Transconductance (gm): Question bank — Hardware deep dive

Every symbol used below is built in the parent note: $g_{m}$ is the slope of output current versus controlling input voltage; $V_{T} \approx 25.9 mV$ is the thermal voltage; $V_{o v} = V_{GS} - V_{t h}$ is the overdrive voltage.

True or false — justify

True or false: Two BJTs biased at the same collector current always have the same

g_{m}

, regardless of their size or

β

True.

g_{m} = I_{C} / V_{T}

depends only on the bias current and temperature — the device-specific

I_{S}

and

β

cancel because

I_{S} e^{V_{B E} / V_{T}}

I_{C}

True or false: Two MOSFETs biased at the same drain current always have the same

g_{m}

False.

g_{m} = 2 k I_{D}

still contains the device parameter

k = μ_{n} C_{o x} W / L

, so a wider (bigger

k

) FET gives more

g_{m}

at the same current. Only the BJT enjoys full device-independence.

True or false:

g_{m}

has units of resistance.

False.

g_{m}

is current-per-voltage (siemens, A/V) — the inverse of resistance. Its reciprocal

1/ g_{m}

is a resistance, which is why the BJT emitter looks like

r_{e} = V_{T} / I_{C} = 1/ g_{m}

True or false: A larger load resistor

R_{L}

changes the transistor's

g_{m}

False.

g_{m}

is set at the input side by the bias point (

I_{C}

V_{o v}

R_{L}

only converts the output current

g_{m} v_{in}

into voltage; it scales the gain

- g_{m} R_{L}

but not

g_{m}

itself.

True or false: Raising temperature raises a BJT's

g_{m}

at fixed collector current.

False (at fixed

I_{C}

g_{m} = I_{C} / V_{T}

and

V_{T} = k T / q

rises with

T

, so

g_{m}

falls if

I_{C}

is truly held constant. (In a real circuit

I_{C}

often drifts too, but the direct dependence lowers

g_{m}

True or false: For a MOSFET, doubling the bias current doubles

g_{m}

False.

g_{m} \propto I_{D}

, so doubling

I_{D}

multiplies

g_{m}

by only

2 \approx 1.41

. Doubling would happen for a BJT, where

g_{m} \propto I_{C}

True or false: The minus sign in

A_{v} = - g_{m} R_{L}

means the amplifier loses power or has negative gain magnitude.

False. The minus is phase inversion — the output swings opposite to the input. The magnitude

g_{m} R_{L}

can be very large; the sign carries no information about loss.

Spot the error

Spot the error: "

g_{m} = I_{C} / V_{B E}

because

g_{m}

is current over the controlling voltage."

g_{m}

is a derivative (local slope), not a DC ratio. The correct slope of the exponential is

I_{C} / V_{T}

; since

V_{T} \approx 26 mV ≪ V_{B E} \approx 0.7 V

, using

V_{B E}

underestimates

g_{m}

by roughly

27 \times

Spot the error: "Increasing

V_{o v}

always increases

g_{m}

, from

g_{m} = k V_{o v}

The formula

g_{m} = k V_{o v}

holds only if you let

I_{D}

rise with

V_{o v}

. For a fixed

I_{D}

, the correct relation is

g_{m} = 2 I_{D} / V_{o v}

, which falls as

V_{o v}

grows. The trap is holding the wrong quantity constant.

Spot the error: "

g_{m} = I_{S} e^{V_{B E} / V_{T}}

, so

g_{m}

depends on the transistor's

I_{S}

That expression equals

I_{C} / V_{T}

after recognising

I_{S} e^{V_{B E} / V_{T}} = I_{C}

. Once written in terms of the bias current,

I_{S}

cancels —

g_{m}

does not depend on

I_{S}

at a given

I_{C}

Spot the error: "The BJT gives more

g_{m}

per amp than a MOSFET, so BJTs are always the better amplifier choice."

The

g_{m} / I

advantage is real, but it ignores input current (BJT base current

I_{B}

), gate leakage-free high input impedance of FETs, noise, and headroom. Higher

g_{m}

efficiency is one factor, not a verdict.

Spot the error: "In saturation

I_{D} = \frac{k}{2} V_{o v}^{2}

, so

g_{m} = \frac{d}{d V _{GS}} \frac{k}{2} V_{o v}^{2} = \frac{k}{2} \cdot 2 V_{o v} = k V_{o v}

— but if I differentiate I should also use the chain rule on

V_{t h}

V_{t h}

is a constant, so

\frac{d V _{o v}}{d V _{GS}} = \frac{d ( V _{GS} - V _{t h} )}{d V _{GS}} = 1

. The chain rule gives exactly

k V_{o v}

; there is no extra term. The "error" is imagining

V_{t h}

varies with

V_{GS}

Spot the error: "Because

g_{m} = 2 k I_{D}

, a MOSFET with zero current still has some

g_{m}

from

k

I_{D} = 0

the square root gives

2 k \cdot 0 = 0

. Equivalently

V_{o v} = 0

means

g_{m} = k V_{o v} = 0

: no channel, no current control, zero transconductance.

Why questions

Why is

g_{m}

defined as a partial derivative evaluated at the operating point Q, rather than a global slope?

The

I

–

V

curve is nonlinear (exponential for BJT, quadratic for FET), so the slope changes everywhere. A small AC signal only experiences the local slope at

Q

, which is why we linearise about the bias point.

Why does differentiating

I_{C} = I_{S} e^{V_{B E} / V_{T}}

bring out a factor

1/ V_{T}

The derivative of

e^{x / V_{T}}

with respect to

x

\frac{1}{V _{T}} e^{x / V_{T}}

— the exponent's inner

1/ V_{T}

falls out by the chain rule. That surviving

1/ V_{T}

is exactly why the tiny

\approx 26 mV

makes

g_{m}

so large.

Why is the BJT

g_{m}

independent of the transistor while the MOSFET

g_{m}

is not?

The BJT's slope re-expresses itself purely as

I_{C} / V_{T}

— device constants vanish. The MOSFET's square law keeps

k = μ_{n} C_{o x} W / L

in the answer (

g_{m} = 2 k I_{D}

), so geometry and process survive.

Why does the common-source / common-emitter gain trace entirely back to

g_{m}

The device's only job in these stages is to turn input voltage into output current,

i_{o u t} = g_{m} v_{in}

. Dropping that current across

R_{L}

gives

A_{v} = - g_{m} R_{L}

, so every volt of gain is

g_{m}

multiplied by the load.

Why do designers often prefer a small overdrive

V_{o v}

in low-power analog design?

At fixed current,

g_{m} = 2 I_{D} / V_{o v}

, so a small

V_{o v}

maximises transconductance per amp of bias current — the best gain efficiency. The cost is reduced voltage headroom and slower speed.

Why does the BJT Small-Signal Model represent the transistor as a dependent current source

g_{m} v_{b e}

rather than a voltage source?

A transistor's controlled quantity is a current (

i_{c}

), governed by input voltage

v_{b e}

. A voltage-controlled current source

g_{m} v_{b e}

captures exactly that — the defining action of transconductance.

Why can't we just measure

g_{m}

once and reuse it as the transistor's current changes?

g_{m}

depends on the bias point:

I_{C} / V_{T}

for BJT and

2 k I_{D}

for FET. Change the current and the slope changes, so

g_{m}

must be re-evaluated at each new

Q

Edge cases

Edge case: What is

g_{m}

for a BJT at

I_{C} = 0

(cut-off)?

Zero.

g_{m} = I_{C} / V_{T} = 0

— with no collector current the exponential slope is flat, so no input wiggle produces output current. The device cannot amplify when off.

Edge case: What happens to MOSFET

g_{m}

V_{GS} \to V_{t h}

from above?

V_{o v} \to 0

, so

g_{m} = k V_{o v} \to 0

. Right at threshold the channel just forms and controls essentially no current; transconductance vanishes at the edge of conduction.

Edge case: If a MOSFET is in the triode (linear) region, does

g_{m} = k V_{o v}

still hold?

No. That form comes from the saturation square law. In triode

I_{D}

depends on both

V_{GS}

and

V_{D S}

, so

\partial I_{D} / \partial V_{GS}

takes a different value (

\propto V_{D S}

). Always confirm the device is in saturation first.

Edge case: As

I_{C} \to \infty

does BJT

g_{m}

grow without bound per the formula?

Mathematically

g_{m} = I_{C} / V_{T}

keeps rising, but physically high-level injection, series resistance, and heating break the ideal exponential law. The formula is a small-signal, moderate-current idealisation, not valid to infinity.

Edge case: Does the output resistance $r_o$ ever change

g_{m}

No —

r_{o}

models the output current's slight dependence on output voltage;

g_{m}

is the input-to-output slope. They are separate parameters that combine as the intrinsic gain

g_{m} r_{o}

, but neither alters the other.

Edge case: A BJT and a MOSFET are both biased at exactly

0 mA

. Compare their

g_{m}

Both are zero.

g_{m} = I_{C} / V_{T} = 0

and

g_{m} = 2 k \cdot 0 = 0

. With no current flowing, neither device has any control action — the "same current, more BJT

g_{m}

" rule only applies at nonzero current.

Edge case: If you halve the temperature (in Kelvin) of a BJT at fixed

I_{C}

, what happens to

g_{m}

V_{T} = k T / q

halves, so

g_{m} = I_{C} / V_{T}

doubles. Colder junctions have a steeper exponential, hence a larger transconductance at the same current (ignoring other temperature effects on

I_{C}

Connections

Thermal Voltage VT — the $V_{T}$ inside every BJT trap above.
Overdrive Voltage Vov — the $V_{o v}$ tradeoff behind the FET traps.
MOSFET Square-Law and Saturation — why $g_{m} = k V_{o v}$ only holds in saturation.
BJT Small-Signal Model / Common-Emitter Amplifier / Common-Source Amplifier — where $g_{m}$ becomes gain.
Early Effect and Output Resistance ro — separate from $g_{m}$ , combines as $g_{m} r_{o}$ .