Current through virtual ground in op-amps

Question

I am currently trying to understand op-amps. I found a video on youtube about I to V convertor and I am wondering since we have a virtual ground, due to $$V_1 = V_2$$ and then there is ground right below the current source (so we have ground on both sides of the source) why is the current through the resistor $$R_F=I_{in}$$ since I would expect the current from the current source to go from the ground below the current source, through the current source and then to the (virtual) ground since the virtual ground, in my mind, short circuits the input terminals of the op-amp.

Answer 1

"bringing" both inputs to the same voltage does not connect them, so no, they are not shorted. The fact that V1 and V2 are the same is an effect of current flowing out or into the Opamps' output, not the effect of connecting them. The feedback structure just makes the opamp produce exactly the amount of current necessary to make the whole circuit act in a way that brings V1 to the same level as V2.

in my mind, short circuits the input terminals of the op-amp.

So, that's the misconception here!

Answer 2

In addition to Marcus Müller`s explanation - here is another one:

As one of the most important properties of negative feedback, any amplifier is forced to assume an operating (quiescent) point within its linear operating range.

That means: The active part of the whole circuit (here: The "naked" opamp unit) acts with its very large open-loop gain Aol (which, unfortunately, is frequency-dependent).

Example: Aol=100dB (1E5). When, for example, the output shows a value of 1 volt, the voltage difference between both opamp input terminals will be 1V/(1E5)=10µV. As you can see, the voltage difference will never be zero - however, it is so small that this tiny difference voltage can be assumed to be zero (with an error that in most cases can be tolerated).