Transconductance (gm): Exercises — Hardware deep dive

Constants used throughout (room temperature, $T = 300$ K): $V_{T} = \frac{k T}{q} \approx 25.9 mV (see [[Thermal Voltage VT]])$

The picture behind every problem on this page

Before any numbers, fix the one image that all of these exercises lean on. A transistor has a curved I–V characteristic: output current versus the controlling input voltage. It is a curve, not a straight line, because the physics is nonlinear (exponential for a BJT, quadratic for a MOSFET). We never operate on the whole curve at once — we pick one DC operating point on it, called the ==bias point $Q$ == (the "quiescent" point, the resting spot with no signal applied).

The figure shows this literally: the blue curve is the BJT's $I_{C}$ vs $V_{B E}$ , the red dashed line is the tangent at $Q$ , and its steepness is $g_{m}$ . Every solution below is really asking "how steep is the tangent here?"

Level 1 — Recognition

Recall Solution L1·Q1

Transconductance is defined as the local slope of output current versus controlling input voltage at the bias point $Q$ : $g_{m} = \frac{\partial I _{o u t}}{\partial V _{in}}_{Q} \approx \frac{Δ I _{o u t}}{Δ V _{in}} = \frac{0.4 \times 1 0 ^{- 3} A}{2 \times 1 0 ^{- 3} V} = 0.2 A/V = 200 mS .$ WHAT we did: replaced the derivative by a small finite ratio (valid because the change is tiny — this is exactly the tangent at $Q$ in the figure above). WHY: $g_{m}$ is the slope; a small $Δ$ approximates the tangent. Units: amps-per-volt $=$ siemens (S).

Recall Solution L1·Q2

(a) BJT: $g_{m} = \frac{I _{C}}{V _{T}}$ — where $I_{C}$ is the DC collector bias current and $V_{T} \approx 25.9$ mV is the thermal voltage. WHY / origin: the base–emitter junction is a forward-biased diode, so $I_{C} = I_{S} e^{V_{B E} / V_{T}}$ (exponential). Differentiating that exponential w.r.t. $V_{B E}$ brings down a factor $1/ V_{T}$ and re-forms $I_{C}$ , giving $g_{m} = I_{C} / V_{T}$ . That is why $β$ and $I_{S}$ vanish — they cancel in the slope. (b) MOSFET (saturation): $g_{m} = k V_{o v} = \frac{2 I _{D}}{V _{o v}} = 2 k I_{D}$ , where $k = μ_{n} C_{o x} \frac{W}{L}$ , $V_{o v} = V_{GS} - V_{t h}$ is the overdrive voltage, and $I_{D}$ is the DC drain bias current. WHY / origin: in saturation the channel charge $\propto (V_{GS} - V_{t h})$ , giving the square law $I_{D} = \frac{k}{2} V_{o v}^{2}$ (see MOSFET Square-Law and Saturation). Differentiating w.r.t. $V_{GS}$ gives $g_{m} = k V_{o v}$ ; substituting $V_{o v}$ from the current gives the other two forms. Key assumption for both: the device is in its active/saturation region so these clean formulas hold — see the "Edge cases" section at the end of the page for exactly where they break.

Level 2 — Application

Recall Solution L2·Q1

$g_{m} = \frac{I _{C}}{V _{T}} = \frac{0.5 \times 1 0 ^{- 3}}{25.9 \times 1 0 ^{- 3}} = 19.3 mS .$ WHY $I_{C} / V_{T}$ and not $I_{C} / V_{B E}$ : $g_{m}$ is the derivative of the exponential $I_{C} = I_{S} e^{V_{B E} / V_{T}}$ ; differentiating brings down $1/ V_{T}$ (see the parent derivation). Geometrically: the tangent to the blue curve at this new, lower $Q$ is gentler than at 1 mA.

Recall Solution L2·Q2

WHY these formulas apply: we assume the device is in saturation, so the square law $I_{D} = \frac{k}{2} V_{o v}^{2}$ and its slope $g_{m} = k V_{o v}$ both hold. Transconductance: $g_{m} = k V_{o v} = 4 \frac{mA}{V ^{2}} \times 0.25 V = 1.0 mS .$ Drain current from the square law: $I_{D} = \frac{k}{2} V_{o v}^{2} = \frac{4}{2} (0.25)^{2} mA = 2 \times 0.0625 = 0.125 mA .$ Cross-check with the current form: $g_{m} = \frac{2 I _{D}}{V _{o v}} = \frac{2 \times 0.125}{0.25} = 1.0$ mS. ✔ (The two forms must agree — they are the same tangent written two ways.)

Recall Solution L2·Q3

$A_{v} = - g_{m} R_{L} = - (1.0 \times 1 0^{- 3}) (20 \times 1 0^{3}) = - 20.$ WHY the minus: more input voltage → more drain current → bigger drop across $R_{L}$ → output node pulled down (inverting). See Common-Source Amplifier.

Level 3 — Analysis

Recall Solution L3·Q1

BJT: $g_{m}^{BJT} = \frac{I _{C}}{V _{T}} = \frac{1 \times 1 0 ^{- 3}}{25.9 \times 1 0 ^{- 3}} = 38.6$ mS. MOSFET: $g_{m}^{MOS} = \frac{2 I _{D}}{V _{o v}} = \frac{2 \times 1 \times 1 0 ^{- 3}}{0.3} = 6.67$ mS. Ratio: $\frac{38.6}{6.67} = 5.79$ . WHY the BJT wins: the BJT's "effective overdrive" is $V_{T} \approx 26$ mV, far smaller than the MOSFET's $V_{o v} = 300$ mV. Since $g_{m} = 2 I / V_{(over)drive}$ -type scaling favours a small denominator, the tiny $V_{T}$ makes the BJT ~6× more efficient here. In the slope figure at the top of the page, the BJT's tangent at 1 mA is far steeper than a MOSFET tangent at the same current would be.

Recall Solution L3·Q2

BJT — $g_{m} \propto I_{C}$ (linear): doubling $I_{C}$ doubles $g_{m}$ , factor $= 2.00$ . MOSFET — $g_{m} = 2 k I_{D} \propto I_{D}$ : doubling $I_{D}$ multiplies $g_{m}$ by $2 = 1.414$ . WHY they differ: the BJT current is exponential in $V_{B E}$ , so its slope $I_{C} / V_{T}$ rides linearly with $I_{C}$ . The MOSFET current is quadratic in $V_{o v}$ , so its slope grows only as a square root of the current. The figure below plots both scaling laws so you can see the linear line pull away from the square-root curve.

How to read the figure below: the horizontal axis is the bias current $I$ (mA); the vertical axis is the resulting $g_{m}$ (mS). The blue straight line is the BJT ( $g_{m} = I_{C} / V_{T}$ , linear); the orange curve is the MOSFET ( $g_{m} = 2 k I_{D}$ , square-root). The two dotted verticals at $1$ mA and $2$ mA mark where we read off the doubling: the blue dots jump by $\times 2$ , the orange dots by only $\times 1.41$ .

Recall Solution L3·Q3

Wrong: $\frac{I _{C}}{V _{B E}} = \frac{1 \times 1 0 ^{- 3}}{0.70} = 1.43$ mS. Correct: $\frac{I _{C}}{V _{T}} = \frac{1 \times 1 0 ^{- 3}}{25.9 \times 1 0 ^{- 3}} = 38.6$ mS. Ratio: $\frac{38.6}{1.43} = \frac{V _{B E}}{V _{T}} = \frac{0.70}{0.0259} = 27.0$ . The student underestimates $g_{m}$ by a factor of $\approx 27$ . Geometrically, $I_{C} / V_{B E}$ is the slope of the chord from the origin to $Q$ , which is far gentler than the tangent at $Q$ .

Level 4 — Synthesis

Recall Solution L4·Q1

Required transconductance from $∣ A_{v} ∣ = g_{m} R_{L}$ : $g_{m} = \frac{∣ A _{v} ∣}{R _{L}} = \frac{100}{4 \times 1 0 ^{3}} = 25 mS .$ Required bias current from $g_{m} = I_{C} / V_{T}$ : $I_{C} = g_{m} V_{T} = (25 \times 1 0^{- 3}) (25.9 \times 1 0^{- 3}) = 6.48 \times 1 0^{- 4} A = 0.648 mA .$ WHY this order: gain fixes $g_{m}$ (through the load), and $g_{m}$ fixes the bias current — the design flows backward from the spec. See Common-Emitter Amplifier.

Recall Solution L4·Q2

From $g_{m} = k V_{o v}$ : $V_{o v} = \frac{g _{m}}{k} = \frac{5 \times 1 0 ^{- 3}}{25 \times 1 0 ^{- 3}} = 0.20 V .$ Since $0.20 V \leq 0.40 V$ , the headroom spec is satisfied. ✔ Drain current: $I_{D} = \frac{k}{2} V_{o v}^{2} = \frac{25 \times 1 0 ^{- 3}}{2} (0.20)^{2} = 12.5 \times 1 0^{- 3} \times 0.04 = 0.5 mA .$ WHY headroom fights $g_{m}$ : for a fixed $I_{D}$ , $g_{m} = 2 I_{D} / V_{o v}$ says a smaller $V_{o v}$ gives more $g_{m}$ — but smaller $V_{o v}$ also means less swing room. Here $k$ was large enough that a modest $V_{o v} = 0.2$ V met both.

Level 5 — Mastery

Recall Solution L5·Q1

(a) $g_{m} = \frac{I _{C}}{V _{T}} = \frac{1 \times 1 0 ^{- 3}}{25.9 \times 1 0 ^{- 3}} = 38.6$ mS. (b) $r_{o} = \frac{V _{A}}{I _{C}} = \frac{50}{1 \times 1 0 ^{- 3}} = 50 \times 1 0^{3} = 50 k Ω.$ (c) Intrinsic gain: $g_{m} r_{o} = \frac{I _{C}}{V _{T}} \cdot \frac{V _{A}}{I _{C}} = \frac{V _{A}}{V _{T}} = \frac{50}{0.0259} = 1931.$ WHY the current cancels: $I_{C}$ appears in $g_{m}$ (numerator) and in $r_{o}$ (denominator), so the intrinsic gain collapses to $V_{A} / V_{T}$ — a pure device figure independent of bias. That is the ceiling on a single BJT stage's gain.

Recall Solution L5·Q2

First find $V_{o v}$ from the square law $I_{D} = \frac{k}{2} V_{o v}^{2}$ : $V_{o v} = \frac{2 I _{D}}{k} = \frac{2 \times 0.5 \times 1 0 ^{- 3}}{1 \times 1 0 ^{- 3}} = 1 = 1.0 V .$ Transconductance (use $g_{m} = 2 k I_{D}$ , or equivalently $k V_{o v}$ ): $g_{m} = 2 k I_{D} = 2 \times 1 \times 1 0^{- 3} \times 0.5 \times 1 0^{- 3} = 1 \times 1 0^{- 6} = 1.0 \times 1 0^{- 3} = 1.0 mS .$ Cross-check: $g_{m} = k V_{o v} = 1 \times 1 0^{- 3} \times 1.0 = 1.0$ mS. ✔ Output resistance: $r_{o} = \frac{1}{λ I _{D}} = \frac{1}{0.02 \times 0.5 \times 1 0 ^{- 3}} = \frac{1}{1 \times 1 0 ^{- 5}} = 100 \times 1 0^{3} = 100 k Ω.$ Intrinsic gain: $g_{m} r_{o} = (1.0 \times 1 0^{- 3}) (100 \times 1 0^{3}) = 100.$ Scaling with current. Write the intrinsic gain symbolically: $g_{m} r_{o} = \frac{2 k I _{D}}{λ I _{D}} = \frac{2 k}{λ I _{D}} \propto \frac{1}{I _{D}} .$ So as $I_{D}$ decreases, the intrinsic gain rises as $1/ I_{D}$ : e.g. cutting $I_{D}$ by 4× raises $g_{m} r_{o}$ by 2×. WHY: $g_{m}$ falls only as $I_{D}$ , but $r_{o}$ climbs as $1/ I_{D}$ — the faster-growing $r_{o}$ wins. This is exactly why MOSFETs are pushed to low currents (toward weak inversion) when maximum gain is the goal.

Recall Solution L5·Q3

From $g_{m} = 2 I_{D} / V_{o v}$ : $V_{o v} = \frac{2 I _{D}}{g _{m}} = \frac{2 \times 0.9 \times 1 0 ^{- 3}}{3 \times 1 0 ^{- 3}} = 0.6 V .$ From $g_{m} = k V_{o v}$ : $k = \frac{g _{m}}{V _{o v}} = \frac{3 \times 1 0 ^{- 3}}{0.6} = 5 \times 1 0^{- 3} = 5 mA/V^{2} .$ Sanity check the current: $I_{D} = \frac{k}{2} V_{o v}^{2} = \frac{5 \times 1 0 ^{- 3}}{2} (0.6)^{2} = 2.5 \times 1 0^{- 3} \times 0.36 = 0.9$ mA. ✔

Edge cases — where these $g_{m}$ formulas do NOT apply

Common mistake "

$g_{m} = k V_{o v}$ or $I_{C} / V_{T}$ works in every region" Why it feels right: they are the formulas the course drills, so students apply them blindly to any bias. The fix — know the valid region for each device:

MOSFET triode (ohmic) region ( $V_{D S} < V_{o v}$ ): the device acts like a voltage-controlled resistor, $I_{D}$ depends on both $V_{GS}$ and $V_{D S}$ , and the saturation slope $g_{m} = k V_{o v}$ no longer holds. You are off the flat part of the curve.
MOSFET subthreshold (weak inversion) ( $V_{GS} < V_{t h}$ , $V_{o v} < 0$ ): current is exponential in $V_{GS}$ , not quadratic, so $g_{m} \approx I_{D} / (n V_{T})$ — a BJT-like law, not $k V_{o v}$ . This is the region ultra-low-power circuits live in.
BJT cutoff / saturation (switch) regions: $g_{m} = I_{C} / V_{T}$ is a forward-active result. In cutoff there is essentially no current (and no useful slope); in deep saturation the exponential law breaks down. Amplifiers must be biased in forward active for $g_{m} = I_{C} / V_{T}$ to mean anything. The one-line rule: a transistor only has the textbook $g_{m}$ inside its amplifying region — BJT forward-active, MOSFET saturation. Check the region before you reach for a formula.

Connections

Parent: Transconductance (gm) — all formulas used here are derived there.
Thermal Voltage VT — the $25.9$ mV in every BJT answer.
MOSFET Square-Law and Saturation — source of $I_{D} = \frac{k}{2} V_{o v}^{2}$ and the saturation/triode boundary.
Overdrive Voltage Vov — the L2/L4 headroom-vs- $g_{m}$ tradeoff.
Common-Emitter Amplifier / Common-Source Amplifier — gain $= - g_{m} R_{L}$ (L2, L4).
BJT Small-Signal Model — where $g_{m} v_{b e}$ lives as a current source.
Early Effect and Output Resistance ro — the L5 intrinsic-gain problems and the $λ \leftrightarrow 1/ V_{A}$ analogy.

Recall One-line answer key

L1: 200 mS ; siemens. L2: 19.3 mS ; 1.0 mS & 0.125 mA ; $A_{v} = - 20$ . L3: 38.6 vs 6.67 mS (×5.79) ; ×2 vs ×1.41 ; wrong by ×27. L4: 25 mS → 0.648 mA ; $V_{o v} = 0.2$ V, $I_{D} = 0.5$ mA. L5: $g_{m} r_{o} = 1931$ ; $g_{m} = 1$ mS, $r_{o} = 100$ kΩ, gain 100 ; $V_{o v} = 0.6$ V, $k = 5$ mA/V².