Temperature dependence of carrier concentration: Visual walkthrough

We are answering ONE question the whole way down:

How many free electrons per cubic centimetre live in the conduction band of a pure crystal at temperature $T$ ?

Call that number $n$ (for pure material we will rename it $n_{i}$ , the " $i$ " meaning intrinsic — coming from the material itself, not from added dopants). See Intrinsic vs extrinsic semiconductors for what "intrinsic" means.

Step 1 — Draw the energy landscape (what $E_{C}$ , $E_{V}$ , $E_{g}$ even are)

WHAT. Every electron in a crystal carries an amount of energy. We draw energy $E$ going up the page (like height). There are two "floors" electrons care about:

$E_{V}$ — the top of the valence band, the crowded downstairs where electrons sit bonded into the lattice.
$E_{C}$ — the bottom of the conduction band, the empty upstairs where an electron is free to move and carry current.
Between them is a forbidden gap of width $E_{g} = E_{C} - E_{V}$ where no electron is allowed to sit.

WHY. A "free carrier" is an electron that made it upstairs to $E_{C}$ . So counting free electrons = counting who reached the upstairs floor. Everything else is bookkeeping about that climb.

PICTURE. The gap is a cliff of height $E_{g}$ . Downstairs is packed; upstairs is empty until heat lifts someone up.

Figure — Temperature dependence of carrier concentration

Step 2 — Occupancy: what is the chance a seat is filled?

WHAT. Suppose there is an allowed seat at energy $E$ . What is the probability it actually holds an electron right now? Nature's answer for electrons is the Fermi–Dirac distribution: $f (E) = \frac{1}{1 + e ^{(E - E_{F}) / k_{B} T}}$

Let us read every symbol, right where it sits:

$f (E)$ — a probability, a number between $0$ and $1$ .
$E_{F}$ — the Fermi level, the energy where the seat is exactly half-full ( $f = \frac{1}{2}$ ). Think "the water line."
$k_{B}$ — Boltzmann's constant, a fixed conversion between temperature and energy.
$k_{B} T$ — the thermal energy budget: how much energy a typical jiggling particle has at temperature $T$ . Bigger $T$ → bigger budget.
$e^{(E - E_{F}) / k_{B} T}$ — the exponential that switches $f$ from "on" (below the water line) to "off" (above it).

WHY this tool and not another? Electrons are fermions: no two can share a state, so their statistics is not the ordinary coin-flip — it is Fermi–Dirac. We must start here; see Fermi-Dirac distribution.

PICTURE. $f (E)$ is a smooth step: near $1$ well below $E_{F}$ , dropping through $\frac{1}{2}$ at $E_{F}$ , near $0$ well above. Raising $T$ softens the step's edge — the water line gets fuzzier.

Step 3 — The Boltzmann shortcut (why the tail is a clean exponential)

WHAT. We only care about seats up in the conduction band, i.e. $E \geq E_{C}$ . Up there, $E$ is far above the Fermi level: $E - E_{F} ≫ k_{B} T$ . When the exponent is large and positive, $e^{(E - E_{F}) / k_{B} T}$ is huge, so the " $+ 1$ " underneath is a rounding error. Drop it: $f (E) \approx e^{- (E - E_{F}) / k_{B} T}$

Term by term:

$- (E - E_{F}) / k_{B} T$ — the higher above the water line, the more negative this is, the tinier the chance.
The whole thing is now a plain decaying exponential in $E$ .

WHY. Two reasons. Physically: high seats are rarely occupied, so we are in the sparse tail. Mathematically: this turns a messy fraction into a clean exponential that we can actually integrate in Step 5. (Valid only for a non-degenerate semiconductor — one not doped so heavily that $E_{F}$ climbs into the band.)

PICTURE. Zoom into the conduction band region of the Fermi–Dirac curve; the true curve and the exponential approximation lie right on top of each other there.

Step 4 — Counting seats: the density of states $g (E)$

WHAT. Probability alone is not a count. We also need how many seats exist in each thin slice of energy. That is the density of states $g (E)$ : the number of allowed states per unit energy per unit volume. Just above the band edge, $g (E) = \frac{( 2 m _{e}^{*} ) ^{3/2}}{2 π ^{2} ℏ ^{3}} E - E_{C}$

Reading it:

$E - E_{C}$ — the star of the show: seats are rare right at the floor $E_{C}$ and grow as $height above the floor$ .
$m_{e}^{*}$ — the electron's effective mass (how heavy it acts inside the crystal); heavier ⇒ more seats.
$ℏ$ — Planck's constant divided by $2 π$ ; it sets the fundamental "size" of one quantum state.
The messy prefactor is just a fixed number packaging those constants.

WHY this shape? In a parabolic band the number of ways to have a given energy grows like $E$ (measured from the edge) — the same square-root you meet for a free particle in 3D. See Density of states in semiconductors.

PICTURE. A sideways parabola-root starting at zero at $E_{C}$ and fanning outward as you go up.

Step 5 — Seats × occupancy, then add them all up (the integral)

WHAT. Electrons in one thin energy slice $d E$ = (seats there) × (chance each is filled) = $g (E) f (E) d E$ . To get the total $n$ we add up all slices from the band floor $E_{C}$ upward. "Add up infinitely many thin slices" is exactly what the integral sign $\int$ means: $n = \int_{E_{C}}^{\infty} g (E) f (E) d E$

$\int_{E_{C}}^{\infty}$ — sum from the conduction floor to infinitely high.
$g (E)$ — the rising of available seats.
$f (E)$ — the falling exponential of occupancy.

WHY an integral and not a sum you can eyeball? Energy is continuous — there is no smallest step between seats — so the natural tool for "total over a continuum" is integration.

PICTURE. $g (E)$ rises, $f (E)$ falls; their product is a little hump. The area under that hump (shaded) is $n$ . Notice: rising × falling gives a peak just above $E_{C}$ , then nothing — so almost all free electrons sit right near the band edge.

Now the arithmetic. Substitute $x = (E - E_{C}) / k_{B} T$ so the exponential becomes clean, and use the standard result $\int_{0}^{\infty} x e^{- x} d x = \frac{π}{2}$ . The $π /2$ and the constants collapse into one bundled symbol $N_{C}$ : $n = N_{C} e^{- (E_{C} - E_{F}) / k_{B} T}, N_{C} = 2 (\frac{2 π m _{e}^{*} k _{B} T}{h ^{2}})^{3/2}$

$N_{C}$ — the effective density of states: pretend all the conduction seats were squeezed into a single level right at $E_{C}$ ; $N_{C}$ is how many that would be. It scales as $T^{3/2}$ (hotter ⇒ more thermally-reachable seats).
$e^{- (E_{C} - E_{F}) / k_{B} T}$ — the chance factor for a seat sitting one gap-fraction above the water line.

By the identical argument downstairs (counting empty seats = holes), the hole count is $p = N_{V} e^{- (E_{F} - E_{V}) / k_{B} T}, N_{V} = 2 (\frac{2 π m _{h}^{*} k _{B} T}{h ^{2}})^{3/2} .$

Step 6 — The multiply trick that kills $E_{F}$ (law of mass action)

WHAT. We have $n$ and $p$ , but both still contain the unknown water line $E_{F}$ . Multiply them: $n p = N_{C} N_{V} e^{- (E_{C} - E_{F}) / k_{B} T} e^{- (E_{F} - E_{V}) / k_{B} T} = N_{C} N_{V} e^{- (E_{C} - E_{V}) / k_{B} T} = N_{C} N_{V} e^{- E_{g} / k_{B} T} .$

Watch the exponents add: $- (E_{C} - E_{F}) - (E_{F} - E_{V}) = - (E_{C} - E_{V}) = - E_{g}$ . The $E_{F}$ cancels exactly.

WHY multiply? Because $+ E_{F}$ and $- E_{F}$ live in the two exponents; multiplying adds the exponents and annihilates $E_{F}$ . This gives the law of mass action: $n p$ depends only on the gap and temperature, not on where the Fermi level (i.e. the doping) sits. See Law of mass action.

PICTURE. Two arrows, $E_{C} - E_{F}$ up and $E_{F} - E_{V}$ down, sliding along the axis: their sum is always the full gap $E_{g}$ no matter where $E_{F}$ floats.

Step 7 — The pure-material shortcut and the square root

WHAT. In a pure crystal every electron promoted upstairs leaves exactly one empty seat downstairs, so electrons and holes come in equal numbers: $n = p = n_{i}$ . Then $n p = n_{i}^{2}$ , and from Step 6: $n_{i}^{2} = N_{C} N_{V} e^{- E_{g} / k_{B} T} ⟹ n_{i} = N_{C} N_{V} e^{- E_{g} /2 k_{B} T}$

The square root hits the exponential too: $e^{- E_{g} / k_{B} T} = e^{- E_{g} /2 k_{B} T}$ . That is where the factor of 2 is born.
$N_{C} N_{V}$ — the geometric-mean seat count, still $\propto T^{3/2}$ .

WHY the square root? Because $n_{i} = n_{i}^{2} = n p$ . There is no new physics — halving the exponent is a mathematical consequence of taking the root of the mass-action product.

PICTURE. A number line of exponents: $n p$ sits at exponent $- E_{g} / k_{B} T$ ; the square root drags it to the midpoint $- E_{g} /2 k_{B} T$ .

Step 8 — Which factor rules? (the exponential is the boss)

WHAT. $n_{i}$ is (slow $T^{3/2}$ prefactor) × (violent exponential). Which wins as $T$ climbs?

WHY it matters. It decides the shape of the famous $ln n_{i}$ vs $1/ T$ plot. Take the log: $ln n_{i} = crawls with l n T ln N_{C} N_{V} - \frac{E _{g}}{2 k _{B}} \cdot \frac{1}{T} .$ Against $1/ T$ this is nearly a straight line of slope $- E_{g} /2 k_{B}$ — the exponential dominates, the prefactor is a barely-visible curve.

PICTURE. Two curves vs $T$ on a log axis: the $T^{3/2}$ prefactor is a gentle slope; the exponential is a rocket. Their product tracks the rocket.

The one-picture summary

Read it left to right: a step of chance ( $f$ ) times a root of seats ( $g$ ) gives a hump whose area is $n$ ; multiplying $n$ and $p$ cancels $E_{F}$ ; the square root halves the exponent into the final law.

Recall Feynman retelling — say it back in plain words

I want to count free electrons upstairs at temperature $T$ . First I draw two floors with a cliff $E_{g}$ between them. For any seat I ask "what's the chance it's filled?" — that's Fermi–Dirac, a soft step at the water line $E_{F}$ . Way up in the conduction band the chance is a tiny decaying exponential, so I use that simpler form. But chance isn't a count, so I also draw how many seats exist at each height — a curve starting at the floor. Multiply chance by seats, add up all the thin slices (that's the integral), and the total collapses into $n = N_{C} e^{- (E_{C} - E_{F}) / k_{B} T}$ , where $N_{C}$ is just "all the seats pretended to sit at the floor." Holes downstairs give a mirror formula with $E_{F}$ flipped. I don't know where the water line is, so I multiply $n$ and $p$ — the $+ E_{F}$ and $- E_{F}$ kill each other, leaving $n p = N_{C} N_{V} e^{- E_{g} / k_{B} T}$ , true no matter how I dope. That's the law of mass action. In pure material $n = p = n_{i}$ , so $n_{i} = n p$ — and the square root halves the exponent, giving $n_{i} = N_{C} N_{V} e^{- E_{g} /2 k_{B} T}$ . The 2 is not physics, it's the square root. The exponential is the boss: plot $ln n_{i}$ vs $1/ T$ and you get a near-straight line of slope $- E_{g} /2 k_{B}$ . At $T \to 0$ carriers vanish; at $T \to \infty$ they saturate at $N_{C} N_{V}$ ; if the gap were zero they'd be everywhere.

For the extrinsic plateau and how mobility rides along, continue with Conductivity and mobility vs temperature and Intrinsic vs extrinsic semiconductors.