Built-in potential of a junction

2.2.6 · Hardware › Doping & PN Junctions

POTENTIAL AATA KYUN HAI?

Intuition Diffusion-vs-drift ki tug of war

N-side electrons se bhari hui hai; p-side unse almost khali hai. Nature concentration gradients se nafrat karta hai → electrons n→p direction mein diffuse karte hain aur holes p→n direction mein diffuse karte hain. Lekin har electron jo n-side chhod ke jaata hai, ek positive donor ion ko peechhe chhodta hai jise woh ab neutralize nahi kar sakta; har hole jo p-side chhod ke jaata hai ek negative acceptor ion ko peechhe chhodta hai. Isse ek aisa region banta hai jo mobile carriers se khali ho jaata hai — depletion region — jisme ek built-in electric field hota hai jo n→p ki taraf point karta hai... ruko, actually positive n-side ions se negative p-side ions ki taraf point karta hai. Woh field carriers ko wapas dhakelta hai (drift). Equilibrium = diffusion current exactly drift current ko cancel karta hai. Jo potential hill net flow ko rokti hai wahi $V_{bi}$ hai.

PEHLE PRINCIPLES SE DERIVE KAISE KAREIN

Hum ise do equivalent tareekon se derive karte hain. Dono ek hi law se shuru hote hain.

Step 1 — Equilibrium par, net current = 0

Electrons ke liye, current ke do parts hain: $J_{n} = drift q μ_{n} n E + diffusion q D_{n} \frac{d n}{d x} = 0$

Yeh step kyun? Equilibrium (bina battery ke) matlab hai kahin bhi net current nahi, isliye dono mechanisms exactly balance karne chahiye.

Step 2 — Einstein relation use karo

Einstein relation diffusion aur mobility ko link karta hai: $D_{n} = \frac{k T}{q} μ_{n}$

Kyun? Diffusion aur drift dono carriers ki same random thermal motion se aate hain, isliye woh independent nahi ho sakte — Einstein's relation unhe ek saath baandhta hai.

Substitute karo aur $μ_{n}$ cancel karo: $q μ_{n} n E + q \frac{k T}{q} μ_{n} \frac{d n}{d x} = 0 \Rightarrow E = - \frac{k T}{q} \frac{1}{n} \frac{d n}{d x}$

Step 3 — $E$ ko potential mein badlo

Kyunki $E = - \frac{d V}{d x}$ : $- \frac{d V}{d x} = - \frac{k T}{q} \frac{1}{n} \frac{d n}{d x} \Rightarrow d V = \frac{k T}{q} \frac{d n}{n}$

Kyun? Hum ek voltage chahte hain (integrate karne laayak), aur $E$ voltage ka slope hai. Yeh cleverly ek position integral ko $n$ par ek bahut aasaan integral mein convert kar deta hai.

Step 4 — Junction ke across integrate karo

P-side (jahan $n = n_{p}$ ) se n-side (jahan $n = n_{n}$ ) tak integrate karo: $V_{bi} = V_{n} - V_{p} = \frac{k T}{q} \int_{n_{p}}^{n_{n}} \frac{d n}{n} = \frac{k T}{q} ln (\frac{n _{n}}{n _{p}})$

Step 5 — Doping values plug in karo

Non-degenerate, fully-ionized doping ke liye:

N-side majority electrons: $n_{n} \approx N_{D}$
P-side minority electrons (mass-action law $n p = n_{i}^{2}$ ): $n_{p} = \frac{n _{i}^{2}}{N _{A}}$

$V_{bi} = \frac{k T}{q} ln (\frac{n _{n}}{n _{p}}) = \frac{k T}{q} ln (\frac{N _{D} \cdot N _{A}}{n _{i}^{2}})$

Yeh step kyun? Humne abstract carrier densities ko un quantities se replace kiya jo fab actually control karta hai: doping levels.

Figure — Built-in potential of a junction

Worked Examples

Common Mistakes (Steel-manned)

Forecast-then-Verify

Recall Compute karne se pehle forecast karo

Q: Agar dono sides $1 0^{16}$ (Si) par doped hain, toh $V_{bi}$ forecast karo: 0.3, 0.7, ya 1.1 V ke kareeb? Verify: $ln (1 0^{32} /1 0^{20}) = ln (1 0^{12}) = 27.6$ ; $\times 0.02585 = 0.714$ V → ~0.7 V. Textbook silicon value ✓.

Flashcards

Equilibrium par built-in field kaun sa physical process rokta hai?

Carriers ka net diffusion — drift exactly diffusion ko cancel karta hai isliye net current = 0.

Built-in potential formula likhiye.

V_{bi} = \frac{k T}{q} ln (\frac{N _{A} N _{D}}{n _{i}^{2}})

Derivation mein

D_{n}

aur

μ_{n}

ko kaun sa relation link karta hai?

Einstein relation,

D_{n} = (k T / q) μ_{n}

Voltmeter se

V_{bi}

se energy extract kyun nahi kar sakte?

Probe junctions par contact potentials use cancel kar dete hain; total loop voltage = 0.

10× doping increase se

V_{bi}

kitna change hota hai?

Lagbhag

+ 60

mV (kyunki

k T / q \cdot ln 10 \approx 59.6

mV).

Ge ka

V_{bi}

Si se kam kyun hota hai?

Ge ka

n_{i}

bahut bada hota hai, jo ratio

N_{A} N_{D} / n_{i}^{2}

aur isliye log ko chhota kar deta hai.

300 K par

k T / q

kya hota hai?

Lagbhag 25.85 mV (thermal voltage

V_{T}

Temperature badhne par

V_{bi}

ka kya hota hai?

Yeh decrease hota hai (Si ke liye ~

- 2

mV/°C) kyunki

n_{i}

tezi se badhta hai, log ratio ko shrink karta hai.

Positive donor ions kis side par exposed hote hain?

Depletion region ke n-side par.

Recall Feynman: 12-saal ke bachche ko explain karo

Socho do bheed wale kamre ek darwaze se jude hue hain. Ek kamra blue balls (electrons) se bhara hai, doosra red balls (holes) se. Darwaza kholo aur balls khali kamre mein gir padti hain. Lekin har ball jo apna ghar chhodti hai, peechhe ek "chipchipa jagah" (charged ion) chhod jaati hai jo balls ko wapas kheenchta hai. Bahut jaldi, darwaze par ek "kheench ki deewar" ban jaati hai jo itni mazboot hoti hai ki koi aur ball cross na kar sake. Us deewar ki takaat, voltage ke roop mein measure ki gayi, built-in potential hai. Kisine battery nahi lagayi — balls ne khud yeh deewar banaayi sirf phailna chahte hue.

Connections

Depletion region width — $V_{bi}$ set karta hai ki carrier-free zone kitni wide hogi.
Mass-action law — $n p = n_{i}^{2}$ , derivation mein $n_{p}$ find karne ke liye use hota hai.
Einstein relation — drift aur diffusion ke beech bridge.
PN junction under bias — external voltage $V_{bi}$ mein add/subtract hota hai.
Diode I-V equation — wahi $k T / q$ thermal voltage exponential ko govern karta hai.
Intrinsic carrier concentration — $n_{i}$ ki temperature dependence $V_{bi} (T)$ ko drive karti hai.
Energy band diagrams — $q V_{bi}$ = junction ke across band bending.

YEH HAI KYA?