0
\$\begingroup\$

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

1) I know that if I have two signals v1 and v2, I can always write them as:

v1 = vcm + vd/2

v2 = vcm - vd/2

Example: v1=7 v2=5

then:

v1 = 6+1

v2 = 6-1

In general, if v1 and v2 are functions of time, I will have a common mode which is a function of time as well (because it's the average of v1 and v2)

2) If I consider the differential stage (two mosfets with sources in common and a current generator under them), I replace v1 and v2 in terms of vcm and vd/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 idealy 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 avarage)? If common mode is a common noise, I then would have:

new v1 = old v1 + noise

new v2 = old v2 + 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 v1 - new v2).

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 v1 and v2 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? v1 and v2 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?

Thank you

\$\endgroup\$
  • \$\begingroup\$ can you provide a link to the book, you are talking about? \$\endgroup\$ – Sir Sy May 20 '18 at 13:19
0
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.