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I can't understand how to apply common mode and differential mode concepts.

  1. I know that if I have two signals \$v_1\$ and \$v_2\$, I can always write them as:

$$v1 = v_{\text{cm}} + v_{\text{d}}/2$$

$$v2 = v_{\text{cm}} - v_{\text{d}}/2$$

Example: \$v_1=7\$, \$v_2=5\$

then:

$$v_1 = 6+1$$

$$v_2 = 6-1$$

In general, if \$v_1\$ and \$v_2\$ are functions of time, I will have a common mode which is a function of time as well (because it's the average of \$v_1\$ and \$v_2\$).

  1. If I consider the differential stage (two MOSFETs with sources in common and a current generator under them), I replace \$v_1\$ and \$v_2\$ in terms of \$v_{\text{cm}}\$ and \$v_{\text{d}}/2\$.

For small signal analysis the circuit is linear, then I can apply superposition principle and analyse the circuit.

This is what I understood, but now there are my doubts.

1st doubt: the Sedra-Smith book analyses the effect of the common mode, and it says that the common mode can be the result of some noise which is common to both terminals of the differential stage. It seems that, if I ideally would be able to remove noise, the common mode would not exist (but in the previous example, the common mode was 6). Here's the doubt: is common mode something which is added by the external environment or does it exist every time I apply two signals (and in this case it is their average)? If common mode is a common noise, I then would have:

new \$v_1\$ = old \$v_1\$ + noise

new \$v_2\$ = old \$v_2\$ + noise

and then I would always be able to compute a new common mode (average of new v1 and new v2) and a differential mode (new \$v_1\$ - new \$v_2\$).

Then, why should I care of noise? It is sufficient to know how the circuit reacts when a common mode is applied (hopefully it should rejects common mode as much as possible)

2nd doubt: When dealing with differential stage, Sedra-Smith always considers a constant common mode voltage, but in general it is the average of two signals which are functions of time, so the common mode voltage in general should be a function of time as well.

3rd doubt: All the books I read consider as \$v_1\$ and \$v_2\$ two sinusoidal signals with same amplitude but with 180° shift. Then of course common mode voltage is 0 and differential signal is the sinusoidal itself. The question is: why consider only this particular case? \$v_1\$ and \$v_2\$ are, in general, generic functions of time.

4th doubt: Is common mode related, in some way, to the feedback which is usually used with op-amps?

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  • \$\begingroup\$ can you provide a link to the book, you are talking about? \$\endgroup\$
    – Sir Sy
    May 20, 2018 at 13:19
  • \$\begingroup\$ Don't think of common mode so much as an average component of your input. Think of it as an unwanted noise that is common to both input lines. e.g. 60Hz noise. The beauty of differential is it (mostly) gains up only signals that are differential and out of phase and (mostly) rejects signals common to both input lines. \$\endgroup\$
    – pat
    Oct 14, 2021 at 19:49

1 Answer 1

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We analyse signals as common mode and differential when it makes sense to do so, when it simplifies the maths, or simplifies the intuition about the circuit. You are correct that for general signals, the concept is not very useful.

So, when is it useful to use differential and common mode concepts?

When analysing a differential input amplifier, like an op-amp, or a long-tailed pair. Such amplifiers usually have radically different gains to differential signals and to common mode signals.

When analysing noise on a signal that has been transmitted differentially. Balanced differential signals are very common in electronics. They are easy to generate with a transformer, or differential output device. They are used as, in general, interference during transmission gets picked up on both conductors, more or less equally, so results in only a common mode signal, with no differential. If they are now received by an amplifier with a good 'common mode rejection ratio' (the normal way the difference in gains is expressed), then the noise will be reduced with respect to the signal.

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