Temperature dependence of carrier concentration: Question bank

First, a shared vocabulary reminder so no term below is unearned.

Definition The words used on this page

Carrier: a mobile charge — a free electron in the conduction band, or a hole in the valence band.
$n$ , $p$ : number of free electrons ( $n$ ) and holes ( $p$ ) per unit volume; $n_{i}$ is their common value in pure material.
$E_{C}$ , $E_{V}$ : the energy of the conduction-band edge (bottom of the electron band) and the valence-band edge (top of the hole band). Picture two horizontal shelves with an empty gap between them.
$E_{g}$ (band gap): the energy height of that empty gap, $E_{g} = E_{C} - E_{V}$ . An electron must be handed at least this much energy to jump from the lower shelf to the upper one.
$E_{F}$ (Fermi level): the energy at which a state has a 50% chance of being occupied; think of it as the "water line" of the electron sea. See Fermi-Dirac distribution.
$E_{D}$ (donor binding energy): the small energy $E_{C} - E_{donor}$ needed to lift a donor's extra electron up into the conduction band. See Doping and dopant ionization.
$k_{B}$ (Boltzmann constant): the conversion between temperature and energy, $k_{B} = 8.617 \times 1 0^{- 5} eV/K$ ; the product $k_{B} T$ is nature's "thermal energy budget" at temperature $T$ .
$N_{C}$ , $N_{V}$ : the effective density of states at each band edge — roughly "how many available seats" sit near $E_{C}$ and $E_{V}$ . Both scale as $N_{C}, N_{V} \propto T^{3/2}$ . See Density of states in semiconductors.
$m_{e}^{*}$ , $m_{h}^{*}$ (effective mass): the electron's and hole's apparent mass inside the crystal, encoding how the lattice bends their motion; they set the size of $N_{C}$ and $N_{V}$ .
$ℏ$ (reduced Planck constant): $ℏ = h /2 π$ , the quantum unit of action that fixes the spacing of allowed energy states.
Ionized donor ( $N_{D}^{+}$ ): a donor atom that has already given up its extra electron to the conduction band.
Freeze-out: low- $T$ regime where donors have not released their electrons — the carrier sits "frozen" on its donor.
Extrinsic / saturation: mid- $T$ regime where all donors are ionized and $n \approx N_{D}$ ; carrier count is flat.
Intrinsic: high- $T$ regime where band-to-band pairs dominate over the doping.
Non-degenerate: doping is light enough that the Fermi level $E_{F}$ stays inside the gap, well away ( $≫ k_{B} T$ ) from the band edge — the condition that lets us swap Fermi–Dirac for the Boltzmann factor.
Degenerate: doping is so heavy that $E_{F}$ pushes into a band (above $E_{C}$ or below $E_{V}$ ); the Boltzmann shortcut then fails and the material behaves almost metallically.

The two figures below are the mental pictures every trap refers back to.

Figure — Temperature dependence of carrier concentration

Why freeze-out carries $E_{D} /2$ , not $E_{D}$ (the missing derivation)

Step 1 — Count what must balance. In freeze-out only donors feed the band, so charge neutrality reads $n \approx N_{D}^{+}$ (ionized donors) and $N_{D}^{+} + n_{D} = N_{D}$ , where $n_{D}$ is the number of donors still holding their electron. Why: every conduction electron came from exactly one emptied donor.

Step 2 — Write occupancy of the donor level. The chance a donor still holds its electron follows Fermi–Dirac at the donor energy $E_{donor} = E_{C} - E_{D}$ : $\frac{n _{D}}{N _{D}} \approx e^{- (E_{F} - E_{donor}) / k_{B} T} = e^{- (E_{F} - E_{C} + E_{D}) / k_{B} T} .$ Why: deep in freeze-out most donors are still occupied, so this small fraction sets the leftover.

Step 3 — Use the band formula for $n$ . From the parent note, $n = N_{C} e^{- (E_{C} - E_{F}) / k_{B} T}$ . Notice $E_{F}$ appears in both Step 2 and here, in opposite signs. Multiplying $n$ by the ionized fraction and demanding $n \approx N_{D}^{+}$ lets us eliminate $E_{F}$ : $n^{2} \approx N_{C} N_{D} e^{- E_{D} / k_{B} T} .$ Why this step: just like the mass-action trick, multiplying two occupancies with opposite $E_{F}$ -signs makes the unknown Fermi level cancel.

Step 4 — Take the square root. Solving the last line, $n \approx N_{C} N_{D} e^{- E_{D} /2 k_{B} T} .$ Why the $/2$ : because $n$ appeared squared in Step 3, the square root halves the exponent — the exact same reason $n_{i}$ carries $E_{g} /2$ . The reservoir (discrete donor level) feeding a continuum always produces this half-exponent.

True or false — justify

Doubling the doping

N_{D}

doubles the intrinsic carrier concentration

n_{i}

False.

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

contains no doping term — it depends only on the host material and temperature. Doubling

N_{D}

raises the extrinsic plateau, not

n_{i}

In the extrinsic (saturation) region, adding heat increases the free-carrier count.

False. Every donor is already ionized, so there is nothing left to release;

n

stays pinned at

N_{D}

until the host bonds start breaking (intrinsic onset).

The law of mass action

n p = n_{i}^{2}

only holds in undoped (intrinsic) material.

False. At thermal equilibrium it holds for any doping, because multiplying

n

and

p

cancels

E_{F}

entirely. Doping shifts

n

and

p

in opposite directions but their product is fixed.

Cooling a semiconductor toward freeze-out lowers its resistance.

False. Carriers stick back onto donors and disappear, so with fewer carriers the material becomes more insulating and resistance rises.

On a

ln n

1/ T

plot, the high-temperature (intrinsic) region is on the right.

False. Large

1/ T

means small

T

(cold), which is on the right; hot is small

1/ T

, on the left. Intrinsic is the steep segment on the left.

The slope of

ln n

1/ T

in the intrinsic region equals

- E_{g} / k_{B}

False. The exponent is

E_{g} /2

, so the slope is

- E_{g} /2 k_{B}

. Forgetting the factor of 2 doubles the band gap you extract.

A larger band gap

E_{g}

pushes the intrinsic-onset temperature higher.

True. A bigger gap makes

n_{i}

smaller at any

T

, so you must heat further before

n_{i}

can overtake

N_{D}

— this is why wide-gap materials tolerate more heat.

In freeze-out the carriers stop moving because their kinetic energy is too low.

False. Freeze-out is about carriers not existing in the band — they've recombined onto donor atoms. It is a population effect, not a mobility effect.

Spot the error

"Since

n = N_{C} e^{- (E_{C} - E_{F}) / k_{B} T}

and this grows with

T

n

always rises with temperature."

The prefactor

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

and the exponent both depend on

T

, but

E_{F}

also moves with

T

. In the extrinsic plateau

E_{F}

drops just enough to keep

n \approx N_{D}

constant — the formula does not read as "monotonically increasing."

"The Boltzmann approximation

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

is exact, so we can use it right at

E = E_{F}

It is an approximation valid only when

E - E_{F} ≫ k_{B} T

. At

E = E_{F}

the true Fermi–Dirac value is exactly

1/2

, while the Boltzmann form gives

1

— a 100% error. See Fermi-Dirac distribution.

"Freeze-out has exponent

E_{D}

because it's just Boltzmann activation over the donor energy."

The correct exponent is

E_{D} /2

. As derived above, multiplying the ionized fraction by the band formula cancels

E_{F}

and leaves

n^{2} \propto e^{- E_{D} / k_{B} T}

; the square root then halves it — a discrete level feeding a continuum, mirroring the

E_{g} /2

of the intrinsic case.

"The density of states

g (E) = \frac{( 2 m _{e}^{*} ) ^{3/2}}{2 π ^{2} ℏ ^{3}} E - E_{C}

shows more carriers appear because more seats open with

T

g (E)

has no temperature in it — the seats are always there. It is the occupancy factor

f (E)

that carries all the temperature dependence. Confusing seats with occupancy is the classic mix-up. See Density of states in semiconductors.

"A student reads a steep slope of

- 13000

K and reports

E_{g} = 1.12

eV using

E_{g} = k_{B} \times slope

They dropped the factor of 2. The correct relation is

E_{g} = 2 k_{B} \times ∣ slope ∣

, so

- 13000

K actually gives

\approx 2.24

eV — or, if

E_{g}

is truly

1.12

eV, the slope should be

\approx - 6500

n = N_{D}^{+} + n_{i}^{'}

means we simply add the doping and intrinsic carrier concentrations everywhere."

The two contributions are never both large at once: in extrinsic

N_{D}^{+}

dominates and

n_{i}^{'}

is negligible; in intrinsic the reverse. Adding them naively double-counts near the crossover and is misleading elsewhere.

Why questions

Why does multiplying

n

and

p

(rather than using either alone) give a doping-independent result?

Because

n \propto e^{+ E_{F} / k_{B} T}

and

p \propto e^{- E_{F} / k_{B} T}

, so the product cancels

E_{F}

— the one quantity that shifts with doping. What survives depends only on the fixed gap

E_{g}

and temperature.

Why is the operating region of real devices the extrinsic plateau and not the intrinsic region?

On the plateau

n \approx N_{D}

is nearly temperature-independent, so the transistor's carrier count (and thus behaviour) is stable against temperature swings. In the intrinsic region carrier count explodes with

T

and the doping loses control.

Why does the exponent in

n_{i}

carry

E_{g} /2

while the plain band formula

n = N_{C} e^{- (E_{C} - E_{F}) / k_{B} T}

carries a full energy?

The single-band formula uses the distance from

E_{F}

to the band edge;

n_{i}

is obtained by taking the square root of the mass-action product

n p \propto e^{- E_{g} / k_{B} T}

, which halves the exponent.

Why can't more heat create carriers in the saturation region even though heat "ionizes things"?

All donors are already ionized — the supply is exhausted. Creating new carriers there would require breaking host lattice bonds, which needs the far larger energy

E_{g}

, not yet available at those temperatures.

Why does the

T^{3/2}

prefactor in

N_{C}

get called a "slow tag-along," and why can we ignore it when reading a slope?

Because

ln n

contains a term

\frac{3}{2} ln T

plus the term

- E_{g} /2 k_{B} \cdot (1/ T)

. Against the

x

-axis variable

1/ T

, the exponential piece is linear and steep, while

\frac{3}{2} ln T

changes by only a factor over a whole doubling of

T

— a gentle curve on top of a strong straight line. Over the narrow window we fit, its contribution to the slope is tiny, so the measured slope reads essentially

- E_{g} /2 k_{B}

Why does mobility decrease with temperature in the extrinsic region even though carrier count is flat?

Hotter lattices vibrate more (more phonons), scattering carriers more often and lowering mobility. Since

n

is constant there, the conductivity actually falls with

T

in the plateau — see Conductivity and mobility vs temperature.

Edge cases

At the exact temperature where

n_{i} = N_{D}

, is the material extrinsic or intrinsic?

Neither cleanly — it is the crossover. Both contributions are comparable, so

n \approx N_{D} + n_{i}

is roughly

2 N_{D}

and the

ln n

1/ T

curve is bending from flat toward steep.

What happens to

n_{i}

T \to 0

The exponential

e^{- E_{g} /2 k_{B} T} \to 0

far faster than the

T^{3/2}

prefactor grows, so

n_{i} \to 0

. A perfect intrinsic semiconductor becomes a perfect insulator at absolute zero.

For a hypothetical zero-gap material (

E_{g} = 0

), what does the intrinsic region look like?

With

E_{g} = 0

the exponential becomes

e^{0} = 1

, so

n_{i} \propto T^{3/2}

with no freeze-out of pairs — carriers exist at any nonzero

T

. There is no steep intrinsic slope because there is no gap to activate over (semimetal-like behaviour).

If the donor binding energy

E_{D}

were essentially zero, would there be a freeze-out region?

No visible freeze-out — donors would ionize even at very low

T

, so the curve would be extrinsic (flat) almost down to absolute zero. Freeze-out only appears because

E_{D} > 0

costs energy to overcome.

In a compensated sample with both donors and acceptors, what pins

n

in the extrinsic region?

The net doping

N_{D} - N_{A}

, not

N_{D}

alone, because acceptors soak up some donor electrons. The plateau sits at

n \approx N_{D} - N_{A}

What does the

ln n

1/ T

curve do for an undoped (truly intrinsic) sample?

It is a single straight line of slope

- E_{g} /2 k_{B}

across all temperatures — no plateau and no freeze-out kink, because there are no dopants to ionize or exhaust.

If doping is pushed so high that

E_{F}

enters the conduction band (degenerate doping), what breaks?

The Boltzmann approximation fails because

E_{C} - E_{F}

is no longer

≫ k_{B} T

— you must keep the full Fermi–Dirac "+1." The formulas

n = N_{C} e^{- (E_{C} - E_{F}) / k_{B} T}

and

n_{i}^{2} = n p

overestimate, the plateau blurs, and the material conducts almost like a metal with little freeze-out.

Right at

E = E_{F}

, what is the Fermi–Dirac occupancy, and why does it matter here?

Exactly

1/2

, independent of temperature. It matters because it marks where the Boltzmann approximation breaks; our carrier integrals are only trustworthy when the band edge sits many

k_{B} T

above

E_{F}

Recall One-line self-test

Name the single quantity whose temperature dependence is the "boss" of $n_{i}$ , and the single reason the extrinsic plateau is flat. ::: The exponential $e^{- E_{g} /2 k_{B} T}$ is the boss; the plateau is flat because every donor is already ionized so there is nothing left to release.