There are several answers on the internet and here on how to calculate the output impedance of the opamp with a given feedback.
My intention, though, is to find this value with an open loop configuration because is considered before feedback connection of the output. Like here, the output resistance saw before the Cgs:
Someone says that is written on the datasheet, but also could not. In this latter case we should need the internal circuitry to estimate it. Is that true? If so, how can I find this value from, for example, an LM324? That impedance seems to be not present inside the datasheet.
EDIT: But I may be wrong, in that circuit I need to model the output of the op amp as a voltage generator+output resistance. I was thinking to open loop only because the oputput of the system in which the feedback takes the signal is not the output of the opamp. But if I'm wrong, I can find the output resistance with the normal closed loop way. Would still be right even if the feedback of my circuit is a drop due to a current and not the output voltage directly from the opamp? I think the answer would be 'yes' because the output is still dependent on the output of the opamp, of course.
Best Answer
Open-loop output impedance of a opamp is rarely, if ever specified. However, the maximum current source and sink capability usually is.
You should therefore go with the datasheet and consider the opamp output a current source within the maximum specified current capability. It's impedance should therefore be considered infinite, since that is the impedance of a perfect current source.
In reality, the impedance won't be infinite, but you don't know what it will be. CMOS output opamps probably look mostly resistive, but the current source model may actually be closer for bipolar output opamps.
In any case, don't try to read into the datasheet what it doesn't say. You are only guaranteed what it guarantees. Assume the current source model when designing the circuit, and everything should work. If you design the circuit right, the open-loop output impedance will be irrelevant anyway. In the closed loop case, the output impedance, like the gain, will be governed by the feedback.