Transconductance and gm/ID

What g_m measures

Raise the gate of a saturated transistor by a few millivolts and its drain current rises; the ratio is the transconductance g_m = ∂I_D/∂V_GS, in siemens. It is the gain of the device itself, because a small signal v_gs on the gate becomes a current g_m·v_gs in the drain.

The square law of E2.3, I_D = ½μC_ox(W/L)V_ov² with the overdrive V_ov = V_GS − V_th, gives three forms of g_m: μC_ox(W/L)V_ov, √(2μC_ox(W/L)I_D) and 2I_D/V_ov. They are the same equation rearranged, so they stand or fall together. The last one needs no process constant, which is why designers reach for it.

Divide by the current and the last form becomes g_m/I_D = 2/V_ov. That ratio is the transconductance efficiency: how much g_m each ampere of bias buys, in V⁻¹. The hook's plot compares that ratio with the simulator. Below about 0.13 V of overdrive the square law promises too much, because its 2/V_ov grows without limit. The device cannot follow: near and below the threshold the channel is weakly inverted and current grows exponentially with V_GS (E2.6), which caps g_m/I_D.

The error also depends on which threshold you subtract. A threshold is a definition (E2.2), and a few tens of millivolts shift every V_ov in the table. A better method needs no threshold at all.

One chart per length: g_m/I_D against current density

Divide the drain current by the width and you get the current density I_D/W, in A/µm. Two devices of the same length at the same density have the same g_m/I_D, whatever their widths. The reason is simple: two identical devices side by side carry twice the current and give twice the g_m, so the ratio and the density stay put.

The chart below sweeps the density itself: an amplifier holds each drain at 0.9 V while a source feeds it 10^x A per µm of width, so x = −6 means 1 µA/µm.

The lab checks the width claim at L = 0.5 µm. W = 8 µm runs 1.4 % to 4.3 % above W = 2 µm, a small edge effect of narrow devices. Close enough to size from one chart, then confirm on the bench.

The flat left end of the chart has a physical meaning. In weak inversion I_D ∝ exp(V_GS/(n·U_T)). U_T = kT/q is the thermal voltage, 25.9 mV at 27 °C. n is the slope factor: only 1/n of a gate change reaches the channel. Then g_m/I_D = 1/(n·U_T), the device's limit; 26.4 V⁻¹ at L = 0.5 µm gives n = 1.46.

Sizing a device with the chart

Now the g_m/I_D method, step by step, for a stage that needs g_m = 100 µS at L = 0.5 µm.

1. Choose g_m/I_D

   High values save current and low values give a smaller, faster device; take 15 V⁻¹, in moderate inversion.

2. Compute the current

   Then I_D = g_m/(g_m/I_D) = 100 µS / 15 V⁻¹ = 6.667 µA.

3. Read the density

   On the L = 0.5 µm curve, g_m/I_D = 15 at 2.77 µA/µm (jd_at_15 in the lab).

4. Compute the width

   W = 6.667 µA / 2.77 µA/µm = 2.41 µm, which rounds to 2.4 µm.

5. Check on the bench

   The current-fed bench measures g_m = 100.3 µS, within 0.3 % of the target, with no threshold and no μC_ox anywhere.

The same method works at any inversion level because the chart is measured, which the square law cannot match. Here both methods size a device for g_m = 100 µS from 5 µA, which needs g_m/I_D = 20.

* 2/Vov = 20 means Vov = 0.1 V;
* W = 3.4 puts 5 uA at about that overdrive
IB vdd d3 DC 5u
XM3 d3 g3 0 0 sky130_fd_pr__nfet_01v8 W=3.4 L=0.5
.control
dc IB 4u 8u 0.01u
let gm3 = 1/deriv(v(g3))
meas dc gm_m3 FIND gm3 AT=5u
.endc

* g_m/I_D = 20 at 0.634 uA/um on the L = 0.5 curve:
* W = 5 uA / 0.634 uA/um = 7.9 um
IB vdd d3 DC 5u
XM3 d3 g3 0 0 sky130_fd_pr__nfet_01v8 W=7.9 L=0.5
.control
dc IB 4u 8u 0.01u
let gm3 = 1/deriv(v(g3))
meas dc gm_m3 FIND gm3 AT=5u
.endc

High g_m/I_D is not free: less density means more width for the same current, so more gate area and more capacitance to charge. The device gets slower; E3.1 measures that speed as f_T. Weak inversion suits low-power and low-frequency stages, strong inversion suits fast ones, and moderate inversion is the usual compromise.

Try it

Run the netlist. The first sweep moves the shared gate of M1 and M2; let turns each drain current into g_m/I_D and into a density, and meas ... WHEN jd1=1u reads g_m/I_D where the density crosses 1 µA/µm. The second sweep moves IB a little around 6.667 µA, so g_m = 1/(dV_GS/dI_D).

Now resize M3 for g_m = 100 µS at g_m/I_D = 10: set IB to 10 µA, read the density at 10 from the Plots tab or add meas dc jd_at_10 FIND jd1 WHEN gmid1=10 FALL=LAST, and compute W. Predict gm_m3 before you run.

Check yourself

* M1 is one device; M2 is four copies of it in parallel (m=4)
.lib sky130 tt
XM1 d1 g 0 0 sky130_fd_pr__nfet_01v8 W=2 L=0.5
XM2 d2 g 0 0 sky130_fd_pr__nfet_01v8 W=2 L=0.5 m=4
VD1 d1 0 0.9
VD2 d2 0 0.9
VGS g 0 0
.control
dc VGS 0 1.2 0.002
let id1 = -i(VD1)
let id2 = -i(VD2)
let ratio = (deriv(id2)/id2)/(deriv(id1)/id1)
meas dc r FIND ratio AT=0.7
.endc
.end