The square law and where SKY130 departs from it

The square law in four steps

The square law comes from a few lines of physics. Each step uses an assumption, named in it.

1. Charge

   Above threshold the gate holds a sheet of mobile charge under the oxide. Per unit area it is C_ox(V_GS − V_th − V(x)), where C_ox is the oxide capacitance per area and V(x) the channel voltage at position x. Assumption: V_th is fixed.

2. Velocity

   The charge drifts with velocity μ·E, where μ is the mobility and E = dV/dx the field along the channel. Assumption: μ is constant and velocity grows in proportion to the field.

3. Integrate

   Current is charge per length times width times velocity. Integrating from source to drain gives I_D = μC_ox(W/L)[(V_GS − V_th)V_DS − V_DS²/2] in triode.

4. Saturate

   The bracket peaks at V_DS = V_GS − V_th. Beyond it I_D = ½·μC_ox·(W/L)·(V_GS − V_th)²: current grows with W/L and with the square of the overdrive.

Take the square root of both sides: √I_D = √(½μC_oxW/L)·(V_GS − V_th). If the law holds, √I_D against V_GS is a straight line that crosses zero at V_th. That makes a simple test.

Testing it with a square-root plot

The lab runs three devices with W/L = 10 at L = 0.15, 0.3 and 1 µm, with V_DS = 1.8 V. The figure plots √I_D in units of √µA, so 30 means 900 µA. The dashed line is fitted to the long device at 1.0 and 1.2 V.

The slope gives the process constant: ½μC_ox·10 = (31.4455 × 10⁻³)² A/V², so μC_ox ≈ 198 µA/V². With that and V_th = 0.526 V, the square law predicts 1.606 mA for every W/L = 10 device at 1.8 V. No length reaches it.

Where the missing current goes

Velocity saturation. In a short channel the field along it is high. Beyond a point, carriers stop speeding up with the field and drift at a nearly fixed velocity. The current then grows more slowly than the square of the overdrive, closer to linearly.

Mobility degradation. A high gate voltage pulls carriers against the oxide surface, where they scatter more and move more slowly. This lowers μ at high V_GS for any length, and it is one reason even the 1 µm device falls 13 % short.

A higher threshold. The short device starts later (E2.2). Its constant-current threshold (1 µA, which is 100 nA × W/L) at V_DS = 1.8 V is 0.587 V against 0.533 V for the long one. Both sit below their 50 mV values of E2.2 because of DIBL.

One test separates the shape from the threshold: the current ratio from V_GS = 1.2 V to 1.8 V. The alpha-power law fits I_D ∝ (V_GS − V_th)^α to each device. Its exponent is 2 for the square law and 1 for a fully velocity-saturated device.

The fix is to size from the device you will use. Scaling the short device's width by the measured current, not by W/L, matches the long one.

* W/L = 10, copied from the 10/1 device
XN1 d1 g 0 0 sky130_fd_pr__nfet_01v8 W=1.5 L=0.15

* W = 1.5 x 1.396 / 0.731 = 2.86, trimmed to 2.81 by a rerun
XN4 d4 g 0 0 sky130_fd_pr__nfet_01v8 W=2.81 L=0.15

The PMOS adds a second gap. At W/L = 2/0.15 µm, |V_GS| = 0.9 V and |V_DS| = 1.8 V the NMOS carries 113.7 µA and the PMOS 39.8 µA, a ratio of 2.85: lower hole mobility and a larger threshold together. As a cross-check, a 7/0.15 µm NMOS at V_GS = V_DS = 1.8 V gives 3.523 mA, and the SKY130 documentation lists 3.512 mA for that size in tt.

Try it

The fit line is built inside the run: s3[500] and s3[600] are the long device's √I_D at 1.0 and 1.2 V, and i_fit squares the line back into a current. Open Plots to see every curve, including fit.

Move the fit points to 0.8 and 0.9 V (points 400 and 450, divide by 0.1) and watch vt_fit and i_sq change. Near threshold the long device is not square-law either, so the "constants" depend on where you fit them.

Check yourself

* four parallel copies of a device against one
.lib sky130 tt
VG g 0 1.2
VD d 0 1.8
VA d d1 0
VB d d4 0
XN1 d1 g 0 0 sky130_fd_pr__nfet_01v8 W=1.5 L=0.15
XN4 d4 g 0 0 sky130_fd_pr__nfet_01v8 W=1.5 L=0.15 m=4
.control
dc VG 1.1 1.3 0.1
let ratio=i(VB)/i(VA)
meas dc ratio FIND ratio AT=1.2
.endc
.end