= Solution
An <ideal operational amplifier> draws no current at either input. With stable <operational-amplifier negative feedback> in its unsaturated linear range, infinite <open-loop gain> also gives a <virtual short>, $V_+=V_-$. The amplifier adjusts its output to enforce that equality; it is not a conducting connection between the inputs. Here $V_+=0$, so the ideal summing node is a <virtual ground>. <Kirchhoff's current law> gives $V_{\rm in}/R_1=-V_{\rm out}/R_2$, hence \b[the ideal gain is $-R_2/R_1$].
For the <finite-input-resistance inverting amplifier> let the inverting-node <voltage> be $v$, and continue to neglect <output impedance>. The controlled output and node-current balance are
$$
V_{\rm out}=-Av,\qquad
\frac{v-V_{\rm in}}{R_1}+\frac{v-V_{\rm out}}{R_2}+\frac{v}{R_{\rm in}}=0.
$$
Eliminating $v$ gives
$$
\boxed{\frac{V_{\rm out}}{V_{\rm in}}
=-\frac{R_2/R_1}{1+\dfrac1A\left(1+\dfrac{R_2}{R_1}+\dfrac{R_2}{R_{\rm in}}\right)}}.
$$
Dividing the denominator by $R_2$ gives the requested resistance-sum form. The ideal result follows whenever $|A|\gg 1+R_2/R_1+R_2/R_{\rm in}$; taking $R_{\rm in}\to\infty$ alone is insufficient if $A$ is small. At finite <frequency> $A$ and the gain are <phasor> ratios, subject to feedback stability.
With equal <resistors> and sufficiently large loop gain, \b[$V_{\rm out}\simeq -V_{\rm in}$]. This gives a controlled sign reversal or 180-degree <phase> inversion, useful for subtraction and opposite-phase signal paths, with amplitude set by a <resistor> ratio rather than uncertain <open-loop gain>. The source still sees approximately $R_1$, so this is not a high-input-impedance <voltage> follower. For the ideal-input, single-pole model, the <noise gain of an operational amplifier> is two and the closed-loop <bandwidth> is approximately half the <gain-bandwidth product>. Output swing, <slew rate> and load constraints must also permit linear operation.
Back to article page
