# What is meant by this statement about op-amps?

I am trying to understand the inverting negative feedback how it stabilizes the op-amp circuit. As the book is explaining this I came across the highlighted sentence:

Thus the voltage vo will not depend on the value of the current that might be supplied to a load impedance connected between terminal 3 and ground.

If I understand correctly what he is trying to say, is that if I have the below circuit, then, io = always regardless of the output resistance. Why?

The book: Microelectronic Circuits SEVENTH EDITION Adel S. Sedra University of Waterloo Kenneth C. Smith University of Toronto ISBN 978–0–19–933913–6

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The op-amp produces an output voltage ($$\v_O\$$) that forces the inverting-input voltage to virtually equal the non-inverting-input voltage. It does this using negative feedback via R2 and R1: -

This happens because the open-loop gain of the op-amp is very, very high; sometimes in excess of 1 million. So, if you think about that sort of gain magnitude, for any reasonable voltage on the output, the negative feedback produces a voltage difference at the inputs to be in the realm of microvolts i.e. they become virtually the same.

But, a normal op-amp can only usually supply 10 to 20 mA of output current so, if you try and draw too much current, the op-amp will fail to make the input voltage difference virtually zero.

It means that the output voltage of the op-amp will not depend on what is connected to it.

In other words, on a perfect op-amp, it means the op-amp will supply whatever current is required to keep the output voltage to what it should be.

If you connect a 1kohm "load" or a 10kohm "load" the output voltage will be the same regardless, the op-amp will supply more current to the 1k load to keep the output voltage at the same level as if it were a 10k load.

That said, as long as the output current is within the limits of the op-amp and the supplies.

In the model shown, output voltage depends on the voltage generated across the internal resistance R, due to the gm of the two input ports.

There is no output resistance show - the output is modelled as an ideal voltage source. Thus, this circuit can (in theory) drive a load of 0.1R (in reality it probably cannot).

A more realistic model would of course have to somehow give the opamp an output resistance, because ideal voltage sources do not exist in the real world.