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In a post I made the other day about common mode signal and op-amps, some exchange in the comments led me to note that the definition of \$V_{\text{CM}}\$ seemed rather arbitrary. The canonical definition of \$V_{\text{CM}}\$ is \$:=\frac{V_1+V_2}{2}\$. Conversely, the (rather natural) definition of the differential mode signal is \$V_{\text{D}}:=V_1-V_2\$. Both of the original voltages \$V_1\$ and \$V_2\$ (as measured against a shared reference) can be reconstructed using \$V_{\text{CM}}\$ and \$V_{\text{D}}\$ in the obvious form:

$$\begin{align}&V_1=V_{\text{CM}}+\frac{1}{2}V_{\text{D}} \\&V_2=V_{\text{CM}}-\frac{1}{2}V_{\text{D}} \quad (\dagger_1)\end{align}$$

In the greater context of op-amp analysis, the common mode and differential signals become relevant because (apparently) the op-amp's gain function acts on them differently. The equation that is used to portray this behavior is shown below:

$$\begin{align}V_{\text{output}}=A_{\text{D}}V_{\text{D}}+A_{\text{CM}}V_{\text{CM}}\end{align}$$

A differential amplifier is then (apparently) designed to minimize \$A_{\text{CM}}\$ and...well...modify \$A_{\text{D}}\$ as needed.

So here is my question. Why is it important to model the \$V_{\text{CM}}\$ term as the arithmetic average of the voltage signals? Consider any arbitrary function \$f\$ that takes as input \$V_1\$ and \$V_2\$...and let \$V_{\text{CM}^*}=f(V_1,V_2)\$. Then, so long as \$V_{\text{D}}\neq0\$, the two \$(\dagger_1)\$ equations can be rewritten as:

$$\begin{align}&V_1=V_{\text{CM}^*}+\alpha V_{\text{D}} \\&V_2=V_{\text{CM}^*}+\beta V_{\text{D}} \quad (\dagger_2)\end{align}$$

, where \$\alpha\$ and \$\beta\$ are, themselves, functions that depend on \$V_1\$ and \$V_2\$ (...and because \$V_D \neq 0\$, we can solve for these unknown values for any given non-equal \$V_1,V_2\$ pair using basic algebra).

I think my confusion may stem from the fact that I don't really understand what the PURPOSE of the \$V_{\text{CM}}\$ term actually is. The only thing I can think of is that \$V_{\text{CM}}\$ is specifically designed to capture the op-amp's noise when \$V_{\text{D}}=0\$...because for this scenario, the original definition of the common mode signal is the only definition that allows \$\dagger\$-like equations to be true (i.e. if \$V_1=V_2\$, then \$(\dagger_1)\$ still holds, where as none of the \$(\dagger_2)\$ equations would...as they are undefined for \$V_1=V_2\$). But if this is so, why not just write the \$V_{\text{output}}\$ equation as:

$$\begin{align}V_{\text{output}}=A_{\text{D}}V_{\text{D}}+\varepsilon \end{align}$$, where \$\varepsilon\$ is just some (measurable) static noise term that is always present in the circuit, regardless of voltage differences? Or is this convention rooted in some empirical finding where the aforementioned \$\varepsilon\$ (apparently) linearly scales with the average of the voltages sitting a little bit outside of the amplifiers input terminals?

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  • \$\begingroup\$ Emphasis to the answers so far: CM (small signal) is most often the unwanted component. It's rejection from the output is a key specification for a circuit or component. Likewise a key parameter for analysis. \$\endgroup\$ Commented Nov 15 at 12:58

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There a couple good answers from a EE perspective. I am going to try to give you a more math based answer in case that is what you are looking for. Hang on, my linear algebra is not very good.

I like to think of common-mode (CM) and differential-mode (DM) as a change of basis. As they are typically defined, the CM-DM transformation creates an orthogonal basis. This is useful because when you care about one basis component and not the other, you can separate them easily.

From the first handwritten line for \$\vec{v_{CM}'}\$, you can see that we need \$\alpha + \beta = 0\$ for \$\vec{v_{CM}'}\$ and \$\vec{v_{DM}'}\$ to be orthogonal. This results in \$\vec{v_{CM}'}\$ being equal to the arithmetic mean of \$v_1\$ and \$v_2\$.

enter image description here

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  • \$\begingroup\$ Perfect. Much appreciated. \$\endgroup\$ Commented Nov 14 at 18:09
  • \$\begingroup\$ @tobalt 's comment combined with this answer is just what I needed. Cheers~ \$\endgroup\$ Commented Nov 14 at 18:19
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The purpose of considering the two modes, common and differential, is that these frequently crop up when dealing with practical systems.

Consider a twisted pair differential line being driven to a remote differential receiver. Perhaps there is some difference in their ground potentials, or a strong varying magnetic field is coupling to the loop made up of the pair of wires going out and the ground returning.

This external interference couples to both lines (more or less) equally. The resulting interference current is common mode, it flows equally down both lines, effectively behaving together as a single line. This is why so many systems use differential mode signalling, the receiver is designed to reject the common mode interference, and only respond to the differential signal.

This is just one example. There are many others where it is easier or more logical to think about the mean signal and the differential signal, rather than the signal on each individual conductor. It's still two degrees of freedom, just viewed differently.

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  • \$\begingroup\$ Hi, Neil_UK. Thank you for the response. I understand everything you are saying and have seen it elsewhere, but I am still struggling to see why the arithmetic average of the signals is the superior way of modeling these systems. That is really what this question is about. As I have stated in my prompt, there are seemingly infinitely many other ways one could model the common mode signal. Why choose the arithmetic mean? \$\endgroup\$ Commented Nov 14 at 16:54
  • \$\begingroup\$ @S.C. Arithmetic "mean" is the "most" simple ... \$\endgroup\$ Commented Nov 14 at 17:03
  • \$\begingroup\$ @Antonio51 is that really how this is motivated? That is...shocking. \$\endgroup\$ Commented Nov 14 at 17:07
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    \$\begingroup\$ @S.C. The mean potential of all conductors is the potential that an unshielded cable will appear to have when viewed in a far field approximation. This is often relevant as mentioned in the answer, e.g. in context of EMI analysis. \$\endgroup\$ Commented Nov 14 at 17:58
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    \$\begingroup\$ @tobalt this is exactly what I was looking for. Thank you very much. \$\endgroup\$ Commented Nov 14 at 17:59
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I think the basic question is not HOW the common mode part of a differential signal is definded, but WHY we split such a signal into two parts: (1) a common mode portion and (2) a differential symmetrical portion.

And the answer to this question is related to the design and analysis of a classical transistor-based differential amplifier.

When analyzing the circuit, you can see that the calculation becomes particularly simple and clear in the case where the same input voltage is applied to both inputs, but with different polarities (V1=-V2, fully diffential). In this case, no signal current flows through the common emitter branch and there is no negative feedback (two simple and equal gain stages).

In practical reality, however, this theoretical special case will occur very rarely. Instead, it will be a matter of evaluating the difference between any two voltages.

And here, mathematics shows us that each of these two voltages V1 and V2 can always be represented by the superposition of two components. Or with other words: When both signals are splitted in two components (with one componenet being the mentioned "fully differential" part) the other part MUST be the arithmetic average of both signals. Othewise, the superposition would not give again the difference between V1 and V2.

  • One component (the common-mode component) is contained in both voltages (V1, V2). This part, of course, is the arithmetic average of both signals.

  • the other component (differential mode) is also part of each input signal - however, with a different sign.

Both components can be treated separately in the calculation, and the amplified output signals are then simply superimposed. As mentioned already, the differential parts of both input voltages are easy to handle (amplification without feedback) - and also amplification of the common mode part of the signals can very easily be found (same gain for each stage with twofold signal feedback).

Finally, it should be emphasized once again that the splitting of V1 and V2 into a common-mode and a differential-mode component is primarily for the purpose of simplifying the calculation (in two steps) of the amplification characteristics of the differential amplifier.

(Question: How could we calculate the amplified values connected with the factors "alpha" resp. "beta", as mentioned in the question ?).

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  • \$\begingroup\$ Thank you for the comments. I would like to answer your question at the end of the paragraph, but I am not sure I am parsing it correctly. Could you clarify what you are asking? Are you inquiring how \$A_{\text{CM}}\$ would be measured? \$\endgroup\$ Commented Nov 14 at 18:03
  • \$\begingroup\$ @ S.C. As I have mentioned: Based on the commonly accepted definitions both input voltages (V1,V2) can be split into Vcm and Vdd=(+/-)(1/2)(V1-V2). And for both parts we have well-known gain formulas for designing/analyzing the circuit. Question to you: What are the gain expressions for (alpha)Vd and (beta)Vd based on the given input signals V1 resp. V2 ? \$\endgroup\$ Commented Nov 15 at 10:56
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Common mode voltage is important in many places.

You mentioned op-amps. As a practical example, op-amp input terminals are usually very close together, so differential voltage is almost zero. But both these voltages must be in certain input range for the op-amp to work properly - for a 5V op-amp, the common mode input voltage range might be from 1V to 4V.

For differential buses like RS-485, it has two wires A and B, and for the sake of argument, let's assume when A is 5V then B is 0V, and when A is 0V then B is 5V. Depending on if transmitting a logic zero or logic one, there is always 5V differential between the wires, but the average voltage is exactly 2.5V in both cases. This is useful as the non-idealities and errors in the system can be measured by analyzing the difference and the average separately. At receiver, only 0.2V of the differential signal is needed to recover the data, and both signals can have DC offset (common mode) from -7V to +12V to allow for noise and differences in the grounds between transmitter and receiver. So if the two data wires are a twisted pair in a shielded cable, the differential voltage fluctuates but on average the voltage of wires is equal with 2.5VDC, the wires have capacitive coupling to each other and to shield, but as both wires have opposite signals, the current that capacitively couples to shield is 0 over total length of the wire, and thus coupled voltage is 0. Only if there is some imbalance that logic 0 and 1 have different DC average, then there is common mode voltage, which does couple to shield.

Another good example are phantom powered XLR microphones. The ampmifier feeds in 48V to both differential audio wires, and ground wire is the 0V power return. Audio is then the difference between two differential wires. If there is some magnetic or capacitive noise or disturbance that couples to both pairs of differential audio, the difference between them does not change so no disturbance is heard in audio, but both wires will have common mode noise in reference to 0V ground.

A more larger scale example is when you rectify AC mains voltage into a capacitor like in almost all power supplies - 230VAC rectified into capacitor is 325VDC, but this DC voltage rides on top of the mains AC voltage which is 325Vpp. So again, the differential is 325VDC but common mode is AC with amplitude of 162V. The fact that both wires fluctuate in reference to 0V earth/ground will capacitively couple to nearby structures.

So it is just useful to understand things as average voltage of two wires (common mode) and equal difference of each wire from the average.

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