**1 Answer**

written 2.4 years ago by |

**Concept of virtual ground :** This means that the differential input voltage $v_{1d}$ between the non inverting and inverting input terminals is essentially zero.

This is obvious because even if output voltage is few volts, due to large open loop gain of op-amp, the difference voltage $v_{id}$ at the input terminals is almost zero.

This means that if output voltage is 10 v and the A i.e. open loop gain is $10^4$ then, we have

$v_{out} = A \ v_1 \ d$

Therefore, we have $v_{1d} = \frac{v_{out}}{A} = \frac{10}{10^4} = l \ m \ v$

Hence, $v_{1d}$ is very small, as $A \rightarrow w$, the difference voltage $v_1 d\rightarrow c$ and realistically assumed to be zero for analyzing the circuits.

$\therefore$ $v_{1d} = \frac{v_{out}}{A}$

$(v_{lnl} - v_{ln2}) = \frac{v_{out}}{\infty } = 0$

$v_{lnl} = v_{ln2}$

Thus, we can say that under linear range of operation, there is virtually short ckt between the two input terminals, in the sense that their voltage are same.