Temperature dependence of carrier concentration: Exercises

Definition Constants and symbols used on this whole page

$k_{B} = 8.617 \times 1 0^{- 5}$ eV/K (Boltzmann constant expressed in electron-volts per kelvin — the "energy-per-degree" conversion).
At $T = 300$ K, $k_{B} T = 0.02585$ eV. This is the room-temperature "thermal energy budget."
For silicon: $E_{g} = 1.12$ eV (the band gap — energy to lift an electron across from valence to conduction band), $N_{C} = 2.8 \times 1 0^{19} cm^{- 3}$ , $N_{V} = 1.04 \times 1 0^{19} cm^{- 3}$ , $n_{i} (300 K) \approx 1.0 \times 1 0^{10} cm^{- 3}$ .
$E_{D} = E_{C} - E_{donor}$ is the donor ionization (binding) energy — the small energy needed to knock the extra electron off a donor atom into the conduction band. $E_{C}$ is the conduction-band edge energy; $E_{donor}$ is the donor level just below it. See Doping and dopant ionization.
$E_{F}$ is the Fermi level — the energy at which a state has a $50%$ chance of being occupied (see Fermi-Dirac distribution). $E_{V}$ is the valence-band edge; $q$ is the elementary charge ( $1.602 \times 1 0^{- 19}$ C), the charge carried by one electron or hole.

Level 1 — Recognition

Can you name the piece and pick the right formula?

Recall Solution L1.1

Region: freeze-out. At very low $T$ the thermal budget $k_{B} T$ is too small to lift electrons off their donor atoms into the conduction band. So the electrons "freeze" — they stick back onto their parent donors. Fewer free electrons ⇒ $n$ falls, and the material becomes more insulating (resistance rises).

Recall Solution L1.2

The intrinsic (high- $T$ ) region is steepest. Its slope is $slope = - \frac{E _{g}}{2 k _{B}} .$ The factor $2$ appears because $n_{i} = n p$ — taking the square root of the mass-action product halves the exponent.

Recall Solution L1.3

The intrinsic carrier formula: $n_{i} = N_{C} N_{V} e^{- E_{g} /2 k_{B} T} .$ Use this whenever the material is pure (no net doping) or when you are computing the intrinsic contribution to compare against $N_{D}$ .

Level 2 — Application

Plug numbers in carefully; watch the factor of 2 and the units.

Recall Solution L2.1

Exponent: $- \frac{E _{g}}{2 k _{B} T} = - \frac{1.12}{2 ( 0.02585 )} = - \frac{1.12}{0.0517} = - 21.66$ . Evaluating carefully, $e^{- 21.66} = 3.93 \times 1 0^{- 10}$ . Prefactor: $N_{C} N_{V} = (2.8 \times 1 0^{19}) (1.04 \times 1 0^{19}) = 2.91 \times 1 0^{38} = 1.71 \times 1 0^{19} cm^{- 3}$ . Product: $n_{i} = 1.71 \times 1 0^{19} \times 3.93 \times 1 0^{- 10} = 6.7 \times 1 0^{9} cm^{- 3} .$ This clean textbook input set gives $6.7 \times 1 0^{9} cm^{- 3}$ ; it is rounded up to the quoted $1.0 \times 1 0^{10} cm^{- 3}$ because the tabulated $N_{C}, N_{V}$ are themselves rounded and a slightly larger effective-mass value nudges the prefactor up. Both numbers are "order $1 0^{10}$ " — the exponential $e^{- 21.66}$ is what dominates, exactly as the parent note stresses.

Recall Solution L2.2

Only the prefactor scales; use the ratio so the constants cancel: $N_{C} (400) = N_{C} (300) (\frac{400}{300})^{3/2} .$ $(\frac{400}{300})^{3/2} = (1.3333)^{1.5} = 1.5396$ . $N_{C} (400) = 2.8 \times 1 0^{19} \times 1.5396 \approx 4.31 \times 1 0^{19} cm^{- 3} .$

Recall Solution L2.3

Extrinsic test: need $N_{D} ≫ n_{i}$ . Here $1 0^{15} ≫ 1 0^{10}$ (five orders of magnitude), so yes, firmly extrinsic, and all donors are ionized: $n \approx N_{D} = 1 0^{15} cm^{- 3} .$ Holes by mass action ( $n p = n_{i}^{2}$ ): $p = \frac{n _{i}^{2}}{n} = \frac{( 1 0 ^{10} ) ^{2}}{1 0 ^{15}} = \frac{1 0 ^{20}}{1 0 ^{15}} = 1 0^{5} cm^{- 3} .$

Level 3 — Analysis

Extract physics from data and graphs.

The figure below is the reference for L3.1–L3.3. It plots $ln n$ (vertical axis) against $1/ T$ (horizontal axis, so cold is on the right, hot on the left). The single red curve is the carrier concentration; its three labelled segments are the intrinsic branch (steep, left), the extrinsic plateau (flat, middle), and the freeze-out branch (gentle, right).

Figure — Temperature dependence of carrier concentration

Recall Solution L3.1

Intrinsic slope $= - \frac{E _{g}}{2 k _{B}}$ , so $E_{g} = - 2 k_{B} \times (slope) = 2 (8.617 \times 1 0^{- 5}) (6500) .$ $E_{g} = 2 \times 8.617 \times 1 0^{- 5} \times 6500 = 1.12 eV .$ That is silicon. Note the 2×: if you forgot it you'd report $0.56$ eV — half the true gap.

Recall Solution L3.2

Freeze-out slope $= - \frac{E _{D}}{2 k _{B}}$ (same square-root/factor-of-2 logic: a discrete donor level feeding the conduction continuum). $E_{D} = 2 k_{B} \times 260 = 2 (8.617 \times 1 0^{- 5}) (260) = 0.0448 eV \approx 45 meV .$ This is a typical shallow-donor energy (e.g. phosphorus in silicon, $\sim 45$ meV). The much smaller slope than the intrinsic branch is exactly why the freeze-out line is far gentler on the graph.

Recall Solution L3.3

Left (small $1/ T$ = high $T$ ), steep down: intrinsic, slope $- E_{g} /2 k_{B}$ .
Middle, flat: extrinsic / saturation, slope $\approx 0$ , $n \approx N_{D}$ .
Right (large $1/ T$ = low $T$ ), gentle down: freeze-out, slope $- E_{D} /2 k_{B}$ (small because $E_{D} ≪ E_{g}$ ).

Level 4 — Synthesis

Combine several ideas into one chain of reasoning.

Recall Solution L4.1

Set $n_{i} = N_{D}$ : $1.71 \times 1 0^{19} e^{- E_{g} /2 k_{B} T} = 1 0^{16} .$ $e^{- E_{g} /2 k_{B} T} = \frac{1 0 ^{16}}{1.71 \times 1 0 ^{19}} = 5.85 \times 1 0^{- 4} .$ Take logs: $- \frac{E _{g}}{2 k _{B} T} = ln (5.85 \times 1 0^{- 4}) = - 7.44$ . $T = \frac{E _{g}}{2 k _{B} ( 7.44 )} = \frac{1.12}{2 ( 8.617 \times 1 0 ^{- 5} ) ( 7.44 )} .$ Denominator $= 2 \times 8.617 \times 1 0^{- 5} \times 7.44 = 1.282 \times 1 0^{- 3}$ . $T = \frac{1.12}{1.282 \times 1 0 ^{- 3}} \approx 874 K .$ A crude estimate (the real value is lower once the prefactor's own $T^{3/2}$ growth is included), but it correctly shows the crossover is far above room temperature — silicon stays comfortably extrinsic at 300 K.

Recall Solution L4.2

Now solve $1.71 \times 1 0^{19} e^{- E_{g} /2 k_{B} T} = 2 \times 1 0^{16}$ . $e^{- E_{g} /2 k_{B} T} = 1.17 \times 1 0^{- 3}, - \frac{E _{g}}{2 k _{B} T} = ln (1.17 \times 1 0^{- 3}) = - 6.75.$ $T = \frac{1.12}{2 ( 8.617 \times 1 0 ^{- 5} ) ( 6.75 )} = \frac{1.12}{1.163 \times 1 0 ^{- 3}} \approx 963 K .$ It rises (874 K → 963 K). Physically: with more donors you need a bigger $n_{i}$ to overtake them, and since $n_{i}$ grows exponentially with $T$ , a slightly higher temperature is needed. This confirms the parent note's forecast: more doping pushes the intrinsic onset up.

Recall Solution L4.3

Freeze-out (low $T$ ): $n$ is tiny and rising fast with $T$ ; the exploding $n$ dominates ⇒ $ρ$ falls as you warm up. (See Conductivity and mobility vs temperature.)
Extrinsic plateau (mid $T$ ): $n \approx N_{D}$ is fixed, so only mobility matters. Phonon scattering makes $μ$ fall with $T$ (roughly $μ \propto T^{- 3/2}$ ), so $ρ$ gently rises.
Intrinsic (high $T$ ): $n_{i}$ explodes exponentially, swamping the mild mobility drop ⇒ $ρ$ plummets. This runaway conduction is device failure — the doping no longer controls the carrier count.

Level 5 — Mastery

One decision — a sign, a factor, a limiting case — settles the whole answer.

Recall Solution L5.1

$2.24 = 2 \times 1.12$ — they used slope $= - E_{g} / k_{B}$ (no factor of 2) instead of the correct slope $= - E_{g} /2 k_{B}$ . $E_{g, correct} = 2 k_{B} \times 6500 = 2 (8.617 \times 1 0^{- 5}) (6500) = 1.12 eV .$ The single missing factor of 2 doubled their band gap. The 2 is not decorative — it comes from $n_{i} = n p$ .

Recall Solution L5.2

If $E_{F}$ is at or above $E_{C}$ , then $E - E_{F}$ is not large compared to $k_{B} T$ near the band edge, so the " $+ 1$ " in Fermi–Dirac is not negligible. The Boltzmann exponential overestimates occupancy, and $n = N_{C} e^{- (E_{C} - E_{F}) / k_{B} T}$ fails. Fix: use the full Fermi-Dirac distribution and the Fermi–Dirac integral $n = N_{C} F_{1/2} (η)$ , where $F_{1/2}$ is the standard integral $\int_{0}^{\infty} \frac{x}{1 + e ^{x - η}} d x$ (times a constant) that replaces the simple exponential, and $η = (E_{F} - E_{C}) / k_{B} T$ measures how far the Fermi level has pushed into the band. This is the degenerate regime — the material behaves metal-like and $n$ saturates near $N_{C}$ regardless of further $E_{F}$ shifts. (See Intrinsic vs extrinsic semiconductors.)

Recall Solution L5.3

As $T \to 0^{+}$ : the exponent $- E_{g} /2 k_{B} T \to - \infty$ , so $e^{- E_{g} /2 k_{B} T} \to 0$ faster than any power of $T$ . The prefactor $N_{C} N_{V} \propto T^{3/2} \to 0$ too, but even if it didn't, the exponential wins: $lim_{T \to 0^{+}} n_{i} = 0.$ Physically: at absolute zero there is no thermal budget, no band-to-band pairs, and (for doped material) donors are fully frozen — a perfect insulator. The exponential is the boss, exactly as the parent note stated.

Recall Solution L5.4

$n p = (1 0^{15}) (1 0^{5}) = 1 0^{20} = (1 0^{10})^{2} = n_{i}^{2} . ✓$ It must hold because in $n p = N_{C} N_{V} e^{- E_{g} / k_{B} T}$ the Fermi level $E_{F}$ cancels when you multiply $n$ and $p$ . So the product depends only on material and temperature, not on how you doped it — the Law of mass action. Doping merely slides $n$ and $p$ along the hyperbola $n p = n_{i}^{2}$ ; their product is locked.

Recall check

Slope of the intrinsic branch in symbols

- E_{g} /2 k_{B}

Slope of the freeze-out branch in symbols

- E_{D} /2 k_{B}

In the extrinsic plateau,

n \approx

N_{D}

(all donors ionized)

Why does doubling

N_{D}

raise the intrinsic-onset temperature?

You need a larger

n_{i}

to overtake the bigger

N_{D}

, and

n_{i}

grows exponentially with

T

, so a higher

T

is required.

What quantity is fixed by the law of mass action regardless of doping?

The product

n p = n_{i}^{2}

at a given

T