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This question already has an answer here:

Let's consider any circuit which contains a differential pair, for example a telescopic cascode single stage OTA:

enter image description here

In every circuit like this one, my professor and every book compute the small signal differential gain by assuming a perfect differential input couple of signals. Thus, as far as regards our example, we have:

enter image description here

and the small signal differential gain, exploiting the symmetry (i.e. the ac grounds on the nodes on the axis of symmetry which allow me to consider just half circuit), becomes:

enter image description here

This procedure is general: every time we have a circuit with a differential input pair, under the hypothesis of differential input signals, we can simplify the computation of the differential gain by exploiting symmetry (half circuit).

The question now is: why do we always compute the differential gain in the particular case of differential signals as input? In the most general case, indeed, we don't have differential signals. Consider for example this basic circuit:

enter image description here

In every book I've read, we pretend that the gain of the op-amp is the differential gain calculated before (i.e. assuming differential input signals). In other words: in order to use the previous expression of the differential gain, we should have the following situation in which the inverting and the non-inverting terminals of the op-amp receive differential signals (in red in the following picture):

enter image description here

But actually those red signals are not present: indeed the non-inverting terminal is fixed to ground, whereas (assuming large differential gain) the inverting terminal is almost zero.

In conclusion: why do we always calculate, in circuits containing a differential pair, the small signal differential gain assuming perfect differential signals as input? Why do we wrongly assume that the gain found in this way is also the differential gain of op-amps used in negative feedback?

Thank you

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marked as duplicate by Elliot Alderson, The Photon, Voltage Spike, Andy aka operational-amplifier Jul 27 at 16:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ From a practical point of view, a factor of 2 doesn't matter because op-amp open-loop gain has such wide tolerance. \$\endgroup\$ – glen_geek Jul 27 at 13:15
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    \$\begingroup\$ why do we rightly assume the gain we have calculated is right? Because that's the way we defined it. Usually the diff gain exceeds common mode gain by so many orders of magnitude, the single ended gain is equal to the diff gain to a very high tolerance. \$\endgroup\$ – Neil_UK Jul 27 at 13:22
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    \$\begingroup\$ How many times are you going to ask the same question? Looking them over, I think your fundamental problem is the meaning of the word differential. This word does not imply equal but opposite sign. \$\endgroup\$ – Elliot Alderson Jul 27 at 13:29
  • \$\begingroup\$ Every time I improve the question, getting more and more precise thanks to the answers users give to me. In every case, you did not answer to my question. You just said that signals at the input terminals of the op amp are not differential. Ok. Why then do we use an expression found under this hypothesis? Remember that I'm here to learn, I could neglect all these doubts and successfully do my exam, but I don't like it. \$\endgroup\$ – Stefanino Jul 27 at 13:37
  • \$\begingroup\$ You must not be reading the answers your getting. In this one I made two main points: 1. Any two input signals can be written as a common-mode part plus a differential part. 2. In an op-amp, the response to the differential part is much stronger, so the response to the differential part determines the overall response. This is all the answer you're going to get to "why do we use an expression for differential gain to determine the overall gain?" \$\endgroup\$ – The Photon Jul 27 at 14:27