MOSFET square-law

The square-law is the drain-current equation of a MOSFET in saturation: the current grows with the square of the Overdrive voltage.

$i_D=\frac{1}{2}{k}_n^{\prime }\frac{W}{L}{V}_{OV}^2=\frac{1}{2}{k}_n(V_{GS}-V_t{)}^2$

Here $V_{OV}=V_{GS}-V_t$ is the overdrive, $V_t$ the Threshold voltage, $k_n^{\prime }={\mu }_n{C}_{ox}$ the process MOSFET transconductance parameter (electron mobility times oxide capacitance per unit area), $W/L$ the channel width-to-length ratio, and $k_n=k_n^{\prime }(W/L)$ the device transconductance parameter. The current depends on $V_{GS}$ (through $V_{OV}$) but not on $V_{DS}$ — that independence is what makes the saturated MOSFET a voltage-controlled current source.

Deriving it from the triode equation

You do not need a separate physical derivation — the square-law falls straight out of the triode equation evaluated at the pinch-off boundary. In triode,

$i_D=k_n^{\prime }\frac{W}{L}\left[V_{OV}{V}_{DS}-\frac{V_{DS}^2}{2}\right]$

Pinch-off (see Channel pinch-off) occurs at $V_{DS}=V_{OV}$. That is the largest $V_{DS}$ for which the triode equation still describes the device; beyond it the current stops growing and stays flat. Substitute $V_{DS}=V_{OV}$:

$i_D=k_n^{\prime }\frac{W}{L}\left[V_{OV}\cdot V_{OV}-\frac{V_{OV}^2}{2}\right]=k_n^{\prime }\frac{W}{L}\left[V_{OV}^2-\frac{V_{OV}^2}{2}\right]=\frac{1}{2}{k}_n^{\prime }\frac{W}{L}{V}_{OV}^2$

That is the square-law. It works because the triode curve is continuous and flattens to zero slope exactly at $V_{DS}=V_{OV}$, so the current “freezes” at this value and holds it through the saturation region. (The flatness is only approximate in a real device — the slow rise with $V_{DS}$ is Channel-length modulation.)

The MOSFET analogue of the BJT exponential

Every transistor’s amplifying behaviour rests on one big-signal control law. For the Bipolar junction transistor it is the exponential BJT collector current $i_C=I_S{e}^{v_{BE}/V_T}$. For the MOSFET it is this square-law. The shapes differ — exponential vs quadratic — which has real consequences: a BJT delivers more transconductance per unit current, but the MOSFET’s square-law and insulated gate make it ideal for dense, low-static-power circuits. In both cases, the small-signal model comes from linearising this big-signal law about a bias point: differentiating the square-law gives the MOSFET transconductance $g_m=k_n{V}_{OV}=2I_D/V_{OV}$.

Worked sanity check

Take $k_n=1{\text{mA/V}}^2$ and $V_t=0.7\text{V}$. Bias at $V_{GS}=1.2\text{V}$, so $V_{OV}=0.5\text{V}$. Then

$I_D=\frac{1}{2}(1{\text{mA/V}}^2)(0.5\text{V}{)}^2=\frac{1}{2}(1)(0.25)\text{mA}=0.125\text{mA}$

Double the overdrive to $V_{OV}=1\text{V}$ ($V_{GS}=1.7\text{V}$): $I_D=\frac{1}{2}(1)(1{)}^2=0.5\text{mA}$ — four times the current for twice the overdrive, the signature of a square law.