D2 · MOSFET

Operating Regions: First a Resistor, Then a Current Source

Once the channel exists, the current that actually flows is decided by the drain voltage. In this unit you raise the drain voltage yourself and see the MOSFET act like a resistor and then, past a certain point, level off its current and turn into a current source.

Raise the drain voltage to see two faces

The curve is the Id-Vds characteristic drawn with the gate switched on. At low drain voltage the current rises straight, almost in proportion, but once the drain voltage reaches the overdrive, the curve bends like a knee and goes flat. Pass that knee yourself, moving from the linear region on the left to the saturation region on the right.

Drain-source voltage VdsVds = 0.4 V

Operating region

Id = k(Vov·Vds - Vds²/2)

Vov = 2.0 V   Id = 0.72

Vds is small. The channel runs evenly from source to drain, so the current is nearly proportional to Vds, and the MOSFET behaves like a resistor whose value is set by voltage.

Linear · like a resistor

Cutoff and linear: an on channel is a resistor

If Vgs is below the threshold there is no channel, and the current is 0: cutoff (D1). Once the channel exists and only a small drain voltage is applied, the channel is a conducting layer running almost evenly from source to drain, and a voltage across it drives a current nearly proportional to Vds, as in Ohm's law. The gate voltage sets the channel's resistance, so the MOSFET is in effect a resistor whose value changes with voltage. The on-resistance of a digital switch and analog switches use exactly this region.

Pinch-off: the channel closes at the drain

As Vds rises, the potential along the channel climbs from the source toward the drain. What holds the channel open is the voltage between gate and channel, and near the drain, where the channel potential is high, that voltage keeps shrinking. When Vds reaches the overdrive Vov = Vgs - Vth, the gate-to-channel voltage at the drain end falls exactly to the threshold, and the channel thickness there becomes 0. This state, in which the channel no longer reaches the drain, is pinch-off.

Saturation: the current flattens

Past pinch-off, raising Vds further only shifts the pinch point slightly inward, and the current the channel carries stays nearly the same. So it becomes the saturation region, where the current levels off; its size is set not by Vds but by the overdrive, following the square law Id ≈ (1/2) k Vov². In real devices the channel gets shorter as the pinch point moves, so the saturation region also tilts slightly upward. This region, where the MOSFET behaves like a current source controlled by Vgs, is the place for amplification, corresponding to the active region of the BJT.

ObserveId ∝ Vds

In the linear region the current is proportional to the drain voltage.

ChooseVds = ?

When the drain voltage reaches the overdrive, the channel pinches off.

Fill inId ≈ (12) k ?

The saturation current follows the square of the overdrive.

On your ownVov = Vgs - ?

Overdrive is the gate voltage beyond the threshold.

Back to the first screen

At low drain voltage the curve rose straight, and over that stretch the MOSFET was a resistor whose value the gate set. Raising the drain voltage up to the overdrive bent the curve at the knee, the moment the channel end at the drain pinched off. Above that lay flat saturation, where the current barely changed however much more voltage you applied, and there the MOSFET became a current source whose size Vgs sets. Even while switched on, one device has two faces, a resistor and a current source, and the boundary between them is whether the drain voltage reaches the overdrive.

Once the channel exists (D1), the current is decided by the drain voltage Vds. When Vds is small, the device sits in the linear (triode) region, with the channel running evenly, and behaves like a voltage-controlled resistor (Id ∝ Vds). When Vds reaches the overdrive Vov = Vgs - Vth, the channel end at the drain pinches off. Above that, in the saturation region, the current stays nearly constant however much Vds rises, and the MOSFET acts as a current source controlled by Vgs (Id ≈ (1/2) k Vov²). For Vgs < Vth it is in cutoff, with Id = 0.

What comes next

In saturation, the MOSFET was a current source controlled by Vgs. As with the BJT, the next unit separates out just the small swings around that operating point and linearizes them. The small drain current a small gate voltage makes, and the constant of proportionality between them, the MOSFET's transconductance gm: you walk the same road to amplification again, this time with the MOSFET.