Channel-length modulation

Channel-length modulation is the slight rise in saturation drain current with $V_{DS}$ caused by the pinch-off point moving back toward the source as $V_{DS}$ increases. It is the reason a real MOSFET in saturation is not a perfect current source but has a finite MOSFET output resistance.

The mechanism

The MOSFET square-law $i_D=\frac{1}{2}{k}_n{V}_{OV}^2$ says the saturation current is independent of $V_{DS}$. That is almost true, not quite. Past Channel pinch-off the channel is pinched at the drain end, and the leftover voltage $V_{DS}-V_{OV}$ drops across a small depletion region between the pinch-off point and the drain. Raise $V_{DS}$ further and that depletion region widens, which pushes the pinch-off point a little way back toward the source. The part of the channel that actually carries current — from the source to the pinch-off point — is now slightly shorter: the effective channel length is $L^{\prime }=L-\Delta L$, with $\Delta L$ growing as $V_{DS}$ grows.

Drain current is inversely proportional to channel length (a shorter resistive channel passes more current for the same overdrive). So as $V_{DS}$ rises, $L^{\prime }$ shrinks and $i_D$ creeps upward instead of staying perfectly flat.

As $V_{DS}$ rises in saturation, pinch-off recedes, $L^{\prime }<L$, $i_D$ rises slightly.

The modified square-law

To a good approximation the effect is captured by a linear factor in $V_{DS}$:

$i_D=\frac{1}{2}{k}_n{V}_{OV}^2(1+\lambda V_{DS})$

where $\lambda$ is the channel-length modulation parameter, a process constant with units of ${\text{V}}^{-1}$, and $V_{OV}=V_{GS}-V_t$ is the Overdrive voltage. Without the $(1+\lambda V_{DS})$ factor this is the ideal square-law; the factor tilts the otherwise-flat saturation curves slightly upward with $V_{DS}$.

The reciprocal of $\lambda$ defines the Early voltage:

$V_A=\frac{1}{\lambda }$

named by direct analogy to the BJT’s Early effect. Channel-length modulation is the MOSFET’s Early effect — same idea (effective base/channel width shrinks with the output voltage, output current rises), same finite-output-resistance consequence.

Why it matters: finite output resistance

A tilted saturation curve means the device has a finite slope $\partial i_D/\partial v_{DS}$ in saturation, i.e. a finite output resistance

$r_o=\frac{1}{\lambda I_D}=\frac{V_A}{I_D}$

(see MOSFET output resistance). This $r_o$ sits in parallel with the drain load in the MOSFET small-signal model, so it directly caps the gain of a Common-source amplifier: $A_v=-g_m(R_D\parallel r_o)$. Typical values: $\lambda \approx 0.01$–$0.05{\text{V}}^{-1}$, so $V_A\approx 20$–$100\text{V}$, giving $r_o$ in the hundreds of kΩ at usual bias currents. Smaller $\lambda$ (larger $V_A$) means flatter curves, larger $r_o$, and higher achievable gain.