As discussed before, the voltage a circuit receives from a source depends on its input impedance as well as the internal impedance of the source, while the voltage it delivers depends on its output impedance as well as the load impedance . It is therefore important to consider these input and output impedances of an amplification circuit as well as its voltage gain.

Here everything inside the red box, including the amplifier as well as and , is treated as the source, while everything inside the blue box, including the amplifier as well as , is treated as the load. Given the amplifier as well as the source and , and the load , we need to find

- Input impedance
- Output impedance
- voltage gain

Consider the typical transistor AC amplification circuit below:

If the capacitances of the coupling capacitors and the emitter by-pass capacitor are large enough with respect to the frequency of the AC signal in the circuit is high enough, these capacitors can all be approximated as short circuit. Moreover, note that the AC voltage of the voltage supply is zero, it can be treated the same as the ground. Now the AC behavior of the transistor amplification circuit can be modeled by the following small signal equivalent circuit:

As shown above, this AC small signal equivalent circuit can be modeled by as an active circuit containing three components:

**AC Input Impedance:**For AC signals, the input of the amplification circuit is shown below, where is the internal resistance of the signal source, and the input impedance of the circuit is the three resistances , and in parallel:

The approximation is due to the fact that typically .**AC Output Impedance:**This is simply the resistance of the resistor

**AC Amplification Gain:**Given the AC input voltage , the base voltage is (voltage divider):

and the base current is

The collector current is and the output voltage is

Here the negative sign indicates the fact that is out of phase with . The voltage gain is therefore

In particular, if the internal resistance of the voltage source is much smaller than input resistance,

and the load resistance is much greater than the output resistance,

then the gain can be approximated as

Note that, as shown before, the resistance of the base-emitter PN-junction is a function of :

which is inversely proportional to . Higher values of and (and therefore ) may be desired to achieve higher voltage gain . However, also note that distortion may also be caused if or is set too high.

**Example 1:**

Here , , (), and the capacitances are large enough so that they can be considered as short circuit for the AC signals. Assume .

- Find base current:

- Find DC load line:
The DC load line is determined by the following two points: ; and ,

- Find DC operating point :
The DC operating point is at

- Find AC load line:
The AC load is . The AC load line is a straight line passing the DC operating point with slope . The intersections of the AC load line with and axes can be found by

- Find input voltage and current:
Assume AC input voltage is and , the overall base voltage is

and the corresponding base current can be found graphically from the input characteristics to be

between 20 and 60 . - Fine

- Find AC output voltage:
The output current is

The output voltage is

- Find the AC voltage gain:

The circuit above can also be analyzed using the small-signal model.

Same as before,
, and we have the following DC variables:

The AC variables:

The voltage gain is:

The input resistance is , the output resistance is .