D3 · MOSFET

The MOSFET Small-Signal Model: The Slope of a Square Curve

In saturation, the MOSFET was a current source controlled by the gate voltage. In this unit you move the operating point yourself and see how the transconductance gm, the slope of the tangent to the square-law curve, changes with the bias.

Raise the operating point and watch the tangent steepen

The curve is the Id-Vgs square law in saturation, starting at the threshold voltage. Increase the overdrive and the operating point Q climbs along the curve, while the tangent drawn to the curve at Q grows steeper. That slope is the transconductance gm, which tells how strongly a small swing of gate voltage swings the drain current.

Bias overdrive VovVov = 0.4 V

Transconductance gm

gm = k Vov = 0.4

Id = 0.08   gm = √(2k Id) = 0.4

The overdrive is small. The operating point sits on the gentle lower part of the curve, so the tangent slope gm is small.

Weak gm

In saturation Id is the square of Vgs

In saturation, the drain current curves upward as the square of the gate voltage: Id = (1/2)k(Vgs - Vth)², the square of the overdrive. It is a curve, but the same idea used for the BJT's small signal (C3) applies: cut out just a very narrow stretch around the operating point and any smooth curve looks like its tangent, a straight line. A small swing of gate voltage, vgs, follows that straight-line relation to make a swing of drain current, id.

The slope gm = k Vov = √(2k Id)

Differentiate the square-law curve at the operating point and the slope of the tangent, the transconductance, comes out as gm = k(Vgs - Vth) = k Vov, directly proportional to the overdrive. Meanwhile, since Id = (1/2)k Vov², writing the same value in terms of current gives gm = √(2k Id). Here a big difference from the BJT shows up: the BJT had gm = Ic/VT, directly proportional to current, but the MOSFET's gm grows only as the square root of current. Doubling gm takes four times the drain current, and at the same current a BJT usually has the larger gm.

The input is open, gate current 0

The BJT's small-signal model had an input resistance rπ, because signal current flowed into the base. A MOSFET's gate, however, is insulated (D1), so practically no signal current goes in; there is no resistance corresponding to rπ on the input side, and the input is an open circuit, an impedance close to infinite. So the small-signal model is very simple: the input is just open, and the output has a single current source proportional to the gate voltage, id = gm vgs . This huge input impedance is what suits MOSFETs to amplifier front ends and digital logic.

Observeid = gm vgs

The output current is the transconductance times the gate voltage.

Choosegm = k ?

The slope is directly proportional to the overdrive.

Fill ingm = √(2 k ?)

In terms of current, gm is proportional to the square root of the drain current.

On your ownig ≈ ?

With an insulated gate, the gate current is almost 0 and the input is open.

Back to the first screen

The more you increased the overdrive, the further the operating point climbed toward the steep part of the square-law curve, and the tangent at Q stood steeper and steeper. The slope of that tangent is the transconductance gm, equal to k Vov, or √(2k Id) in terms of current. There lay the difference: where the BJT's gm was directly proportional to current, the MOSFET's gm grows only as the square root of current. And because no current enters the insulated gate, the model has no rπ, leaving only the output current source gm vgs . Next comes fitting this compact linear box into a circuit.

To a small signal around the saturation operating point (D2), a MOSFET is a linear element. Differentiating the square law Id = (1/2)k Vov² at the Q point gives the slope gm = k Vov = √(2k Id) (the transconductance). Unlike the BJT (gm = Ic/VT), gm grows only as the square root of current. The gate is insulated, so no signal current enters and the input is practically an open (infinite) impedance, leaving a small-signal model with an open input and a single output current source id = gm vgs .

What comes next

You now hold a simple linear box with an open input and a single current source at the output. In the next unit you fit this box into a common-source amplifier. A small signal at the gate swings the drain current, and that current flows through the drain resistor to make a large output voltage, giving a voltage gain Av = -gm Rd of exactly the same form as the common emitter (C4).