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I often see op-amp's modelled with a finite output impedance or resistance as follows:

enter image description here

Let's assume that the op-amp above is implemented with the following structure: enter image description here

From the transistor level diagram, how can the model say that there is an Rout in series? Is it referring to the small-signal output impedance of M5 in parallel with Mb4 rout?

In short, what is the origin of this Rout in the model diagram from a circuit perspective?

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    \$\begingroup\$ M5 and Mb4....? \$\endgroup\$
    – user253751
    Oct 29, 2022 at 0:35

2 Answers 2

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From the transistor level diagram, how can the model say that there is an Rout in series?

That model is a generic one, based on the fact that it is (A) practically impossible to build any amplifier with zero output impedance, and (B) op-amps are generally not designed for low open-loop output impedance.

The entire point of an op-amp is that the things that matter in closed-loop -- being the gain, the gain-bandwidth product, the input offset, etc., are optimized, and everything else is pretty much left to find its own value. Output impedance is one of those things that get left to find its own value.

Is it referring to the small-signal output impedance of M5 in parallel with Mb4 rout?

In that particular schematic, yes. But in practical terms, all op-amps have fairly high output impedances because it generally doesn't matter. Sometimes it does, and then you have to pay attention. But usually it doesn't (or you can work around it) and then you can ignore it.

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  • \$\begingroup\$ Got it. But how come the output impedance appears in series? If you think back to MOSFET small signal models, the output impedance was always to ground in parallel with the VCVS. \$\endgroup\$ Oct 29, 2022 at 11:57
  • \$\begingroup\$ It's a Thevanin equivalent circuit. Whatever the actual circuitry is doing, it'll act like the model under enough circumstances to make the model useful. \$\endgroup\$
    – TimWescott
    Oct 29, 2022 at 14:26
  • \$\begingroup\$ Ah! I get it now. That makes sense. Looking back at it now, whenever we used to calculate the output impedance of a transistor level small-signal circuit, all independent sources were zeroed anyways, that was just the thevenin equivalent output resistance! It's clear now. \$\endgroup\$ Oct 29, 2022 at 14:42
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The series output resistance is the Thevenin equivalent resistance looking into the output, essentially, the drain-source resistance of the output FETs.

If the FETs are modeled correctly, then the op-amp output resistance will also be modeled correctly.

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