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enter image description here

In the above illustration the common mode noise is hitting to both + and - wires.

Here is my understanding of the "common mode noise":

In a Single-ended system: the - wire is GND and + wire is carrying the actual signal, so the noise will only affect the + wire but not the - wire since it is GND. Noise then will appear at the output (difference of the + and - wires).

In a Differential-ended system(differential signalling):* the - wire and the + wire both are carrying signals which are mirrors of each other, so the same noise will simultaneously be added to the + wire and the - wire. Noise will not appear at the output.

If my understanding is correct; does that mean if one talks about "common mode noise", he is talking about a single-ended system?

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    \$\begingroup\$ Differential systems reduce the effects of common mode noise, but don't remove it completely. Differential systems usually have some rating that tells you how much common mode noise will be reduced, and how much common mode noise they can handle. \$\endgroup\$ – JRE Nov 27 '16 at 10:13
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If my understanding is correct; does that mean if one talks about "common mode noise", he is talking about a single-ended system?

A differential receiver has to "manage" the common mode noise and although a perfect receiver will certainly achieve this, the real world receivers don't. For instance (just an example), an op-amp may be specified as having 100 dB common mode rejection but the fine detail in the data sheet tells you that this might only be at 100 Hz and at (say) 10 kHz, this has degraded to 60 dB and maybe 40 dB at 100 kHz etc..

So no, common mode noise is a big issue for both types of system.

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  • \$\begingroup\$ I think you mean differential ended signalling is more immune to noise comparing to single ended but it is still not perfect. Some op-amp inputs are not mirrored(no differential signalling). Are you talking about op-amps which have differential inputs? They cannot subtract Vin+ and Vin- perfectly and that quality depends on amplitude of Vin+ and Vin- wrt GND and how fast the signals are varying? \$\endgroup\$ – floppy380 Nov 27 '16 at 10:42
  • \$\begingroup\$ I'm using op-amps as an example of a chip that suffers from common mode problems as frequency rises. The same will be true for devices intended as differential receivers of analogue or digital signals. \$\endgroup\$ – Andy aka Nov 27 '16 at 10:45
  • \$\begingroup\$ I just wondered what it means "differential receiver". Does it mean they are designed to used for mirrored inputs or does it mean they subtract the inputs? \$\endgroup\$ – floppy380 Nov 27 '16 at 10:48
  • \$\begingroup\$ @doncarlos all differential receivers subtract inputs and all have to have "mirrored" inputs so that the subtraction becomes signal addition and common-mode cancellation. \$\endgroup\$ – Andy aka Nov 27 '16 at 11:20
  • \$\begingroup\$ Worth noting: deferentially signaled devices run into common mode noise issues only because, once we realized they can drastically reduce CM noise, we just pushed them further until the noise became a problem again! =) \$\endgroup\$ – Cort Ammon Nov 27 '16 at 18:45
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No -- common-mode noise can affect a differential signal if the source impedances aren't balanced

Compare the two circuits below:

schematic

simulate this circuit – Schematic created using CircuitLab

schematic

simulate this circuit

In both circuits, we have a 1.1kHz, 1V differential mode signal superimposed on a 90Hz, 500V (!) common mode signal, and an ideal (infinite CMRR) subtractor as the receiver, with its input impedance set to 1MΩ by the resistors at the bottom of the bridge. However, in the first circuit, the source impedances are perfectly balanced, while in the second circuit, they are imbalanced by a rather extreme ratio of 10 to 1. The simulations of the first circuit show a near-perfect 1.1kHz sine wave at the output of the subtractor, while on the second circuit, the subtractor output contains several dozen mV of 90Hz. Oops!

The only fix to this (other than fixing the bad signal source) is to increase the input impedance that common-mode signals "see", as in the circuit below.

schematic

simulate this circuit

Now, the common mode signal has been reduced to sub-millivolt levels, but at the cost of having to greatly increase the input impedance of the receiver, which becomes impractical beyond a certain point.

In practical circuits, transformers can achieve high common mode impedances, but have their own drawbacks as they're bulky, can pick up magnetic noise, and are hard to achieve wide bandwidth with; in the solid-state world, advanced bootstrapping techniques (look up the "Whitlock bootstrap" if you want to know more, but keep in mind it's patented) are used.

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