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.

In the first figure, 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 the following three parameters so that the red and blue boxes in the first figure can be modeled by the corresponding boxes in the second figure:

- 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 Voltage Gain:**Given the AC input voltage , the base voltage and current are

The collector current is and collect voltage is

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

If

then the gain can be approximated as

To achieve higher gain , we want to have smaller and greater . However, this also means the input resistance is small and the output resistance is large, neither is desirable.Note that is not constant. As shown before, of the base-emitter PN-junction is approximately inversely proportional to .

Also note that and affects the DC operating point. Distortion may be caused if or is set properly.

**Example 1:**

, , , and . We further assume , and the capacitances are large enough so that they can be considered as short circuit for AC signals.

- Find base current:

- Find DC load line determined by the following two points:

- Find DC operating point :

- 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 . - Find

- 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 .

**Example 2:**

Consider the circuit below with its AC small-signal model:

We can find the voltage gain, the input and output resistances when and ( ).

- AC voltage gain:
First, . Then apply KVL to the collector to get

Solving this we get and then find the AC voltage gain:

We see that when , i.e., there is not negative feedback, the gain becomes the same as the result before without feedback. - AC input resistance:
This is in parallel with the resistence of the circuit to the right of the base. First convert the current source in parallel with to a voltage source in series with and then we get the current into this circuit to be:

and the resistance is

and the total input resistance is:

- AC output resistance:
- Open-circuit voltage with is simply

- Short-circuit current with :
Short-circuit current by alone is ,
Short-circuit current by current source alone is
. The total short-circuit current is

- Output resistance:

- Open-circuit voltage with is simply

**Example 3:**

Consider both the DC operating point and its AC small signal model of the circuit below:

Apply KVL to get

and

We assume , and consider the following for the DC operating point to be in the middle of the linear region:

- Given
, we need to find :

But

- Alternatively, given
, we can find :

Solving to get .

Next consider the AC equivalent circuit based on small-signal model of the transistor in the dashed line box:

As is significantly greater than all resistors in the circuit, it can
be ignored in the AC analysis. Apply KCL to the emitter to get

Solving for :

- AC input resistance:

- AC voltage gain:

- AC output resistance:
- Open-circuit voltage ():
- Short-ciruit current ():
- Output resistance: