Av, Vd is equivalent Thevenin voltage source and $R_0$ is Thevenin eq. resistance looking back into o/p terminal of OP-AMP.

The value of i/p resistance $R_i$ is finite but very high here ad that of o/p resistance $R_0$ is non-zero because the OP-AMP is non – ideal.

The circuit is made of differential i/p non – ideal/

The circuit is made of differential i/p resistance $R_i$ ,

The voltage gain Av and output $R_0$

Parameters:

$R_i$, Av and $R_0⇒$ Open-loop parameters.

$Vd = (V_1-V_2)$ is called differential i/p voltage and Av is called open loop gain.

Hence o/p voltage $V_0=$

$V_0=Av \times Vd=Av(V_1-V_2)$

$V_1$ and $V_2$ Is voltage at the non – inverting and inverting i/p terminals of op-amp, w.r.t grid Since both i/p terminals are allowed to be connected to independent potentials, w.r.t grid, the i/p side of OP-AMP is said to be double ended type. But the o/p is single ended type.

OP-AMP responds only to differential i/p voltageVd. That means it produces o/p voltage which is ∝ to difference between the i/p voltages and not to individual voltages. Hence OP-AMP are also called differential amps.

$∴V_0=Av(V_1-V_2)$ This shows, polarity of o/p voltage depends on polarity of differential i/ps signal Vd.

The differential i/p voltage Vd is $Vd = V_0/Av$

It is used of the value of only differential i/p voltage Vd and does not give individual i/p`s voltages $V_1,V_2$

Open loop voltage gain Av is for very large value. Hence the value of Vd even for maximum o/p voltage is extremely small. Eg. , $V_{0(\max)} =10 V$ at 741 OPAM needs, $Vd - 10/2 \times 10^5=50 μV$ . Thus we need a very small differential i/p voltage Vd to obtain the maximum possible o/p voltage.