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. enter image description here

As a result, the transfer function of this opamp:

enter image description here

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

enter image description here

***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?

  • \$\begingroup\$ It's not clear to me. But are you thinking about something like this? \$\endgroup\$ Commented Dec 12, 2023 at 15:59
  • \$\begingroup\$ "Common mode" voltage is coefficient of (V1+V2)/2 ... \$\endgroup\$
    – Antonio51
    Commented Dec 12, 2023 at 16:58

1 Answer 1


If you plug my bottom four equations into my top equation and then multiply out you should find that four of the terms cancel out leaving your original (top) equation.


  • \$\begingroup\$ Posts consisting of mostly images are not very flexible, have limited readability, do not format according to site style, etc.. Have you taken a look at Markdown and LaTeX formatting? \$\endgroup\$ Commented Dec 12, 2023 at 23:46
  • \$\begingroup\$ @user350400 Where do the top and second formulas come from? \$\endgroup\$
    – TSLee
    Commented Dec 14, 2023 at 16:49
  • \$\begingroup\$ @TSLee Can't remember where I first saw them, just something I've learned along the way. \$\endgroup\$
    – user350400
    Commented Dec 14, 2023 at 17:15

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