Editing Differential amplifier (section)

=== Symmetrical feedback network eliminates common-mode gain and common-mode bias ===
[[File:Op-Amp Differential Amplifier input impedence and common bias.svg|thumb|280px|Figure 6: Differential amplifier with non-ideal op-amp: input bias current and differential input impedance]]
In case the operational amplifier's (non-ideal) input bias current or differential input impedance are a significant effect, one can select a feedback network that improves the effect of common-mode input signal and bias. In Figure&nbsp;6, current generators model the input bias current at each terminal; ''I''<sup>+</sup><sub>b</sub> and ''I''<sup>−</sup><sub>b</sub> represent the input bias current at terminals ''V''<sup>+</sup> and ''V''<sup>−</sup> respectively.

The [[Thévenin's theorem|Thévenin equivalent]] for the network driving the ''V''<sup>+</sup> terminal has a voltage ''V''<sup>+</sup>' and impedance ''R''<sup>+</sup>':
: <math>{V^+}' = V^+_\text{in} R^+_\parallel / R^+_\text{i} - I^+_\text{b} R^+_\parallel; \quad \text{where} \quad {R^+}' = R^+_\parallel  = R^+_\text{i} \parallel R^+_\text{f},</math>
while for the network driving the ''V''<sup>−</sup> terminal:
: <math>{V^-}' =  V^-_\text{in} R^-_\parallel / R^-_\text{i} + V_\text{out} R^-_\parallel / R^-_\text{f} - I^-_\text{b} R^-_\parallel; \quad \text{where} \quad {R^-}' = R^-_\parallel = R^-_\text{i} \parallel R^-_\text{f}.</math>
The output of the op-amp is just the open-loop gain ''A''<sub>ol</sub> times the differential input current ''i'' times the differential input impedance 2''R''<sub>d</sub>, therefore
: <math> V_\text{out} = A_\text{ol} \cdot 2 R_\text{d} \frac{{V^+}' - {V^-}'}{2R_\parallel + 2R_\text{d}} = ({V^+}' - {V^-}') A_\text{ol} R_\parallel / (R_\parallel \parallel R_\text{d}),</math>
where ''R''<sub>||</sub> is the average of ''R''<sup>+</sup><sub>||</sub> and ''R''<sup>−</sup><sub>||</sub>.

These equations undergo a great simplification if 
: <math>R^+_\text{i} = R^-_\text{i}, \quad R^+_\text{f} = R^-_\text{f},</math>
resulting in the relation
: <math>V^+_\text{in} - V^-_\text{in} - R_\text{i} I^\Delta_\text{b} = V_\text{out} \left[ \frac{R_\text{i}}{R_\text{f}} + \frac{1}{A_\text{ol} \frac{R_\text{i}}{R_\text{i} \parallel R_\text{f} \parallel R_\text{d}}}\right],</math>
which implies that the closed-loop gain for the differential signal is ''V''<sup>+</sup><sub>in</sub>&nbsp;−&nbsp;''V''<sup>−</sup><sub>in</sub>, but the common-mode gain is identically zero. 

It also implies that the common-mode input bias current has cancelled out, leaving only the input offset current ''I''<sup>Δ</sup><sub>b</sub> = ''I''<sup>+</sup><sub>b</sub>&nbsp;−&nbsp;''I''<sup>−</sup><sub>b</sub> still present, and with a coefficient of ''R''<sub>i</sub>. It is as if the input offset current is equivalent to an input offset voltage acting across an input resistance ''R''<sub>i</sub>, which is the source resistance of the feedback network into the input terminals.

Finally, as long as the open-loop voltage gain ''A''<sub>ol</sub> is much larger than unity, the closed-loop voltage gain is ''R''<sub>f</sub>/''R''<sub>i</sub>, the value one would obtain through the rule-of-thumb analysis known as "virtual ground".<ref 
group="nb">For the closed-loop common-mode gain to be zero only requires that the ratio of resistances ''R''<sub>f</sub> / ''R''<sub>i</sub> be matched in the inverting and non-inverting legs. For the input bias currents to cancel, the stricter relation given here must obtain.</ref>