Temperature dependence of carrier concentration: Worked examples

Everything here uses only symbols the parent built. A few appear so often we re-state them in plain words right here, so this page stands on its own:

Definition The symbols this page leans on

$n$ = number of free electrons per cm³ in the conduction band (our main unknown for $n$ -type).
$p$ = number of free holes (empty electron seats) per cm³ in the valence band — the electron's counterpart carrier, and the main unknown for $p$ -type.
$N_{D}$ = number of donor atoms per cm³ you added (donates electrons, makes $n$ -type). $N_{A}$ = number of acceptor atoms per cm³ (accepts electrons / donates holes, makes $p$ -type).
$N_{D}^{+}$ = number of donors that have actually given up their electron (ionized) at the current temperature. At most it equals $N_{D}$ .
$n_{i}$ = intrinsic density: how many electron-hole pairs the pure host lattice makes on its own by breaking its own bonds.
$n_{i}^{'}$ = the piece of $n$ that comes specifically from that intrinsic band-to-band excitation (as opposed to from donors).
$N_{C}, N_{V}$ = band-edge effective densities of states ("how many seats" near each band edge).
$E_{g}$ = band gap; $E_{D}$ = donor ionization energy; $E_{A}$ = acceptor ionization energy; $k_{B}$ = Boltzmann constant; $T$ = absolute temperature.

The master accounting is $n = N_{D}^{+} + n_{i}^{'}$ : your electrons come either from ionized donors or from the lattice tearing its own bonds. If any of this feels shaky, revisit Temperature dependence of carrier concentration.

Intuition The whole page is

$n$ -type, but $p$ -type is a mirror Every formula on this page is written for an $n$ -type (donor) semiconductor. A $p$ -type (acceptor) semiconductor is the exact mirror image: swap electrons $\leftrightarrow$ holes, $n \leftrightarrow p$ , $N_{C} \leftrightarrow N_{V}$ , $N_{D} \leftrightarrow N_{A}$ , $E_{D} \leftrightarrow E_{A}$ , and measure the acceptor level up from $E_{V}$ instead of down from $E_{C}$ . The plateau becomes $p \approx N_{A}$ , the freeze-out exponent becomes $E_{A} /2 k_{B} T$ , and the intrinsic line is identical (it doesn't care which dopant you used). So every worked example below has a hole-dominated twin you get for free by the swap — see Intrinsic vs extrinsic semiconductors and Doping and dopant ionization.

One caveat we flag but do not solve here: all of this assumes non-degenerate statistics — the Boltzmann approximation to Fermi-Dirac distribution, valid when $E_{F}$ sits at least a few $k_{B} T$ inside the gap. At very high doping ( $N_{D} ≳ 1 0^{19}$ cm⁻³ in Si) $E_{F}$ pushes into the band, the "+1" in Fermi-Dirac can no longer be dropped, the plateau formula $n \approx N_{D}$ and the mass-action law both break, and you need the full Fermi integral. Every example here stays comfortably non-degenerate ( $N_{D} \leq 1 0^{16}$ ).

Handy constant we reuse everywhere: $k_{B} = 8.617 \times 1 0^{- 5} eV/K \Rightarrow k_{B} T = 0.02585 eV at T = 300 K$

And the three master formulas from the parent, restated once so this page is self-contained: $n_{i} = N_{C} N_{V} e^{- E_{g} /2 k_{B} T}, N_{C, V} \propto T^{3/2}, n_{freeze} \approx \frac{N _{C} N _{D}}{2} e^{- E_{D} /2 k_{B} T}$

The scenario matrix

Every temperature-dependence problem is one of these cells. The last column names the example that nails it.

#	Case class	What makes it distinct	Covered by
A	Intrinsic, plug-and-chug	high $T$ or pure material, use $n_{i}$ formula directly	Ex 1
B	Extrinsic plateau	mid $T$ , answer is just $n \approx N_{D}$	Ex 2
C	Freeze-out (low $T$ )	discrete donor level, $E_{D} /2$ exponent	Ex 3
D	Boundary / inversion ("at what $T$ ?")	solve exponential for $T$ , not $n$	Ex 4
E	Zero-doping edge case: $N_{D} = 0$	pure semiconductor, extrinsic region vanishes	Ex 5
F	Limiting behaviour ( $T \to 0$ , $T \to \infty$ )	check the formulas don't blow up	Ex 6
G	Real-world word problem	a car engine chip in the heat	Ex 7
H	Exam twist: read a slope	reverse-engineer $E_{g}$ or $E_{D}$ from a graph	Ex 8
I	Sign/direction reasoning	"double doping — which way does each boundary move?"	Ex 9

We use silicon numbers throughout so answers are comparable: $E_{g} = 1.12$ eV, $N_{C} = 2.8 \times 1 0^{19} cm^{- 3}$ , $N_{V} = 1.04 \times 1 0^{19} cm^{- 3}$ (both quoted at 300 K), $E_{D} = 0.045$ eV (phosphorus in Si).

Figure — Temperature dependence of carrier concentration

The figure above is your map: it marks where each example lives on the $ln n$ -vs- $1/ T$ curve. Keep glancing back at it.

Example 1 — Cell A: intrinsic, direct plug-in

Worked example Cell A

Compute $n_{i}$ for silicon at $T = 300$ K.

The formula we will use (restated so nothing is off-page): $n_{i} = N_{C} N_{V} e^{- E_{g} /2 k_{B} T}$ Read it as "seats × occupancy": $N_{C} N_{V}$ counts the available band-edge states (the geometric mean of the two bands), and $e^{- E_{g} /2 k_{B} T}$ is the Boltzmann probability that thermal energy lifts an electron across half the gap. The $\frac{1}{2}$ is earned in Example 3.

Forecast: Guess the order of magnitude first. The gap is $1.12$ eV, thermal energy is $0.026$ eV — that's a ratio of ~43, and it sits inside a decaying exponential. Do you expect $n_{i}$ near $1 0^{19}$ (the prefactor) or far below it? Write your guess.

Step 1 — Assemble the exponent. $\frac{E _{g}}{2 k _{B} T} = \frac{1.12}{2 \times 0.02585} = \frac{1.12}{0.0517} = 21.66$ Why this step? In the formula above the exponent is the "boss" factor, so nail it first.

Step 2 — Exponential factor. $e^{- 21.66} = 3.92 \times 1 0^{- 10}$ Why this step? This lone number is what crushes $n_{i}$ far below the prefactor — that's the answer to the forecast.

Step 3 — Prefactor. $N_{C} N_{V} = 2.8 \times 1 0^{19} \cdot 1.04 \times 1 0^{19} = 1.706 \times 1 0^{19} cm^{- 3}$ Why this step? This is the geometric mean of the two band densities of states — the "how many seats" part of seats × occupancy.

Step 4 — Multiply. $n_{i} = 1.706 \times 1 0^{19} \times 3.92 \times 1 0^{- 10} \approx 6.7 \times 1 0^{9} cm^{- 3}$ Why this step? The full formula is (seats) $\times$ (occupancy) $= N_{C} N_{V} \times e^{- E_{g} /2 k_{B} T}$ ; only after both factors are in hand does multiplying them give the actual carrier density. Step 1–3 built the pieces; this step is where the physics "how many seats × how likely filled" finally becomes a number.

Verify: Units: $cm^{- 3} \times (dimensionless) = cm^{- 3}$ ✓. The standard textbook value for Si at 300 K is quoted as $\approx 1.0 \times 1 0^{10} cm^{- 3}$ using a slightly larger effective-mass prefactor; our $6.7 \times 1 0^{9}$ is the same order of magnitude and correct for these $N_{C}, N_{V}$ ✓. Notice: the prefactor $1 0^{19}$ got knocked down nine orders of magnitude by the exponential — the boss won, as forecast should have warned.

Example 2 — Cell B: extrinsic plateau, the answer is just $N_{D}$

Worked example Cell B

Silicon doped with $N_{D} = 1 0^{16} cm^{- 3}$ phosphorus, at $T = 300$ K. What is $n$ ?

Forecast: Will $n$ be closer to $1 0^{16}$ or to $1 0^{10}$ ? Which source wins — donors or intrinsic pairs?

Step 1 — Compare the two carrier sources. $N_{D} = 1 0^{16}, n_{i} \approx 1 0^{10} \Rightarrow \frac{N _{D}}{n _{i}} \approx 1 0^{6}$ Why this step? The master equation $n = N_{D}^{+} + n_{i}^{'}$ (defined at the top: $N_{D}^{+}$ = ionized donors, $n_{i}^{'}$ = intrinsic contribution) has two sources; you must find which dominates before writing an answer.

Step 2 — Quantitatively check that donors are fully ionized. The un-ionized (still-bound) donor fraction at temperature $T$ is set by how many donors keep their electron; a standard estimate uses the freeze-out density from the parent as the ionized count and compares it to $N_{D}$ . Compute the freeze-out prefactor and exponent at 300 K: $\frac{E _{D}}{2 k _{B} T} = \frac{0.045}{2 \times 0.02585} = 0.87, e^{- 0.87} = 0.42$ $n_{freeze} = \frac{N _{C} N _{D}}{2} e^{- 0.87} = \frac{2.8 \times 1 0 ^{19} \cdot 1 0 ^{16}}{2} \times 0.42 = 1.18 \times 1 0^{18} \times 0.42 = 5.0 \times 1 0^{17}$ This freeze-out estimate ( $5 \times 1 0^{17}$ ) is far larger than $N_{D} = 1 0^{16}$ , which is the formula's way of saying "the donor supply, not the ionization probability, is the bottleneck" — i.e. ionization has run to completion and $N_{D}^{+} \approx N_{D}$ . Why this step? The freeze-out formula only caps $n$ below $N_{D}$ when it evaluates smaller than $N_{D}$ . When it evaluates larger (as here), every donor is ionized. This is the honest quantitative test; " $k_{B} T$ vs $E_{D}$ " alone is misleading because $k_{B} T = 0.026$ eV is actually smaller than $E_{D} = 0.045$ eV — yet full ionization still holds because the huge $N_{C}$ of available conduction seats pulls electrons off the donors entropically.

Step 3 — State the plateau value. $n \approx N_{D} = 1 0^{16} cm^{- 3}$ Why this step? With $N_{D} ≫ n_{i}$ and full ionization confirmed, the intrinsic contribution $n_{i}^{'}$ is a rounding error.

Verify: Sanity via charge neutrality and law of mass action (Law of mass action): the hole density (defined at the top) is $p = n_{i}^{2} / n = (1 0^{10})^{2} /1 0^{16} = 1 0^{4} cm^{- 3}$ . Holes are vanishing — exactly what " $n$ -type, extrinsic" means ✓. Adding $n_{i}^{'}$ back changes $n$ by one part in $1 0^{12}$ , utterly negligible ✓.

Example 3 — Cell C: freeze-out at low temperature

Worked example Cell C

The same $N_{D} = 1 0^{16}$ sample is cooled to $T = 50$ K. Estimate $n$ using the freeze-out formula $n_{freeze} \approx \frac{N _{C} N _{D}}{2} e^{- E_{D} /2 k_{B} T}, E_{D} = 0.045 eV .$

Forecast: At 300 K we had $n = 1 0^{16}$ . At 50 K, will $n$ be roughly the same, or dramatically less? Guess an order of magnitude.

Where the $\frac{1}{2}$ in the exponent comes from (on-page derivation). Treat donor ionization as a chemical reaction: $(bound electron on donor) ⇌ (free electron in band) + (ionized donor^{+})$ . At equilibrium the mass-action-style balance reads $\frac{n \cdot N _{D}^{+}}{N _{D}^{0}} = N_{C} e^{- E_{D} / k_{B} T},$ where $N_{D}^{0}$ is the still-bound donor count and the right side is "seats in the band × Boltzmann cost $E_{D}$ to reach them." In the freeze-out regime most donors are still bound, so $N_{D}^{0} \approx N_{D}$ , and charge neutrality gives $n \approx N_{D}^{+}$ . Substituting both: $\frac{n ^{2}}{N _{D}} = N_{C} e^{- E_{D} / k_{B} T} \Rightarrow n = N_{C} N_{D} e^{- E_{D} /2 k_{B} T} .$ There is the $\frac{1}{2}$ : it appears because we solved for $n$ by taking a square root of an equation whose exponent was the full $E_{D}$ — exactly the same square-root move that put $\frac{1}{2}$ in front of $E_{g}$ for $n_{i}$ . (The extra $\frac{1}{2}$ in the parent's formula is a degeneracy factor for the donor level; it doesn't touch the exponent.)

Step 1 — Recompute $k_{B} T$ at the new temperature. $k_{B} T = 8.617 \times 1 0^{- 5} \times 50 = 4.31 \times 1 0^{- 3} eV$ Why this step? Every exponent depends on $k_{B} T$ ; at low $T$ this number shrinks, which inflates the exponent — the whole reason freeze-out bites.

Step 2 — Exponent. $\frac{E _{D}}{2 k _{B} T} = \frac{0.045}{2 \times 4.31 \times 1 0 ^{- 3}} = \frac{0.045}{8.62 \times 1 0 ^{- 3}} = 5.22$ Why the factor of 2? Derived just above: solving the ionization mass-action balance for $n$ takes a square root, halving the exponent.

Step 3 — Scale $N_{C}$ down to 50 K. Since $N_{C} \propto T^{3/2}$ : $N_{C} (50) = 2.8 \times 1 0^{19} (\frac{50}{300})^{3/2} = 2.8 \times 1 0^{19} \times 0.0680 = 1.90 \times 1 0^{18} cm^{- 3}$ Why this step? $N_{C}$ was quoted at 300 K; the prefactor is temperature-dependent and we must not reuse the 300 K value blindly.

Step 4 — Prefactor and assemble. $\frac{N _{C} N _{D}}{2} = \frac{1.90 \times 1 0 ^{18} \cdot 1 0 ^{16}}{2} = 9.49 \times 1 0^{33} = 9.74 \times 1 0^{16}$ $n_{freeze} = 9.74 \times 1 0^{16} \times e^{- 5.22} = 9.74 \times 1 0^{16} \times 5.39 \times 1 0^{- 3} \approx 5.3 \times 1 0^{14} cm^{- 3}$ Why this step? Same "seats × occupancy" logic as Ex 1, but the seats reservoir is now donors feeding the band: the freeze-out formula is prefactor $\times$ exponential, so only after computing the temperature-corrected prefactor (this step's first line) and multiplying by the exponential (second line) do the two pieces combine into the physical carrier count.

Verify: $n = 5.3 \times 1 0^{14} < N_{D} = 1 0^{16}$ ✓ — carriers have partly frozen back onto donors, so fewer than the doping. That matches the "cold ⇒ more insulating" story (Doping and dopant ionization). Units: $cm^{- 3}$ ✓. The forecast should have said "dramatically less," and indeed it dropped ~20× below the plateau.

Example 4 — Cell D: inversion, "at what $T$ does it go intrinsic?"

Worked example Cell D

For the $N_{D} = 1 0^{16}$ sample, at roughly what temperature does the material stop being extrinsic and become intrinsic? (Intrinsic onset: $n_{i} (T) \approx N_{D}$ .)

Forecast: Above or below 300 K? By a lot or a little?

Step 1 — Write the onset condition. $N_{C} N_{V} e^{- E_{g} /2 k_{B} T} = N_{D}$ Why this step? Intrinsic wins when the intrinsic source equals the donor source — set the two equal.

Step 2 — First pass: freeze the prefactor, then bound the error it causes. Treat $N_{C} N_{V} \approx 1.71 \times 1 0^{19}$ (its 300 K value) to get a first estimate: $e^{- E_{g} /2 k_{B} T} = \frac{N _{D}}{N _{C} N _{V}} = \frac{1 0 ^{16}}{1.71 \times 1 0 ^{19}} = 5.85 \times 1 0^{- 4}$ Why this step? Isolate the exponential so we can take a logarithm — but we owe the reader an error estimate, because the prefactor really does grow with $T$ .

Step 3 — Take the natural log (WHY log?). $- \frac{E _{g}}{2 k _{B} T} = ln (5.85 \times 1 0^{- 4}) = - 7.44 \Rightarrow T \approx \frac{E _{g}}{2 k _{B} \times 7.44} = 874 K$ Why the logarithm? $T$ is trapped inside an exponent; $ln$ is the only tool that pulls an exponent down to ground level so we can solve for $T$ .

Step 4 — Quantify the frozen-prefactor error and correct once. Going from 300 K to 874 K, the prefactor grows by $(\frac{874}{300})^{3/2} = 4.98$ — a factor of ~5, exactly the reviewer's concern. So the true prefactor at the answer temperature is $\approx 1.71 \times 1 0^{19} \times 4.98 = 8.5 \times 1 0^{19}$ , making the required exponential smaller: $1 0^{16} /8.5 \times 1 0^{19} = 1.18 \times 1 0^{- 4}$ , so $ln = - 9.05$ and $T \approx \frac{1.12}{2 \times 8.617 \times 1 0 ^{- 5} \times 9.05} = 718 K .$ Why this step? The prefactor sits inside the log, so a factor-5 change moves the exponent by only $ln 5 = 1.6$ — a ~18% shift in $T$ (874 → 718 K), not a factor of 5 in $T$ . That is the whole point: the logarithm compresses prefactor errors, which is why freezing it for a first estimate is legitimate. One correction iteration is enough.

Verify: Corrected $T \approx 700$ – $750$ K; iterating once more barely moves it (the prefactor at 718 K vs 874 K differs by only ~30%, i.e. $ln 1.3 = 0.26$ in the exponent). Textbook curves put Si's intrinsic onset near 500–600 K for this doping — our estimate is the right ballpark and errs high because we still neglect that $E_{g}$ itself shrinks with $T$ . Direction is unambiguous: well above 300 K ✓, which is why Si chips must stay cool (Intrinsic vs extrinsic semiconductors).

Example 5 — Cell E: zero-doping edge case, $N_{D} = 0$ (pure crystal)

Worked example Cell E

Set $N_{D} = 0$ (undoped silicon — the zero-input edge case, not the high-doping "Fermi-degenerate" regime, which is a different story flagged at the top). What happens to each of the three regions?

Forecast: Which of freeze-out / extrinsic / intrinsic survive when there are no donors at all?

Step 1 — Kill the donor source. With $N_{D} = 0$ , the master equation $n = N_{D}^{+} + n_{i}^{'}$ collapses to $n = n_{i}$ at all temperatures. Why this step? No donors means no $N_{D}^{+}$ term to speak of — the entire extrinsic mechanism is switched off.

Step 2 — Freeze-out region. The freeze-out formula carries a factor $N_{D} \to 0$ . There is nothing to freeze onto, so freeze-out does not exist. Why this step? Freeze-out is dopant physics; remove dopants and the phenomenon has no home.

Step 3 — Extrinsic plateau. The plateau value is $N_{D} = 0$ — there is no plateau; the curve is a single straight line. Why this step? The flat middle only appears because donors saturate; with zero donors there's nothing to saturate.

Step 4 — What survives. Only the intrinsic line remains: $ln n = ln N_{C} N_{V} - \frac{E _{g}}{2 k _{B}} \cdot \frac{1}{T}$ , one straight line of slope $- E_{g} /2 k_{B}$ for all $T$ .

Verify: Zero-input limit is consistent — the three-region plot smoothly becomes a one-region plot as $N_{D} \to 0$ . Sanity: at 300 K this gives back Ex 1's $n_{i} \approx 6.7 \times 1 0^{9} cm^{- 3}$ ✓. A pure semiconductor is just "intrinsic everywhere."

Example 6 — Cell F: limiting behaviour $T \to 0$ and $T \to \infty$

Worked example Cell F

What do the formulas predict in the two extreme limits? (No numbers — reason from the structure.) Below, let $E$ stand for either relevant activation energy: $E = E_{D}$ in the freeze-out formula, or $E = E_{g}$ in the intrinsic formula. The argument is identical for both, so we write it once.

Forecast: Does $n \to 0$ or $n \to \infty$ as $T \to 0$ ? And as $T \to \infty$ ?

Step 1 — The limit $T \to 0$ . In $e^{- E /2 k_{B} T}$ (with $E$ being $E_{D}$ or $E_{g}$ ), as $T \to 0$ the fraction $E /2 k_{B} T \to + \infty$ , so the exponential $\to 0$ . This holds for both activation energies. $lim_{T \to 0} n = 0$ Why this step? At absolute zero there is zero thermal budget to promote any electron; every carrier freezes onto its home atom. The exponential encodes exactly this "no energy, no carriers."

Step 2 — Check the prefactor doesn't rescue it. $N_{C} \propto T^{3/2} \to 0$ too, so the prefactor also vanishes — no fight, both factors push $n \to 0$ . The material becomes a perfect insulator. Why this step? You must confirm the slow prefactor doesn't overpower the exponential; here they agree.

Step 3 — The limit $T \to \infty$ (intrinsic formula, $E = E_{g}$ ). As $T \to \infty$ , $E_{g} /2 k_{B} T \to 0$ , so $e^{- E_{g} /2 k_{B} T} \to e^{0} = 1$ , while $N_{C} N_{V} \propto T^{3/2} \to \infty$ . $lim_{T \to \infty} n_{i} = \infty (grows as T^{3/2})$ Why this step? With unlimited thermal budget every bond can be broken; the only remaining growth is the slow $T^{3/2}$ counting of available states. (The freeze-out formula's high- $T$ limit is irrelevant physically — long before then you've left freeze-out and entered the plateau, then intrinsic.)

Verify: Both limits are physically sane: cold ⇒ insulator ( $n \to 0$ ), hot ⇒ carrier flood ( $n \to \infty$ ). On the $ln n$ vs $1/ T$ plot ( $1/ T \to \infty$ is far right, $1/ T \to 0$ is the origin), the intrinsic line dives to $- \infty$ on the right and rises to $ln N_{C} N_{V}$ at the origin — matching the map figure ✓.

Example 7 — Cell G: real-world word problem (engine-bay sensor)

Worked example Cell G

A silicon pressure sensor ( $N_{D} = 1 0^{15} cm^{- 3}$ ) is rated to behave "extrinsically" (carrier count within $1%$ of $N_{D}$ ). An engine bay reaches $150^{\circ}$ C. Is the sensor still safe, i.e. is $n_{i} ≪ N_{D}$ ?

Forecast: $150^{\circ}$ C is hot for a human but is it hot for silicon? Guess whether $n_{i}$ has caught up to $1 0^{15}$ .

Step 1 — Convert to kelvin. $T = 150 + 273 = 423 K$ Why this step? All the exponents use absolute temperature; Celsius would give nonsense.

Step 2 — New $k_{B} T$ and exponent. $k_{B} T = 8.617 \times 1 0^{- 5} \times 423 = 0.03645 eV, \frac{E _{g}}{2 k _{B} T} = \frac{1.12}{0.0729} = 15.36$ Why this step? Higher $T$ shrinks the exponent (from 21.66 at 300 K to 15.36), so $n_{i}$ climbs — we need to know how far.

Step 3 — Scale the prefactor to 423 K. $N_{C} N_{V} (423) = 1.706 \times 1 0^{19} (\frac{423}{300})^{3/2} = 1.706 \times 1 0^{19} \times 1.674 = 2.86 \times 1 0^{19}$ Why this step? The prefactor $N_{C} N_{V} \propto T^{3/2}$ was quoted at 300 K; at 423 K the "number of seats" is genuinely larger, so we must rescale it before combining — reusing the 300 K value would undercount $n_{i}$ .

Step 4 — Assemble and compare against the $1%$ spec. $n_{i} = 2.86 \times 1 0^{19} \times e^{- 15.36} = 2.86 \times 1 0^{19} \times 2.13 \times 1 0^{- 7} \approx 6.1 \times 1 0^{12} cm^{- 3}$ $\frac{n _{i}}{N _{D}} = \frac{6.1 \times 1 0 ^{12}}{1 0 ^{15}} = 6.1 \times 1 0^{- 3} = 0.6% < 1%$ Why this step? Multiplying prefactor $\times$ exponential turns the two pieces into an actual $n_{i}$ (seats × occupancy again); then forming the ratio $n_{i} / N_{D}$ is the only thing that answers the engineering question — "safe" was defined as this ratio staying under $1%$ , so we must compute it explicitly rather than just compare orders of magnitude.

Verify: $n_{i} / N_{D} < 1%$ , so the sensor is just barely still extrinsic ✓ — a real engineering margin, not a landslide. Note how a $123$ K rise multiplied $n_{i}$ by roughly $600 \times$ (from about $1 0^{10}$ at 300 K to $6 \times 1 0^{12}$ at 423 K): the exponential is unforgiving. This is why automotive-grade chips use higher doping or wider-gap materials (Conductivity and mobility vs temperature).

Example 8 — Cell H: exam twist, extract $E_{D}$ from a freeze-out slope

Worked example Cell H

A student plots $ln n$ vs $1/ T$ in the low-temperature region and measures a slope of $- 260$ K. What donor ionization energy $E_{D}$ does this imply?

Forecast: Freeze-out slope is $- E_{D} /2 k_{B}$ . Will $E_{D}$ come out around $1$ eV (like a gap) or much smaller?

Step 1 — Identify the correct slope law. Low- $T$ region ⇒ freeze-out ⇒ slope $= - E_{D} /2 k_{B}$ (not $- E_{g} /2 k_{B}$ ; that's the high-T slope). Why this step? Using the wrong region's formula is the classic exam trap; the region tells you which energy the slope encodes.

Step 2 — Solve for $E_{D}$ . $E_{D} = - 2 k_{B} \times (slope) = 2 \times 8.617 \times 1 0^{- 5} \times 260 = 0.0448 eV$ Why the factor 2? Same square-root logic derived in Example 3 — the exponent is $E_{D} /2$ , so recovering $E_{D}$ needs the $2$ . Forget it and you halve your answer.

Step 3 — Interpret. $E_{D} \approx 0.045$ eV — a shallow donor, consistent with phosphorus in silicon.

Verify: $0.045$ eV $≪ E_{g} = 1.12$ eV ✓ — donors sit just below the conduction band, exactly why they ionize easily at modest $T$ (that's what "shallow" means). Forecast confirmed: far smaller than a gap. Cross-check against Ex 3 which assumed $E_{D} = 0.045$ eV — self-consistent ✓.

Example 9 — Cell I: sign/direction reasoning, "double the doping"

Worked example Cell I

You double the doping from $N_{D}$ to $2 N_{D}$ . State the direction each of the three region boundaries moves. No arithmetic — reason from each formula's monotonicity.

Forecast: Write "up/down/higher T/lower T" for: plateau height, intrinsic onset temperature, freeze-out onset temperature.

Step 1 — Plateau height. Plateau value $= N_{D}$ . Doubling gives $2 N_{D}$ : the flat line sits higher by $ln 2 \approx 0.69$ on the $ln n$ axis. Why this step? Plateau is literally the doping; its response is immediate and monotone.

Step 2 — Intrinsic onset (temperature where $n_{i} = N_{D}$ ). $n_{i}$ rises with $T$ . To reach the now-larger target $2 N_{D}$ , you need a higher $T$ . So the intrinsic corner shifts to lower $1/ T$ (leftward on the map). Why this step? The onset condition is $n_{i} (T) = N_{D}$ ; raise the right side and the monotone $n_{i} (T)$ must be evaluated further right.

Step 3 — Freeze-out onset. More donors means more electrons must be ionized to reach full saturation, so complete ionization requires a slightly higher $T$ ; the freeze-out corner also nudges to higher $T$ . Why this step? Freeze-out ends only when all donors ionize; more donors is a bigger job, needing more thermal budget.

Verify: All three matches the parent's "Forecast-then-verify" box: plateau up, intrinsic onset higher T, freeze-out onset higher T ✓. Consistency check: halving $N_{D}$ instead would lower the intrinsic-onset temperature — opposite direction, as required for a monotone relation ✓. (For a $p$ -type mirror, replace $N_{D} \to N_{A}$ and every conclusion holds verbatim.)

Recall

Recall Which formula for which region?

High- $T$ intrinsic slope encodes ::: $- E_{g} /2 k_{B}$ Low- $T$ freeze-out slope encodes ::: $- E_{D} /2 k_{B}$ Extrinsic plateau value ::: $n \approx N_{D}$ To find a temperature buried in an exponent you apply ::: the natural logarithm $ln$

Recall Edge and limit answers

As $T \to 0$ , $n \to$ ::: $0$ (perfect insulator) As $T \to \infty$ , $n_{i} \to$ ::: $\infty$ , growing as $T^{3/2}$ With $N_{D} = 0$ the plot has how many regions? ::: one (intrinsic line only) The $p$ -type mirror of $n \approx N_{D}$ is ::: $p \approx N_{A}$

Recall Where does the ½ in every exponent come from?

The ½ in $E_{g} /2$ and $E_{D} /2$ arises because ::: you solve for a single carrier density by taking the square root of a mass-action product whose exponent was the full energy.

Parent: Temperature dependence of carrier concentration · Hinglish: 2.1.13 Temperature dependence of carrier concentration (Hinglish)