Finite-input-resistance inverting amplifier ($G=-\frac{R_2/R_1}{1+(1+R_2/R_1+R_2/R_{\mathrm{in}})/A}$)

With the noninverting input grounded, a differential input impedance $R_{\mathrm{in}}$ draws current from the summing node. For zero output impedance, $V_o=-Av$ and Kirchhoff's current law gives $(v-V_i)/R_1+(v-V_o)/R_2+v/R_{\mathrm{in}}=0$. Eliminating $v$ yields the displayed voltage gain. The resistor-ratio result requires $|A|\gg 1+R_2/R_1+R_2/R_{\mathrm{in}}$ as well as stable, unsaturated operation. The same expression applies to small-signal phasors with complex $A(f)$.