Reverse saturation current — Hinglish notes

2.2.13 · Hardware › Doping & PN Junctions

YEH HAI KYA?

"Saturation" kyun? Kyunki jaise hi tum thoda sa bhi reverse-bias karo, har minority carrier jo depletion edge tak pahunchta hai woh sweep ho jaata hai. Tum junction tak pahunchne wale minority carriers ki supply se zyada sweep nahi kar sakte — isliye $∣ V ∣$ badhane se current nahi badhta. Yeh $I_{S}$ par saturate ho jaata hai.

"Minority carriers" kyun? Depletion-region field n → p ki taraf point karti hai. Yeh field electrons ko n ki taraf aur holes ko p ki taraf push karti hai — majority carriers ke liye galat direction hai, lekin minority carriers ko cross karaane ke liye bilkul sahi direction. Isliye $I_{S}$ ek minority-carrier show hai.

KOI CURRENT FLOW KYUN KARTA HAI?

Kisi bhi $T > 0$ par, thermal energy bonds todti hai aur har jagah electron–hole pairs banati hai. Neutral p-region mein kuch electrons (minorities) lagaataar create hote hain; kuch depletion edge tak diffuse hote hain, field feel karte hain, aur n-side mein sweep ho jaate hain. N-side par holes ke liye bhi aisa hi hota hai. Yahi trickle $I_{S}$ hai.

Recall Feynman: 12-saal ke bacche ko samjhao

Ek hill (junction) imagine karo jisme ek taraf slide jaati hai. Zyaatar bacche (majority carriers) galat side par hain slide use karne ke liye, isliye almost koi cross nahi karta. Lekin kabhi kabhi garmi ek rare bacche (minority carrier) ko slide ke top par "kick" karti hai, aur whoosh woh ek dam slide ho jaata hai. Koi fark nahi padta slide kitni steep ho — slide karne wale bacchon ki ginti is baat par depend karti hai ki kitne top tak kick hote hain, steepness par nahi. Bacchon ki wahi trickle reverse saturation current hai.

$I_{S}$ KO SCRATCH SE DERIVE KAISE KAREIN

Hum ise minority-carrier diffusion se banate hain. N-side consider karo (wahan holes minorities hain).

Step 1 — Equilibrium minority concentration. Mass-action law: $n p = n_{i}^{2}$ . N-side par, majority electrons $\approx N_{D}$ , toh equilibrium minority holes: $p_{n 0} = \frac{n _{i}^{2}}{N _{D}}$ Yeh step kyun? Yeh un rare carriers ki "supply" set karta hai jo current feed karenge.

Step 2 — Diffusion length. N-region mein inject hua ek minority hole recombine hone se pehle average ek khaas distance diffuse karta hai. Steady-state diffusion equation solve karne par $D_{p} \frac{d ^{2} p}{d x ^{2}} = \frac{p - p _{n 0}}{τ _{p}}$ ek exponential decay milta hai characteristic length ke saath $L_{p} = D_{p} τ_{p}$ Yeh step kyun? $L_{p}$ par gradient yeh set karta hai ki carriers junction tak kitni tezi se diffuse karte hain.

Step 3 — Depletion edge par diffusion current density. Reverse bias ke under depletion edge par minority concentration ~0 ho jaati hai, isliye excess hai $Δ p \approx - p_{n 0}$ length $L_{p}$ par. Fick's law $J_{p} = - q D_{p} d p / d x$ se magnitude milti hai: $J_{p} = \frac{q D _{p} p _{n 0}}{L _{p}} = \frac{q D _{p} n _{i}^{2}}{L _{p} N _{D}}$

Step 4 — Symmetrically electron contribution (p-side se) add karo: $J_{n} = \frac{q D _{n} n _{p 0}}{L _{n}} = \frac{q D _{n} n _{i}^{2}}{L _{n} N _{A}}$

Step 5 — Total. Total current density ko junction area $A$ se multiply karo:

$n_{i}^{2}$ kyun sabse zyada matter karta hai. $n_{i} \propto T^{3/2} e^{- E_{g} /2 k T}$ ke saath, hume milta hai $n_{i}^{2} \propto T^{3} e^{- E_{g} / k T}$ . Isliye $I_{S}$ extremely temperature-sensitive hai — silicon mein yeh roughly har ~10 °C mein double ho jaata hai. Yahi asli wajah hai ki diodes/transistors temperature ke saath drift karte hain.

Worked examples

Common mistakes

Flashcards

Reverse saturation current ke liye kaun se carriers responsible hain?

Thermally generated minority carriers jo junction field se sweep hote hain.

I_{S}

ko "saturation" current kyun kehte hain?

Kyunki yeh minority carriers ki supply se limited hai, isliye yeh ~constant rehta hai chahe reverse voltage kitna bhi bada ho.

I_{S}

ka poora expression likho.

I_{S} = q A n_{i}^{2} (\frac{D _{p}}{L _{p} N _{D}} + \frac{D _{n}}{L _{n} N _{A}})

I_{S}

intrinsic concentration ke saath kaise scale karta hai?

I_{S} \propto n_{i}^{2}

(

n_{i}

nahi), kyunki minority concentration

= n_{i}^{2} / N

hai.

I_{S}

temperature par kaise depend karta hai aur roughly kitna?

\propto T^{3} e^{- E_{g} / k T}

; silicon mein roughly har ~10 °C mein double ho jaata hai.

Diode equation mein, strong reverse bias ke under

I

kya approach karta hai?

I \to - I_{S}

(exponential term → 0).

Reverse bias ke under minority carriers kyun cross karte hain, majority kyun nahi?

Depletion field direction majority carriers ko door push karta hai lekin minorities ko cross karaa deta hai.

N-region mein equilibrium minority hole concentration kya hai?

p_{n 0} = n_{i}^{2} / N_{D}

(mass-action

n p = n_{i}^{2}

se).

Diffusion length

L_{p}

kya hai aur yeh kyun appear karta hai?

L_{p} = D_{p} τ_{p}

; yeh concentration gradient set karta hai jo diffusion current feed karta hai.

Zyada heavily doped side

I_{S}

mein kam contribute kyun karti hai?

Zyada doping → kam minority carriers (

n_{i}^{2} / N

chota) → chota diffusion current.

Connections

PN Junction Diode Equation — $I_{S}$ iska pre-factor hai.
Depletion Region & Built-in Field — woh field jo minorities ko sweep karti hai.
Minority & Majority Carriers — $I_{S}$ ka supply source.
Mass-Action Law (np = ni²) — $p_{n 0}, n_{p 0}$ deta hai.
Intrinsic Carrier Concentration ni(T) — $T$ dependence drive karta hai.
Diffusion Current & Fick's Law — transport mechanism.
Reverse Breakdown (Zener/Avalanche) — jahan flat curve finally break hoti hai.