Depletion region and space charge — Hinglish notes

2.2.5 · Hardware › Doping & PN Junctions

KYA ho raha hai?

YEH KYUN banta hai?

Free electrons n-side par dense hain, holes p-side par dense hain.
Concentration gradient → diffusion. Electrons p-side mein diffuse karte hain, holes n-side mein.
Woh recombine karte hain, boundary ke paas mobile carriers ko khatam kar dete hain.
Ab uncovered fixed dopant ions hil nahi sakte — woh space charge ke roop mein wahin rehte hain.
Yeh charge ek E-field n→p direction mein banata hai jo carriers ko wapas push karta hai (drift).
Equilibrium = drift exactly diffusion ko cancel karta hai. Net current = 0.

Figure — Depletion region and space charge

KAISE derive karte hain field, potential, aur width ko

Hum Poisson's equation 1D mein use karte hain. Yeh master tool hai — baaki sab kuch isi se nikalti hai.

YEH equation KYUN? Gauss's law kehta hai ki charge ek diverging field create karta hai ( $\nabla \cdot E = ρ / ε$ ). 1D mein yeh hai $d E / d x = ρ / ε_{s}$ . Aur $E = - d ϕ / d x$ potential ki definition se.

Step 1 — Depletion approximation

Maano ki space charge ek clean box hai: andar fully depleted, bahar neutral. $ρ (x) = {- q N_{A} + q N_{D} - x_{p} < x < 0 (p-side) 0 < x < x_{n} (n-side)$

Yeh step kyun? Real carrier profiles smooth hote hain, lekin edges ko sharp treat karne se messy integrals simple ho jaati hain — aur yeh accurate hai kyunki carriers jaldi vanish ho jaate hain (ek Debye length ke andar).

Step 2 — Charge neutrality

Poora crystal neutral hai, isliye total positive = total negative charge: $q N_{A} x_{p} = q N_{D} x_{n} \Rightarrow == N_{A} x_{p} = N_{D} x_{n} ==$

Yeh step kyun? Koi net charge create nahi hua — humne sirf use rearrange kiya. Isliye exposed positive aur negative charge balance hona chahiye. Consequence: depletion region lightly-doped side mein zyada gehri jaati hai.

Step 3 — E-field ke liye integrate karo

$d E / d x = ρ / ε_{s}$ ko integrate karo. P-side par ( $ρ = - q N_{A}$ ), $E = 0$ ke saath $x = - x_{p}$ par: $E (x) = - \frac{q N _{A}}{ε _{s}} (x + x_{p}), - x_{p} < x < 0$ N-side par ( $ρ = + q N_{D}$ ), $E = 0$ ke saath $x = x_{n}$ par: $E (x) = - \frac{q N _{D}}{ε _{s}} (x_{n} - x), 0 < x < x_{n}$

Field junction par maximum hoti hai ( $x = 0$ ), shape mein triangular: $E_{ma x} = - \frac{q N _{A} x _{p}}{ε _{s}} = - \frac{q N _{D} x _{n}}{ε _{s}}$

Yeh step kyun? Field edges par zero se shuru hoti hai (neutral region), har charged slab se linearly badhti hai, aur metallurgical junction par peak karti hai. Dono expressions $x = 0$ par match karte hain kyunki Step 2 ki neutrality ki wajah se.

Step 4 — Potential ke liye integrate karo (built-in voltage)

$V_{bi}$ $∣ E ∣$ vs $x$ curve ke neeche ka area hai (ek triangle): $V_{bi} = - \int_{- x_{p}}^{x_{n}} E d x = \frac{1}{2} ∣ E_{ma x} ∣ W$ jahan $W = x_{p} + x_{n}$ total width hai.

Step 5 — Total width solve karo

$x_{p}, x_{n}$ (neutrality se) ko $V_{bi} = \frac{1}{2} ∣ E_{ma x} ∣ W$ mein substitute karo aur solve karo: $W = \frac{2 ε _{s} V _{bi}}{q} (\frac{1}{N _{A}} + \frac{1}{N _{D}})$

Yeh form kyun? Zyada $V_{bi}$ → wider region (field banane ke liye zyada charge chahiye). Zyada doping → narrower region (same charge expose karne ke liye kam ions chahiye).

Worked Examples

Common Mistakes

Active Recall

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho ek kamra beech se do hisso mein banta hai. Baayein half mein red balls (holes) bhari hain aur daayein half mein blue balls (electrons). Beech ka darwaza kholo: balls khali side ki taraf bhagti hain aur pair up karke gayab ho jaati hain. Lekin jo ball jaati hai woh ek fixed magnet uncover kar deti hai jo zameen par chipki hai (dopant ion) — red balls ke jaane se negative magnets bachte hain, blue balls ke jaane se positive wale. Yeh chipke magnets ek invisible "haawa" banate hain jo itni tez ho jaati hai ki koi aur ball cross nahi kar sakti. Woh shaant khaali patli chhiddar beech mein, chipke magnets se bhari lekin free balls ke bina, yahi depletion region hai.

Flashcards

Depletion region kya hai?

Junction ka ek zone jo mobile carriers se khaali hai, jisme sirf fixed ionized dopant ions (space charge) hain.

Depletion region ki n-side par kaun sa charge rehta hai?

Positive donor ions (

N_{D}^{+}

P-side par kaun sa charge rehta hai?

Negative acceptor ions (

N_{A}^{-}

Charge-neutrality condition kya hai?

N_{A} x_{p} = N_{D} x_{n}

Depletion region kaun si side mein zyada gehri jaati hai?

Lightly-doped side mein.

Junction mein E-field aur potential derive karne ke liye kaun si starting equation hai?

Poisson's equation

d^{2} ϕ / d x^{2} = - ρ / ε_{s}

E-field vs position ki shape kaisi hoti hai?

Triangular — x mein linear, metallurgical junction par maximum.

Built-in potential ka formula?

V_{bi} = (k T / q) ln (N_{A} N_{D} / n_{i}^{2})

Total depletion width ka formula?

W = (2 ε_{s} V_{bi} / q) (1/ N_{A} + 1/ N_{D})

Reverse bias width ko kaise affect karta hai?

Use wider banata hai;

V_{bi}

ki jagah

V_{bi} + V_{R}

use karo, isliye

W \propto V_{bi} + V_{R}

Forward bias width ko kaise affect karta hai?

Use narrow banata hai;

V_{bi} - V_{F}

use karo.

Equilibrium par junction se net current kitni hoti hai?

Zero — diffusion current exactly drift current ko cancel karta hai.

Depletion width heavy doping ke saath kyun shrink karti hai?

Same charge expose karne ke liye kam ions chahiye hote hain, isliye region patli hoti hai (

W \propto 1/ N

effect).

Connections

Doping (Donors and Acceptors) — $N_{A}$ , $N_{D}$ ions supply karta hai jo space charge bante hain.
Built-in Potential and Fermi Level Alignment — $V_{bi}$ ka thermodynamic origin.
Poisson's Equation and Gauss's Law — is derivation ka mathematical engine.
Diode I-V Characteristics — forward/reverse bias ke under depletion behavior.
Junction Capacitance and Varactors — $C_{j} = ε_{s} A / W$ , seedha width se.
Drift and Diffusion Currents — equilibrium par balance karne wale do currents.