The small-signal model

A transistor around its operating point

Bias a transistor at an operating point: fixed DC voltages and drain current, here V_GS = 0.7815 V and I_D = 25 µA. Wiggle the gate by a few millivolts and the drain current wiggles almost in proportion. For small changes the curved device acts like a straight line.

Write small changes in lower case (v_gs, i_d). Then i_d = g_m·v_gs + g_mb·v_bs + v_ds/r_o: the small-signal model.

1. g_m, the transconductance

   g_m = ∂I_D/∂V_GS, the slope of the I_D-V_GS curve at the bias (E2.5 Transconductance and g_m/I_D).

2. g_mb, the body transconductance

   The body acts as a second, weaker gate. When the source rises above it, the threshold rises too (the body effect). Its transconductance is g_mb = ∂I_D/∂V_BS.

3. r_o, the output resistance

   r_o = 1/g_ds, with g_ds = ∂I_D/∂V_DS (E2.4 Channel-length modulation and output resistance).

ngspice reports these numbers per device. A SKY130 transistor is a subcircuit, so the name has a path: @m.xm1.msky130_fd_pr__nfet_01v8[gm] is g_m of the transistor inside XM1. The lab saves them during its input sweep and reads them where v(out) crosses 0.9 V.

The model, drawn and rebuilt

Every gain in this module is this circuit plus the resistors around it. In the common-source stage R_D goes to VDD, which is ground for the signal, so v_out = −g_m·(R_D ‖ r_o)·v_in. The minus sign means it inverts.

The lab builds the model from parts. A G element is a voltage-controlled current source: G1 mout 0 in 0 322.83u pushes 322.83 µS × v(in) from mout through itself to ground. AC analysis (E1.4 AC analysis and the Bode plot) builds the same linear circuit, so the gains should agree.

XM1 out in 0 0 sky130_fd_pr__nfet_01v8 W=2 L=0.15
G1 mout 0 in 0 322.83u
RO mout 0 62.64k
RDM mout 0 36k

XM1 out in 0 0 sky130_fd_pr__nfet_01v8 W=2 L=0.15 nrs=0 nrd=0
G1 mout 0 in 0 322.85u
RO mout 0 62.71k
RDM mout 0 36k

Why 0.9 %? The SKY130 subcircuit puts a small resistance in series with source and drain: a sheet resistance times nrs or nrd squares, about 34 Ω here. The printed g_m belongs to the transistor inside. E3.4 shows why a source resistor lowers the gain.

How small is small

Back to the hook. The model is a straight line; the device is not. Near this bias the current grows faster than linearly with V_GS, so the output dips further than it rises: 0.70 V down, 0.61 V up.

At 1 mV the swing matches the AC gain to four digits. At 10 mV it is 0.1 % low, at 100 mV 10 % low. So a small signal is small against what bends the curve. In strong inversion that is the overdrive V_GS − V_th. Near threshold, where this device sits, it is a few times kT/q (26 mV).

Capacitances and the transit frequency

At high frequency the gate is no longer an open circuit: charge in the channel and in the gate's overlap of the diffusions forms capacitors. The lab measures them on copies of M1 at the same bias.

Junction capacitance grows with the drain's area and perimeter (AD, AS, PD, PS); leave them out and it is much smaller. C_gd is small but matters most: it joins the input to an output that moves 7.3 times as much the other way, so it draws about 8 times its own current (E6).

How fast can a transistor work? Feed AC current into the gate and hold the drain still. As frequency rises C_gg takes more current, i_g = 2πf·C_gg·v_gs, so the current gain |i_d/i_g| falls as 1/f. The transit frequency f_T is where it reaches 1, about g_m/(2π·C_gg).

f_T rises with overdrive and falls steeply with length: 38 times lower at L = 1 µm than at 0.15 µm. The square law's 1/L² predicts 44; E2.3 showed why a 0.15 µm device departs from it. The estimate lands 4 to 6 % low. At f_T part of the gate's capacitive current also reaches the drain, through C_gd and the channel, and g_m does not count it.

Try it

Run the netlist. The dc sweep finds the input that puts v(out) at 0.9 V and reads the device parameters there; alter moves every gate to that input; ac then measures the transistor stage, the hand-built model and the capacitances in one run.

Now add nrs=0 nrd=0 to XM1, run, copy the new gm_b and ro_b into G1 and RO, and run again: the two gains agree to four digits. Then change L=0.15 to L=0.5 on M1 only and predict which way ro_b and the gain move before you run.

Check yourself

* a voltage-controlled current source of 1 mS into 10 kohm
V1 in 0 dc 0 ac 1
G1 out 0 in 0 1m
R1 out 0 10k
.control
ac lin 3 1k 3k
meas ac a FIND vr(out) AT=1k
.endc
.end