What does common mode mean? - Page 1
Forum section: electronics / beginners

author: z9Rct4RJ2lutJ64XxSqH
author_rank: Newbie
timestamp: October 25, 2015, 09:52:05 pm
content: What does common mode mean in simple terms? I heard the words "common mode" thrown around, but can't seem to figure out what it means (google searches haven't helped). Can you help out someone that may have a degree with the word "mechanical" in it? thanks.

author: max_torque
author_rank: Super Contributor
timestamp: October 25, 2015, 10:11:27 pm
content: "common mode" is used when refering to a signal that is present on two conductors equally.So common mode noise is noise that is present on say both +ve and -ve power wires of a device, or common mode voltage is a voltage equally applied to those two wires.For example, take a 9v battery. It's positive terminal is 9v (nominally) higher than it's negative terminal, but if say we connected the -ve terminal to say the mains electricity, both terminals would have that voltage as "common mode voltage" on them. (when referenced to mains earth)for example, with a differential signal:

author: SL4P
author_rank: Super Contributor
timestamp: October 25, 2015, 10:40:34 pm
content: If that is hard to 'get', read about balanced or differential signalling e.g. RS422/485 or in a more specific case LVDS.The objective is to increase noise immunity of the transmitted information over the link.

author: z9Rct4RJ2lutJ64XxSqH
author_rank: Newbie
timestamp: October 26, 2015, 12:19:23 am
content: Is balanced audio based on a similar concept?https://en.m.wikipedia.org/wiki/Balanced_audio

author: SL4P
author_rank: Super Contributor
timestamp: October 26, 2015, 12:36:32 am
content: Yep... exactly.Good pick up!Remember for balanced lines / common mode noise rejection, that neither leg should be tied to ground or an absolute voltage rail.

author: TimFox
author_rank: Super Contributor
timestamp: October 26, 2015, 04:28:57 pm
content: If you have two voltages, A and B, with respect to a defined ground or common point, then by simple algebra you can defineV(differential) = A - BV(common mode) = (A + B)/2.The inverse,A = Vcm + (Vdiff)/2B = Vcm - (Vdiff)/2This is true for any arbitrary voltages A and B, including time-varying voltages.For many amplifiers, the output response to Vdiff is much stronger than to Vcm.This is called "common-mode rejection" and is very useful.See Wikipedia:https://en.wikipedia.org/wiki/Common-mode_rejection_ratio

author: dom0
author_rank: Super Contributor
timestamp: October 26, 2015, 07:40:34 pm
content: And while we are talking about differential signaling we can get things straight at the beginning: A signal is not made differential by voltage levels alone, what mattersmostis the impedance to ground/common. Tada. If you want to know why, draw up a differential (op-amp-) amplifier and calculate differential and common mode gains for different source impedances on each input.

author: yashrk
author_rank: Frequent Contributor
timestamp: October 26, 2015, 08:36:23 pm
content: Basically think of it as same thing (signal) is provided to two different pins is called common mode (signal)

author: SL4P
author_rank: Super Contributor
timestamp: October 26, 2015, 09:46:44 pm
content: ^^^^^^ two pins with same signal that is phase inverted with respect to each other!

author: dom0
author_rank: Super Contributor
timestamp: October 26, 2015, 09:49:54 pm
content: Quote from: SL4P on October 26, 2015, 09:46:44 pmThat's differential, not common, mode.

author: SL4P
author_rank: Super Contributor
timestamp: October 27, 2015, 02:19:38 am
content: Quote from: dom0 on October 26, 2015, 09:49:54 pmCommon mode as a concept is largely irrelevant in any context other than for differential signals...?There is no means of reducing /cancelling common mode noise without differential signals to subtract.

author: Arjan Emm
author_rank: Regular Contributor
timestamp: October 27, 2015, 05:32:01 pm
content: Quote from: SL4P on October 27, 2015, 02:19:38 amLike a common mode filter. Basically a lossy 1:1 transformer with the primary in the phase/+ and the secondary in the return/0/-/ground. And connected such, that the differential signals cancel each other out magnetically. so no impedance. The common mode signals both magnetize the core and see an inductive impedance and so blocking them to a certain degree. Present in many devices you plug in the mains.

author: fivefish
author_rank: Frequent Contributor
timestamp: October 27, 2015, 11:42:43 pm
content: not phase inverted.... polarity inverted. (phase has to do with time delay)

author: SL4P
author_rank: Super Contributor
timestamp: October 28, 2015, 07:26:50 am
content: <pedantic>What are two signals that are 180-degrees out of phase with each other?Inverted.If we take that all the way, then the term inversion is 'correct' until one leg is notperfectlysymmetrical with the other - which will make the CM receiver work harder! (one wire slightly longer than the other?)Being ~180 out of phase doesn't make any assumption about absolute amplitude of the signals - just theirrelativeamplitude to each other.(One could be 0+3V, while the other 0-1V)...If the phase offset is +/- some degrees - then they are not 'inverted', but the differential receiver still has a chance at recovering the difference until the amplitude is below the 'slicing' level of the receiver.Nominally square signalling will have a far better chance of recovery than a signal with slow edges, due to the overlap vs the reduced difference window of the receiver.</pedantic>

author: dom0
author_rank: Super Contributor
timestamp: October 28, 2015, 08:05:36 am
content: I think there is some confusion about phase inversion vs. phase shift (of 180°).For a certain class of signals - that is symmetric periodic signals -they are the same. For almost every other kind of signalthey are not; the reason why a inversion in e.g. a transfer function (like a bode plot) can be observed by a 180° phase shift is simply that in frequency domain analysis we always concern ourselves with a number of spot frequencies, i.e. sine stimuli. Sines are both symmetrical and periodic, thus phase shift and inversion become indistinguishable to them.

author: SL4P
author_rank: Super Contributor
timestamp: October 28, 2015, 03:30:10 pm
content: Well phrased.In my last comment, perhaps I could have simplified the statements by saying the differential transmitter is simply generating a 'complementary' signal pair for use by a receiver after any common mode 'noise' has been discounted.