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

This is your self-testing station for the parent topic. Each problem is graded by cognitive level. Read the problem, try it on paper, then open the Solution. After each level a [!mistake] callout shows the trap that feels right — and why it isn't.

Before you start, meet every symbol once so none is a stranger.

Definition The quantities we keep reusing

$n$ = number of free electrons per cm³ (charge carriers in the conduction band, "CB").
$p$ = number of holes per cm³ (empty bond-spots in the valence band, "VB", that act like positive carriers).
$n_{i}$ = the intrinsic carrier count — how many electrons and holes a pure crystal makes on its own from heat alone. In pure material $n = p = n_{i}$ .
$N_{D}$ = donor density: number of pentavalent dopant atoms (e.g. P) added per cm³. Each can donate one electron.
$N_{A}$ = acceptor density: number of trivalent dopant atoms (e.g. B) added per cm³. Each can accept one electron, i.e. release one hole.
$E_{g}$ = band gap in electron-volts (eV): the energy a bound electron must gain to jump from the top of the VB to the bottom of the CB and become free.
$E_{D}$ = donor-level energy, measured on the same energy axis as the bands. A pentavalent dopant creates this shallow level just below the CB edge. The energy needed to free its spare electron is the gap from $E_{D}$ up to the CB edge, written $E_{C B} - E_{D} \approx 0.045 eV$ for P in Si — tiny compared with the full $E_{g} = 1.1 eV$ . (So $E_{D}$ is measured from the same zero as $E_{C B}$ ; the ionisation cost is the small difference $E_{C B} - E_{D}$ , not $E_{D}$ itself.)
$k_{B}$ = Boltzmann's constant $= 1.38 \times 1 0^{- 23} J/K = 8.617 \times 1 0^{- 5} eV/K$ . It converts a temperature into an energy.
$k_{B} T$ = thermal energy budget available at temperature $T$ . At room temperature ( $300 K$ ), $k_{B} T \approx 0.026 eV$ — remember this number, it decides what "easy" means.

Intuition Where

$σ \propto e^{- E_{g} /2 k_{B} T}$ comes from (the ½ factor, built from scratch) We want the number of electrons that heat manages to lift across the gap. Two ingredients multiply.

How many seats are available near the CB edge — the density of states $g (E)$ . Near a band edge this grows like $g (E) \propto E - E_{C B}$ (seats fan out just above the edge).
How likely each seat is occupied — the Fermi factor. For energies well above the Fermi level, that factor simplifies to the Boltzmann tail $e^{- (E - E_{F}) / k_{B} T}$ : the chance of finding an electron drops off exponentially the higher you look.

Multiplying "seats × chance" and adding up (integrating) over all energies above the CB edge gives $n \propto e^{- (E_{C B} - E_{F}) / k_{B} T} .$ In a pure crystal the Fermi level $E_{F}$ sits halfway up the gap, so $E_{C B} - E_{F} = \frac{1}{2} E_{g}$ . That halving is exactly where the ½ is born: creating a free electron also leaves a hole, and the gap energy is shared equally between the two. Substituting $E_{C B} - E_{F} = E_{g} /2$ : $n \propto e^{- E_{g} /2 k_{B} T} \Rightarrow σ = σ_{0} e^{- E_{g} /2 k_{B} T} .$ Here $σ_{0}$ collects everything that varies slowly with $T$ (the density-of-states constants, the charge $e$ , the mobilities). It has the units of conductivity, S/m, and is not the conductivity at any real temperature — it is the extrapolated value the formula would give if the exponent were zero.

Level 1 — Recognition

L1·Q1 — Classify by band gap

Given band gaps: Diamond $5.5 eV$ , Germanium $0.7 eV$ , Copper $0 eV$ (bands overlap), Quartz SiO₂ $9 eV$ . Label each as conductor, semiconductor, or insulator.

Recall Solution

The rule: conductor $E_{g} = 0$ · semiconductor $0.1$ – $3 eV$ · insulator $> 3 eV$ .

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

What to notice in the figure: four vertical energy ladders. The cyan block is the filled valence band, the faint white block is the empty conduction band, and the amber double-arrow is the band gap you must clear. Scan left to right: the amber arrow grows — zero for copper (blocks touch), a short hop for germanium, a tall wall for diamond and quartz. The dashed white line marks the $3 eV$ "insulator" convention; anything whose amber arrow pokes above it is an insulator.

Where do the boundaries come from? They follow from comparing $E_{g}$ with the thermal budget $k_{B} T \approx 0.026 eV$ .

If $E_{g} = 0$ the bands overlap: empty states sit right beside filled ones, so electrons move under any field → conductor.
If $E_{g}$ is a few tenths of an eV up to about $3 eV$ , the Boltzmann factor $e^{- E_{g} /2 k_{B} T}$ still lets a usable trickle of electrons cross at room temperature (and heating helps a lot) → semiconductor. Below ~ $0.1 eV$ the material behaves almost metallic; the $0.1 eV$ end is a soft lower edge, not a hard wall.
Once $E_{g} > 3 eV$ , $e^{- E_{g} /2 k_{B} T}$ is astronomically small at $300 K$ → virtually no carriers → insulator. The $3 eV$ line is a convention, not a cliff: borderline materials like GaN ( $3.4 eV$ ) or SiC ( $3.3 eV$ ) are called "wide-gap semiconductors," and zero-gap materials like graphene ( $E_{g} = 0$ but bands only touch) are semimetals — reminders that the classes shade into one another.

Applying the rule:

Copper: $E_{g} = 0$ → conductor (valence and conduction bands overlap).
Germanium: $0.7 eV$ , inside $0.1$ – $3$ → semiconductor.
Diamond: $5.5 eV > 3$ → insulator.
Quartz: $9 eV > 3$ → insulator (the champion — that huge gap is why glass doesn't conduct).

L1·Q2 — Name the carrier

For n-type Si (doped with phosphorus) and p-type Si (doped with boron), name the majority carrier in each and state whether the dopant is pentavalent or trivalent.

Recall Solution

n-type: dopant P has 5 valence electrons (pentavalent). One electron is left over → majority carrier = electrons. ("n" = negative.)
p-type: dopant B has 3 valence electrons (trivalent). One bond is missing an electron → a hole → majority carrier = holes. ("p" = positive.)

Level 2 — Application

L2·Q1 — Dopant number density

Silicon has $N_{S i} = 5 \times 1 0^{22}$ atoms/cm³. It is doped with phosphorus at 1 part per million by atom count. Find the donor density $N_{D}$ .

Recall Solution

"1 part per million" means a fraction $1 0^{- 6}$ of the atoms are dopant. $N_{D} = 1 0^{- 6} \times 5 \times 1 0^{22} = 5 \times 1 0^{16} cm^{- 3} .$ Why this matters: each of these becomes one free electron once ionised — the pipeline of new carriers.

L2·Q2 — Electron & hole count after doping

Take the doped Si from L2·Q1 with $N_{D} = 5 \times 1 0^{16} cm^{- 3}$ and $n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ at $300 K$ . Assume complete ionisation. Find $n$ and $p$ .

Recall Solution

Step 1 — free electrons. The donor's ionisation cost is only $E_{C B} - E_{D} \approx 0.045 eV$ . This is a bit larger than the average thermal quantum $k_{B} T = 0.026 eV$ , so a single "typical" kick doesn't guarantee release. But ionisation isn't governed by the average — it's governed by the spread of thermal energies. The fraction of donors still holding their electron goes like $e^{- (E_{C B} - E_{D}) / k_{B} T}$ -type Boltzmann statistics, and with $0.045/0.026 \approx 1.7$ in the exponent that bound fraction is small: at $300 K$ well over 99 % of donors are ionised (a full calculation, not a hand-wave — donor levels are only $\sim 0.045 eV$ deep, versus the $1.1 eV$ full gap the intrinsic electrons must clear). So the "complete ionisation" assumption is a good approximation, not an exact statement: $n \approx N_{D} = 5 \times 1 0^{16} cm^{- 3} .$ The right mental picture: $0.045 eV$ is comparable to $k_{B} T$ , so the ladder is easy to climb; the full $1.1 eV$ gap is $\sim 42 k_{B} T$ , so climbing that is astronomically rarer — which is why doping wins. Step 2 — holes via mass action. The see-saw law $n p = n_{i}^{2}$ gives $p = \frac{n _{i}^{2}}{n} = \frac{( 1.5 \times 1 0 ^{10} ) ^{2}}{5 \times 1 0 ^{16}} = \frac{2.25 \times 1 0 ^{20}}{5 \times 1 0 ^{16}} = 4.5 \times 1 0^{3} cm^{- 3} .$ So electrons rose ~ $1 0^{6} \times$ ; holes dropped ~ $1 0^{6} \times$ . Electrons are the majority carrier → n-type confirmed.

Edge warning — when does $n p = n_{i}^{2}$ break? The mass-action law assumes a non-degenerate semiconductor: the Fermi level stays inside the gap and the Boltzmann tail approximation holds. At very heavy doping (roughly $N_{D} ≳ 1 0^{19} cm^{- 3}$ , "degenerate" doping) the Fermi level pushes into the conduction band, the material behaves almost metallic, and $n p = n_{i}^{2}$ no longer holds. Our $5 \times 1 0^{16}$ is far below that, so we are safe.

Level 3 — Analysis

L3·Q1 — Compare n-type vs p-type conductivity

Two Si samples are doped to the same level $N_{D} = N_{A} = 5 \times 1 0^{16} cm^{- 3}$ . Mobilities: $μ_{e} = 1350$ , $μ_{h} = 450 cm^{2} / (V\cdots)$ . Compute $σ_{n}$ , $σ_{p}$ and their ratio.

Recall Solution

Why we may drop the minority term. The full conductivity is $σ = e (n μ_{e} + p μ_{h})$ . For the n-type sample, $n = 5 \times 1 0^{16}$ while (from mass action) $p = n_{i}^{2} / n = 4.5 \times 1 0^{3}$ . So the hole term is smaller than the electron term by a factor $\frac{p μ _{h}}{n μ _{e}} = \frac{( 4.5 \times 1 0 ^{3} ) ( 450 )}{( 5 \times 1 0 ^{16} ) ( 1350 )} \approx 3 \times 1 0^{- 14} .$ That is fourteen orders of magnitude — utterly negligible. So keep only the majority carrier: $σ_{n} = e n μ_{e} = (1.6 \times 1 0^{- 19}) (5 \times 1 0^{16}) (1350) = 10.8 S/cm .$ $σ_{p} = e p μ_{h} = (1.6 \times 1 0^{- 19}) (5 \times 1 0^{16}) (450) = 3.6 S/cm .$ $\frac{σ _{n}}{σ _{p}} = \frac{μ _{e}}{μ _{h}} = \frac{1350}{450} = 3.$ Why 3×? Same carrier count, but electrons in the conduction band have smaller effective mass than holes → they accelerate more easily → higher mobility. The carrier type sets a factor of 3; the doping itself set the factor of $1 0^{6}$ .

L3·Q2 — Band gap from a slope (why the $ln σ$ vs $1/ T$ trick works)

An intrinsic semiconductor obeys $σ = σ_{0} e^{- E_{g} /2 k_{B} T}$ (the prefactor $σ_{0}$ , defined in the tools block above, has units S/m and collects the slowly-varying density-of-states, charge and mobility factors). Two measurements: at $T_{1} = 300 K$ , $σ_{1} = 1.0 S/m$ ; at $T_{2} = 350 K$ , $σ_{2} = 8.0 S/m$ . Find $E_{g}$ in eV. Use $k_{B} = 8.617 \times 1 0^{- 5} eV/K$ .

Recall Solution

Why take a ratio? The unknown $σ_{0}$ cancels when we divide, leaving only $E_{g}$ . $\frac{σ _{2}}{σ _{1}} = e^{- \frac{E _{g}}{2 k _{B}} (\frac{1}{T _{2}} - \frac{1}{T _{1}})} .$ Take $ln$ : $ln \frac{σ _{2}}{σ _{1}} = - \frac{E _{g}}{2 k _{B}} (\frac{1}{T _{2}} - \frac{1}{T _{1}}) .$ Note $\frac{1}{T _{2}} - \frac{1}{T _{1}} = \frac{1}{350} - \frac{1}{300} = - 4.762 \times 1 0^{- 4} K^{- 1}$ (negative, since $T_{2} > T_{1}$ ). $ln 8 = 2.079.$ $E_{g} = \frac{- 2 k _{B} l n ( σ _{2} / σ _{1} )}{\frac{1}{T _{2}} - \frac{1}{T _{1}}} = \frac{- 2 ( 8.617 \times 1 0 ^{- 5} ) ( 2.079 )}{- 4.762 \times 1 0 ^{- 4}} \approx 0.75 eV .$ That is germanium-like.

What to notice in the figure: the cyan curve is a perfectly straight line when we plot $ln σ$ against $1/ T$ . That straightness is the whole point — it confirms a single activation energy. The two amber dots are our measurements; the slope between them is $- E_{g} /2 k_{B}$ , so the steepness of the line is the band gap (times $- 1/2 k_{B}$ ). Read a steeper line → bigger gap.

Level 4 — Synthesis

L4·Q1 — Where does the extra conductivity come from? (build it end to end)

Pure Si at $300 K$ has $n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ , $μ_{e} = 1350$ , $μ_{h} = 450$ . Compute (a) intrinsic conductivity $σ_{i}$ ; (b) conductivity after n-doping to $N_{D} = 5 \times 1 0^{16}$ ; (c) the enhancement factor.

Recall Solution

(a) Intrinsic. Both carriers present at $n = p = n_{i}$ : $σ_{i} = e n_{i} (μ_{e} + μ_{h}) = (1.6 \times 1 0^{- 19}) (1.5 \times 1 0^{10}) (1350 + 450) .$ $σ_{i} = (1.6 \times 1 0^{- 19}) (1.5 \times 1 0^{10}) (1800) = 4.32 \times 1 0^{- 6} S/cm .$ (b) n-doped. Now $n \approx N_{D} = 5 \times 1 0^{16}$ , holes negligible: $σ_{n} = e N_{D} μ_{e} = (1.6 \times 1 0^{- 19}) (5 \times 1 0^{16}) (1350) = 10.8 S/cm .$ (c) Enhancement. $\frac{σ _{n}}{σ _{i}} = \frac{10.8}{4.32 \times 1 0 ^{- 6}} = 2.5 \times 1 0^{6} .$ Just one dopant atom per million silicon atoms multiplied conductivity by 2.5 million — the whole reason semiconductor electronics exists.

Edge warning: this $1 0^{6}$ boost holds in the extrinsic plateau around room temperature. Cool the crystal far enough and donors freeze out (carriers vanish); heat it far enough and intrinsic generation takes over — both are explored in Level 5.

L4·Q2 — A compensated crystal (both dopants at once)

A Si sample is doped with both phosphorus ( $N_{D} = 8 \times 1 0^{16}$ ) and boron ( $N_{A} = 3 \times 1 0^{16}$ ) cm⁻³ (recall $N_{A}$ = acceptor density from the tools block). Is it n-type or p-type? Find the net majority carrier concentration and then the minority one ( $n_{i} = 1.5 \times 1 0^{10}$ ).

Recall Solution

Idea — donors and acceptors cancel. Each acceptor hole swallows one donor electron. What survives is the difference: $n \approx N_{D} - N_{A} = 8 \times 1 0^{16} - 3 \times 1 0^{16} = 5 \times 1 0^{16} cm^{- 3} .$ Since donors win, the sample is n-type. Minority holes via mass action: $p = \frac{n _{i}^{2}}{n} = \frac{( 1.5 \times 1 0 ^{10} ) ^{2}}{5 \times 1 0 ^{16}} = 4.5 \times 1 0^{3} cm^{- 3} .$

Level 5 — Mastery

L5·Q1 — Freeze-out: when "complete ionisation" fails

We keep saying "assume complete ionisation." Show it breaks at low temperature. For P in Si the donor sits an ionisation cost $E_{C B} - E_{D} = 0.045 eV$ below the conduction-band edge (see tools block for how $E_{D}$ is measured). Compare the thermal budget $k_{B} T$ at $300 K$ , $77 K$ (liquid N₂) and $20 K$ . At which temperatures is $0.045 eV$ "easy"? Use $k_{B} = 8.617 \times 1 0^{- 5} eV/K$ .

Recall Solution

Compute the thermal budget at each temperature: $k_{B} T (300) = 0.0259 eV, k_{B} T (77) = 0.00663 eV, k_{B} T (20) = 0.00172 eV .$ Compare each with the ionisation cost $0.045 eV$ :

$300 K$ : $0.045/0.0259 \approx 1.7$ — thermal energy is the same order as the cost, so the Boltzmann spread ionises nearly all donors → complete ionisation OK.
$77 K$ : $0.045/0.00663 \approx 6.8$ — cost is many times the budget → many electrons fall back onto the donor → partial freeze-out.
$20 K$ : $0.045/0.00172 \approx 26$ — the budget is tiny → electrons are frozen onto donors → carriers vanish, conductivity collapses.

What to notice in the figure: carrier count (log axis) traces three zones left to right. At low $T$ (amber band) the curve climbs — donors are waking up out of freeze-out. Then a long flat plateau (cyan band) where the curve hugs the dashed $n = N_{D}$ line — every donor is ionised and nothing changes. Finally at high $T$ (white band) the curve shoots up as heat rips electrons straight across the full $1.1 eV$ gap: the intrinsic region. The flat middle is where devices are designed to live.

L5·Q2 — When does intrinsic generation overtake the doping?

For the n-type sample with $N_{D} = 5 \times 1 0^{16}$ , doping dominates only while $n_{i} ≪ N_{D}$ . The intrinsic count grows with temperature as $n_{i} (T) = A e^{- E_{g} /2 k_{B} T},$ where $A$ is a temperature prefactor (units cm⁻³) that collects the density-of-states constants; for our purposes we only need that it is roughly constant over the range, so it will cancel. Given $n_{i} = 1.5 \times 1 0^{10}$ at $300 K$ and $E_{g} = 1.1 eV$ , estimate the temperature $T_{⋆}$ at which $n_{i}$ rises to equal $N_{D}$ (the onset of the intrinsic region).

Recall Solution

Why a ratio again? Divide $n_{i} (T_{⋆}) = N_{D}$ by $n_{i} (300) = 1.5 \times 1 0^{10}$ to cancel the prefactor $A$ : $\frac{N _{D}}{n _{i} ( 300 )} = exp [- \frac{E _{g}}{2 k _{B}} (\frac{1}{T _{⋆}} - \frac{1}{300})] .$ $\frac{5 \times 1 0 ^{16}}{1.5 \times 1 0 ^{10}} = 3.33 \times 1 0^{6} .$ Take $ln$ : $ln (3.33 \times 1 0^{6}) = 15.02$ . $15.02 = - \frac{1.1}{2 ( 8.617 \times 1 0 ^{- 5} )} (\frac{1}{T _{⋆}} - \frac{1}{300}) = - 6383 (\frac{1}{T _{⋆}} - \frac{1}{300}) .$ Solve for the bracket: $\frac{1}{T _{⋆}} - \frac{1}{300} = - \frac{15.02}{6383} = - 2.353 \times 1 0^{- 3} K^{- 1} .$ Add $\frac{1}{300} = 3.333 \times 1 0^{- 3}$ : $\frac{1}{T _{⋆}} = 3.333 \times 1 0^{- 3} - 2.353 \times 1 0^{- 3} = 0.980 \times 1 0^{- 3} K^{- 1} .$ Invert: $T_{⋆} = \frac{1}{0.980 \times 1 0 ^{- 3}} \approx 1020 K .$ Interpretation: below $\sim 1020 K$ the deliberate donors ( $N_{D} = 5 \times 1 0^{16}$ ) outnumber the heat-generated intrinsic carriers, so the sample stays reliably n-type and the device behaves. Above it, intrinsic carriers swamp the doping, $n$ and $p$ climb together, the mass-action-based "majority carrier" picture blurs, and a transistor built here stops switching. (The real prefactor $A$ drifts with $T$ , shifting $T_{⋆}$ somewhat, but the order — many hundreds of kelvin — is robust.)

Related deep threads: Fermi-Dirac Distribution (why $k_{B} T$ sets ionisation and where the Boltzmann tail comes from), p-n Junction Diodes (put n and p side by side), Thermistors & Hall Effect (measuring what you computed here), Photovoltaic Cells and Transistors and Integrated Circuits (why the $1 0^{6}$ enhancement matters).

Quick self-check

Net carriers in compensated crystal (

N_{D} = 8 \times 1 0^{16}

N_{A} = 3 \times 1 0^{16}

)

n \approx 5 \times 1 0^{16} cm^{- 3}

, n-type

n-type vs p-type conductivity ratio at equal doping

σ_{n} / σ_{p} = μ_{e} / μ_{h} = 3

Why the 2 in

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

promoting one electron makes an electron–hole pair;

E_{F}

sits mid-gap so

E_{C B} - E_{F} = E_{g} /2

Minority hole count when

n = 5 \times 1 0^{16}

n_{i} = 1.5 \times 1 0^{10}

p = n_{i}^{2} / n = 4.5 \times 1 0^{3} cm^{- 3}