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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.

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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?

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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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    – Transistor
    Jan 20 at 10:14
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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: -

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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.

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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.

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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.

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