Small-signal models

Transistors are nonlinear: the BJT's collector current depends exponentially on v_BE and the MOSFET's drain current on the square of its overdrive. Yet amplifiers are expected to be linear, so that a voice or a sensor signal comes out larger but with the same shape. The resolution is the small-signal idea. Bias the transistor at a DC operating point, then add a signal small enough that the curve looks straight over the range it covers. For that small signal, the transistor behaves like a linear circuit of resistors and controlled sources, and all the linear circuit tools of ELE-202 apply. This lesson builds those linear models; lesson 11 uses them to design amplifiers.

The accessible pedestrian signal's microphone produces only a few millivolts at normal street noise levels. Amara must amplify that by a factor of about a hundred before the envelope detector from lesson 4 can measure it. To predict the gain, she needs a small-signal model of the input transistor.

Separating DC and signal

In an amplifier, each voltage and current has a DC part, set by the bias, and a signal part. The convention uses capital letters with capital subscripts for DC values (I_C, V_BE), lower-case letters with lower-case subscripts for signals (i_c, v_be), and lower-case letters with capital subscripts for the total (i_C = I_C + i_c). The analysis is done in two steps: first the DC circuit finds the Q-point; then the small-signal circuit, with every DC source set to zero, finds the signal behavior. Setting a DC voltage source to zero makes it a short circuit to signals, which is why a power-supply rail is called signal ground. Setting a DC current source to zero makes it an open circuit. Coupling and bypass capacitors are chosen large enough to act as short circuits at signal frequencies.

Linearizing the BJT

Write v_BE = V_BE + v_be in the exponential law and expand:

If v_be is much smaller than V_T, the squared and higher terms are negligible and the signal current is proportional to the signal voltage, i_c = (I_C/V_T) v_be. The proportionality constant is the transconductance:

At I_C = 1 mA, g_m ≈ 40 mA/V, also written 40 mS: every millivolt at the base–emitter junction changes the collector current by 40 µA. The transconductance depends only on the bias current, not on the transistor's size or type, which makes it very predictable. The small-signal condition is usually taken as v_be below about 10 mV; at that level, the ratio of the second-order term to the linear term, v_be/(2V_T), is about 20 percent at the peak, which is a significant distortion for audio but acceptable for many sensor jobs. For low distortion, signals at the base–emitter junction must be kept to a few millivolts, or negative feedback must be used, as lesson 11 shows.

The hybrid-π model

The base current is i_C/β, so a base–emitter signal also produces a base signal current i_b = i_c/β = (g_m/β) v_be. Looking into the base, the signal therefore sees a resistance

Adding the output resistance r_o = V_A/I_C from the Early effect gives the hybrid-π model: between base and emitter, a resistor r_π; between collector and emitter, a voltage-controlled current source g_m v_be in parallel with r_o. This model is the workhorse of BJT amplifier analysis. At 1 mA with β = 150 and V_A = 100 V: g_m = 40 mS, r_π = 3.75 kΩ and r_o = 100 kΩ. An equivalent T model places a resistance r_e = α/g_m ≈ V_T/I_E, about 25 Ω at 1 mA, in the emitter lead, with the controlled source from collector to base. The T model is often easier to use when there is a resistor in the emitter.

The MOSFET small-signal model

For a MOSFET in saturation, i_D = ½k_n'(W/L)(V_OV + v_gs)². Expanding in the same way gives a linear term and a square term; the linear term defines the MOSFET transconductance,

and the small-signal condition is v_gs much smaller than 2V_OV. The gate draws no current, so the model has an open circuit at the input: a voltage-controlled current source g_m v_gs from drain to source in parallel with r_o. Each of the three forms of g_m is useful. The middle one is the most revealing: at a given current, g_m is set by the overdrive, and a MOSFET at V_OV = 0.2 V has g_m = I_D/0.1 V, while a BJT at the same current has g_m = I_C/0.025 V, four times larger. Equivalently, a MOSFET would need an overdrive of 2V_T ≈ 50 mV to match the BJT, and at such low overdrives the square law is no longer accurate.

Small-signal analysis step by step

Every small-signal problem follows the same recipe. First, solve the DC circuit for the Q-point. Second, compute the small-signal parameters from the bias: g_m, r_π (or r_e) and r_o. Third, draw the small-signal circuit: replace DC supplies by ground, DC current sources by open circuits, coupling and bypass capacitors by short circuits, and the transistor by its model. Fourth, solve that linear circuit for the gain, input resistance and output resistance using nodal analysis or the voltage-divider and parallel-resistance shortcuts from ELE-202. The result is valid only for small signals at mid-band frequencies, where the capacitors are short circuits and the transistor's internal capacitances are still open circuits; lesson 12 deals with the frequencies at which those assumptions fail.

import math
VT, IC = 0.025, 0.9e-3
gm = IC / VT
RC = 5100.0
# Compare the exact exponential response with the linear small-signal prediction
print(" vbe peak   linear vc    exact +peak   exact -peak   2nd harmonic")
for vbe_pk in (0.001, 0.005, 0.010, 0.020):
    lin = gm * vbe_pk * RC
    up = IC * (math.exp(vbe_pk / VT) - 1) * RC
    dn = IC * (1 - math.exp(-vbe_pk / VT)) * RC
    h2 = vbe_pk / (4 * VT) * 100           # 2nd-harmonic distortion estimate, percent
    print(f"{vbe_pk*1000:6.0f} mV   {lin:8.3f} V   {up:10.3f} V   {dn:10.3f} V     {h2:5.1f} %")

 vbe peak   linear vc    exact +peak   exact -peak   2nd harmonic
     1 mV      0.184 V        0.187 V        0.180 V       1.0 %
     5 mV      0.918 V        1.016 V        0.832 V       5.0 %
    10 mV      1.836 V        2.257 V        1.513 V      10.0 %
    20 mV      3.672 V        5.625 V        2.528 V      20.0 %

The output shows the price of an amplifier with no feedback. With 10 mV at the base, the output's positive peak is about half as large again as its negative peak, which a listener would hear as harsh distortion. Real microphone amplifiers therefore use an unbypassed emitter resistor or overall feedback, trading some gain for linearity, as lesson 11 explains.

Input resistance, output resistance and loading

Gain alone does not describe an amplifier. Any real signal source has its own internal resistance, and any real load draws current, so an amplifier stage is characterized by three numbers: its voltage gain with no load, A_vo; its input resistance R_in, the resistance the source sees looking into the input; and its output resistance R_out, the Thévenin resistance seen looking back into the output with the input signal set to zero. With these three, the stage becomes the simple voltage-amplifier model from ELE-202, and the overall gain from source to load follows from two voltage dividers:

For the common-emitter stage in the last example, R_in is the bias resistors in parallel with r_π, a few kilohms, and R_out is R_C ∥ r_o, about 4.9 kΩ. Amara's electret microphone capsule has an output resistance of about 2 kΩ, comparable to R_in, so roughly a third of the microphone signal is lost at the input before any gain is applied; and if the stage drove a 5 kΩ load directly, half the output would be lost again. Recognizing these divider losses early is what drives the choice of amplifier configuration in lesson 11: a stage with high input resistance at the front, one with low output resistance at the back, and gain in the middle.

Where the models come from in practice

Datasheets for discrete transistors often give small-signal parameters directly as h-parameters, measured at a stated bias: h_fe is the small-signal current gain (close to β), h_ie is the input resistance (close to r_π), and h_oe is the output conductance (1/r_o). Circuit simulators compute all of these automatically from the device model at the DC operating point they find first, exactly the two-step process described here. Hand analysis with simple models is still essential, because it shows which components set the gain and how to change it, which a simulator's numbers alone do not.