# BJT Small Signal Modelling

concept
When using the BJT as an amplifier we'll use what's called the "small signal model" of the BJT. This model is a circuit segment which will replace the symbol for the BJT in our circuits and makes the analysis of the circuit much simpler. The key is to figure out what combination of components will operate the same as a transistor.
fact
The small signal model we're going to use is the following:
So you can see that the base connects to the emitter through a resistor labelled $$r_{\pi}$$ and the collector is connected to the emitter with a dependent current source which multiplies the voltage drop across $$r_{\pi}$$. This small signal model is used to calculate how AC signals (which our input signals always are) make the circuit react. To do a small signal analysis of a circuit we need to replace EVERYTHING in the circuit with its small signal equivalent. So the BJT becomes the circuit shown above, but also constant voltage sources become ground and constant current sources become open circuits.
fact
The following picture shows how to convert a simple circuit to its complete small signal analysis.
You can see that the battery became a short to ground, the AC signal remained, and the capacitor became a short circuit. You'll get plenty of practice at this as we go on. Now we're going to figure out what values $$r_{\pi}$$, $$r_o$$ and $$g_m$$ will take. These are determined by the DC voltages and currents around the transistor's terminals.
fact
$$g_m = \frac{I_C}{V_T}$$ Where $$V_T$$ is the usual thermal voltage $$\approx 26$$mV at room temperature. $$r_{\pi} = \frac{\beta}{g_m}$$ $$r_o = \frac{V_A + V_{CE}}{I_C} \approx \frac{V_A}{I_C}$$ Where $$V_A$$ is something called the "Early Voltage" and is determined by the transistor. It's typically in the range of 50V - 200V. We use the simplification $$\frac{V_A}{I_C}$$ since in general $$V_A \gt\gt V_{CE}$$
derivation
To find why $$g_m = \frac{I_C}{V_T}$$ we're going to start by seeing that to get the correct collector voltage we must have: $$g_m = \frac{dI_C}{dV_{BE}}$$ That is, $$g_m$$ is a measure of how much the collector current will change when we change the base-emitter voltage by some small amount. Expanding the derivative we have: $$\frac{d}{dV_{BE}}I_S\exp(\frac{V_{BE}}{V_T}) = \frac{1}{V_T}I_S\exp(\frac{V_{BE}}{V_T}) = \frac{I_C}{V_T}$$. Just like I said.
derivation
Now for why $$r_{\pi} = \frac{\beta}{g_m}$$. For some given $$V_{BE}$$ we're going to have some $$I_C$$ which means that given a fixed and known $$\beta$$ we can calculate $$I_B$$. Our job here is to pick $$r_{\pi}$$ so that our small signal model has the correct $$I_B$$ for a given $$V_{BE}$$. So we know that $$I_B = \frac{1}{\beta}I_C = \frac{1}{\beta}g_m v_{\pi}$$ And $$v_{\pi} = r_{\pi}I_B$$ so: $$\frac{1}{\beta}g_mr_{\pi} = 1 \implies r_{\pi} = \frac{\beta}{g_m}$$

## Input Impedance

An important part of an amplifier circuit is its input and output impedances. This means the impedance that is seen by the input signal and the impedance seen by the output signal respectively. We want a high input impedance and a low output impedance.
fact
The input impedance is the impedance between the input signal voltage and ground. We want it to be large in order to avoid loss of signal to voltage division.
example

Find the input impedance of the circuit shown.

Our first step is to convert the circuit into its small signal equivalent. Here DC voltages become ground and the coupling capacitors become short circuits. Now we can see that the input impedance is given by: $$Z_{in} = R_{B1}||R_{B2}||r_{\pi}$$
When there is a resistor between the emitter and ground the equation becomes a little bit more complex, if you haven't seen it before the equivalent resistance of two resistors with different currents can be a little weird. In the case of our BJT emitter resistor we have to multiply the emitter resistor's resistance by $$\beta + 1$$ because that's how much more current is running through that resistor.
fact
The equivalent resistance of two resistors with different currents running through them like below is given by: $$Z_{eq} = R_1 + (\beta + 1)R_E$$
example

Find the input resistance of the circuit below.

Again we first convert the circuit to its small signal equivalent. Now we use our above fact about equivalent resistances with different currents to find $$Z_{in}$$: $$Z_{in} = R_{B1}||R_{B2}||(r_{\pi} + (\beta + 1)R_E)$$
When dealing with BJTs (and MOSFETs) you'll make heavy use of the above technique of equivalent resistances of resistors with different currents to make sure you understand the process and how to use it.
practice problems