11.4 Small-signal models and analysis

Transistor Small-Signal Models

Small-signal models let you analyze how a BJT responds to tiny AC signals riding on top of its DC bias. Instead of dealing with the transistor's full nonlinear behavior, you replace it with a linear circuit made of resistors and controlled sources. This only works when the AC signal is small enough that the transistor stays in its active region, close to the Q-point you've already set with DC biasing.

Hybrid-π Model and T-Model

The hybrid-π model is the one you'll use most often in this course. It replaces the transistor with:

• A voltage-controlled current source: $g_m \cdot v_{be}$
• An input resistance between base and emitter: $r_\pi$
• An output resistance between collector and emitter: $r_o$

You connect these elements between the transistor's three terminals, and the rest of the circuit stays the same. This model works well for low- and mid-frequency analysis, which is where most intro-level problems live.

The T-model is an alternative that looks at things from the emitter's perspective. It uses:

• A resistance $r_e$ in the emitter leg (where $r_e = \frac{V_T}{I_C}$, typically a small value)
• A current-controlled current source $\alpha \cdot i_e$

The T-model can be more intuitive for certain circuit configurations (like common-base), but both models describe the same transistor behavior. For most problems in an intro course, the hybrid-π model is your go-to.

AC Equivalent Circuit and Small-Signal Parameters

To actually do small-signal analysis, you build an AC equivalent circuit. Here's the process:

1. Find the DC operating point (Q-point). Solve for $I_C$, $V_{CE}$, and $I_B$ using your DC biasing techniques.
2. Calculate the small-signal parameters ($g_m$, $r_\pi$, $r_o$) from the Q-point values.
3. Replace the transistor with its small-signal model (hybrid-π or T-model).
4. Short all DC voltage sources and open all DC current sources, since they don't contribute to the AC signal.
5. Analyze the resulting linear circuit using standard techniques (KVL, KCL, voltage dividers, etc.) to find voltage gain, input resistance, or whatever the problem asks for.

The three core small-signal parameters you need to know:

• Transconductance $g_m$: How effectively the transistor converts a small change in $V_{BE}$ into a change in $I_C$.

$g_m = \frac{I_C}{V_T}$

• Base-emitter resistance $r_\pi$: The small-signal resistance looking into the base.

$r_\pi = \frac{V_T}{I_B} = \frac{\beta}{g_m}$

• Collector-emitter resistance $r_o$: The output resistance of the transistor, which accounts for the slight dependence of $I_C$ on $V_{CE}$.

$r_o = \frac{V_A}{I_C}$

In all of these, $V_T$ is the thermal voltage, approximately 25–26 mV at room temperature, and $V_A$ is the Early voltage, a device parameter typically in the range of 50–200 V.

Key Transistor Parameters

Transconductance ($g_m$)

Transconductance tells you how much the collector current changes for a small change in base-emitter voltage. A higher $g_m$ means the transistor is a more sensitive amplifier.

$g_m = \frac{I_C}{V_T}$

Since $V_T \approx 26 \text{ mV}$ at room temperature, a transistor biased at $I_C = 1 \text{ mA}$ has:

$g_m = \frac{1 \text{ mA}}{26 \text{ mV}} \approx 38.5 \text{ mA/V}$

Notice that $g_m$ is directly proportional to the collector current. If you double $I_C$, you double $g_m$. This is why the choice of Q-point directly affects your amplifier's gain.

Base-Emitter Resistance ($r_\pi$) and Collector-Emitter Resistance ($r_o$)

$r_\pi$ is the dynamic resistance seen looking into the base terminal. It relates to $g_m$ through $\beta$:

$r_\pi = \frac{\beta}{g_m} = \frac{\beta \cdot V_T}{I_C}$

For the same 1 mA bias point with $\beta = 100$:

$r_\pi = \frac{100}{38.5 \text{ mA/V}} \approx 2.6 \text{ k}\Omega$

Higher collector current means lower $r_\pi$, which means the base draws more AC current from whatever is driving it.

$r_o$ models the fact that $I_C$ isn't perfectly constant as $V_{CE}$ changes (this is the Early effect). It acts as a large resistance in parallel with the current source in the hybrid-π model:

$r_o = \frac{V_A}{I_C}$

With $V_A = 100 \text{ V}$ and $I_C = 1 \text{ mA}$, you get $r_o = 100 \text{ k}\Omega$. Because $r_o$ is usually much larger than the load resistance, it's often ignored in first-pass calculations. But for more accurate analysis or high-gain designs, it matters.

Small-Signal Analysis

Frequency Response

Everything above assumes the signal frequency is low enough that internal capacitances don't matter. At higher frequencies, two parasitic capacitances inside the BJT start to affect performance:

• $C_\pi$ (base-emitter capacitance): This is the larger of the two and sits in parallel with $r_\pi$ in the hybrid-π model.
• $C_\mu$ (base-collector capacitance): Smaller in value, but its effect gets multiplied by the voltage gain of the stage (this is the Miller effect, which you may encounter later).

At low frequencies, these capacitances act like open circuits and have negligible effect. As frequency increases, they provide low-impedance paths that shunt signal current away from the controlled source, reducing the transistor's gain.

The unity-gain frequency ($f_T$) is the frequency where the transistor's short-circuit current gain drops to 1:

$f_T = \frac{g_m}{2\pi(C_\pi + C_\mu)}$

A transistor with a higher $f_T$ can amplify at higher frequencies. For a typical small-signal BJT, $f_T$ might be several hundred MHz to a few GHz. If your signal frequency is well below $f_T$, you can safely ignore the capacitances and use the simpler resistor-and-current-source model.