I know the rule that states that the negative feedback will keep the difference in potential between the inverting input and the non-inverting input (almost) zero.

Now I just wanted to take a moment and imagine how the circuit does that with the inverting amplifier and I think I get myself very confused.

First I tried to imagine the voltages and current when the Vin rises a little bit above ground potential (as always) this will drive the output terminal towards the (-) supply by some amount.

And here it comes, Now the negative terminal should compensate this rise in Vin but when the Vout is driven negative now the potential difference across Rf increased (I imagine the current now is from Vin to Vout) which means that the V- will also increase and now I am in a loop the inverting terminal is now behaving like a non-inverting one (in my head.)

I don't know what I got wrong or what I didn't notice I think it's very trivial but I am stuck I would appreciate any clarification.

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1 Answer 1


...( I imagine the current now is from Vin to Vout ) which means that the V- will also increase...

OK, current does flow from Vin, through \$R_{in}\$, and then through \$R_F\$ to \$V_{out}\$, because for this opamp, "+"input and "-"input are assumed to consume no current. Right so far...

Why should V- increase?
When Vin is at a positive voltage, and Vout is at a negative voltage...somewhere in between there exists a zero voltage. Guess where this happens?...at the opamp's inverting input pin.

Opamp's inverting pin in this circuit must always sit near zero volts. We call this point a virtual ground, because it mimics the real ground on the opamp's non-inverting input. You might think of this virtual ground as the fulcrum on a see-saw: when one end goes up, the other end goes down.


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