Previously we considered a simple circuit composed of a source of and and a load of . Now we generalize this idea to a more complicated circuit containing an amplification circuit shown on top of the figure below, where everything, including the amplifier as well as and , inside the red box is treated as the source, while everything, including the amplifier as well as , inside the blue box is treated as the load. Given the amplifier as well as the source and , and the load , we need to find , and also the voltage gain .

Based on the previous discussion, a single transistor AC amplification circuit is given as shown in the figure.

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 .**AC Output Impedance:**This is simply the resistance of the resistor

**AC Amplification Gain:**Given the AC input voltage , the base voltage can be found (voltage divider) to be

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 , as

The voltage gain is therefore

In particular, if the input resistance is much larger than the internal resistance of the voltage source, i.e.,

and the output resistance is much smaller than the load resistance, i.e.,

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 capacitors are large enough so that they can be considered as short circuit for the AC signals.

- Find base current:

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

- Find DC operating point :
The intersection of the DC load line and the curve corresponding to is the DC operating point , and .

- 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

This can be obtained graphically from the output i-v characteristics, based on , to be

Note that the output is in opposite phase (180 phase shift) with the input.

- Find the 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 .