Editing Negative-feedback amplifier (section)

===Loaded open-loop gain===
Figure 3 indicates the output node, but not the choice of output variable. A useful choice is the short-circuit current output of the amplifier (leading to the short-circuit current gain). Because this variable leads simply to any of the other choices (for example, load voltage or load current), the short-circuit current gain is found below.

First the loaded '''open-loop gain''' is found. The feedback is turned off by setting ''g''<sub>12</sub> = ''g''<sub>21</sub> = 0. The idea is to find how much the amplifier gain is changed because of the resistors in the feedback network by themselves, with the feedback turned off. This calculation is pretty easy because ''R''<sub>11</sub>, ''R''<sub>B</sub>, and ''r''<sub>π1</sub> all are in parallel and ''v''<sub>1</sub> = ''v''<sub>π</sub>. Let ''R''<sub>1</sub> = ''R''<sub>11</sub> || ''R''<sub>B</sub> || ''r''<sub>π1</sub>. In addition, ''i''<sub>2</sub> = −(β+1) ''i''<sub>B</sub>. The result for the open-loop current gain ''A''<sub>OL</sub> is:

::<math> A_\mathrm{OL} = \frac { \beta i_\mathrm{B} } {i_\mathrm{S}} = g_m R_\mathrm{C} \left( \frac { \beta }{ \beta +1} \right) 
\left( 
\frac {R_1} {R_{22} + 
\frac {r_{ \pi 2} + R_\mathrm{C} } {\beta + 1 } } \right)   \ . </math>