As far as I know, voltage is a defined with respect to a common ground in circuits. So if there is a source and a receiver/amplifier; then they should have a common ground if we want to talk meaningfully about the voltages at any node.

Below illustration showing a differential amplifier used for both single-ended and differential inputs:

enter image description here

I can see in single-ended input case the signal ground of the source is tied to one input of the diff. amplifier. So they share a common ground.

But where is the common ground for the differential input case? Is the source's ground is tied to the ground of the diff. amp? Where is the differential source's ground? Where is the differential amplifier's ground? Can someone draw a more clear example diagram which also shows how and where their grounds are tied?


I tried to make the confusion more clear with this illustration below:

enter image description here

We have on the left a differential signalling circuit and it has its own ground GNDA.

We have on the right a differential amplifier and it has its own ground GNDB. Its rails are supplied with +15/-15V wrt GNDB.

Imagine for a moment in time the signal source is outputting +1V and -1V with respect to its own ground GNDA.

One might say here the differential amplifier will take the difference which is +1 - (-1) = 2V and this 2V will be the voltage with respect to GNDB.

Now lets say we measured GNDA = GNDB + 100V.

What does that mean? It doesn't matter or does it?

And here I tried to simulate the situation with LTspice:

enter image description here

It seems if GNDA - GNDB is greater than some voltages the output corrupts.

What is this about?

  • \$\begingroup\$ voltage is a defined with respect to a common ground in circuits But what if I have two such point where I measure the voltage with respect to common ground. But then I measure between those two points. Ha! Not using common ground. \$\endgroup\$ Commented Sep 19, 2017 at 13:46
  • \$\begingroup\$ Do you mean something like : I'm feeding two signals comming from sources with separate grounds, what will be the outcome? If that's the case try to add the op-amp supply in your drawing. \$\endgroup\$
    – judoka_acl
    Commented Sep 19, 2017 at 13:53
  • \$\begingroup\$ pls see my edit \$\endgroup\$
    – GNZ
    Commented Sep 19, 2017 at 15:19

2 Answers 2


But where is the common ground for the differential input case?

It can be anywhere! Assume that the average voltage of the two input voltages is at common ground, so 0 Volt. Now what is the differential input voltage ? It is:

\$V_{in,diff} = (V_{inp} - 0) - (V_{inn} - 0) = V_{inp}-V_{inn}\$

Now I use a different common ground which is at + 12.3 V, what do we get now:

\$V_{in,diff} = (V_{inp} - 12.3) - (V_{inn} - 12.3) = V_{inp}-V_{inn}\$

See, no difference!

Since it is the difference between \$V_{inp}\$ and \$V_{inp}\$ what counts, whatever the common ground is, it is added and subtracted so the net result is zero.

The common ground is irrelevant. Also they do not need to be tied although in practice they often are. Ethernet (network) cables for example use differential mode signals. The ground does not need to be connected. On both sides of the Ethernet cables small isolation transformers are used to re-define the ground level so that it suits the local circuit.

Also in practice the voltage between inputs and the ground of the circuit will be limited for example by the input voltage range of the amplifier.

  • \$\begingroup\$ I read your answer thanks. But please see my edit I tried to make my confusion more clear. \$\endgroup\$
    – GNZ
    Commented Sep 19, 2017 at 15:18
  • \$\begingroup\$ Now lets say we measured GNDA = GNDB + 100V. What does that mean? It doesn't matter or does it? As I said, it doesn't matter, do the calculation I show. The differential voltage is what is important. Shifting grounds 100 V is irrelevant, it does not change the difference that the opamp circuit sees. Have you read: en.wikipedia.org/wiki/Differential_signaling ? \$\endgroup\$ Commented Sep 19, 2017 at 17:38
  • 1
    \$\begingroup\$ I see what matters is the difference but there is also something called "Common Mode Range Extends to Negative Supply" in the data sheet of LM324: onsemi.com/pub/Collateral/LM324-D.PDF Isnt 100V in my case "common mode voltage"? \$\endgroup\$
    – GNZ
    Commented Sep 19, 2017 at 17:51
  • \$\begingroup\$ See the last sentence in my answer: the input voltage range can of course not be exceeded. I was assuming you still did not understand that the ground levels can be different. Of course obviously a 100 V ground voltage difference will fry an LM324. \$\endgroup\$ Commented Sep 19, 2017 at 18:38
  • \$\begingroup\$ The reason I asked you because I wasnt sure if this i.sstatic.net/UkiR8.png illustration can really be modelled like this i.sstatic.net/jJL7q.png . Because in LTspice thereis a path for current caused by common mode voltage but in reality there is air. I mean if there is no common ground I just dont get where would the 100V common mode voltage's current return, via which path(current needs a loop)? Thanks for answers \$\endgroup\$
    – GNZ
    Commented Sep 19, 2017 at 18:41

If you have gndA=gndB+100V it doesn't matter, the differential amplifier will read the difference between the 2 signals so it will have an input voltage of 2V. Let me explain it with the simplest differential amplifier:

enter image description here

Vo = R2/R1 ⋅(V1− V2 ) if r2/r1 = r4/r3, the only important thing is that Vo must be smaller than the positive (or larger than the negative if Vo is negative) rail of the amplifier or it will saturate.


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