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:
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:
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
This is simply the resistance of the resistor .
Given the AC input voltage , the base voltage and current are
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.
, , , and . We further assume , and the capacitances are large enough so that they can be considered as short circuit for AC signals.
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
Assume AC input voltage is
, the overall base voltage is
The output current is
The circuit above can also be analyzed using the small-signal model.
Same as before,
, and we have the following DC variables:
Consider the circuit below with its AC small-signal model:
We can find the voltage gain, the input and output resistances when and ( ).
. Then apply KVL to the collector to get
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:
Consider both the DC operating point and its AC small signal model of the circuit below:
Apply KVL 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