D2 · MOSFET

Operating Regions: First a Resistor, Then a Current Source

Once the channel exists, the actual current is decided by the drain voltage. Raise the drain voltage yourself and see the MOSFET behave like a resistor, then past a point the current flattens and it turns into a current source.

Raise the drain voltage and see two faces

The curve is the Id-Vds characteristic drawn with the gate held on. When the drain voltage Vds is small, the current rises almost in proportion. But once Vds reaches the overdrive, the curve bends like a knee and goes flat. Cross that knee yourself.

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. The MOSFET behaves like a voltage-controlled resistor.

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 zero: cutoff (D1). Once the channel exists, apply only a small drain voltage and the channel is a nearly uniform conducting layer from source to drain. Put a voltage across that layer and current flows as by Ohm’s law, nearly proportional to Vds. The resistance of the channel here is set by the gate voltage. So the MOSFET is a resistor whose value you change with voltage, and the on-resistance of a digital switch or an analog switch uses exactly this region.

Pinch-off: the channel closes at the drain

As Vds rises, the potential along the channel climbs from source toward drain. The force that opens the channel is the voltage between gate and channel, and near the drain the channel potential is high, so this value shrinks. When Vds reaches the overdrive Vov = Vgs - Vth, at the drain end the gate-to-channel voltage falls exactly to the threshold and the channel thickness there goes to zero. This state, where the channel no longer reaches the drain, is pinch-off.

Saturation: the current flattens

Beyond pinch-off, raising Vds further only nudges the pinch point slightly inward, while the current the channel carries stays nearly the same. So the current flattens: the saturation region. Here the size of the current is set not by Vds but by the overdrive, following the square law Id ≈ (1/2) k Vov². This region, where the MOSFET behaves like a current source controlled by Vgs, is the seat of 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 - ?

The overdrive is the gate voltage in excess of the threshold.

Back to the first screen

When the drain voltage was low the curve rose straight. In that stretch the MOSFET was a resistor whose value the gate set. As you raised the drain voltage to the overdrive, the curve bent at the knee, the moment the channel end at the drain pinched off. Above it lay a flat saturation where more voltage barely changed the current, and there the MOSFET became a current source whose size Vgs set. One device, even while on, has two faces: a resistor and a current source. What drew the boundary was whether the drain voltage reached the overdrive.

Once the channel exists (D1), the current is decided by the drain voltage Vds. With small Vds the channel runs evenly and the device acts like a voltage-controlled resistor in the linear (triode) region (Id ∝ Vds). When Vds reaches the overdrive Vov = Vgs - Vth, the channel end at the drain pinches off. Above it, in the saturation region, raising Vds barely changes the current, so the MOSFET acts like a current source controlled by Vgs (Id ≈ (1/2) k Vov²). If Vgs < Vth it is in cutoff, Id = 0.

What comes next

In saturation the MOSFET was a current source controlled by Vgs. The next unit, as with the BJT, peels off only the small wobble about that operating point and linearizes it. A small gate voltage makes a small drain current, and that constant of proportion is the MOSFET’s transconductance gm. We walk the same road to amplification again, now with the MOSFET.