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I was reading a text about instrumentation amplifiers. I couldn't find any easy explanation what really common mode voltage means and its importance.

  • 1
    \$\begingroup\$ In an IA, the resistors must be matched as closely as possible to avoid errors, and the phrase common mode refers to when a signal appears that is common to both inputs, essentially when they are tied together. For instance in your diagram, two sources are depicted. The output from the sensor is one source, connected obviously to only 1 terminal of the IA. The second source, the "common mode" voltage source, represents any signals that might be common to both inputs. \$\endgroup\$
    – krb686
    Commented Nov 24, 2013 at 23:15
  • \$\begingroup\$ Taken from wikipedia on common mode rejection ratio:"For example, when measuring the resistance of a thermocouple in a noisy environment, the noise from the environment appears as an offset on both input leads, making it a common-mode voltage signal. The CMRR of the measurement instrument determines the attenuation applied to the offset or noise." \$\endgroup\$
    – krb686
    Commented Nov 24, 2013 at 23:18
  • 1
    \$\begingroup\$ Well common because the signal is literally common (appearing on both) inputs. As far as mode I don't know, because it doesn't refer to a "mode" that the IA operates in or anything like that. Take a look at this picture. m.eet.com/media/1138273/17407-figure_4.pdf It does a good job of explaining it. There are 3 types of "common mode" signals. There's eLC, a common AC noise. eGD, where the ground is floating, or Eos, where the driver is offset a certain voltage. IAs need good CMRR, or common mode rejection ratio, to avoid errors associated with such common signals. \$\endgroup\$
    – krb686
    Commented Nov 24, 2013 at 23:41
  • \$\begingroup\$ Yes if you look at your diagram, all it's doing is amplifying the difference from the sensor outputs. So say this circuit was in a mobile device, where there is no real local ground connection, then the entire sensor and circuit might be floating above real ground, shown by Vcm. \$\endgroup\$
    – krb686
    Commented Nov 25, 2013 at 0:20
  • 1
    \$\begingroup\$ And again, please read the page I already linked on CMRR from wikipedia. You already know what a common mode voltage is, and that explains why they are important. Amplifiers are not perfect. They don't just multiply the difference of the inputs, they multiply the difference of the inputs plus the average times the common mode gain. So in a situation where say the inputs are at 14V and 16V, and the gain is 100, the output won't simply be (16-14)*100 = 200. It will be (16-14)*100 + (16+14)/2 *Acm \$\endgroup\$
    – krb686
    Commented Nov 25, 2013 at 0:51

3 Answers 3


The common mode voltage is a voltage offset that is "common" to both the inverting and noninverting (i.e. "+" and "-") inputs of the instrumentation amp. An instrumentation amplifier is set up as a difference amplifier, so it measures the difference between these two inputs and so rejects any voltage that is common to the two. In other words, if you have two signals v1(t) and v2(t) on the two inputs:

v1(t) = f1(t) + Vcm(t)

v2(t) = f2(t) + Vcm(t)

what the instrumentation amp will measure is:

vo(t) = v1(t) - v2(t) = (f1(t) + Vcm(t)) - (f2(t) + Vcm(t)) = f1(t) - f2(t)

Note that Vcm(t) (the common mode voltage that appears in both input signals) is cancelled out. Also note that this doesn't have to be a DC signal, but can vary with time.

Now why do we care about common mode voltage when selecting a difference amplifier? As other folks have said, there are two key characteristics of the amplifier to consider, the common-mode rejection ratio (CMRR) and the common mode range.

The CMRR is important because the instrumentation amplifier is not an ideal difference amplifier. An ideal difference amplifier would reject 100% of the common mode voltage in the input signals, and would only measure the difference between the two signals. In a real-world instrument amp, this is not the case, and there is a measurable (although typically very very small) amount of the common-mode voltage on the input that gets into the output.

The common-mode range is important, because it limits how far away from ground the measured input signals can be. This is a limit because typically you can't measure signals outside the supply voltages (often referred to as "rails) of the amplifier. There are exceptions to this, but in general the voltage of each input signal must remain within the supply rails of the amplifier. So if you are supplying your amplifier with rails of +/-12V, you may be unable to measure the difference between two signals with a common-mode offset of 15V, even if the difference between the two signals is only 20mV. For example, if your two signals are completely DC and are:

V1 = 15 + 0.010

V2 = 15 - 0.010

Vo = V1 - V2 = 0.020

You would not be able to measure these if your instrumentation amplifier had a common-mode range of +/-12V.

  • 1
    \$\begingroup\$ Your definitions only match the usual way of defining the common mode voltage if \$f_1(t)=-f_2(t)\$. \$\endgroup\$
    – The Photon
    Commented Nov 25, 2013 at 1:21
  • \$\begingroup\$ @Robert Ussery that was the most clear explanation until now i found \$\endgroup\$
    – user16307
    Commented Nov 25, 2013 at 10:16
  • 1
    \$\begingroup\$ We are actually talking of vector space representation. We are after a 2-dimensional space defined by the two input voltages <v1,v2> . Switching to <vcm,f1,f2> base means using a 3-elements base to represent a 2-dimension space and hence involves that rank of <vcm,f1,f2> is 2 anyway and its elements are not linearly indipendent (i.e. orthogonal) anylonger. Really not a great idea, being orthogonal really simplifies a lot calculations. @ThePhoton 's base <vcm,vd> is what has to be used instead: it's 2-dimension and linearly indipendent \$\endgroup\$
    – carloc
    Commented Dec 26, 2017 at 23:17

Say a circuit has two inputs, \$v_1(t)\$ and \$v_2(t)\$, we can mathematically decompose this into a common-mode and differential part, making the two circuits below equivalent:


simulate this circuit – Schematic created using CircuitLab

For these circuits to be equivalent, we need to have

\$V_{cm} = \frac{V_1+V_2}{2}\$

\$V_d = V_1 - V_2\$.

And we call \$V_{cm}\$ the common mode voltage, and we call \$V_d\$ the differential voltage.

Why is it important?

When talking about instrumentation amps we prefer to express the input in terms of common mode and differential because in-amps are designed to have high gain for differential signals and ideally no response to common-mode signals.

That is

\$V_{o-d} = A V_{i-d}\$

where \$V_{o-d}\$ is the differential signal at the output, \$V_{i-d}\$ is the differential signal at the input, and A is the gain of the amplifier.


\$V_{o-cm} = V\$

where V is some voltage not related to the inputs.

  • \$\begingroup\$ What is VD and VD/2 in your figure? \$\endgroup\$
    – user16307
    Commented Nov 25, 2013 at 10:15
  • \$\begingroup\$ VD is the differential signal or voltage. VD/2 is half of VD. \$\endgroup\$
    – The Photon
    Commented Nov 25, 2013 at 17:13
  • \$\begingroup\$ u wrote "When talking about instrumentation amps we prefer to express the input in terms of common mode and differential because in-amps are designed to have high gain for differential signals and ideally no response to common-mode signals.". but i dont understand. in reality there is only one signal at a moment to each input. what do u mean they have no response to cm voltages? \$\endgroup\$
    – user16307
    Commented Dec 2, 2013 at 15:13
  • \$\begingroup\$ "no response to common mode signals" means that if + and - inputs change the same way then the output shouldn't change. For example if both inputs go up 10 mV, the output shouldn't change. If both inputs go down 5 mV, the output shouldn't change \$\endgroup\$
    – The Photon
    Commented Dec 2, 2013 at 16:56
  • \$\begingroup\$ I may be wrong, but your answer looks incorrect and guilty of the same assumption you mentioned under the top answer (that the differential components are equal and opposite). The differential parts can be any signal riding on a voltage common to both inputs and not necessarily Vd / 2. Just like inputs of -5V on 15V and 2V on 15V will yield the same output as inputs of -4V on 15V and 3V on 15V. So I don't get how your second op-amp diagram and first equation are valid. Unless we're only concerned with an assumed Vcm as the real Vcm can only be determined empirically(?). \$\endgroup\$ Commented Mar 31, 2017 at 23:59

common mode voltage is nothing but the offset @ which the diff signal is travelling above a common reference i.e. Ground. So, CM voltage has significance from op amp's operation point of view but it doesn't make any impact on the diff signal being interpreted @ the receiver because receiver just measures the difference between the two signals.

  • \$\begingroup\$ RE: "doesn't make any impact". That's only true if your receiver has perfect common mode rejection. \$\endgroup\$
    – The Photon
    Commented Jun 3, 2015 at 4:42

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