Why Common Mode Voltage is the Average of Inputs in Op-Amp CMRR

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Common mode voltage in op-amps is calculated as the average of the input voltages, reflecting the average level of the signals applied to the non-inverting and inverting terminals. This average is significant because, while ideal op-amps amplify only differential signals, real-world op-amps also amplify common mode signals, albeit to a lesser extent. The concept of common mode voltage is crucial for understanding how variations in the average level can affect output in non-ideal amplifiers. In circuits with pseudo differential signals, the common mode level can shift, impacting performance, which is why fully differential signals are preferred for better consistency. Ultimately, considering the average voltage as a reference level rather than a fixed ground can lead to more accurate circuit designs.
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An ideal op-amp amplifies the differential input signal which means that its CMRR tends to infinity. However in real applications common mode signals are also slightly amplified.

Suppose you apply a 10V voltage to the non-inverting terminal of the op-amp and a 7V voltage to the inverting one... It is said that the common mode voltage is (10+7)/2 = 8.5V

My question is why do we take the average of the inputs when measuring the common mode voltage? In this situation shouldn't 7V be the common signal?
 
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Keep in mind the difference between open loop amplifier and a closed loop amplifier, ie with or without feedback.

Open loop amplifier amplifies differential signals, but by using feedback it can handle single ended signals. Since an opamp is fundamentally a differential circuit you need to consider the difference between the two inputs. As you point out, you consider the open loop gain which is ideally infinite for differential signals and 0 for common mode signals.

For fully differential input signal, each side of the input swings equally but in opposite directions. The average of the two signals is always the same. That is the common mode voltage.

For non ideal amplifiers the output voltage will have a small dependence on where the average level sits. So if you changed the common mode from 8.5 to 6.5 you might get say 1 mV of change in the output.

You maybe thinking along the lines of what would be called a psuedo differential signal. There one input is kept at a constant voltage (7 V) in your example and the other input varies with respect to it. With a psuedo differential signal, the common mode level constantly changes so its contribution to the output will also vary. That's one reason why fully differential signals offer better performance.

For analog circuits you need to avoid thinking of 0V or ground as some universal reference point. In some cases it makes much more sense to consider the average level of the signal as the reference level.

A voltage/potential is always between 2 things and what you call 0 V could be much different than what I call 0 V in another circuit due to a difference in our "grounds" or if the circuits where floating with respect to ground (like a battery powered circuit). With different 0 V levels you can still get the correct amplifier output because the differential signal is the same for you and me.
 
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