# Differential gain of an op-amp and feedback

In an op-amp, we know from basic electronic courses that

vo = A(v1 - v2)

Assuming large differential gain and finite output, we get that (v1-v2) is very very small, ideally zero.

In a university course I'm studying what's inside of an op-amp. The first stage is a differential stage (2 mosfets with sources in common and a current generator under them). We're studying it by using small signal analysis, I suppose because in future there will be negative feedback and thus a very very small differential input.

Now the question, assuming that the following reasoning is correct: negative feedback reduces the differential input (it becomes a small signal) and the small signal differential gain coincides with the A gain of basic electronic courses (that "A" we use when op-amp is simply a 3 terminal object). But when the circuit "switches on", feedback is still not working, I mean: in the exact moment the circuit switches on, some signal (not necessarily a small signal) arrives to inverting and non-inverting terminals of the op-amp; the op-amp amplifies this signal (the signal is not yet small, so the gain is not yet A) and only now some output signal is brought to the input, thus making the differential input small. Which is the expression of the gain when the signal is not yet small?

• ittc.ku.edu/~jstiles/412/handouts/… May 21, 2018 at 14:03
• Impossible to say because in that fleeting moment that the op-amp is being powered, sub-sections of the op-amp may do strange things or very little and gain may increase or be low. It's totally dependant on the specific op-amp and not a theorizable situation. May 21, 2018 at 14:07
• You're over-thinking this. There is no "exact moment" when a circuit switches on -- there is always some nonzero amount of time during which the power supplies ramp up. The behavior of a circuit during this time very much depends on the details of the design and the speed of the transition. May 21, 2018 at 14:47