MOSFET Drain Current in the Linear (Triode) Region

Below pinch-off — that is, with the drain-source voltage smaller than the overdrive — the inversion layer runs continuously from source to drain and the transistor behaves like a resistor whose value the gate sets. The current has two terms. The first, $(V_{GS}-V_{th})V_{DS}$, is plain Ohm's law through a channel of fixed conductance. The second, $-V_{DS}^2/2$, corrects for the fact that the channel is not uniform: the drain end sits at a higher potential, so the gate-to-channel voltage there is smaller and the layer is thinner.

When $V_{DS}$ is small compared with the overdrive, the squared term is negligible and the device is a clean linear resistor of value $1/[k_n (W/L)(V_{GS}-V_{th})]$. This is the operating point of every pass gate, transmission gate, analogue switch and power MOSFET, and it is where the datasheet parameter $R_{DS(on)}$ comes from. The formula also says what to do about it: on-resistance falls with gate drive and with width, which is why a power MOSFET is enormous and why driving its gate hard enough matters more than most of its other specifications.

The boundary between this region and saturation is not an arbitrary line. Set $V_{DS} = V_{GS}-V_{th}$ in this equation and it gives $\tfrac{1}{2}k_n(W/L)(V_{GS}-V_{th})^2$, which is exactly the saturation formula. The two expressions meet with the same value and the same slope, which is the condition the model was constructed to satisfy. Past that point this equation turns over and starts predicting a FALLING current, which no transistor does — it is the model leaving its domain, not a physical effect, and a solver that quietly evaluates it there is lying.

The practical warning for switching applications is about the trip through the triode region. A MOSFET used as a switch spends most of its time either fully off or deep in triode where it dissipates $I^2 R_{DS(on)}$, and both states are cheap. The expensive part is the transit: during a transition the device passes through saturation carrying substantial current with substantial voltage across it, and that product is the switching loss. It is why gate drive strength matters so much in power electronics, and it is the same physics as the short-circuit power term in a CMOS logic gate, seen at a scale where it burns your fingers instead of a milliwatt.