Temperature dependence of carrier concentration

2.1.13 · Hardware › Band Theory & Carrier Physics

Intuition Badi picture

Ek doped semiconductor mein dopant atoms ki ek fixed number hoti hai, lekin jo free carriers tumhe actually milte hain woh temperature ke saath change hote hain. Kyun? Kyunki temperature woh "energy budget" hai jo nature electrons ko promote karne mein kharchti hai.

Bahut thanda → dopants ko ionize karne ke liye itni thermal energy nahi → carriers apne donor atoms pe freeze ho jaate hain (freeze-out).
Bilkul sahi → saare dopants ionized hain, lekin host lattice ki bonds todne ke liye itni energy nahi → carrier count flat rehta hai aur doping ke barabar hota hai (extrinsic / saturation).
Bahut garam → thermal energy host bonds ko tod deti hai, itni tezi se electron-hole pairs banate hue ki woh dopants ko daba dete hain (intrinsic).

Toh $ln n$ vs $1/ T$ ka plot teen regions mein hota hai, ek mein nahi.

HUM KYA count kar rahe hain?

Teen quantities jo humein chahiye:

KAISE: first principles se intrinsic $n_{i}$ derive karna

Step 1 — Occupancy. Electrons Fermi–Dirac statistics follow karte hain: $f (E) = \frac{1}{1 + e ^{(E - E_{F}) / k_{B} T}}$ Kyun? Electrons fermions hain; yeh probability hai ki energy $E$ par ek available state occupied hai.

Step 2 — Boltzmann approximation. $E - E_{F} ≫ k_{B} T$ ke liye (non-degenerate semiconductor ki conduction band mein yeh sach hai), "+1" negligible hai: $f (E) \approx e^{- (E - E_{F}) / k_{B} T}$ Yeh step kyun? Yeh ek ugly integral ko ek Gaussian-type integral mein convert karta hai jo hum exactly kar sakte hain.

Step 3 — Seats (density of states). Band edge $E_{C}$ ke paas: $g (E) = \frac{( 2 m _{e}^{*} ) ^{3/2}}{2 π ^{2} ℏ ^{3}} E - E_{C}$ Kyun? Parabolic band → free-electron-like $g (E) \propto E$ band edge se measure kiya gaya.

Step 4 — Integrate karo. $n = \int_{E_{C}}^{\infty} g (E) f (E) d E$ $x = (E - E_{C}) / k_{B} T$ substitute karo; standard integral $\int_{0}^{\infty} x e^{- x} d x = \frac{π}{2}$ deta hai $n = N_{C} e^{- (E_{C} - E_{F}) / k_{B} T}, N_{C} = 2 (\frac{2 π m _{e}^{*} k _{B} T}{h ^{2}})^{3/2}$ Similarly holes ke liye: $p = N_{V} e^{- (E_{F} - E_{V}) / k_{B} T}$ .

Step 5 — Intrinsic case. Pure material mein $n = p = n_{i}$ . Multiply karo: $n p = N_{C} N_{V} e^{- (E_{C} - E_{V}) / k_{B} T} = N_{C} N_{V} e^{- E_{g} / k_{B} T}$ Yeh step kyun? Multiply karne se $E_{F}$ cancel ho jaata hai — "law of mass action." Square root lo: $n_{i} = N_{C} N_{V} e^{- E_{g} /2 k_{B} T}$

Teen regions (shape kyun aisi hai)

Figure — Temperature dependence of carrier concentration

$ln n$ y-axis par aur $1/ T$ x-axis par plot karo (toh "hot" left pe hai).

Region	Temperature	Kya ho raha hai	$ln n$ vs $1/ T$ ka slope
Freeze-out	bahut low	donors abhi ionized nahi hue	$- E_{D} /2 k_{B}$ (chota)
Extrinsic / saturation	middle	saare donors ionized, $n \approx N_{D}$	$\approx 0$ (flat)
Intrinsic	high	band-to-band pairs dominate karte hain	$- E_{g} /2 k_{B}$ (steep)

Worked examples

Classic mistakes ko steel-man karna

Pehle forecast karo, phir verify karo

Flashcards

n_{i}

mein

e^{- E_{g} /2 k_{B} T}

kyun hai aur

e^{- E_{g} / k_{B} T}

kyun nahi?

Kyunki

n_{i} = n p = N_{C} N_{V} e^{- E_{g} /2 k_{B} T}

; mass-action product ki square root lene se exponent half ho jaata hai.

Carrier concentration ke teen temperature regions kya hain?

Freeze-out (low

T

), extrinsic/saturation (mid

T

n \approx N_{D}

), intrinsic (high

T

Extrinsic region mein

n

approximately kiske barabar hota hai?

Donor concentration

N_{D}

(saare donors ionized).

Extrinsic region mein

ln n

1/ T

flat kyun hota hai?

Saare dopants ionized ho gaye hain (koi bacha nahi) aur host bonds todne ke liye bahut thanda hai, toh carrier count

N_{D}

par saturate ho jaata hai.

Intrinsic region mein

ln n

1/ T

ka slope kya hai?

- E_{g} /2 k_{B}

Law of mass action kya hai?

n p = N_{C} N_{V} e^{- E_{g} / k_{B} T} = n_{i}^{2}

, doping se independent (equilibrium mein).

N_{C}

physically kya hai aur yeh

T

ke saath kaise scale karta hai?

Conduction band mein effective density of states,

N_{C} \propto T^{3/2}

Hum Fermi-Dirac ke liye Boltzmann approximation kyun use karte hain?

Kyunki non-degenerate semiconductor mein

E - E_{F} ≫ k_{B} T

, toh

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

, jo integral solvable bana deta hai.

Freeze-out ke dauran resistance ka kya hota hai?

Yeh badhti hai — carriers waapas donors pe chipak jaate hain, toh material zyada insulating ho jaata hai.

300 K par Si ka roughly

n_{i}

kitna hai?

Lagbhag

1.0 \times 1 0^{10}

^{- 3}

High temperature par silicon devices kyun fail ho jaate hain?

Woh intrinsic region mein jaate hain (

n_{i} ≳ N_{D}

), toh doping carrier type/count ko control nahi kar paata.

Recall Feynman: 12-saal ke bachche ko explain karo

Ek school imagine karo jisme "helper" bacchon ki ek fixed number hai (dopants) jo haath uthaake volunteer ho sakte hain (free electrons ban jaate hain).

Thandi subah (freeze-out): sabko neend aa rahi hai, bahut kam haath uthte hain. Bahut kam volunteers.
Garam din (extrinsic): har helper ne haath utha diya hai. Aur zyada volunteers nahi mil sakte — helpers itne hi hain! Number flat rehti hai.
Super hot (intrinsic): ab itna garam hai ki regular non-helper bacche bhi uchhalkar khade hone lagte hain. Achanak original helpers se kahin zyada volunteers hain, aur helpers ab matter nahi karte. Isliye temperature ke saath volunteers ki number pehle badhti hai, phir flat hoti hai, phir explode karti hai.

Connections

Fermi-Dirac distribution — occupancy $f (E)$ jo humne approximate ki.
Density of states in semiconductors — $g (E) \propto E - E_{C}$ "seats."
Law of mass action — jahan se $n p = n_{i}^{2}$ aata hai.
Doping and dopant ionization — $N_{D}$ aur $E_{D}$ set karta hai.
Intrinsic vs extrinsic semiconductors — yahan name kiye gaye regimes.
Conductivity and mobility vs temperature — poori $σ (T)$ ke liye $n (T)$ ko $μ (T)$ ke saath combine karta hai.