Electrical properties — conductors, semiconductors, insulators; doping (n-type, p-type): Worked examples

This page is the practice engine for the parent topic. We take every rule the parent built — band gaps, the $σ = σ_{0} e^{- E_{g} /2 k_{B} T}$ law, doping, the law of mass action — and hit it from every angle a problem can come at you. Nothing here uses a symbol we haven't already earned, but let us re-anchor the ones we lean on hardest.

Definition The quantities we will keep using

$n$ = number of free electrons per cubic centimetre (cm⁻³). Picture tiny negative dots roaming the top floor of the building (the conduction band).
$p$ = number of holes per cm⁻³. A hole is an empty seat on the ground floor (valence band) that behaves like a positive dot.
$n_{i}$ = the intrinsic value both $n$ and $p$ take in a pure crystal, where every free electron left exactly one hole behind, so $n = p = n_{i}$ .
$N_{D}$ = the donor concentration — how many pentavalent dopant atoms (P, As, Sb) we packed in per cm³. Each one is ready to donate one loosely-held electron.
$N_{A}$ = the acceptor concentration — how many trivalent dopant atoms (B, Al, Ga) we packed in per cm³. Each one is ready to accept one electron, leaving a mobile hole behind.
$e$ = the elementary charge, the fixed size of the charge on one electron (negative) or one hole (positive): $e = 1.602 \times 1 0^{- 19} coulomb$ . It is the "unit price" of charge — every carrier carries exactly one $e$ .

Formula The two engines that drive every example

$σ = σ_{0} e^{- E_{g} /2 k_{B} T} (how heat wakes up carriers)$ $n \cdot p = n_{i}^{2} (the law of mass action — a fixed handshake)$ Here $σ$ is conductivity (units S/cm or S/m), $E_{g}$ the band gap in eV, $k_{B} = 8.617 \times 1 0^{- 5}$ eV/K is Boltzmann's constant (the "thermal energy per kelvin" ruler), and $T$ the absolute temperature in kelvin. (Note: the $e$ inside $e^{...}$ is Euler's number $\approx 2.718$ , the base of natural growth — not the charge $e$ above. Same letter, different job; the exponent tells them apart.)

The prefactor $σ_{0}$ is the conductivity ceiling — the value $σ$ would approach if every carrier that could ever be excited were already free (the $T \to \infty$ limit, where the exponential becomes $1$ ). Physically it bundles together the carrier charge, mobility, and the total density of available states; it does not depend on temperature. That is exactly why, in every band-gap example below, we take a ratio of two conductivities — it makes the unknown $σ_{0}$ cancel out.

Intuition Where the ½ in the exponent is born

Why $E_{g} / 2 k_{B} T$ and not $E_{g} / k_{B} T$ ? Freeing one electron does two things at once: it lifts a negative dot up to the conduction band and leaves a positive hole behind — a pair. Boltzmann statistics say the number of pairs is set by how many electron states at the top and hole states at the bottom line up thermally, and the density-of-states counting shares the gap energy equally between the two members of the pair. Concretely, the Fermi level of a pure crystal sits in the middle of the gap (by electron–hole symmetry), so each carrier is only $E_{g} /2$ away from its band edge, not the full $E_{g}$ . Multiply "electrons at $E_{g} /2$ " by "holes at $E_{g} /2$ " and the two halves add back to a single $E_{g}$ in the product — but each individual carrier density carries the tell-tale $e^{- E_{g} /2 k_{B} T}$ . That surviving $\frac{1}{2}$ is the fingerprint you extract as a slope in Ex 5 and mustn't drop in Ex 9.

The scenario matrix

Every problem this topic throws is one of the cells below. The last column names the worked example that nails that cell.

#	Case class	The twist it tests	Nailed by
C1	Pure (intrinsic) crystal	$n = p = n_{i}$ , sign symmetry	Ex 1
C2	n-type, complete ionisation	$n \approx N_{D}$ , find minority $p$	Ex 2
C3	p-type, complete ionisation	$p \approx N_{A}$ , find minority $n$	Ex 3
C4	Compensated crystal (both dopants)	which wins? net $N_{D} - N_{A}$	Ex 4
C5	Band-gap extraction (two-temperature)	use $ln σ$ vs $1/ T$ slope	Ex 5
C6	Degenerate / limiting inputs	$T \to 0$ , $E_{g} \to 0$ , $T \to \infty$	Ex 6
C7	Metal vs semiconductor sign of $d σ / d T$	conductivity goes down with heat	Ex 7
C8	Real-world word problem	a thermistor thermometer	Ex 8
C9	Exam twist — trap in the "2"	forget the factor of 2 → wrong $E_{g}$	Ex 9
C10	Freeze-out edge case (cold dopants)	ionisation incomplete at low $T$	Ex 10

Ex 1 — Pure silicon (cell C1)

Forecast: guess before reading — will there be more electrons or more holes in a pure crystal?

In a pure crystal, every excited electron leaves one hole. So $n = p = n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ . Why this step? No dopant means the only carriers come in electron–hole pairs; they must match one-for-one.
Product: $n \cdot p = n_{i}^{2} = (1.5 \times 1 0^{10})^{2} = 2.25 \times 1 0^{20} cm^{- 6}$ . Why this step? This number is the constant "handshake value" we reuse in every doped example below.
Neutrality: negative charge density from electrons $= - e n$ , positive from holes $= + e p$ , where $e = 1.602 \times 1 0^{- 19}$ C is the elementary charge defined above. Since $n = p$ , these cancel — total charge $0$ . Why this step? Confirms our forecast: pure = perfectly balanced, no majority carrier.

Recall Verify

Check $n_{i}^{2}$ ::: $(1.5 \times 1 0^{10})^{2} = 2.25 \times 1 0^{20} cm^{- 6}$ , and $- e n + e p = e (p - n) = 0$ . ✓

Ex 2 — Phosphorus-doped n-type (cell C2)

Forecast: the product $n p$ is locked at $2.25 \times 1 0^{20}$ . If $n$ shoots up, what must $p$ do?

Complete ionisation gives each P atom one free electron, so $n \approx N_{D} = 5 \times 1 0^{16} cm^{- 3}$ . Why? $E_{D} \approx 0.045$ eV $≪ k_{B} T \approx 0.026$ eV — thermal energy tears the 5th electron loose almost always. (Ex 10 shows what happens when this assumption breaks.)
Apply mass action: $p = \frac{n _{i}^{2}}{n} = \frac{2.25 \times 1 0 ^{20}}{5 \times 1 0 ^{16}} = 4.5 \times 1 0^{3} cm^{- 3}$ . Why? Generation/recombination pins the product; if $n$ rose $\times 1 0^{6}$ , $p$ must fall $\times 1 0^{6}$ .
Identify carriers: electrons ( $5 \times 1 0^{16}$ ) vastly outnumber holes ( $4.5 \times 1 0^{3}$ ) → n-type confirmed.

Recall Verify

$n \cdot p$ back-check ::: $(5 \times 1 0^{16}) (4.5 \times 1 0^{3}) = 2.25 \times 1 0^{20} = n_{i}^{2}$ . ✓ The handshake holds.

Ex 3 — Boron-doped p-type (cell C3)

Forecast: boron is trivalent — will majority carriers be electrons or holes this time?

Complete ionisation: each B accepts one electron from the valence band, creating one hole. So $p \approx N_{A} = 2 \times 1 0^{16} cm^{- 3}$ . Why? Acceptor level sits $\approx 0.045$ eV above the valence band — easy to fill at $300$ K.
Mass action for the minority: $n = \frac{n _{i}^{2}}{p} = \frac{2.25 \times 1 0 ^{20}}{2 \times 1 0 ^{16}} = 1.125 \times 1 0^{4} cm^{- 3}$ . Why? Same locked product; here the holes are boosted so electrons are suppressed.
Carrier type: holes ( $2 \times 1 0^{16}$ ) ≫ electrons ( $1.125 \times 1 0^{4}$ ) → p-type confirmed.

Recall Verify

$p \cdot n$ ::: $(2 \times 1 0^{16}) (1.125 \times 1 0^{4}) = 2.25 \times 1 0^{20}$ . ✓

Ex 4 — Compensated crystal (cell C4)

Forecast: the electrons from donors will fall into the acceptor holes first. Who is left standing?

Compensation subtraction: net donors $= N_{D} - N_{A} = 8 \times 1 0^{16} - 3 \times 1 0^{16} = 5 \times 1 0^{16} cm^{- 3}$ . Why? Each acceptor "eats" one donor electron. Only the surplus donates freely. Sign of $(N_{D} - N_{A})$ decides the type — positive ⇒ n-type.
Majority electrons: $n \approx N_{D} - N_{A} = 5 \times 1 0^{16} cm^{- 3}$ (net donor wins → n-type).
Minority holes: $p = \frac{n _{i}^{2}}{n} = \frac{2.25 \times 1 0 ^{20}}{5 \times 1 0 ^{16}} = 4.5 \times 1 0^{3} cm^{- 3}$ . Why? The mass-action product ignores how $n$ got its value; only the final $n$ matters.

Recall Verify

Net + product ::: $8 - 3 = 5$ (× $1 0^{16}$ ); $p = 2.25 \times 1 0^{20} /5 \times 1 0^{16} = 4.5 \times 1 0^{3}$ . ✓

Ex 5 — Extracting the band gap from two temperatures (cell C5)

Forecast: heating raised $σ$ eightfold — do you expect a band gap around $0.1$ eV or around $1$ eV?

We use $ln σ = ln σ_{0} - \frac{E _{g}}{2 k _{B} T}$ . Taking the difference at two temperatures kills the unknown $σ_{0}$ (that conductivity ceiling from the opening box) — that is why we choose two data points instead of one.

Subtract the two log-equations: $ln \frac{σ _{2}}{σ _{1}} = \frac{E _{g}}{2 k _{B}} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$ Why? The $ln σ_{0}$ term is identical at both $T$ , so it cancels — leaving only $E_{g}$ .
Solve for $E_{g}$ : $E_{g} = \frac{2 k _{B} l n ( σ _{2} / σ _{1} )}{\frac{1}{T _{1}} - \frac{1}{T _{2}}}$
Plug numbers. $ln (32/4) = ln 8 = 2.0794$ ; $\frac{1}{300} - \frac{1}{400} = 8.333 \times 1 0^{- 4} K^{- 1}$ ; $2 k_{B} = 1.7234 \times 1 0^{- 4}$ eV/K. $E_{g} = \frac{( 1.7234 \times 1 0 ^{- 4} ) ( 2.0794 )}{8.333 \times 1 0 ^{- 4}} \approx 0.430 eV .$

The figure below drives this: it is the $ln σ$ -vs- $1/ T$ line, with its axes labelled directly on the plot — the vertical axis $ln σ$ , the horizontal axis $1/ T$ in units of $1 0^{- 3} K^{- 1}$ . Because $ln σ = ln σ_{0} - \frac{E _{g}}{2 k _{B}} \cdot \frac{1}{T}$ , the data fall on a straight line (violet) whose slope, annotated right on the figure as $- E_{g} /2 k_{B}$ , is the band gap. The two coloured dots are our $T_{1} = 300$ K (magenta) and $T_{2} = 400$ K (orange) measurements — the very points we subtracted in step 1. Slide from one dot to the other: the vertical drop over the horizontal run is the slope you convert to $E_{g}$ .

Figure — Electrical properties — conductors, semiconductors, insulators; doping (n-type, p-type)

Recall Verify

$E_{g}$ value ::: $1.7234 \times 1 0^{- 4} \times 2.0794/8.333 \times 1 0^{- 4} = 0.430$ eV. ✓ (a plausible semiconductor gap)

Ex 6 — Degenerate and limiting inputs (cell C6)

Forecast: which case makes a semiconductor behave like a perfect insulator, and which like a metal?

(a) $T \to 0$ : the exponent $- E_{g} /2 k_{B} T \to - \infty$ , so $e^{- \infty} = 0$ . Thus $n \to 0$ , $σ \to 0$ . Why? No thermal energy ⇒ no electron can jump the gap ⇒ a cold semiconductor is a perfect insulator.
(b) $T \to \infty$ : exponent $\to 0$ , so $e^{0} = 1$ and $σ \to σ_{0}$ (a finite ceiling — the very meaning of $σ_{0}$ ). Why? Once heat is abundant, essentially all carriers are excited; conductivity saturates, it does not blow up.
(c) $E_{g} \to 0$ : exponent $\to 0$ for any $T$ , so $σ \to σ_{0}$ even when cold. Why? A zero gap means valence and conduction bands overlap — that is exactly a conductor (a metal-like limit).

The figure below carries the argument: on it the vertical axis is $σ / σ_{0}$ (a fraction of the ceiling) and the horizontal axis is temperature $T$ in kelvin, both labelled on the plot, with the ceiling $σ_{0}$ drawn as a dotted navy line and each limit arrow-annotated. Trace the magenta curve ( $E_{g} = 1.1$ eV): near $T \to 0$ on the left it hugs zero (the insulator limit of case a), and it climbs toward the dotted ceiling on the right (case b). The violet curve ( $E_{g} \to 0$ ) starts already near the ceiling even when cold — the metal-like limit of case c.

Recall Verify

Limits ::: $e^{- \infty} = 0$ , $e^{0} = 1$ . So $σ (T \to 0) = 0$ , $σ (T \to \infty) = σ_{0}$ , $σ (E_{g} \to 0) = σ_{0}$ . ✓

Ex 7 — Metal vs semiconductor: the sign of $d σ / d T$ (cell C7)

Forecast: a metal cools its own conduction as it warms; does a semiconductor gain or lose carriers per degree?

Metal: more heat ⇒ more lattice vibrations ⇒ electrons scatter more ⇒ $σ$ decreases. Sign of $d σ / d T$ is negative. Why? Metals already have a full conduction band; heat only adds obstacles, never carriers.
Semiconductor: differentiate $ln σ = ln σ_{0} - \frac{E _{g}}{2 k _{B} T}$ : $\frac{1}{σ} \frac{d σ}{d T} = \frac{E _{g}}{2 k _{B} T ^{2}} .$ Why differentiate the log? It hands us the fractional change directly — cleaner than differentiating $σ$ itself.
Plug in $E_{g} = 1.1$ , $2 k_{B} = 1.7234 \times 1 0^{- 4}$ , $T = 300$ : $\frac{1}{σ} \frac{d σ}{d T} = \frac{1.1}{1.7234 \times 1 0 ^{- 4} \times 30 0 ^{2}} \approx 0.0709 K^{- 1} (+ 7.1% per K) .$ Positive sign ⇒ conductivity climbs steeply with heat — opposite to copper.

The figure below contrasts the two signs directly: vertical axis is relative conductivity (normalised to $1$ at the dotted $300$ K reference line), horizontal axis is temperature $T$ in kelvin, both labelled on the plot. The orange curve (copper) falls as $\propto 1/ T$ — heat scatters its already-free electrons. The magenta curve (silicon) rises exponentially — heat frees brand-new carriers across the gap. Their opposite slopes are the whole point.

Recall Verify

Fractional slope ::: $1.1/ (1.7234 \times 1 0^{- 4} \times 90000) = 0.0709 K^{- 1}$ . ✓

Ex 8 — Real-world word problem: a thermistor thermometer (cell C8)

Forecast: a $10$ K rise — will conductivity change by a boring $3%$ or by a surprising double-digit percentage?

Ratio form (kills $σ_{0}$ again): $\frac{σ _{2}}{σ _{1}} = exp [\frac{E _{g}}{2 k _{B}} (\frac{1}{T _{1}} - \frac{1}{T _{2}})] .$ Why ratio? We know $σ_{1}$ ; we want $σ_{2}$ without ever needing the ceiling $σ_{0}$ .
Compute the bracket: $\frac{1}{300} - \frac{1}{310} = 1.0753 \times 1 0^{- 4} K^{- 1}$ ; and $\frac{E _{g}}{2 k _{B}} = \frac{0.67}{1.7234 \times 1 0 ^{- 4}} = 3887.7 K$ . Why? This isolates the pure number inside the exponential before we exponentiate.
Multiply them: exponent $= 3887.7 \times 1.0753 \times 1 0^{- 4} = 0.4180$ . Why? The two pieces from step 2 combine into the single dimensionless argument the $exp$ needs.
Exponentiate: $σ_{2} = 1.00 \times e^{0.4180} = 1.519 S/m$ . Why it matters: a mere $10$ K gave a $52%$ conductivity jump — that steep sensitivity is precisely why thermistors make good temperature sensors.

Verify: units check — the bracket is $K^{- 1}$ , $E_{g} /2 k_{B}$ is $eV / (eV/K) = K$ , product is dimensionless ✓ (an exponent must be a pure number). Sanity: heating a semiconductor raises $σ$ , and indeed $1.519 > 1.00$ ✓.

Recall Verify

New conductivity ::: $exp (0.67/ (1.7234 \times 1 0^{- 4}) \times (1/300 - 1/310)) = 1.519$ S/m. ✓

Ex 9 — Exam twist: the factor-of-2 trap (cell C9)

Forecast: where did the missing factor hide — in $k_{B}$ , or in the physics of pair creation?

Correct relation: slope $= - \frac{E _{g}}{2 k _{B}}$ , so $E_{g} = 2 k_{B} \times ∣ slope ∣$ . Why the 2? Exciting one electron creates a pair (electron + hole); as the opening intuition box showed, the density-of-states derivation splits $E_{g}$ across both, putting a $2$ under the gap.
Correct value: $E_{g} = 1.7234 \times 1 0^{- 4} \times 6380 = 1.099 \approx 1.10 eV$ .
Diagnose: the student used $E_{g} = k_{B} \times ∣ slope ∣$ , forgetting the $2$ — halving the true gap. This is silicon's actual $\approx 1.1$ eV gap, confirming the corrected answer.

Recall Verify

Correct $E_{g}$ ::: $2 \times 8.617 \times 1 0^{- 5} \times 6380 = 1.099$ eV (vs the wrong $0.55$ ). ✓

Ex 10 — Freeze-out: when ionisation is incomplete (cell C10)

Forecast: at $300$ K nearly every donor gave up its electron. At one-tenth the temperature, do you expect all, most, or few donors still ionised?

Compute the thermal ruler at 30 K: $k_{B} T = 8.617 \times 1 0^{- 5} \times 30 = 2.585 \times 1 0^{- 3} eV = 0.00259 eV$ . Why this step? $k_{B} T$ is the typical energy a jiggling lattice can hand a bound electron. It is the yardstick every ionisation is measured against.
Compare to the donor binding energy: $\frac{E _{D}}{k _{B} T} = \frac{0.045}{0.00259} = 17.4$ . Why? The Boltzmann factor $e^{- E_{D} / k_{B} T}$ governs the fraction of donors that can shake loose an electron. A ratio of $17$ means the exponent is a large negative number.
Evaluate the fraction: $e^{- 17.4} = 2.8 \times 1 0^{- 8}$ — essentially none of the donors are ionised. Why? At $300$ K the ratio was only $0.045/0.026 = 1.7$ (easy to overcome); at $30$ K it ballooned to $17$ , and the exponential of a big negative number collapses to near zero. The donor electrons are frozen back onto their P atoms.
Conclusion — the edge case: the "complete ionisation" assumption fails. Here $n ≪ N_{D}$ ; the free-electron count drops as the crystal cools, a regime called carrier freeze-out. So a doped semiconductor cooled far enough starts to behave like an insulator again — the exact opposite of the "always $n \approx N_{D}$ " reflex from Ex 2–4.

Recall Verify

Freeze-out check ::: $k_{B} T (30 K) = 0.00259$ eV; $E_{D} / k_{B} T = 17.4$ ; $e^{- 17.4} = 2.8 \times 1 0^{- 8} \Rightarrow$ negligible ionisation. ✓

Recall

Majority carrier of P-doped Si

electrons (n-type)

Law of mass action

n \cdot p = n_{i}^{2}

, fixed by generation–recombination equilibrium

Sign of

d σ / d T

for a metal

negative (heat adds scattering, no new carriers)

Slope of

ln σ

1/ T

- E_{g} /2 k_{B}

Compensated crystal rule

subtract

N_{A}

from

N_{D}

; sign of the survivor sets the type

What

σ_{0}

means

the temperature-independent conductivity ceiling reached as

T \to \infty

Why the ½ in the exponent

freeing one electron makes an electron–hole pair; the gap splits equally, each carrier sits

E_{g} /2

from its band edge

When complete ionisation fails

at low

T

where

k_{B} T < E_{D}

— carrier freeze-out,

n ≪ N_{D}

Value of elementary charge

e

1.602 \times 1 0^{- 19}

C, the charge on one electron or one hole