Mass action law (np = ni²): Exercises — Hardware deep dive

Intuition How to use this page

Each problem has a fully worked solution hidden inside a collapsible [!recall]- box. On screen it appears as a grey bar with a small triangle ▸ on its left and the word "Solution" — click that bar (or the triangle) and the full solution unfolds downward; click again to hide it. The trailing minus in [!recall]- is what makes it start collapsed, so try each problem on paper first, then click to reveal. Problems climb from L1 Recognition (do you know the rule?) through L4 Synthesis, up to L5 Mastery (can you fuse it with everything else?). After each level, a [!mistake] box shows the trap most people fall into there.

Everything here rests on the parent law [[Mass action law (np = ni²) (index 2.1.7)| $n p = n_{i}^{2}$ ]] and the constants from [[Effective density of states $N_{C}$ , $N_{V}$ ]], Intrinsic carrier concentration $n_i$, and Band gap $E_g$ and its temperature dependence.

Definition Constants and symbols used everywhere below (silicon,

$T = 300 K$ unless stated)

$n_{i} = 1.0 \times 1 0^{10} cm^{- 3}$ (silicon, room temperature)
$k T = 0.02585 eV \approx 0.0259 eV$ at $300 K$
Boltzmann constant $k = 8.617 \times 1 0^{- 5} eV/K$
$E_{g} (Si) = 1.12 eV$
$N_{C}$ , $N_{V}$ = the effective density of states (a plain-language term, not an equation — the double-equals here is just this vault's way of highlighting a key term you should be able to recall). $N_{C}$ is the effective number of available electron "seats" bunched at the conduction-band edge, and $N_{V}$ the same for hole "seats" at the valence-band edge; both are measured in $cm^{- 3}$ . Physically they replace the whole spread-out band by one equivalent count of states sitting right at the edge, so that $n = N_{C} e^{- (E_{C} - E_{F}) / k T}$ works with a single number. For silicon both are $\sim 1 0^{19} cm^{- 3}$ . Full treatment: Effective density of states $N_C$, $N_V$.
"per cm³" is written $cm^{- 3}$ ; all concentrations are per cm³.

A quick reminder of what each letter means: $n$ = free electrons in the conduction band, $p$ = free holes in the valence band, $N_{D}$ = donor doping (each donates one electron), $N_{A}$ = acceptor doping (each accepts one electron, i.e. creates one hole). "Fully ionized" means every dopant has given up / grabbed its carrier.

Level 1 — Recognition

Can you recall the rule and plug numbers into it?

L1.1 — Is this equilibrium claim allowed?

A student writes: "In a silicon bar sitting in the dark with no wires attached, $n p = n_{i}^{2}$ ." True or false, and why?

Recall Solution

True. The three conditions for the law are: (1) thermal equilibrium, (2) no illumination, (3) no current / applied voltage. "Dark, no wires" satisfies all three. The law is a statement about a system that has settled — generation of electron–hole pairs exactly balances recombination, which pins the product $n p$ to the material constant $n_{i}^{2}$ .

L1.2 — Minority carrier from majority

An n-type silicon sample has electron concentration $n = 1 0^{15} cm^{- 3}$ . Find the hole concentration $p$ .

Recall Solution

Use $n p = n_{i}^{2}$ directly, solving for $p$ : $p = \frac{n _{i}^{2}}{n} = \frac{( 1 0 ^{10} ) ^{2}}{1 0 ^{15}} = \frac{1 0 ^{20}}{1 0 ^{15}} = 1 0^{5} cm^{- 3} .$ What we did: divided the constant $n_{i}^{2}$ by the known majority carrier. Why: the product is fixed, so knowing one carrier fixes the other. The minority (holes here) is tiny — $1 0^{5}$ vs $1 0^{15}$ , ten orders of magnitude smaller.

L1.3 — State the constant

Write $n_{i}^{2}$ in terms of band-structure quantities, and $n_{i}$ itself. Explain the factor of 2.

Recall Solution

Recall (from the definition box above) that $N_{C}$ and $N_{V}$ are the effective densities of states at the conduction- and valence-band edges — the "number of seats" available to electrons and holes — each $\sim 1 0^{19} cm^{- 3}$ in silicon, and $E_{g}$ is the band gap. Then: $n_{i}^{2} = N_{C} N_{V} e^{- E_{g} / k T}, n_{i} = N_{C} N_{V} e^{- E_{g} /2 k T} .$ Taking the square root of the whole right side square-roots the prefactor ( $N_{C} N_{V}$ ) and halves the exponent (because $e^{- E_{g} / k T} = e^{- E_{g} /2 k T}$ ). The gap appears as $E_{g} /2$ in $n_{i}$ but as $E_{g}$ in $n_{i}^{2}$ .

Level 2 — Application

Combine the law with charge neutrality to solve real doped samples.

L2.1 — Standard n-type

Silicon is doped with $N_{D} = 5 \times 1 0^{16} cm^{- 3}$ donors (fully ionized), $N_{A} = 0$ . Find $n$ and $p$ .

Recall Solution

Since $N_{D} = 5 \times 1 0^{16} ≫ n_{i} = 1 0^{10}$ , the donors swamp the intrinsic carriers, so charge neutrality $n = p + N_{D}$ gives $n \approx N_{D} = 5 \times 1 0^{16} cm^{- 3} .$ Then from mass action: $p = \frac{n _{i}^{2}}{n} = \frac{1 0 ^{20}}{5 \times 1 0 ^{16}} = 2 \times 1 0^{3} cm^{- 3} .$ Check the approximation: we assumed $p ≪ N_{D}$ . Indeed $2 \times 1 0^{3} ≪ 5 \times 1 0^{16}$ ✅, so $n \approx N_{D}$ was justified.

L2.2 — Standard p-type

Silicon is doped with $N_{A} = 2 \times 1 0^{17} cm^{- 3}$ acceptors, $N_{D} = 0$ . Find $p$ and $n$ .

Recall Solution

Acceptors create holes. With $N_{A} ≫ n_{i}$ , neutrality $p = n + N_{A}$ gives $p \approx N_{A} = 2 \times 1 0^{17} cm^{- 3} .$ $n = \frac{n _{i}^{2}}{p} = \frac{1 0 ^{20}}{2 \times 1 0 ^{17}} = 5 \times 1 0^{2} cm^{- 3} .$ Here holes are the majority and electrons the minority — mirror image of L2.1.

L2.3 — Compensated sample (both dopants)

Silicon has $N_{D} = 8 \times 1 0^{16}$ and $N_{A} = 3 \times 1 0^{16} cm^{- 3}$ , both fully ionized. Find $n$ and $p$ .

Recall Solution

Donors and acceptors partially cancel. The net doping is $N = N_{D} - N_{A} = 8 \times 1 0^{16} - 3 \times 1 0^{16} = 5 \times 1 0^{16} cm^{- 3} (net donor) .$ Since $N ≫ n_{i}$ , the sample is n-type with $n \approx N = 5 \times 1 0^{16} cm^{- 3}, p = \frac{n _{i}^{2}}{n} = \frac{1 0 ^{20}}{5 \times 1 0 ^{16}} = 2 \times 1 0^{3} cm^{- 3} .$ What matters is the difference, not each dopant alone: the $3 \times 1 0^{16}$ acceptors "eat" $3 \times 1 0^{16}$ of the donor electrons, leaving a net $5 \times 1 0^{16}$ .

Level 3 — Analysis

When doping is comparable to $n_{i}$ , the shortcuts break — use the full quadratic.

L3.1 — Lightly doped, near-intrinsic regime

A material has $n_{i} = 2 \times 1 0^{13} cm^{- 3}$ (say, a narrow-gap semiconductor at $300 K$ ) and net donor doping $N = N_{D} - N_{A} = 1 \times 1 0^{13} cm^{- 3}$ . Find $n$ and $p$ exactly — the shortcut $n \approx N$ is NOT valid here (why?).

Recall Solution

Why the shortcut fails: $N = 1 0^{13}$ is not $≫ n_{i} = 2 \times 1 0^{13}$ ; in fact $N < n_{i}$ . The intrinsic carriers are not negligible, so we must solve the full system. Why we may still write $p = n_{i}^{2} / n$ : mass action $n p = n_{i}^{2}$ holds at any doping so long as we are at thermal equilibrium (the $E_{F}$ -cancellation derivation never assumed the material was pure — only equilibrium). So even here, doped, $p = n_{i}^{2} / n$ is exact. Substitute it into charge neutrality $n - p = N$ : $n - \frac{n _{i}^{2}}{n} = N \Rightarrow n^{2} - N n - n_{i}^{2} = 0.$ The physically meaningful (positive) root is $n = \frac{N}{2} + (\frac{N}{2})^{2} + n_{i}^{2} .$ Plug in $N = 1 \times 1 0^{13}$ , $n_{i} = 2 \times 1 0^{13}$ : $\frac{N}{2} = 0.5 \times 1 0^{13}, (\frac{N}{2})^{2} = 0.25 \times 1 0^{26}, n_{i}^{2} = 4 \times 1 0^{26} .$ $n = 0.5 \times 1 0^{13} + (0.25 + 4) \times 1 0^{26} = 0.5 \times 1 0^{13} + 4.25 \times 1 0^{13} .$ $4.25 = 2.0616 \Rightarrow n = (0.5 + 2.0616) \times 1 0^{13} = 2.5616 \times 1 0^{13} cm^{- 3} .$ Then $p = n - N = 2.5616 \times 1 0^{13} - 1.0 \times 1 0^{13} = 1.5616 \times 1 0^{13} cm^{- 3} .$ Sanity check the product: $n p = (2.5616) (1.5616) \times 1 0^{26} = 4.000 \times 1 0^{26} = n_{i}^{2}$ ✅. Note $n$ and $p$ are comparable — the sample is only weakly n-type, exactly what you'd expect when $N ≲ n_{i}$ .

The figure below plots this whole story. Read the axes first: the horizontal axis is the net doping $N = N_{D} - N_{A}$ and the vertical axis is the resulting electron concentration $n$ , both on a log scale (cm⁻³). Three lines are drawn:

blue solid = the exact quadratic solution $n = \frac{N}{2} + (N /2)^{2} + n_{i}^{2}$ ;

orange dashed = the heavy-doping asymptote $n \approx N$ (the L2 shortcut), which the blue curve hugs on the right;

green dotted = the near-intrinsic floor $n \to n_{i}$ (the L3.2a limit), which the blue curve flattens onto at the far left.

The red dot marks this exact L3.1 point ( $N = 1 0^{13}$ , $n = 2.56 \times 1 0^{13}$ ), sitting in the bend where neither asymptote alone is accurate — which is precisely why the shortcut fails here.

L3.2 — Limiting behaviour of the quadratic (both signs of $N$ )

From $n = \frac{N}{2} + (N /2)^{2} + n_{i}^{2}$ , show what happens (a) when $N \to 0$ , (b) when $N ≫ n_{i}$ , and (c) when $N < 0$ (net acceptor, p-type). Interpret each.

Recall Solution

(a) $N \to 0$ (pure material): $n = 0 + 0 + n_{i}^{2} = n_{i} .$ With $N = 0$ there is no net doping, so $n = p = n_{i}$ — the intrinsic case. The formula correctly reduces to it. What it looks like: the blue curve in the figure lands on the horizontal $n_{i}$ line at the far left.

(b) $N ≫ n_{i}$ : the $n_{i}^{2}$ under the root is negligible, so $n \approx \frac{N}{2} + (\frac{N}{2})^{2} = \frac{N}{2} + \frac{N}{2} = N .$ Fully doped n-type limit: $n \approx N$ , the L2 shortcut. What it looks like: the curve becomes the straight line $n = N$ (the dashed orange asymptote) on the right.

(c) $N < 0$ (net acceptor doping, p-type): the same formula still applies if we keep the sign of $N$ . Write $N = - ∣ N ∣$ . Then $n = - \frac{∣ N ∣}{2} + (\frac{∣ N ∣}{2})^{2} + n_{i}^{2} .$ This is a small positive number (the root slightly exceeds $∣ N ∣/2$ ), so $n$ becomes the minority carrier. For $∣ N ∣ ≫ n_{i}$ it reduces to $n \approx n_{i}^{2} /∣ N ∣$ (expand the root), exactly the minority-electron formula in p-type material (compare L2.2). And $p = n - N = n + ∣ N ∣ \approx ∣ N ∣$ — the majority holes. So one signed formula covers n-type ( $N > 0$ ), intrinsic ( $N = 0$ ) and p-type ( $N < 0$ ). Always solve for the carrier tied to the positive root and get the other from $p = n - N$ .

Recall Bonus corner case: degenerate (very heavy) doping

All the arithmetic above used the non-degenerate (Boltzmann) band populations $n = N_{C} e^{- (E_{C} - E_{F}) / k T}$ , valid while $E_{F}$ stays inside the gap, a few $k T$ below $E_{C}$ (or above $E_{V}$ ). If you dope so heavily that $E_{F}$ crosses into the conduction band (roughly $N_{D} ≳ N_{C} \sim 1 0^{19} cm^{- 3}$ in silicon), the material is degenerate: you must use full Fermi–Dirac statistics (Fermi level & Fermi–Dirac statistics), the simple $n p = n_{i}^{2}$ acquires corrections, and $n$ no longer tracks $N_{D}$ linearly. Qualitatively, the seesaw stiffens — this page's formulas assume you stay below that regime.

Level 4 — Synthesis

Bring in temperature, quasi-Fermi levels, and multiple effects at once.

L4.1 — Temperature doubling of $n_{i}$

For silicon, $n_{i}^{2} = N_{C} N_{V} e^{- E_{g} / k T}$ . By what factor does $n_{i}$ change when temperature rises from $300 K$ to $310 K$ ? Use $E_{g} = 1.12 eV$ , $k = 8.617 \times 1 0^{- 5} eV/K$ .

Recall Solution

First, is it fair to treat the prefactor as constant? The prefactor $N_{C} N_{V}$ actually carries a mild $T^{3/2}$ each, so $N_{C} N_{V} \propto T^{3/2}$ , and $E_{g}$ drifts by only a few meV over $10 K$ . Let's quantify the prefactor drift: $(\frac{310}{300})^{3/2} = (1.0333)^{1.5} = 1.050$ — a 5% rise. Compare that to the exponential's effect, computed below, which is a factor of ~2 (100%). So the prefactor drift is $\sim 20 \times$ smaller than the exponential swing and is legitimately negligible for a "by what factor / roughly doubles" question; we drop it, keeping only the dominant $e^{- E_{g} /2 k T}$ .

$n_{i} \propto e^{- E_{g} /2 k T}$ . The ratio is $\frac{n _{i} ( 310 )}{n _{i} ( 300 )} = exp [- \frac{E _{g}}{2 k} (\frac{1}{310} - \frac{1}{300})] .$ Compute the bracket. $\frac{1}{310} - \frac{1}{300} = \frac{300 - 310}{300 \cdot 310} = \frac{- 10}{93000} = - 1.0753 \times 1 0^{- 4} K^{- 1}$ . $\frac{E _{g}}{2 k} = \frac{1.12}{2 \times 8.617 \times 1 0 ^{- 5}} = \frac{1.12}{1.7234 \times 1 0 ^{- 4}} = 6499.9 K .$ Exponent $= - 6499.9 \times (- 1.0753 \times 1 0^{- 4}) = + 0.6990$ . $\frac{n _{i} ( 310 )}{n _{i} ( 300 )} = e^{0.6990} = 2.012.$ So $n_{i}$ roughly doubles over a $10 K$ rise — this is the famous extreme temperature sensitivity (and the neglected $5%$ prefactor rise would nudge this to $\approx 2.11$ , not changing the "doubles" conclusion). Why exponential and not linear? Because carriers must be thermally lifted across the gap; the population of "lucky" high-energy states follows the Boltzmann factor $e^{- E_{g} /2 k T}$ , which changes explosively with $T$ .

L4.2 — Non-equilibrium: injection breaks the law

Under illumination, a silicon sample in steady state has electron and hole quasi-Fermi levels split by $F_{n} - F_{p} = 0.30 eV$ at $300 K$ . Find the ratio $n p / n_{i}^{2}$ , and evaluate $n p$ .

Recall Solution

What are $F_{n}$ and $F_{p}$ ? At equilibrium a single Fermi level $E_{F}$ sets both carrier populations. Under injection (light or current) electrons and holes fall out of mutual equilibrium, so we give each its own quasi-Fermi level: $F_{n}$ = electron quasi-Fermi level (sets $n$ ) and $F_{p}$ = hole quasi-Fermi level (sets $p$ ). Their split $F_{n} - F_{p}$ measures how far from equilibrium the sample is; at equilibrium $F_{n} = F_{p} = E_{F}$ . See Quasi-Fermi levels & non-equilibrium carriers. Away from equilibrium the law generalizes: $n p = n_{i}^{2} e^{(F_{n} - F_{p}) / k T} .$ With $F_{n} - F_{p} = 0.30 eV$ and $k T = 0.02585 eV$ : $\frac{n p}{n _{i}^{2}} = e^{0.30/0.02585} = e^{11.605} = 1.096 \times 1 0^{5} .$ Now multiply through by $n_{i}^{2} = (1 0^{10})^{2} = 1 0^{20} cm^{- 6}$ : $n p = n_{i}^{2} \times 1.096 \times 1 0^{5} = (1 0^{20}) (1.096 \times 1 0^{5}) = 1.096 \times 1 0^{25} cm^{- 6} .$ Interpretation: light injects excess pairs, so $n p$ jumps five orders of magnitude above equilibrium ( $n p / n_{i}^{2} \approx 1.1 \times 1 0^{5}$ , i.e. $n p \approx 1.1 \times 1 0^{25}$ ). The equilibrium law $n p = n_{i}^{2}$ is the special case $F_{n} = F_{p}$ (a single Fermi level), giving $e^{0} = 1$ .

Level 5 — Mastery

Everything at once: derive, generalize, and reason about a designed device.

L5.1 — Design a target minority scenario (carrier ratio)

You need an n-type silicon region where the minority hole concentration is exactly $p = 1 0^{6} cm^{- 3}$ at $300 K$ . What net donor doping $N_{D}$ (assume $N_{A} = 0$ , $N_{D} ≫ n_{i}$ ) achieves this? Then confirm the majority electron concentration.

Recall Solution

From mass action, $p = n_{i}^{2} / n$ , so $n = n_{i}^{2} / p$ : $n = \frac{( 1 0 ^{10} ) ^{2}}{1 0 ^{6}} = \frac{1 0 ^{20}}{1 0 ^{6}} = 1 0^{14} cm^{- 3} .$ Since $N_{D} ≫ n_{i}$ was assumed, $N_{D} \approx n = 1 0^{14} cm^{- 3}$ . Confirm: with $n = 1 0^{14} ≫ n_{i} = 1 0^{10}$ ✅ the assumption holds, and $n p = 1 0^{14} \times 1 0^{6} = 1 0^{20} = n_{i}^{2}$ ✅. So doping to $N_{D} = 1 0^{14} cm^{- 3}$ gives exactly the required $p = 1 0^{6} cm^{- 3}$ .

L5.2 — Full derivation chain, from generation–recombination to a number

Starting from the generation–recombination balance $G (T) = r n p$ , and the band populations, prove $n p = n_{i}^{2}$ and then evaluate $n_{i}$ for a hypothetical material with $N_{C} = N_{V} = 2.5 \times 1 0^{19} cm^{- 3}$ , $E_{g} = 1.00 eV$ , $T = 300 K$ .

Recall Solution

Symbols recalled first (self-contained):

$G (T)$ = thermal generation rate of electron–hole pairs (pairs per cm³ per second); depends only on temperature and gap, not on doping.
$r$ = recombination coefficient (a constant of proportionality); recombination rate is $r n p$ because it needs one electron and one hole to meet — a two-body event.
$E_{C}$ = conduction-band edge energy, $E_{V}$ = valence-band edge energy, so the gap is $E_{g} = E_{C} - E_{V}$ .
$E_{F}$ = Fermi level, the reference energy that sets equilibrium carrier populations (see Fermi level & Fermi–Dirac statistics).
$N_{C}, N_{V}$ = effective densities of states at the two band edges (defined in the top box).

Step 1 — the product is constant. At equilibrium generation equals recombination: $G (T) = r n p \Rightarrow n p = \frac{G ( T )}{r} = const (T) .$ Step 2 — evaluate the constant in pure material. In an undoped crystal $n = p = n_{i}$ , so the constant is $n_{i}^{2}$ . Hence for any doping (still at equilibrium) $n p = n_{i}^{2}$ . Step 3 — express $n_{i}^{2}$ via band statistics. With $n = N_{C} e^{- (E_{C} - E_{F}) / k T}$ and $p = N_{V} e^{- (E_{F} - E_{V}) / k T}$ , multiplying cancels $E_{F}$ : $n p = N_{C} N_{V} e^{- (E_{C} - E_{V}) / k T} = N_{C} N_{V} e^{- E_{g} / k T} = n_{i}^{2} .$ Step 4 — number. $k T = 8.617 \times 1 0^{- 5} \times 300 = 0.025851 eV$ . $n_{i} = N_{C} N_{V} e^{- E_{g} /2 k T} = 2.5 \times 1 0^{19} \times e^{- 1.00/ (2 \times 0.025851)} .$ Exponent $= - \frac{1.00}{0.051702} = - 19.342$ , so $e^{- 19.342} = 3.985 \times 1 0^{- 9}$ . $n_{i} = 2.5 \times 1 0^{19} \times 3.985 \times 1 0^{- 9} = 9.96 \times 1 0^{10} cm^{- 3} .$ So this material has $n_{i} \approx 1.0 \times 1 0^{11} cm^{- 3}$ — close to silicon's, as expected for a slightly smaller gap partly offset by our chosen prefactor.

L5.3 — Sensitivity synthesis: how much doping error from a temperature error?

Continue L5.2's material. If your process temperature is off by $+ 5 K$ (so $T = 305 K$ instead of $300$ ), by what factor does $n_{i}^{2}$ (and hence the minority carrier $p = n_{i}^{2} / N_{D}$ at fixed $N_{D}$ ) change? Use $E_{g} = 1.00 eV$ .

Recall Solution

$n_{i}^{2} \propto e^{- E_{g} / k T}$ (ignore prefactor drift). Ratio: $\frac{n _{i}^{2} ( 305 )}{n _{i}^{2} ( 300 )} = exp [- \frac{E _{g}}{k} (\frac{1}{305} - \frac{1}{300})] .$ $\frac{1}{305} - \frac{1}{300} = \frac{300 - 305}{305 \cdot 300} = \frac{- 5}{91500} = - 5.4645 \times 1 0^{- 5} K^{- 1}$ . $\frac{E _{g}}{k} = \frac{1.00}{8.617 \times 1 0 ^{- 5}} = 11605.0 K$ . Exponent $= - 11605.0 \times (- 5.4645 \times 1 0^{- 5}) = + 0.6342$ . $\frac{n _{i}^{2} ( 305 )}{n _{i}^{2} ( 300 )} = e^{0.6342} = 1.886.$ So a mere $+ 5 K$ error nearly doubles the minority carrier concentration (factor $1.89$ ) at fixed doping. Since minority carriers drive diode leakage, this is why thermal control matters enormously in device physics — the exponential in $n_{i}^{2}$ amplifies tiny temperature errors.

Recall One-line self-test before you leave

Every answer above either (a) plugged into $n p = n_{i}^{2}$ , (b) added charge neutrality $n + N_{A} = p + N_{D}$ , (c) took the positive root of $n^{2} - N n - n_{i}^{2} = 0$ , or (d) generalized to $n p = n_{i}^{2} e^{(F_{n} - F_{p}) / k T}$ off equilibrium. If you can say which of these four each problem needed, you own the topic.

Connections

Intrinsic carrier concentration $n_i$
Fermi level & Fermi–Dirac statistics
Effective density of states $N_C$, $N_V$
Charge neutrality condition
Doping: donors and acceptors
Quasi-Fermi levels & non-equilibrium carriers
Band gap $E_g$ and its temperature dependence

Exercises — Mass action law (np = ni²)

Level 1 — Recognition

L1.1 — Is this equilibrium claim allowed?

L1.2 — Minority carrier from majority

L1.3 — State the constant

Level 2 — Application

L2.1 — Standard n-type

L2.2 — Standard p-type

L2.3 — Compensated sample (both dopants)

Level 3 — Analysis

L3.1 — Lightly doped, near-intrinsic regime

L3.2 — Limiting behaviour of the quadratic (both signs of $N$ )

Level 4 — Synthesis

L4.1 — Temperature doubling of $n_{i}$

L4.2 — Non-equilibrium: injection breaks the law

Level 5 — Mastery

L5.1 — Design a target minority scenario (carrier ratio)

L5.2 — Full derivation chain, from generation–recombination to a number

L5.3 — Sensitivity synthesis: how much doping error from a temperature error?

Connections

Concept Map