Reverse bias behavior — Hinglish notes

2.2.8 · Hardware › Doping & PN Junctions

80/20 core: Reverse bias depletion region ko choda kar deta hai, potential barrier ko badhata hai, aur sirf ek chhoti si temperature-dependent saturation current $I_{S}$ ko flow karne deta hai — jab tak field itni strong na ho jaye ki junction break down kar jaye.

Kya ho raha hai?

Current kaise aata hai: $I$ ko reverse bias mein derive karna

Shockley diode equation se shuru karo (carrier injection + diffusion se derive ki gayi):

$I = I_{S} (e^{V / (n V_{T})} - 1)$

jahan $V_{T} = \frac{k T}{q}$ thermal voltage hai ( $\approx 25.85 mV$ at $300 K$ ) aur $n$ ideality factor hai.

Zyada voltage se zyada current kyun nahi milta? Reverse current minority carriers se aata hai jo junction ke paas thermally generate hote hain aur phir field se sweep hoke cross karte hain. Jis rate pe woh generate hote hain woh temperature par depend karta hai, field strength par nahi. Jab field itni strong ho jaaye ki har generated carrier ko sweep kar sake, tab aur zyada voltage lagane se current nahi badhega — carriers ki supply bottleneck hai.

Depletion width badhti hai: derivation

Depletion region charge store karta hai; ise Poisson's equation ke zariye ek one-sided problem ki tarah treat karo. Junction ke across total voltage $V_{bi} - V$ hai (reverse bias ke liye $V < 0$ , isliye yeh badhta hai):

$W = \frac{2 ε}{q} (\frac{1}{N _{A}} + \frac{1}{N _{D}}) (V_{bi} - V)$

Kyunki $W$ badhta hai, junction ek parallel-plate capacitor ki tarah act karta hai jiska gap badhta ja raha hai, isliye junction capacitance ghatti hai:

$C_{j} = \frac{ε A}{W} = \frac{C _{j 0}}{1 - V / V _{bi}}$

Iska faayda varactor diodes mein uthaya jaata hai (voltage-controlled capacitor).

Breakdown: current hamesha ke liye nahi jaata

Ek critical reverse voltage $V_{B R}$ par current achanak shoot up karta hai (yeh koi defect nahi — yeh reversible hai agar power limited ho):

Worked examples

Common mistakes

Forecast-then-Verify

Recall Pehle predict karo, phir check karo

Q: $V = - 10 V$ vs $V = - 2 V$ par (breakdown se neeche), kiska $∣ I ∣$ zyada hai? Forecast karo… phir verify karo: dono $\approx I_{S}$ hain — practically identical, kyunki current saturated hai.

Flashcards

Reverse bias kaun se terminals connect karta hai?

+ to N-side, − to P-side (isliye

V < 0

Reverse bias mein depletion width ka kya hota hai?

Yeh choda hota hai (

W \propto V_{bi} - V

Breakdown se door reverse current ki magnitude kya hoti hai?

Saturation current

I_{S}

(chhota, roughly constant).

Reverse current voltage par independent kyun hota hai?

Yeh minority carriers ki thermal generation se limited hai, field se nahi.

Temperature ke saath

I_{S}

kaise change hoti hai?

Roughly har 10 °C mein double hoti hai (

I_{S} \propto n_{i}^{2} \propto e^{- E_{g} / k T}

Shockley eq se reverse-bias diode current jab

∣ V ∣ ≫ V_{T}

ho?

I \approx - I_{S}

300 K par thermal voltage

V_{T}

kya hai?

k T / q \approx 25.85

mV.

Reverse bias se junction capacitance kyun ghatti hai?

Choda depletion region = bada capacitor gap,

C_{j} = ε A / W

Zener breakdown mechanism kya hai?

Electrons ka direct field tunneling bonds se bahar (thin, low-V junctions).

Avalanche breakdown mechanism kya hai?

Carriers ki impact ionization chain reaction (wide, high-V junctions).

Temp coefficient sign: Zener vs avalanche

V_{B R}

Zener T ke saath ghatta hai; avalanche T ke saath badhta hai.

Recall Feynman: 12-saal ke bachche ko samjhao

Socho ek nadi (current) jisme ek chhota sa gate hai. Forward bias gate ko khol deta hai — bahut saara paani behta hai. Reverse bias gate ko band kar deta hai aur reet ke thele bhi laga deta hai (choda depletion region), isliye almost koi paani nahi nikal sakta. Jo thodi si trickle nikal jaati hai woh gate ke paas seedhi barish se aati hai (garmi kuch charges bana deti hai) — aur garam din matlab zyada barish, isliye zyada trickle. Gate par bahut zyada force lagao aur woh ek dum toot jaata hai — wahi breakdown hai.

Connections

Forward bias behavior — iska ulta case (barrier ghatta hai, current exponentially badhta hai)
Shockley diode equation — dono regimes ke liye parent formula
Depletion region & built-in potential — $W$ aur $V_{bi}$ ka source
Zener diodes — deliberately breakdown mein operate kiye jaate hain
Varactor diodes — $C_{j} (V)$ voltage-dependence exploit karte hain
Doping concentration N_A N_D — $W$ , $I_{S}$ , aur breakdown type set karta hai