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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.

Ground the non-inverting input and take the ideal operational amplifier in stable, unsaturated negative feedback. Its inverting input is a virtual ground and draws no current. Kirchhoff's current law there givesThe minus sign is the phase inversion of the inverting amplifier.

An external source sees the input resistor between its terminal and the virtual ground. The input current is $I_i=V_i/R_i$, soThe infinite input resistance of the ideal operational amplifier itself is distinct from the finite input resistance of the complete inverting amplifier. Negative feedback makes the output approach an ideal voltage source; a real amplifier retains a small nonzero output resistance.

In the linear inverting amplifier, the non-inverting terminal is grounded and negative feedback connects the output to the inverting terminal. For large open-loop gain $A$, $V_o=A(V_{+}-V_{-})$; a finite unsaturated output therefore requires $V_{-}\simeq V_{+}=0$. ThusThis is a virtual ground: it has approximately ground potential, but it is not a conducting connection to ground. The ideal operational amplifier draws no input current, so input current flows through the feedback resistor. Saturation, loss of feedback, or insufficient bandwidth invalidates the approximation.

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]