Transconductance (gm): Visual walkthrough — Hardware deep dive

Step 1 — A knob and a hose: what are we even measuring?

WHAT. We draw the relationship as a curve: knob voltage on the horizontal axis, output current on the vertical axis.

WHY. Everything about "how much control the input has over the output" is hidden in the shape of this curve. Before any algebra, we just look at it.

PICTURE. Look at the curve below. The horizontal axis is the input voltage (the knob). The vertical axis is the output current (the hose). Notice it is not a straight line — it bends upward. That bend is the whole story.

Step 2 — Why a slope and not a ratio

WHAT. We pick one place on the curve — the operating point (bias point) labelled $Q$ — and draw the little wedge that hugs the curve there.

WHY this tool, the slope? The transistor sits parked at $Q$ (set by the DC power supply). A tiny AC signal only nudges it left and right by a hair around $Q$ . Over that hair-width the curve looks straight, and the current it produces is governed by the local steepness — the slope. A ratio like $I_{C} / V_{B E}$ (the height of $Q$ divided by its horizontal position) is the slope of the dashed line from the origin to $Q$ — a totally different, much gentler line. The signal never feels that line.

PICTURE. Below: the amber wedge is the true local slope at $Q$ . The faint white dashed line from the origin is the ratio $I_{C} / V_{B E}$ . See how much steeper the amber wedge is — that gap is why " $g_{m} = I_{C} / V_{B E}$ " is the classic mistake (it underestimates by ~27×).

Step 3 — The BJT curve is an exponential

WHAT. We name the exact shape of Step 1's curve: an exponential in $V_{B E}$ .

WHY. The base–emitter junction is a forward-biased diode. The number of charge carriers pushed across it grows exponentially with the voltage — this is physics, not a modelling choice.

PICTURE. The curve below is the same one from Step 1, now labelled with the law. Every time $V_{B E}$ climbs by $V_{T} \approx 26 mV$ , the current multiplies by $e \approx 2.718$ . That relentless multiplication is why the curve rockets upward.

Step 4 — Take the slope of the exponential

WHAT. We compute the slope of the exponential curve.

WHY the $1/ V_{T}$ ? The exponential has a magic property: its slope is a copy of itself. But our exponent is $V_{B E} / V_{T}$ , not just $V_{B E}$ , so the "chain rule" tacks on the rate at which the exponent changes — which is $1/ V_{T}$ . Small $V_{T}$ ⇒ large $1/ V_{T}$ ⇒ steep curve ⇒ big $g_{m}$ .

PICTURE. Below, three tangent wedges at three heights. As we slide $Q$ up the curve, the wedge gets steeper in exact proportion to the height. The slope literally tracks the current.

Step 5 — The beautiful cancellation

WHAT. We recognise that the whole exponential factor is just the current we already have, and substitute.

WHY it matters. Two different BJTs at the same $I_{C}$ have the same $g_{m}$ . The slope of the curve at height $I_{C}$ is always $I_{C} / V_{T}$ — like every exponential, the steepness is fixed by the height and the horizontal scale, nothing else.

PICTURE. Same current line drawn across two different- $I_{S}$ curves; the tangent slope where each crosses is identical. Bias at $1 mA$ → $g_{m} = 1 mA /25.9 mV \approx 38.6 mS$ , every single time.

Step 6 — The MOSFET is a parabola, not an exponential

WHAT. We plot the MOSFET's curve: flat zero until $V_{GS}$ reaches $V_{t h}$ , then a parabola rising as the square of the overdrive.

WHY square, not exponential? The channel charge grows linearly with overdrive and the push moving it grows with voltage too — two linear effects multiplied give a square. Different physics, different curve, different scaling.

PICTURE. The parabola below. Note the dead-flat region for $V_{GS} < V_{t h}$ (the degenerate "off" case: $g_{m} = 0$ , no slope, no gain — the transistor is a closed tap).

Step 7 — Slope of the parabola, and its three faces

WHAT. We measure the parabola's steepness at the bias point.

WHY three forms? Each answers a different design question. Have $V_{o v}$ ? Use $k V_{o v}$ . Fixed the current $I_{D}$ and want to see the trade-off? Use $2 I_{D} / V_{o v}$ — it shows that at fixed current, more overdrive means less $g_{m}$ . Want the scaling with current alone? $2 k I_{D}$ shows $g_{m} \propto I_{D}$ .

PICTURE. Below: BJT (exponential) vs MOSFET (parabola) slopes plotted against bias current. The BJT line is straight ( $g_{m} \propto I_{C}$ ); the MOSFET is a bending square-root ( $g_{m} \propto I_{D}$ ). Per amp of bias, the BJT wins.

Step 8 — From slope to gain

WHAT. We connect $g_{m}$ to the thing engineers actually buy: voltage gain.

WHY. All amplifier gain traces back to this one slope. Bigger $g_{m}$ ⇒ bigger gain for the same load. Everything on this page exists to make that number as large as usefully possible.

PICTURE. The summary below.

The one-picture summary

Recall Feynman: retell the whole walkthrough in plain words

We started by drawing a picture of a transistor as a magic tap: twist the handle (input voltage), water gushes (output current). The picture was a curve, and it was bent — not a straight line. So we asked: how much extra water for a tiny twist right where the tap is set? That's the slope, and we called it $g_{m}$ . We were careful: the slope at that spot is much steeper than the lazy line drawn from "no water" up to where we're parked — that lazy line is the ratio trap.

For the BJT the curve is an exponential (a diode grows its current explosively with voltage). Exponentials have a party trick: their slope is a copy of themselves, scaled by $1/ V_{T}$ . So the slope became $I_{C} \times \frac{1}{V _{T}}$ — and all the messy transistor details cancelled. Any BJT at the same current has the same $g_{m}$ : about $38.6 mS$ at a milliamp.

For the MOSFET the curve is a parabola that stays flat until the gate crosses threshold, then rises as the square of the overdrive. The slope of a parabola grows linearly, giving $g_{m} = k V_{o v}$ , which we rewrote three ways. Per amp of current the BJT (linear) beats the MOSFET (square-root).

Finally, why bother: feed a whisper of signal into the handle, the slope turns it into a current wiggle, and dropping that across a resistor makes an amplified, upside-down copy — gain $= - g_{m} R_{L}$ . Every bit of amplifier gain is really just the steepness of that one bent curve.

Recall Numbers to be able to reproduce

BJT at $1 mA$ , $V_{T} = 25.9 mV$ ::: $g_{m} = 1/0.0259 \approx 38.6 mS$ MOSFET $k = 2 mA/V^{2}$ , $V_{o v} = 1 V$ ::: $g_{m} = k V_{o v} = 2 mS$ , and check $I_{D} = \frac{k}{2} V_{o v}^{2} = 1 mA$ Same $1 mA$ , MOSFET $V_{o v} = 0.2 V$ ::: $g_{m} = 2 I_{D} / V_{o v} = 10 mS$ (BJT's $38.6$ still wins)

Connections

Transconductance (gm) — the parent topic this walkthrough visualises.
Thermal Voltage VT — the $25.9 mV$ that sets the exponential's scale.
BJT Small-Signal Model — where $g_{m} v_{b e}$ becomes a current source.
MOSFET Square-Law and Saturation — the parabola of Step 6.
Overdrive Voltage Vov — the $V_{o v}$ knob of the FET trade-off.
Common-Emitter Amplifier / Common-Source Amplifier — gain $= - g_{m} R_{L}$ .
Early Effect and Output Resistance ro — pairs with $g_{m}$ for intrinsic gain $g_{m} r_{o}$ .