# Transfer Function of Differential opamp with common mode voltage

When we consider the relationship between Vout and Vin of the differential amplifier, we connect one of its inputs each time to the ground to ignore that input voltage in the calculation, in order to calculate the transfer function between another input and Vout. After we get all the transfer functions between each input and Vout, we can sum up all of them to get a complete output response by applying the superposition theorem.

As a result, the transfer function of this opamp:

What I don't get is this transfer function considering common-mode voltage:

***V2 now is the common mode voltage in the third equation.

This article, https://masteringelectronicsdesign.com/the-differential-amplifier-common-mode-error-part-1/, ignores the contribution of another input (the real voltage input 2) to the output. How could this be possible? What exactly is the subtraction of V1 and common voltage, "V1 - V2" in the first term?

• It's not clear to me. But are you thinking about something like this? Commented Dec 12, 2023 at 15:59
• "Common mode" voltage is coefficient of (V1+V2)/2 ... Commented Dec 12, 2023 at 16:58