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?

  • \$\begingroup\$ ittc.ku.edu/~jstiles/412/handouts/… \$\endgroup\$
    – BeB00
    Commented May 21, 2018 at 14:03
  • 2
    \$\begingroup\$ 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. \$\endgroup\$
    – Andy aka
    Commented May 21, 2018 at 14:07
  • \$\begingroup\$ 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. \$\endgroup\$
    – Dave Tweed
    Commented May 21, 2018 at 14:47

1 Answer 1


A is unimportant. For an actual opamp the A can vary heavily. 9000 vs 11000 so basing your small signal gain on this number is a bad idea.

in an actual circuit A just needs to be sufficiently large so that we can make the valid assumptions in order to calculate the small signal gain based on the negative feedback applied. Here is how you do that. http://www.radio-electronics.com/info/circuits/opamp_basics/operational-amplifier-gain.php

As far as startup conditions, You haven't gotten that far in your class to see the full opamp circuit which looks more like this. https://i.sstatic.net/U9Gr4.png As you can see on startup the output is (close to) high but really its startup value is unimportant. This initial value then propagates through the negative feedback path and presents its input to the opamp. then the differential (v1 - v2) is amplified by A presenting a new output which is then propagated through the feedback path again. Parasitic capacitance keeps the output from changing instantaneously which prevents oscillation from occurring with a linear feedback network, such as a resistor network.

  • \$\begingroup\$ I'm not sure what you mean in that last sentence. Stability has nothing to do with the feedback network being linear. In fact, delays make stability worse. \$\endgroup\$
    – Sven B
    Commented May 25, 2018 at 17:51

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