Could you physically clarify why we use virtual ground in op amps?
I have seen the argument that since A(open loop gain) tends to infinity, we need the difference in inputs to tend to zero(to maintain a finite output.) Now, this is confusing because aren't the inputs under our control? What would happen if I gave one very large positive and one very large negative input to the op amp?
Electronic – Operational amplifiers and virtual ground
operational-amplifiervirtual-ground
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Best Answer
The term "virtual earth" is a reflection of the fact that if one of the inputs of an op amp, say \$V_-\$ is connected to the zero volt (earth) line then if the op amp is to work in the non-saturation region the potential difference between the two inputs is going to be very much less than the supply voltage.
So the voltage of the other input \$V_+\$ is going to be very close to zero volts - virtually at earth potential.
Say that an op amp has an open loop gain \$A=10^5\$ and the supply voltage is \$\pm 10\,\rm V\$ then the input-output characteristic of the op amp might look something like this.
You will see from that characteristic that as long as you restrict the potential difference between \$V_+\$ and \$V_-\$ to less that about \$\pm 10^{-4}\,\rm V\$ the output voltage will be proportional to the potential difference across the two inputs.
\$V_{\rm out} = 10^5(V_+ - V_-)\$
If you wanted to you could have a much larger potential difference across the two inputs but then the op amp output potential would be either $\pm10\,\rm V$ and that output voltage would not change if the input voltages were changed unless the potential difference between the two inputs dropped to below \$10^{-4}\,\rm V\$.
In my example \$V_- = 0\,\rm V\$ (earth) then \$V_+\$ cannot differ in potential by more than \$10^{-4}\,\rm V\$ ie it is virtually at earth potential.
You will note that as the open loop gain of an op amp increases the approximation that the two inputs are at the same potential gets better and better.