Small-Signal Analysis
Replace a thoroughly non-linear device with a straight line that is wrong everywhere except near one point — and get the whole of linear circuit theory in exchange.
Skip to the animationSmall-signal analysis replaces a non-linear device with its tangent at the operating point, making the circuit linear — which separates the DC and AC problems entirely and buys the whole of linear circuit theory, at the price of a validity range of a few millivolts.
The problem
Collector current depends exponentially on base-emitter voltage: 60 mV multiplies it tenfold. Superposition, impedance and transfer functions all assume linearity, so none of them legitimately applies to a device like that.
The move
Over a small enough region, a smooth curve is indistinguishable from its tangent. Replace the device with its slope at the Q-point and the circuit becomes linear. Formally it is a Taylor expansion truncated after the first-order term.
This is why the biasing topic comes first. The tangent is taken at the operating point, so the slope — and therefore the gain — depends on where you biased the stage.
Two circuits from one schematic
| DC analysis | AC analysis | |
|---|---|---|
| Capacitors | Open circuit | Short circuit |
| Supply rails | At their supply values | Grounded |
| Device model | The real curves, or a simple model | The small-signal model |
| Finds | The Q-point | Gain, impedances, frequency response |
Grounding the supply in the AC circuit feels wrong and is right: a fixed voltage carries no signal, so as far as the signal is concerned it is a short to ground. This is also why supply decoupling matters — a rail that is not stiff stops being a signal ground.
Transconductance
Differentiating the exponential gives g_m = I_C/V_T, with V_T ≈ 26 mV at room temperature. At 1 mA that is 38 mA/V — and it depends on nothing about the device, only on the bias current.
A common-emitter stage's gain is then A_v = −g_m·R_C, which rearranges to the DC voltage across R_C divided by 26 mV. So the achievable gain is set by how much supply headroom you spent on the load resistor.
MOSFETs are different
Drain current follows a square law rather than an exponential, giving g_m = 2I_D/(V_GS − V_th). At equal current that is typically five to ten times smaller than a BJT's.
This is the concrete reason bipolar devices persist in precision analogue front ends despite CMOS dominating everywhere else. More gain per unit of bias current, and better matching.
The validity range
The tangent is a good fit only near the Q-point. For a BJT, "small" means roughly under 10 mV peak; beyond that the curvature produces harmonic distortion that the linear model cannot predict at all.
Power amplifiers therefore need large-signal analysis, and distortion becomes a measured specification rather than an assumption. Negative feedback reduces the distortion by the loop gain, which is one of the main reasons to use it.
What it buys
- Gain and input/output impedances by inspection, with ordinary resistor arithmetic.
- Poles, Bode plots and stability analysis.
- Cascading stages by multiplying gains — which is what makes a multi-stage amplifier designable rather than something discovered on the bench.
The numbers you will be asked for
- Transconductance, BJT
g_m = I_C / V_T
V_T ≈ 26 mV
- Transconductance, MOSFET
g_m = 2 I_D / (V_GS − V_th)
- Input resistance
r_π = β / g_m
- CE voltage gain
A_v = −g_m · R_C
- Gain in terms of headroom
A_v = −V_RC / V_T
- Emitter resistance
r_e = V_T / I_E ≈ 26 / I_E(mA) ohms
Watch it work
Check yourself
question 1 / 4
One question at a time. Pick an answer to see why it is right or wrong, then move on — there is no score to keep and nothing is saved.