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?
Electronic – the common mode in common mode input voltage range
common-mode
Related Solutions
It's an output common-mode voltage, not input. So, the output is differential around whatever voltage you apply to the common-mode pin (assuming you have enough headroom for it to still act like an opamp).
For common mode signals you get two wires: one with a signal and one with the inverse of that signal. If you add them you'll get zero, if you'd take the difference you'll get the signal.
What's the cool thing about this. If the two wires (the pair of wires) picks up some noise. Than the noise will be added on both wires. If you add the two signals you only keep the noise.. But if you take the difference you'll get the signal without the noise.
Signal is S (without noise)
Cable a => S
Cable b => -S
The Difference is S - (-S) = 2*S
Signal is S (with noise)
Cable a => S + noise
Cable b => -S + noise
The Difference is (S+noise) - (-S+noise) = S+S+noise-noise = 2S
So the noise will be cancelled.
example image:
When Common mode signals are used they are mostly transferred with twisted pair cables like this:
Related Topic
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Best Answer
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.