Question: Explain the concept of virtual ground in operational amplifiers?
0 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.

renu • 35 views
 modified 5 days ago  • written 25 days ago by RB • 100