Editing Negative-feedback amplifier (section)

====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>