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

[unseensoul]

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?

 

[davidrit]

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.

 

[oswaler]

The common mode voltage is the average of the inputs in op-amp CMRR because it represents the voltage level that is common to both inputs. In the given example, the common mode voltage is 8.5V because it is the voltage that is present on both the non-inverting and inverting terminals of the op-amp. This is important because the op-amp is designed to amplify the difference between the two inputs, not the common voltage level. Therefore, when measuring the common mode voltage, we take the average of the inputs to accurately represent the common voltage level that is present in both inputs. This average value is used in the calculation of the common mode rejection ratio (CMRR) of the op-amp, which is a measure of its ability to reject common mode signals.

In an ideal op-amp, the CMRR tends to infinity, meaning that it amplifies only the differential input signal and completely rejects any common mode signals. However, in real applications, there are slight imperfections and limitations that cause the op-amp to amplify common mode signals to some extent. This is why the common mode voltage is also slightly amplified in real op-amps.

In summary, the common mode voltage is the average of the inputs in op-amp CMRR because it represents the common voltage level present in both inputs and is used in the calculation of the op-amp's ability to reject common mode signals.