Temperature dependence of carrier concentration

WHAT are we counting?

The three quantities we need:

HOW: deriving intrinsic $n_i$ from first principles

Step 1 — Occupancy. Electrons obey Fermi–Dirac statistics: $f(E) = \frac{1}{1+e^{(E-E_F)/k_BT}}$ Why? Electrons are fermions; this is the probability an available state at energy $E$ is occupied.

Step 2 — Boltzmann approximation. For $E - E_F \gg k_BT$ (true in the conduction band of a non-degenerate semiconductor), the "+1" is negligible: $f(E) \approx e^{-(E-E_F)/k_BT}$ Why this step? It converts an ugly integral into a Gaussian-type one we can do exactly.

Step 3 — Seats (density of states). Near the band edge $E_C$ : $g(E) = \frac{(2m_e^*)^{3/2}}{2\pi^2\hbar^3}\sqrt{E-E_C}$ Why? Parabolic band → free-electron-like $g(E)\propto\sqrt{E}$ measured from the band edge.

Step 4 — Integrate. $n = \int_{E_C}^{\infty} g(E) f(E)\,dE$ Substitute $x=(E-E_C)/k_BT$ ; the standard integral $\int_0^\infty \sqrt{x}\,e^{-x}dx=\frac{\sqrt\pi}{2}$ gives $\boxed{n = N_C\, e^{-(E_C-E_F)/k_BT}}, \quad N_C = 2\left(\frac{2\pi m_e^*k_BT}{h^2}\right)^{3/2}$ Similarly for holes: $p = N_V\, e^{-(E_F-E_V)/k_BT}$ .

Step 5 — Intrinsic case. In pure material $n=p=n_i$ . Multiply: $np = N_C N_V\, e^{-(E_C-E_V)/k_BT} = N_C N_V\, e^{-E_g/k_BT}$ Why this step? Multiplying makes $E_F$ cancel — the "law of mass action." Take the square root: $\boxed{n_i = \sqrt{N_C N_V}\; e^{-E_g/2k_BT}}$

The three regions (WHY the shape)

Figure — Temperature dependence of carrier concentration

Plot $\ln n$ on the y-axis vs $1/T$ on the x-axis (so "hot" is on the left).

Region	Temperature	What's happening	Slope of $\ln n$ vs $1/T$
Freeze-out	very low	donors not yet ionized	$-E_D/2k_B$ (small)
Extrinsic / saturation	middle	all donors ionized, $n\approx N_D$	$\approx 0$ (flat)
Intrinsic	high	band-to-band pairs dominate	$-E_g/2k_B$ (steep)

Worked examples

Worked example 1.

$n_i$ of silicon at 300 K Given $E_g=1.12$ eV, $N_C=2.8\times10^{19}$ , $N_V=1.04\times10^{19}$ cm $^{-3}$ , $k_BT=0.02585$ eV. $n_i=\sqrt{2.8\times10^{19}\cdot1.04\times10^{19}}\;e^{-1.12/(2\cdot0.02585)}$ Why the square root? Law of mass action from Step 5. Exponent $=-1.12/0.0517=-21.66$ , so $e^{-21.66}=3.9\times10^{-10}$ . $\sqrt{2.91\times10^{38}}=1.71\times10^{19}$ . $n_i \approx 1.71\times10^{19}\times3.9\times10^{-10} \approx 1.0\times10^{10}\text{ cm}^{-3}$ . ✓ (textbook value)

Steel-manning the classic mistakes

Forecast-then-verify

Flashcards

Why does

n_i

have

e^{-E_g/2k_BT}

and not

e^{-E_g/k_BT}

Because

n_i=\sqrt{np}=\sqrt{N_CN_V}e^{-E_g/2k_BT}

; taking the square root of the mass-action product halves the exponent.

What are the three temperature regions of carrier concentration?

Freeze-out (low

T

), extrinsic/saturation (mid

T

n\approx N_D

), intrinsic (high

T

In the extrinsic region, what does

n

approximately equal?

The donor concentration

N_D

(all donors ionized).

Why is

\ln n

1/T

flat in the extrinsic region?

All dopants are ionized (none left) and it's too cold to break host bonds, so carrier count saturates at

N_D

What is the slope of

\ln n

1/T

in the intrinsic region?

-E_g/2k_B

What is the law of mass action?

np = N_C N_V e^{-E_g/k_BT} = n_i^2

, independent of doping (at equilibrium).

What is

N_C

physically and how does it scale with

T

Effective density of states in the conduction band,

N_C\propto T^{3/2}

Why do we use the Boltzmann approximation to Fermi-Dirac?

Because in a non-degenerate semiconductor

E-E_F\gg k_BT

, so

f(E)\approx e^{-(E-E_F)/k_BT}

, making the integral solvable.

What happens to resistance during freeze-out?

It rises — carriers stick back onto donors, so the material becomes more insulating.

Roughly

n_i

of Si at 300 K?

About

1.0\times10^{10}

^{-3}

Why do silicon devices fail at high temperature?

They enter the intrinsic region (

n_i\gtrsim N_D

), so doping no longer controls carrier type/count.

Recall Feynman: explain to a 12-year-old

Imagine a school with a fixed number of "helper" kids (dopants) who can raise their hands to volunteer (become free electrons).

Cold morning (freeze-out): everyone's sleepy, few hands go up. Very few volunteers.
Warm day (extrinsic): every helper has raised their hand. You can't get more volunteers — there are only so many helpers! The number stays flat.
Super hot (intrinsic): now it's SO hot that even the regular non-helper kids start jumping up. Suddenly there are way more volunteers than the original helpers, and the helpers don't matter anymore. The number of volunteers vs temperature therefore rises, flattens, then explodes.

Connections

Fermi-Dirac distribution — the occupancy $f(E)$ we approximated.
Density of states in semiconductors — the $g(E)\propto\sqrt{E-E_C}$ "seats."
Law of mass action — where $np=n_i^2$ comes from.
Doping and dopant ionization — sets $N_D$ and $E_D$ .
Intrinsic vs extrinsic semiconductors — the regimes named here.
Conductivity and mobility vs temperature — combines $n(T)$ with $\mu(T)$ for full $\sigma(T)$ .

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, ek doped semiconductor mein dopant atoms ki sankhya fix hoti hai, lekin actual free carriers temperature ke saath change karte hain. Iski wajah simple hai: temperature ka matlab hai "energy budget" jo nature electrons ko upar promote karne pe kharch karti hai. Bahut thanda ho to donor apne electron ko release hi nahi karta — isko freeze-out kehte hain, carriers gayab ho jaate hain aur resistance badh jaata hai.

Beech ke temperature pe, saare donors ionize ho chuke hote hain, par lattice ke bonds todne ke liye energy kam hai — isliye $n\approx N_D$ ho ke flat ho jaata hai. Isko extrinsic/saturation region bolte hain, aur yahi wo zone hai jahan hum real devices chalate hain, kyunki carrier count temperature ke saath stable rehta hai. Jab bahut zyada garam kar do, tab host bonds toot ke electron-hole pairs banne lagte hain itni tezi se ki doping ka effect dab jaata hai — ye intrinsic region hai, aur yahi wajah hai ki overheat hone pe Si device fail ho jaata hai.

Sabse important formula: $n_i=\sqrt{N_CN_V}\,e^{-E_g/2k_BT}$ . Yaad rakho — exponent mein $E_g/2$ hai, poora $E_g$ nahi, kyunki humne $np$ ka square root liya (law of mass action, jahan $E_F$ cancel ho jaata hai). Agar tum $\ln n$ vs $1/T$ ka graph banao to teen slope dikhenge: thoda tilt (freeze-out), flat (extrinsic), aur steep $-E_g/2k_B$ (intrinsic). Bas isi teen-region picture ko samajh lo, poora topic clear ho jaata hai. Mnemonic: "Frozen Extras Ignite".

Temperature dependence of carrier concentration

WHAT are we counting?

HOW: deriving intrinsic $n_i$ from first principles

The three regions (WHY the shape)

Worked examples

Steel-manning the classic mistakes

Forecast-then-verify

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Band Theory & Carrier Physics

Connections