MOSFET cut-off region

The cut-off region is the MOSFET state with no channel and no current — the device is off. The condition is simply

$V_{GS}<V_t$

the gate-to-source voltage is below the Threshold voltage. With $V_{GS}$ below $V_t$ the gate field is not strong enough to invert the surface, so no channel forms (see channel formation). With no channel there is no conducting path between source and drain, and $i_D=0$ regardless of $V_{DS}$.

What it means in a circuit

Cut-off is the open-switch state. In a Common-source amplifier or any resistor-loaded stage, no drain current means no drop across the drain resistor $R_D$, so the output sits all the way up at the supply:

$V_{DS}=V_{DD}(V_{GS}<V_t)$

This is the flat top-left plateau of the Voltage-transfer characteristic: as long as the input is below threshold the output is pinned at $V_{DD}$ and nothing happens. Driving the input further below $V_t$ changes nothing — the device is already fully off. For amplification you must bias the input above $V_t$ and into the saturation region; cut-off is the wrong side of the curve for linear signals, but it is exactly the off-state you want when using the MOSFET as a switch.

Cut-off is one of the three MOSFET regions of operation, alongside the MOSFET triode region and MOSFET saturation region. For a p-MOSFET the off condition is the magnitude version: cut-off when $|V_{GS}|<|V_{tp}|$ (with $V_{tp}<0$).