential component of the input signals, the common component of the input signals (which is the average value of the two input voltages) will also appear at the output. In practice, this common mode component will cause an error in the measurement of the signals.
 
 

To eliminate the effect of the common mode component, we can either

                    (i)    make the input common mode component equal to zero, i.e. make V2 = -V1
                           such that the average value of the two input signals equal to zero
or

(ii) choose the resistor values of R1 to R4 in such a way that Acm is zero.
 
(i) is usually not possible in practice due to the constraint of the measuring circuitry used to produce V1 and V2 (e.g. the Bridge circuit).

(ii) can be achieved theoretically by making R1 = R2 and R3 = R4. However, this is not feasible in practice due to the tolerance of the resistors used.

Because of this imperfection, a figure of merit used to describe differential amplifier is the Common Mode Rejection Ratio (CMRR), which is defined as

                   CMRR = 20 log (Ad/Acm)

For a perfect differential amplifier, the CMRR is equal to ¥, as Acm is zero.

In practice, a CMRR in excess of 80dB to 100dB will be needed for high accuracy measuring system (e.g. a microcomputer data acquisition system). This is very difficult to achieve if the differential amplifier uses discrete resistors for R1 to R4.