Editing Negative-feedback amplifier (section)

====Background on resistance determination====
Figure 6 shows an equivalent circuit for finding the input resistance of a feedback voltage amplifier (left) and for a feedback current amplifier (right). These arrangements are typical [[Miller theorem#Applications|Miller theorem applications]].

In the case of the voltage amplifier, the output voltage β''V''<sub>out</sub> of the feedback network is applied in series and with an opposite polarity to the input voltage ''V<sub>x</sub>'' travelling over the loop (but in respect to ground, the polarities are the same). As a result, the effective voltage across and the current through the amplifier input resistance ''R''<sub>in</sub> decrease so that the circuit input resistance increases (one might say that ''R''<sub>in</sub> apparently increases). Its new value can be calculated by applying [[Miller theorem#Miller theorem (for voltages)|Miller theorem]] (for voltages) or the basic circuit laws. Thus [[Kirchhoff's circuit laws|Kirchhoff's voltage law]] provides:

::<math> V_x = I_x R_\mathrm{in} + \beta v_\mathrm{out} \ , </math>

where ''v''<sub>out</sub> = ''A''<sub>v</sub> ''v''<sub>in</sub> = ''A''<sub>v</sub> ''I''<sub>x</sub> ''R''<sub>in</sub>. Substituting this result in the above equation and solving for the input resistance of the feedback amplifier, the result is:

::<math> R_\mathrm{in}(fb) = \frac {V_x} {I_x} = \left( 1 + \beta A_v \right ) R_\mathrm{in} \ . </math>

The general conclusion from this example and a similar example for the output resistance case is:
''A series feedback connection at the input (output) increases the input (output) resistance by a factor ( 1 + β ''A''<sub>OL</sub> )'', where ''A''<sub>OL</sub> = open loop gain.

On the other hand, for the current amplifier, the output current β''I''<sub>out</sub> of the feedback network is applied in parallel and with an opposite direction to the input current ''I<sub>x</sub>''. As a result, the total current flowing through the circuit input (not only through the input resistance ''R''<sub>in</sub>) increases and the voltage across it decreases so that the circuit input resistance decreases (''R''<sub>in</sub> apparently decreases). Its new value can be calculated by applying the [[Miller theorem#Dual Miller theorem (for currents)|dual Miller theorem]] (for currents) or the basic Kirchhoff's laws:

::<math> I_x = \frac {V_\mathrm{in}} {R_\mathrm{in}} + \beta i_\mathrm{out} \ . </math>

where ''i''<sub>out</sub> = ''A''<sub>i</sub> ''i''<sub>in</sub> = ''A''<sub>i</sub> ''V''<sub>x</sub> / ''R''<sub>in</sub>. Substituting this result in the above equation and solving for the input resistance of the feedback amplifier, the result is:

::<math> R_\mathrm{in}(fb) = \frac {V_x} {I_x} = \frac { R_\mathrm{in}  } { \left( 1 + \beta A_i \right ) } \ . </math>

The general conclusion from this example and a similar example for the output resistance case is:
''A parallel feedback connection at the input (output) decreases the input (output) resistance by a factor ( 1 + β ''A''<sub>OL</sub> )'', where ''A''<sub>OL</sub> = open loop gain.

These conclusions can be generalized to treat cases with arbitrary [[Norton's theorem|Norton]] or [[Thevenin's theorem|Thévenin]] drives, arbitrary loads, and general [[two-port network|two-port feedback networks]]. However, the results do depend upon the main amplifier having a representation as a two-port – that is, the results depend on the ''same'' current entering and leaving the input terminals, and likewise, the same current that leaves one output terminal must enter the other output terminal.

A broader conclusion, independent of the quantitative details, is that feedback can be used to increase or to decrease the input and output impedance.