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. Move the operating point yourself and see how the transconductance gm, the slope of the tangent to the square curve, changes with the bias.

Raise the operating point and watch the tangent steepen

The curve is the Id-Vgs square law of the saturation region. Raise the overdrive to move the operating point Q up the curve. The tangent drawn to the curve at Q grows steeper and steeper. That slope is the transconductance gm, the ratio of drain current to the small gate voltage that makes it.

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 bends upward as the square of the gate voltage: Id = (1/2)k(Vgs - Vth)², that is, the square of the overdrive. It is a curve, but we use the very same idea as in the BJT small-signal case (C3). Take only a very narrow stretch around the operating point and any smooth curve looks like its tangent, a straight line. A small wobble of the gate voltage vgs makes, along that straight-line relation, a wobble of drain current id.

The slope gm = k Vov = √(2k Id)

Differentiating the square curve at the operating point gives the slope of the tangent, the transconductance, as gm = k(Vgs - Vth) = k Vov. It is 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 appears. The BJT had gm = Ic/VT, directly proportional to current, but the MOSFET’s gm grows only as the square root of the current. To double gm you must spend four times the drain current.

The input is open, gate current zero

The BJT small-signal model had an input resistance rπ, because a signal current flowed into the base. But the MOSFET gate is insulated (D1), so essentially no signal current enters. There is therefore no resistance like rπ on the input side; the input is an open circuit, an impedance close to infinity. The small-signal model is thus very simple: the input is just open, and the output has a single current source id = gm vgs proportional to the gate voltage. This huge input impedance is what makes the MOSFET suited to the front of amplifiers and to 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 the square root of the drain current.

On your ownig ≈ ?

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

Back to the first screen

The more you raised the overdrive, the higher the operating point climbed on the square curve, and the tangent at Q stood steeper and steeper. The slope of that tangent is the transconductance gm, whose value was k Vov, or √(2k Id) in terms of current. Unlike the BJT, which had a gm directly proportional to current, the MOSFET’s gm grows only as the square root of current, and that difference lived right there. Moreover, thanks to the insulated gate, no current enters the input, so the model has no rπ and only the output current source gm vgs remains. It is time to drop this lean linear box into the next circuit.

To a small signal about a saturation operating point (D2), the MOSFET is a linear device. 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 the current. The gate is insulated, so no signal current enters and the input is essentially an open (infinite) impedance. The small-signal model is thus just 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 only one output current source. The next unit fits this box into the common-source amplifier. A small signal at the gate wobbles the drain current, and that current flows through a drain resistor to make a large output voltage. In exactly the same form as the common-emitter (C4), a voltage gain Av = -gm Rd comes out.