Transconductance of MOSFET

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In short: A MOSFET in saturation has gm = 2·Id/Vov (equivalently μCox·(W/L)·Vov), while a BJT has gm = Ic/VT, so at the same current the BJT delivers several times more gain per milliamp because its gm is pinned to the thermal voltage rather than to a design-chosen overdrive.

Why transconductance is the number analog designers care about

Almost every analog stage is a transconductor wearing a load. The transistor converts a small input voltage into a small output current, and the load turns that current back into voltage. The conversion factor is gm = ∂Id/∂Vgs, measured at the bias point. Voltage gain of a common-source stage is roughly −gm·(ro‖RD), the unity-gain bandwidth of an amplifier is gm/(2π·CL), and thermal noise referred to the input scales as 1/gm. If you can estimate gm quickly and correctly, you can size half an op-amp on the back of an envelope, which is exactly what interviewers at analog houses test.

This article derives the MOSFET result from the square-law model, shows where the BJT formula comes from, and then explains the practical consequences using real bias numbers. For the full equivalent circuit that gm plugs into, see our companion piece on the MOSFET small-signal model.

Deriving gm for a long-channel MOSFET

In saturation, with Vov = Vgs − Vth and channel-length modulation ignored, the drain current of an NMOS is:

Id = (1/2) · μn·Cox · (W/L) · (Vgs − Vth)^2
   = (1/2) · k' · (W/L) · Vov^2

Differentiate with respect to Vgs while holding Vds fixed. Vth is constant, so dVov/dVgs = 1 and the power rule does the rest:

gm = dId/dVgs = k' · (W/L) · Vov

That is already a usable result, but it hides the design trade. Substitute k'·(W/L) = 2·Id / Vov^2 from the current equation and the formula collapses to the form everybody memorises:

gm = 2·Id / Vov        also   gm = sqrt( 2·k'·(W/L)·Id )

Key fact: The three forms are the same number. Use 2Id/Vov when you know the bias, μCox·(W/L)·Vov when you know the geometry, and √(2k'(W/L)Id) when you want to see how gm grows only as the square root of current.

A worked example: a device with k' = 200 µA/V², W/L = 20, biased at Vov = 0.2 V. Then Id = 0.5 · 200 µ · 20 · 0.04 = 80 µA and gm = 200 µ · 20 · 0.2 = 0.8 mS. Check with 2Id/Vov = 160 µA / 0.2 V = 0.8 mS. Same answer, as it must be.

The square-law curve and its tangent at the bias point. A steeper tangent means more current change per volt of gate swing.

Where the BJT formula comes from

A bipolar transistor in forward-active mode follows the exponential Ebers–Moll relation Ic = Is·exp(Vbe/VT), with thermal voltage VT = kT/q ≈ 25.9 mV at 300 K. Differentiating an exponential returns itself divided by the exponent's scale:

gm = dIc/dVbe = Ic / VT

There is no process parameter and no geometry in that result. At Ic = 1 mA, gm ≈ 38.6 mS regardless of whether the device is a SiGe HBT or a discrete 2N3904. The only knobs are current and temperature (VT rises about 0.33% per kelvin, so gm at fixed current falls as the die heats up).

Head-to-head: same current, very different gm

Put both devices at Id = Ic = 1 mA. The BJT gives 38.6 mS no matter what. The MOSFET gives 2 mA/V ÷ Vov, so the answer depends on how hard you overdrive the gate:

Device	Bias	gm at 1 mA	gm/Id
BJT	Ic = 1 mA	38.6 mS	≈ 38.6 V⁻¹
MOSFET	Vov = 0.4 V	5 mS	5 V⁻¹
MOSFET	Vov = 0.2 V	10 mS	10 V⁻¹
MOSFET	Vov = 0.1 V	20 mS	20 V⁻¹

Key fact: gm/Id for a BJT is 1/VT, a fixed ceiling near 38.6 V⁻¹. A MOSFET can only approach that ceiling by dropping Vov toward zero, which costs it width, speed and linearity.

The BJT's exponential law gives a fixed high gm per amp; the MOSFET's gm is tunable and only competes at very small overdrive.

Watch it move: gm as a gate-swing amplifier

As the gate voltage sweeps right, the operating point climbs the curve and the drain-current bar grows faster than linearly, the square law at work.

Why the two devices behave so differently

The difference is physical. In a BJT, the base-emitter voltage changes the barrier for injected carriers exponentially, so every 60 mV at room temperature changes Ic by a decade. In a strongly inverted MOSFET, Vgs sets the density of charge in the channel linearly, and drift current scales with that charge and with how fast it moves, giving a squared dependence. Square law is gentler than exponential, which is why gm only grows as √Id at fixed W/L.

There is a twist that interviewers love. Reduce Vov far enough and the MOSFET enters weak inversion (subthreshold), where current becomes exponential too: Id ∝ exp(Vgs/(n·VT)) with slope factor n of roughly 1.2 to 1.5. There gm = Id/(n·VT), so the MOSFET's gm/Id saturates near 1/(n·VT), always slightly below the BJT limit. Moderate inversion is the smooth bridge between the two regimes, and it is where many low-power designs live.

Practical consequences for design

Power efficiency. For a given gm target, BJTs need less current. A 10 mS input stage costs 260 µA in a BJT but 1 mA in a MOSFET at Vov = 0.2 V.
Input impedance. The MOSFET gate draws no DC current, so rin is effectively infinite; a BJT has rπ = β/gm, a few kilohms at 1 mA.
Intrinsic gain. gm·ro for a MOSFET is 2/(λ·Vov), which improves at low Vov but is limited by short-channel effects; for a BJT it is VA/VT, around 2,000 for VA = 50 V.
Linearity and headroom. Larger Vov gives better linearity and swing tolerance but lowers gm/Id. This is the central MOSFET trade.
Short-channel reality. In 28 nm and below, velocity saturation makes Id closer to linear in Vov, so gm approaches Cox·W·vsat and the 2Id/Vov rule overestimates gm. Always confirm in SPICE.

In practice nobody trusts hand formulas for final sizing. Designers sweep gm/Id versus Id/W with a DC simulation in Cadence Spectre, Synopsys HSPICE or Siemens Eldo and pick a point from the lookup curve. We cover that workflow in gm/Id design methodology. Teams that need production analog blocks sized and verified can also engage Avecas design services.

Common interview traps

First, forgetting that the three gm formulas assume saturation and a fixed Vds; in triode, gm = k'(W/L)·Vds instead. Second, quoting gm = Ic/VT for a MOSFET or 2Id/Vov for a BJT. Third, ignoring that gm at fixed Id depends on temperature through mobility for MOSFETs and VT for BJTs. A clean answer states the regime, the formula, and a number: "At 1 mA, a BJT gives about 39 mS; a MOSFET at 0.2 V overdrive gives 10 mS."

If you want a structured route into this material, our CourseTron analog courses pair derivations with SPICE labs, and ChipXpert's VLSI training programs add placement-oriented interview practice.

Ready to go deeper? Join the CourseTron analog design course to practise gm/Id sizing, build amplifiers in SPICE and get interview-ready on transistor-level fundamentals.

FAQ

Why does gm = 2Id/Vov make a MOSFET look better at small Vov?

Because for a fixed current, a smaller overdrive means a wider device, and a wider device is a stronger transconductor. The cost is bigger capacitance, lower speed and poorer linearity, so designers rarely go below about 0.1 V overdrive in strong inversion.

Is BJT gm really independent of device size?

Yes, at a given collector current and temperature. Area changes Is, which shifts the required Vbe, but the slope Ic/VT is unchanged. High-level injection at very large currents is the exception.

What is gm/Id and why is it used?

It is transconductance efficiency, gm divided by drain current. It captures how much gain you get per unit of power and maps cleanly onto inversion level, so it works across weak, moderate and strong inversion where the square law fails.

Does the 2Id/Vov formula hold in modern FinFET nodes?

Only approximately. Velocity saturation and mobility degradation flatten the Id–Vgs curve, so extracted gm values from foundry models differ from the hand result. Use the formula for intuition and the simulator for sign-off.