Nope, the model does not need to change. You get a NaN result because R1 appears at the denominator of the transfer function. What happens, physically, when you make R1 zero? R2 does not matter any more, and the output takes a value that corresponds to the current given by the difference between V1 and V2 divided by R3. So, there is a solution.

You get a NaN because the computer cannot divide by zero. It is the server response to a division by zero. If you want to find the solution, all you have to do is to make R1 very small, like 0.01 ohms. In that case the solution is Vout = 14.06V and Vocm = 12V (when the rest of the parameters are R3 = 1k, R2 = 20k, R4 = 20k, tolerance = 1%, V1 = 12.10V, V2 = 12V).

I am curious. Why would you make R1 zero in a differential amplifier and achieve such a high imbalance? Do you have a practical application?

Regards, Dana.

Hi Ian,

common mode voltage error IS dependent on gain, but indeed not proportionally to it. If you increase the gain by a factor 10, to common mode voltage error will increase much less than a factor 10.
So if you need a differential amplfier consisting of two gain stages, it is best to make as much gain as possible with the first amplifier to obtain a common mode voltage error on the output of the second stage that is as small as possible.

Ben.

i really enjoyed this common mode voltage error calculator..
BUT maybe you should add that the Vocm can be positive or negative,
ciao
ian