i have a question and i hope is not that strange. i spend some time asking myself how people came out with the mathematical model for differential signals and why is this better in this way?!

to be clear i'm asking about this model: vcm = (v1+v2)/2 and vdiff = v1 - v2!

i can understand why the equation for vdiff, that is obvious, but what about vcm?

in the simplest cases vcm is an offset voltage that is the same for both signals v1 and v2. and now i get why is this common and how the equation for vcm works. but what about signals that have on their common mode also noise and they also have different offset voltage?

how the equation works now and why is this still useful?

  • \$\begingroup\$ Does this equation set better describe the effects of line-to-line imbalances? \$\endgroup\$ Nov 21, 2018 at 13:07

2 Answers 2


I will answer your question using - as an example - the classical differential amplifier (with an ohmic resistor RE in the common emitter path).

This circuit reacts upon both input voltages V1 and V2 at both input nodes. That means: We are interested in the output voltage Vo at one or at both signal voltages at the output node(s).

Of course, for calculating the corresponding gains we could start from the beginning (without using well-known formulas) on the basis of Kirchhoffs laws (KCL and KVL). However, it is much more simple and elegant to split the calculation into two parts which are very simple to treat.

For this purpose the two arbitrary input signal voltages V1 and V2 are split into the following signals:

Arithm. mean value: Vcm=(V1+V2)/2

Differential (symmetric): Vdd=(V1-V2)/2=Vd/2.

As we can see: V1=Vcm+Vdd and V2=Vcm-Vdd.

(Because Vcm appears in both parts, it is a common mode value. As a general rule: Two arbitrary voltages - positiv or negative - can always be split into a common mode part Vcm and symmetric diff. part Vdd. The amplified signals of both parts can easily be found and superimposed at the output).

Now - it is a simple task to find the gain for both parts and to calcuate the output voltages using superposition.

(1) Common mode gain: Gcm=Vocm/Vcm=-gmRc/(1+2gmRE)

(2RE because for common mode signals we have two equal emitter currents through RE)

(2) Symm. diff. gain: Gdd=Vodd/Vdd=(+-)gmRc

(We have used the formula for an emitter stage without RE-feedback, because both emitter signal currents have opposite signs and cancel each other).

Using these gain expressions, we can find the output signals (with |Gd|=gmRc/2):

(1) Vo1=-Vcm*|Gcm|-Vdd*|Gdd|=-Vcm*|Gcm|-Vd*|Gd|

(2) Vo2=-Vcm*|Gcm|+Vdd*|Gdd|=-Vcm*|Gcm|+Vd*|Gd|

Comment 1: Please note, that we have a common mode input voltage Vcm also in case one of the input is grounded !)

Comment 2: Since we want that the diff. amplifiers reacts upon differential signals Vd only, we want to have a very small common mode output. Therefore, we select a very large resistor RE or - as a better solution - we are using the large dynamic output resistance of a third transistor in the common leg of the circuit.

  • \$\begingroup\$ Just adding the fact that common mode signals can be injected on inputs terminals by "electromagnetic" influence. See how electrocardiography or electroenphalography is done. \$\endgroup\$
    – user288518
    Jun 15, 2021 at 8:14

what about signals that have on their common mode also noise There's no actual "line" that carries the common mode voltage \$V_{cm}\$. That is just the "theoretical" middle voltage between the differential signal pair.

Being a theoretical value on the transmitter side, no, it doesn't have noise.

and they also have different offset voltage?

Read your equation twice: if \$V_1\$ and \$V_2\$ have an offset, no matter which one, you just shift \$V_{cm}\$.


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