Fermi level and Fermi-Dirac distribution — Hinglish notes

2.1.4 · Hardware › Band Theory & Carrier Physics

WHY karte hain hum yeh? (Zaroorat kya hai?)

WHAT problem solve kar rahe hain? Hamare paas billions electrons hain aur allowed energy states (bands) ka ek continuum hai. Hum har electron ko track nahi kar sakte. Hume ek statistical rule chahiye: energy $E$ par kisi state ke liye, chance kya hai ki wahan ek electron baitha ho?

WHY classical Boltzmann statistics use nahi karte? Kyunki electrons Pauli exclusion principle follow karte hain: ek quantum state mein zyada se zyada ek hi electron. Classical particles distinguishable hote hain aur states freely share kar sakte hain. Yeh ek constraint poori distribution badal deta hai.

Fermi–Dirac distribution scratch se derive karna

Hum $f (E)$ grand-canonical idea se derive karte hain: ek single state temperature $T$ aur chemical potential $μ$ wale ek bade reservoir ke saath electrons exchange karta hai.

Step 1 — Ek state mein sirf 0 ya 1 electron ho sakta hai. Yeh step kyon? Pauli exclusion. Occupation number $n \in {0, 1}$ hi possible hai.

Step 2 — Har option ki grand-canonical probability. Ek state mein energy $E$ aur $n$ electrons hone ki probability Gibbs factor $e^{- (E n - μ n) / k_{B} T}$ ke proportional hoti hai. Yeh step kyon? $μ$ (chemical potential) reservoir se ek particle add karne ka energy cost/gain hai; $- μ n$ term particle exchange ko account karta hai.

$n = 0$ : weight $= e^{0} = 1$
$n = 1$ : weight $= e^{- (E - μ) / k_{B} T}$

Step 3 — Average occupation paane ke liye normalize karo.

f (E) = ⟨ n ⟩ = \frac{0 \cdot 1 + 1 \cdot e ^{- (E - μ) / k_{B} T}}{1 + e ^{- (E - μ) / k_{B} T}}

Yeh step kyon? Average occupation = (sum of $n \times$ weight) / (sum of weights).

Step 4 — Top aur bottom dono ko numerator ke exponential se divide karo taaki clean form mile:

f (E) = \frac{1}{1 + e ^{(E - μ) / k_{B} T}}

$T = 0$ par, $μ \to E_{F}$ , aur finite $T$ par bhi semiconductors mein hum usually $μ \approx E_{F}$ likhte hain. Toh:

Figure — Fermi level and Fermi-Dirac distribution

$f (E)$ ka behavior — curve padhna

$T = 0$ K par (step function):

$E < E_{F}$ : $e^{(E - E_{F}) / k_{B} T} \to e^{- \infty} = 0 \Rightarrow f = 1$ (sab states below filled hain).
$E > E_{F}$ : $e^{+ \infty} \to \infty \Rightarrow f = 0$ (sab states above empty hain).

$T > 0$ par: yeh sharp step kuch $k_{B} T$ ki width mein smear ho jaata hai. Kuch electrons thermally $E_{F}$ ke upar kick ho jaate hain, aur neeche empty states (holes) chhod jaate hain.

Boltzmann approximation (80/20 shortcut)

Jab $E - E_{F} ≫ k_{B} T$ ho (states $E_{F}$ se kaafi upar, jaise semiconductor mein conduction band), toh denominator mein $1$ negligible ho jaata hai:

f (E) \approx e^{- (E - E_{F}) / k_{B} T}

Yeh kyon important hai: yeh classical Boltzmann tail hai — isi wajah se hum non-degenerate semiconductors mein carrier concentrations easily compute kar sakte hain. Yeh ek approximation ~80% device calculations cover karta hai.

Worked examples

Common mistakes (Steel-manned)

Recall Feynman: ek 12 saal ke bachche ko samjhao

Ek stadium imagine karo jisme seats ground floor se upar tak stacked hain, aur ek rule hai: ek seat mein sirf ek hi person. Log lazy hote hain aur sabse neeche ki seats chahte hain. Jab bahut thandi hai ( $T = 0$ ), sab log jitna neeche ho sake baith jaate hain — har neeche ki seat li gayi, har upar ki seat khaali. Full aur empty seats ke beech ki "water line" Fermi level hai. Jab garam hota hai, kuch log line ke paas wale excited ho ke upar ki seats par kood jaate hain, aur kuch neeche ki seats khaali ho jaati hain. Fermi–Dirac curve bas yeh batata hai ki kisi bhi diye gaye seat mein koi baitha hone ki chance kya hai — line se bahut neeche 1, bahut upar 0, aur line par fuzzy $\frac{1}{2}$ .

Active-recall flashcards

Fermi–Dirac distribution formula kya hai?

f (E) = \frac{1}{1 + e ^{(E - E_{F}) / k_{B} T}}

Definition ke anusar,

E = E_{F}

par

f (E)

kya hai?

Exactly

\frac{1}{2}

(50%), kisi bhi temperature par.

Fermi level physically kya hai?

Electrons ka electrochemical potential; woh energy jahan occupation probability = 1/2 ho; equilibrium mein poore system mein constant rehta hai.

Electrons ke liye Boltzmann ki jagah Fermi–Dirac kyon use karte hain?

Electrons Pauli exclusion principle follow karte hain (ek state mein max ek), isliye occupation 1 se cap hoti hai; Boltzmann mein aisa koi cap nahi.

T = 0

K par

f (E)

kaisa dikhta hai?

Ek step function:

E < E_{F}

ke liye

f = 1

E > E_{F}

ke liye

f = 0

f (E)

ka Boltzmann approximation kya hai aur kab valid hai?

f \approx e^{- (E - E_{F}) / k_{B} T}

, valid jab

E - E_{F} ≫ k_{B} T

ho (non-degenerate,

E_{F}

se kaafi upar).

E_{F}

Δ

neeche koi state EMPTY hone ki probability kya hai?

f (E_{F} + Δ)

— same as probability ki

E_{F}

Δ

upar koi state filled hai (antisymmetry:

f (E_{F} + Δ) + f (E_{F} - Δ) = 1

Room temperature (300 K) par

k_{B} T

ki value kya hai?

Lagbhag 0.0259 eV (≈ 26 meV).

Kya

f (E)

electrons ka number deta hai?

Nahi — yeh ek probability hai (0–1). Electron count ke liye

n = \int g (E) f (E) d E

chahiye.

Inversion: diya

f

, toh

E - E_{F}

kya hai?

E - E_{F} = k_{B} T ln (\frac{1 - f}{f})

Connections

Density of states — carrier counts paane ke liye $f (E)$ ko $g (E)$ ke saath pair karo.
Intrinsic and extrinsic semiconductors — doping $E_{F}$ ko bands ki taraf shift karta hai.
Carrier concentration n and p — $n = \int g_{c} (E) f (E) d E$ , $p = \int g_{v} (E) (1 - f) d E$ .
Pauli exclusion principle — denominator mein "1 +" ka physical origin.
Band gap and conduction/valence bands — $E_{F}$ bands ke relative kahan hota hai.
Chemical potential and equilibrium — flat $E_{F}$ = koi net electron flow nahi.