Transconductance (gm) — Hinglish notes

2.4.13 · Hardware › Transistors: BJT & FET

Transconductance HAI KYA?

WHY ek slope (derivative) na ki ratio? Kyunki transistor nonlinear hota hai. $I_{C}$ vs $V_{B E}$ ek exponential curve hai. Small signals ke liye hume sirf bias point $Q$ par local slope ki parwah hai — yahi woh effective gain hai jo tiny AC signal actually experience karta hai.

First principles se $g_{m}$ KAISE derive karein

BJT case (Ebers–Moll / diode law se)

MOSFET case (square law se)

Hume kyun parwah hai? — gain link

Worked examples

Common mistakes (steel-manned)

Active recall

Recall Quick self-test (answers chhupao)

$g_{m}$ words mein kya hai? → bias point par output current vs input control voltage ka slope.
BJT formula? → $I_{C} / V_{T}$ .
MOSFET ke teen forms? → $k V_{o v} = 2 I_{D} / V_{o v} = 2 k I_{D}$ .
$I_{C} / V_{T}$ kyun, $I_{C} / V_{B E}$ kyun nahi? → $I_{C} = I_{S} e^{V_{B E} / V_{T}}$ ka derivative $1/ V_{T}$ neeche le aata hai.
CE/CS stage ka gain? → $- g_{m} R_{L}$ .

Recall Feynman: 12-saal ke bachche ko samjhao

Socho ek garden ka naal (tap). Handle thoda sa ghoomana input voltage hai. Kitna extra paani aata hai woh output current hai. Transconductance hai handle thoda ghoomane par paani kitna uchhalata hai. Ek "sensitive" naal (bada $g_{m}$ ) halke se ghoomane par bahut paani uchhaalta hai — yahi accha amplifier hai. Transistor ek jaadu ka naal hai jahan handle par ek phusphusahat ek fire-hose ko control karti hai, aur $g_{m}$ naapata hai ki woh jaadu kitna sensitive hai.

Connections

BJT Small-Signal Model — $g_{m}$ dependent current source $g_{m} v_{b e}$ hai.
Thermal Voltage VT — $25.9$ mV kahan se aata hai ( $k T / q$ ).
MOSFET Square-Law and Saturation — $I_{D} = \frac{k}{2} V_{o v}^{2}$ ka source.
Common-Emitter Amplifier / Common-Source Amplifier — gain $= - g_{m} R_{L}$ .
Overdrive Voltage Vov — FETs mein $g_{m}$ /current tradeoff control karta hai.
Early Effect and Output Resistance ro — $g_{m}$ ke saath combine hokar intrinsic gain $g_{m} r_{o}$ deta hai.

Transconductance

g_{m}

define karo

Slope

\partial I_{o u t} / \partial V_{in}

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

BJT transconductance formula

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

, jahan

V_{T} = k T / q \approx 25.9

mV.

BJT

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

kyun hai,

I_{C} / V_{B E}

kyun nahi?

g_{m}

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

ka derivative hai; differentiate karne par

1/ V_{T}

neeche aata hai, aur exponential wapas

I_{C}

ban jaata hai.

MOSFET

g_{m}

(teen equivalent forms)

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

, jahan

k = μ_{n} C_{o x} W / L

V_{o v} = V_{GS} - V_{t h}

g_{m}

bias current ke saath kaise scale karta hai?

BJT: linear (

\propto I_{C}

). MOSFET: square-root (

\propto I_{D}

CE/CS stage ka small-signal voltage gain

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

(inverting).

Fixed

I_{D}

par,

V_{o v}

badhane ka

g_{m}

par kya effect hota hai?

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

ghatta hai — zyada overdrive se constant current par transconductance kam hoti hai.

Bias ke har amp par zyada

g_{m}

kaun deta hai, BJT ya FET?

BJT (denominator mein tiny

V_{T} \approx 26

mV ki wajah se).