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The closed-loop gain from an equivalent differential input-voltage disturbance to the output is the noise gain. For an ideal-input inverting amplifier, $\beta =R_1/(R_1+R_2)$ and $G_{\mathrm{n}}=1+R_2/R_1$, although the voltage gain is $-R_2/R_1$. A dominant-pole amplifier has approximate closed-loop bandwidth $f_T/G_{\mathrm{n}}$ when loop gain is large at low frequency and the circuit is stable.

The ideal voltage-feedback Operational amplifier has zero input currents and, while stable negative feedback holds it in its linear operating range, equal input voltages: $i_{+}=i_{-}=0$ and $v_{+}=v_{-}$. The first rule follows from infinite input impedance; the second is the large-open-loop-gain limit of $v_o=A(v_{+}-v_{-})$. Equal input electric potential is a virtual short, not an electrically conducting connection.

For an inverting amplifier with the noninverting input grounded, $v_{-}=0$ and node-current balance gives $v_i/R_i=-v_o/R_f$, hence $v_o=-R_f{v}_i/R_i$. For a noninverting amplifier, the feedback divider gives $v_{-}=v_o{R}_g/(R_f+R_g)=v_i$, hence $v_o=(1+R_f/R_g)v_i$.

One should check output swing, input common-mode range, slew rate, Gain-bandwidth product and stability before using these limits in a real circuit. Saturation, positive feedback or operation beyond bandwidth invalidates the equal-voltage rule; finite input bias currents and offset voltage can also matter.^[ref]