The common mode voltage at the input of a difference amplifier is defined as $$\dfrac{V_1+V_2}{2}$$ and common mode input voltage range is the range of voltages at the input for which the opamp works as it's supposed to. But what is common mode about that?

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    \$\begingroup\$ I'd rephrase your opening statement to "the common mode voltage at the input of a difference amplifier is...". "Common mode" could mean "common mode rejection" and this is different. \$\endgroup\$
    – Andy aka
    Commented Apr 1, 2014 at 21:27

3 Answers 3


For an op-amp in normal operation, V1 ~= V2, so it collapses to just the input voltage.

The range of acceptable input voltages is the common mode input voltage range.


Common mode voltage is distinguised from differential mode voltage.

The differential mode voltage for two inputs is how different they are. That is \$V_1-V_2\$.

The common mode voltage is the part of the voltage that is the same for both, that is, the part that they have in common. As you say, the formula is \$\dfrac{V_1+V_2}{2}\$.

We can make this more mathematical by noticing that with these definitions

\$V_1 = V_c + V_d / 2\$


\$V_2 = V_c - V_d / 2\$.

You can see that the common-mode term is common between these two equations.

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    \$\begingroup\$ What about the case that an opamp is used as a comperator for binary signals? Where high is equal to Vdd and low is equal to ground and the threshold is Vdd/2 and the common mode input range is 1/6*Vcc to 5/6*Vcc. The Vc will be somewhere in the range of 1/4*Vdd to 3/4*Vdd and thus within range, but will it operate properly? The input voltage does exceed the common mode input range. \$\endgroup\$
    – Henk
    Commented Apr 2, 2014 at 9:08
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    \$\begingroup\$ @Henk, you should not use an op-amp with it's common mode input outside the specified range. Maybe the only effect will be reduced bandwidth, or maybe it will cause large currents to flow in the input pins, disturbing the rest of the circuit. Some op-amps also have issues when the common mode input is "large" (more than a few mV). Choose another op-amp for your application or, even better, use a part designed to be a comparator and not an op-amp. \$\endgroup\$
    – The Photon
    Commented Apr 2, 2014 at 16:21

Fundamentally, the term common mode implies that the signal at the two input terminals of a differential amplifier is identical in both magnitude and phase.

When signals V1 and V2 are applied as input we can spilt them into a combination of common mode and differential mode signals in the following manner

V1 = (V1 + V2)/2 + (V1 - V2)/2

V2 = (V1 + V2)/2 + {-(V1 - V2)}/2

thus, (V1 + V2)/2 is the common part and (V1 - V2)/2 is the differential part of the signal.

One of the signals that is always applied in the common mode is the DC Bias. It is this common mode signal that the Input Common Mode Range is associated with. Therefore, the ICMR is the range of input DC (or Common Mode) voltage range for which the circuit is in saturation (or as intended to be).

From the previous computation if we consider V1 and V2 to have both ac and dc components, such as (V0 +/- A sin wt) then V0 would be the DC common mode signal applied to the amplifier.


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