Editing Negative-feedback amplifier (section)

===Input and output resistances===
[[Image:Feedback amplifier input resistance.PNG|thumb|500px|Figure 6: Circuit set-up for finding feedback amplifier input resistance]]
Feedback is used to better match signal sources to their loads. For example, a direct connection of a voltage source to a resistive load may result in signal loss due to [[voltage division]], but interjecting a negative feedback amplifier can increase the apparent load seen by the source, and reduce the apparent driver impedance seen by the load, avoiding signal attenuation by voltage division. This advantage is not restricted to voltage amplifiers, but analogous improvements in matching can be arranged for current amplifiers, transconductance amplifiers and transresistance amplifiers.

To explain these effects of feedback upon impedances, first a digression on how two-port theory approaches resistance determination, and then its application to the amplifier at hand.

====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.

====Application to the example amplifier====
These resistance results now are applied to the amplifier of Figure 3 and Figure 5. The ''improvement factor'' that reduces the gain, namely ( 1 + β<sub>FB</sub> A<sub>OL</sub>), directly decides the effect of feedback upon the input and output resistances of the amplifier. In the case of a shunt connection, the input impedance is reduced by this factor; and in the case of series connection, the impedance is multiplied by this factor. However, the impedance that is modified by feedback is the impedance of the amplifier in Figure 5 with the feedback turned off, and does include the modifications to impedance caused by the resistors of the feedback network.

Therefore, the input impedance seen by the source with feedback turned off is ''R''<sub>in</sub> = ''R''<sub>1</sub> = ''R''<sub>11</sub> || ''R''<sub>B</sub> || ''r''<sub>π1</sub>, and with the feedback turned on (but no feedforward)

::<math> R_\mathrm{in} = \frac {R_1} {1 + { \beta }_\mathrm{FB} A_\mathrm{OL} } \ , </math>

where ''division'' is used because the input connection is ''shunt'': the feedback two-port is in parallel with the signal source at the input side of the amplifier. A reminder: ''A''<sub>OL</sub> is the  ''loaded'' open loop gain [[Negative feedback amplifier#Loaded open-loop gain|found above]], as modified by the resistors of the feedback network.

The impedance seen by the load needs further discussion. The load in Figure 5 is connected to the collector of the output transistor, and therefore is separated from the body of the amplifier by the infinite impedance of the output current source. Therefore, feedback has no effect on the output impedance, which remains simply ''R''<sub>C2</sub> as seen by the load resistor ''R''<sub>L</sub> in Figure 3.<ref>The use of the improvement factor ( 1 + β<sub>FB</sub> A<sub>OL</sub>) requires care, particularly for the case of output impedance using series feedback. See Jaeger, note below.</ref><ref name=Jaeger>{{cite book | title = Microelectronic Circuit Design |author1=R.C. Jaeger  |author2=T.N. Blalock   |name-list-style=amp | publisher = McGraw-Hill Professional | year = 2006 |edition=Third |page=Example 17.3 pp. 1092–1096| isbn = 978-0-07-319163-8 | url = http://worldcat.org/isbn/978-0-07-319163-8 | no-pp = true }}</ref>

If instead we wanted to find the impedance presented at the ''emitter'' of the output transistor (instead of its collector), which is series connected to the feedback network, feedback would increase this resistance by the improvement factor ( 1 + β<sub>FB</sub> A<sub>OL</sub>).<ref>That is, the impedance found by turning off the signal source ''I''<sub>S</sub> = 0, inserting a test current in the emitter lead ''I<sub>x</sub>'', finding the voltage across the test source ''V<sub>x</sub>'', and finding ''R''<sub>out</sub> = ''V<sub>x</sub> / I<sub>x</sub>''.</ref>