Transconductance (gm) — Hardware notes

WHAT is transconductance?

WHY a slope (derivative) and not a ratio? Because the transistor is nonlinear. $I_C$ vs $V_{BE}$ is an exponential curve. For small signals we only care about the local slope at the bias point $Q$ — that's the effective gain the tiny AC signal actually experiences.

HOW to derive $g_m$ from first principles

BJT case (from the Ebers–Moll / diode law)

Formula Deriving BJT

$g_m$ Step 1 — start with the physics. The collector current follows the exponential junction law: $I_C = I_S\,e^{V_{BE}/V_T}$ Why this step? The base–emitter junction is a forward-biased diode; carrier injection is exponential in $V_{BE}$ . $V_T=kT/q\approx 25.9\text{ mV}$ at room temperature.

Step 2 — differentiate w.r.t. the controlling voltage. $g_m=\frac{dI_C}{dV_{BE}} = I_S\,e^{V_{BE}/V_T}\cdot\frac{1}{V_T}$ Why this step? $g_m$ is the slope by definition. The derivative of $e^{x/V_T}$ brings down a factor $1/V_T$ .

Step 3 — recognise the exponential is just $I_C$ again. $\boxed{g_m=\frac{I_C}{V_T}}$ Why this step? Because $I_S e^{V_{BE}/V_T}=I_C$ . This is the beautiful result: $g_m$ depends only on the bias current, not on the individual transistor's $I_S$ or $\beta$ .

MOSFET case (from the square law)

Formula Deriving MOSFET

$g_m$ Step 1 — saturation drain current (square law): $I_D=\frac{1}{2}\mu_n C_{ox}\frac{W}{L}\,(V_{GS}-V_{th})^2 = \frac{k}{2}(V_{GS}-V_{th})^2$ where $k=\mu_n C_{ox}\frac{W}{L}$ and $V_{ov}=V_{GS}-V_{th}$ is the overdrive voltage. Why this step? Above threshold, the channel charge $\propto (V_{GS}-V_{th})$ and carrier velocity $\propto V_{DS}$ ; integrating gives a square dependence in saturation.

Step 2 — differentiate w.r.t. $V_{GS}$ : $g_m=\frac{dI_D}{dV_{GS}}=k\,(V_{GS}-V_{th})=\boxed{k\,V_{ov}}$

Step 3 — rewrite in useful equivalent forms. Using $I_D=\tfrac{k}{2}V_{ov}^2$ : $g_m=\frac{2I_D}{V_{ov}}=\sqrt{2k\,I_D}$ Why this step? Solve $I_D$ for $V_{ov}$ and substitute — each form is handy in different design situations.

WHY do we care? — gain link

Worked examples

Common mistakes (steel-manned)

Active recall

Recall Quick self-test (hide answers)

What is $g_m$ in words? → slope of output current vs input control voltage at the bias point.
BJT formula? → $I_C/V_T$ .
MOSFET three forms? → $kV_{ov}=2I_D/V_{ov}=\sqrt{2kI_D}$ .
Why $I_C/V_T$ and not $I_C/V_{BE}$ ? → derivative of exponential brings $1/V_T$ .
Gain of CE/CS stage? → $-g_mR_L$ .

Recall Feynman: explain to a 12-year-old

Imagine a garden tap. Turning the handle a tiny bit is the input voltage. How much extra water gushes out is the output current. Transconductance is how much the water jumps for a small twist of the handle. A "sensitive" tap (big $g_m$ ) blasts a lot of water for a gentle twist — that's a good amplifier. A transistor is a magic tap where a whisper on the handle controls a firehose, and $g_m$ measures how sensitive that magic is.

Connections

BJT Small-Signal Model — $g_m$ is the dependent current source $g_m v_{be}$ .
Thermal Voltage VT — where the $25.9$ mV comes from ( $kT/q$ ).
MOSFET Square-Law and Saturation — source of $I_D=\tfrac{k}{2}V_{ov}^2$ .
Common-Emitter Amplifier / Common-Source Amplifier — gain $=-g_mR_L$ .
Overdrive Voltage Vov — controls the $g_m$ /current tradeoff in FETs.
Early Effect and Output Resistance ro — combines with $g_m$ for intrinsic gain $g_m r_o$ .

Define transconductance

g_m

The slope

\partial I_{out}/\partial V_{in}

at the bias point — change in output current per unit change in controlling input voltage. Units: siemens (A/V).

BJT transconductance formula

g_m = I_C/V_T

, with

V_T=kT/q\approx25.9

mV.

Why is BJT

g_m=I_C/V_T

and not

I_C/V_{BE}

g_m

is the derivative of

I_C=I_S e^{V_{BE}/V_T}

; differentiating brings down

1/V_T

, and the exponential re-becomes

I_C

MOSFET

g_m

(three equivalent forms)

g_m = kV_{ov} = 2I_D/V_{ov} = \sqrt{2kI_D}

, where

k=\mu_nC_{ox}W/L

V_{ov}=V_{GS}-V_{th}

How does

g_m

scale with bias current?

BJT: linear (

\propto I_C

). MOSFET: square-root (

\propto\sqrt{I_D}

Small-signal voltage gain of CE/CS stage

A_v=-g_mR_L

(inverting).

For fixed

I_D

, effect of raising

V_{ov}

g_m

g_m=2I_D/V_{ov}

decreases — higher overdrive lowers transconductance at constant current.

Which gives more

g_m

per amp of bias, BJT or FET?

BJT (thanks to the tiny

V_T\approx26

mV in the denominator).

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, transconductance ( $g_m$ ) ka matlab hai — transistor ke input voltage ka output current par kitna strong control hai. Formula ki jagah pehle intuition samjho: agar main $V_{BE}$ (BJT me) ya $V_{GS}$ (FET me) thoda sa hilaun, to output current $I_C$ ya $I_D$ kitna jump karta hai? Yeh jump kitna bada hai — usko $g_m$ measure karta hai. Isliye $g_m$ hi transistor ke amplification ka "asli engine" hai.

BJT ke liye derivation simple hai: $I_C = I_S e^{V_{BE}/V_T}$ , aur is curve ka slope lo (derivative), to milta hai $g_m = I_C/V_T$ . Yaha $V_T\approx26$ mV thermal voltage hai. Kamaal ki baat — $g_m$ sirf bias current par depend karta hai, transistor ke specific $\beta$ ya $I_S$ par nahi! MOSFET me square law se derive karo: $I_D=\frac{k}{2}V_{ov}^2$ , slope lo to $g_m=kV_{ov}=2I_D/V_{ov}=\sqrt{2kI_D}$ .

Common galti: log $g_m$ ko $I_C/V_{BE}$ samajh lete hain — galat! Woh ratio nahi, slope hai. Kyunki curve exponential hai, slope me $V_T$ (26 mV) aata hai, na ki $V_{BE}$ (0.7 V). Isse 27 guna difference aa jata hai. Ek aur point: BJT ka $g_m$ current ke saath linear badhta hai, lekin FET ka sirf square-root — isliye same current par BJT zyada $g_m$ deta hai.

Kyun important hai? Kyunki amplifier ka gain seedha $g_m$ se aata hai: common-emitter/common-source stage ka gain $A_v = -g_m R_L$ . Zyada $g_m$ matlab zyada gain. Exam aur real design dono me, pehle bias set karke $g_m$ nikalo, phir gain — yahi 80/20 hai.

Go deeper — visual, from zero

Connections