Body effect and substrate bias — Hinglish notes

2.4.16 · Hardware › Transistors: BJT & FET

Body effect KYA hai?

Controlling quantity hai $V_{S B}$ (nMOS ke liye, source-to-bulk). Wo parameter jo set karta hai ki $V_{T}$ kitni strongly respond karta hai, wo hai body-effect coefficient $γ$ (gamma).

YEH KYUN hota hai? (first-principles story)

Gate charge $Q_{G}$ ko semiconductor ke andar do cheezein balance karni padti hain:

Inversion charge $Q_{in v}$ — actual channel electrons jo hum chahte hain, aur
Depletion charge $Q_{d e p}$ — channel ke neeche depleted region mein immobile ionised acceptor ions.

$Q_{G} = Q_{in v} + Q_{d e p}$

Transistor turn on karne ke liye hume itna gate voltage chahiye ki (a) depletion region create ho aur (b) surface invert ho. Toh threshold roughly hota hai

$V_{T} = V_{F B} + 2 ϕ_{F} + \frac{Q _{d e p}}{C _{o x}}$

jahan $V_{F B}$ flat-band voltage hai, $2 ϕ_{F}$ strong inversion ke liye zaruri surface potential hai, aur $Q_{d e p} / C_{o x}$ wo gate voltage hai jo depletion charge support karne mein "use up" ho jaata hai.

Depletion charge KO KAISE derive karte hain?

Yeh step kyun? Poisson's equation $\frac{d ^{2} ψ}{d x ^{2}} = - \frac{ρ}{ε}$ , constant charge density $ρ = - q N_{A}$ ke saath, do baar integrate karne par ek parabola deta hai; edge se surface tak evaluate karne par $ψ \propto W^{2}$ milta hai. $W$ ke liye invert karo aur $q N_{A}$ se multiply karo toh square-root law milta hai — depletion charge sirf $ψ$ ki tarah badhta hai, linearly nahi.

Ab threshold par zaroori band bending. Bina back-bias ke surface ko $2 ϕ_{F}$ tak bend karna hoga. Reverse source-body bias $V_{S B}$ ke saath, wo total potential jo depletion region ko sustain karna hai, wo hai $2 ϕ_{F} + V_{S B}$ :

$Q_{d e p} (V_{S B}) = 2 q ε_{s i} N_{A} (2 ϕ_{F} + V_{S B})$

$V_{T}$ mein plug karo aur $V_{S B} = 0$ baseline se increase ko alag karo:

$2 ϕ_{F}$ subtract kyun karte hain? Kyunki $V_{T 0}$ (threshold at $V_{S B} = 0$ ) already band bending $2 ϕ_{F}$ ke liye depletion charge include karta hai. Hum sirf back-bias ki wajah se badhne wala extra charge add karte hain, isliye baseline term subtract karte hain. Warna double-count ho jaata.

Worked examples

Common mistakes

Active recall

Recall 12-saal ke bachche ko samjhao (Feynman)

Socho channel ek tiny paani ki pipe hai aur gate ek haath hai jo use kholne ke liye press karta hai. Pipe ke neeche soft mud hai (mobile carriers). Agar aap source ko ground se upar khiichte ho (yahi back-bias hai), toh aap soft mud ko kheench lete ho aur hard, dry, stuck rocks expose ho jaate hain (wo fixed ions hain). Ab tumhara haath (gate) rocks ko clear karne ke liye sirf pipe kholne se pehle zyada zyada push karna padta hai. Toh "turn on karne ke liye minimum push" — threshold — badhta hai. Aur kyunki rocks clear karna har extra scoop ke saath aasaan hota jaata hai, extra push ek square root ki tarah badhta hai, straight line ki tarah nahi.

Flashcards

Body effect kya hai?

MOSFET threshold voltage

V_{T}

ki source-to-body voltage

V_{S B}

par dependence; reverse bias depletion wide karta hai,

V_{T}

badhata hai.

Body-effect threshold equation batao.

V_{T} = V_{T 0} + γ (2 ϕ_{F} + V_{S B} - 2 ϕ_{F})

Body-effect coefficient

γ

define karo.

γ = 2 q ε_{s i} N_{A} / C_{o x}

, units

V

V_{T}

V_{S B}

ke saath kyun badhta hai?

Reverse bias depletion region wide karta hai, zyada fixed acceptor ions expose karta hai, isliye

Q_{d e p}

badhta hai aur zyada gate voltage ki zarurat hoti hai.

Kya

V_{T}

–

V_{S B}

relation linear hai?

Nahi — square-root, kyunki

Q_{d e p} \propto 2 ϕ_{F} + V_{S B}

Poisson's equation se aata hai.

High channel doping

N_{A}

γ

kyun badhata hai?

Volume mein zyada ionised acceptors matlab per unit band bending mein zyada depletion charge, isliye

V_{T}

zyada strongly respond karta hai.

Thick oxide

γ

kyun badhata hai?

C_{o x} = ε_{o x} / t_{o x}

shrink hota hai, isliye depletion charge ki har unit zyada gate voltage cost karti hai.

Stacked (series) transistors mein body effect kyun hota hai?

Upar wale devices ka source ground ke upar lift hota hai,

V_{S B} > 0

deta hai, unka

V_{T}

badhata hai aur drive current kam karta hai.

Q_{d e p}

vs band bending

ψ

derive karo.

Poisson deta hai

ψ = q N_{A} W^{2} /2 ε_{s i}

, isliye

W = 2 ε_{s i} ψ / q N_{A}

aur

Q_{d e p} = q N_{A} W = 2 q ε_{s i} N_{A} ψ

Formula mein

2 ϕ_{F}

subtract kyun karte hain?

Kyunki

V_{T 0}

mein already band bending

2 ϕ_{F}

ke liye depletion charge hai; hum sirf back-bias ki wajah se badhne wala extra add karte hain.

Connections

MOSFET threshold voltage — body effect $V_{T 0}$ ko modify karta hai.
Depletion region and Poisson's equation — √ law ka source.
Oxide capacitance Cox — $γ$ scale set karta hai.
CMOS NAND gate delay — series stacks ke zariye practical impact.
Channel-length modulation — ek aur " $V_{T}$ /current constant nahi hai" effect.
Flat-band voltage and work function — $V_{T}$ mein $V_{F B}$ term.