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

This page is the drill hall for the mass action law. The parent note told you what the law is and why it holds. Here we hit it with every kind of input it can face — heavy doping, light doping, exact-balance doping, compensated doping, zero net doping, illumination, and a change of temperature — and we solve each one to a number, checking that number every single time.

Before we start, one promise: we use only these tools, all defined already in the parent —

$n$ = electron concentration (per cm³), $p$ = hole concentration (per cm³),
$n_{i}$ = intrinsic carrier concentration ( $n = p$ in pure material),
$n p = n_{i}^{2}$ at thermal equilibrium (see [[Intrinsic carrier concentration $n_{i}$ ]]),
charge neutrality: $n + N_{A} = p + N_{D}$ with fully ionized dopants (see Charge neutrality condition and Doping: donors and acceptors).

Everything below is just these two equations, applied over and over.

The scenario matrix

Every problem the mass action law throws at you is one cell in this table. The "control knob" is the net doping $N \equiv N_{D} - N_{A} (donors minus acceptors, both fully ionized) .$ The behaviour is decided entirely by how $∣ N ∣$ compares to $n_{i}$ , and by the sign of $N$ .

Cell	Regime (input)	What dominates	Formula shortcut	Example
A	$N > 0$ , $N ≫ n_{i}$ (heavy n-type)	electrons	$n \approx N$ , $p = n_{i}^{2} / N$	Ex 1
B	$N < 0$ , $∣ N ∣ ≫ n_{i}$ (heavy p-type)	holes	$p \approx ∣ N ∣$ , $n = n_{i}^{2} / ∣ N ∣$	Ex 2
C	$N = 0$ (undoped OR exactly compensated)	neither	$n = p = n_{i}$	Ex 3
D	$N \sim n_{i}$ (lightly doped, comparable)	mixed — need quadratic	$n = \frac{N}{2} + (N /2)^{2} + n_{i}^{2}$	Ex 4
E	$N_{D}, N_{A}$ both large but nearly equal (compensation)	net $N$ small	use net $N$ in D	Ex 5
F	Non-equilibrium (light/injection)	excess carriers	$n p = n_{i}^{2} e^{(F_{n} - F_{p}) / k T} > n_{i}^{2}$	Ex 6
G	Temperature changed	$n_{i} (T)$ moves	recompute $n_{i}$ , redo A/B	Ex 7
H	Word problem / exam twist	pick the right cell	identify regime first	Ex 8

Throughout, unless a problem says otherwise, we use silicon at room temperature: $n_{i} = 1.0 \times 1 0^{10} cm^{- 3}, k T = 0.02585 eV (T = 300 K) .$

Example 1 — Cell A: heavy n-type

Worked example Heavy donor doping

Silicon, $N_{D} = 1 \times 1 0^{16} cm^{- 3}$ , no acceptors. Find $n$ and $p$ .

Forecast: guess before reading — will $p$ be bigger or smaller than $n_{i}$ ? By how many orders of magnitude?

Compute net doping. $N = N_{D} - N_{A} = 1 0^{16} - 0 = 1 0^{16}$ . Why this step? The whole problem is decided by $N$ vs $n_{i}$ ; get $N$ first.
Compare to $n_{i}$ . $N = 1 0^{16} ≫ n_{i} = 1 0^{10}$ (six orders of magnitude). We are firmly in cell A, so the shortcut is safe. Why this step? Only when $N ≫ n_{i}$ can we drop the quadratic's tiny correction term.
Majority carrier. $n \approx N = 1 0^{16} cm^{- 3}$ . Why this step? Neutrality $n - p = N$ ; since $p$ is going to be minuscule, $n \approx N$ .
Minority carrier via mass action. $p = \frac{n _{i}^{2}}{n} = \frac{( 1 0 ^{10} ) ^{2}}{1 0 ^{16}} = \frac{1 0 ^{20}}{1 0 ^{16}} = 1 0^{4} cm^{- 3}$ . Why this step? Mass action pins the product; once $n$ is fixed, $p$ is forced.

Verify: product $n p = 1 0^{16} \cdot 1 0^{4} = 1 0^{20} = (1 0^{10})^{2} = n_{i}^{2}$ ✅. And $p = 1 0^{4} ≪ n_{i} = 1 0^{10}$ — the minority carrier collapsed far below the intrinsic value. Units: (cm⁻³)(cm⁻³)=cm⁻⁶, matches $n_{i}^{2}$ .

Example 2 — Cell B: heavy p-type (the mirror image)

Worked example Heavy acceptor doping

Silicon, $N_{A} = 5 \times 1 0^{15} cm^{- 3}$ , no donors. Find $p$ and $n$ .

Forecast: which carrier is now the majority? Which formula gets the square?

Net doping. $N = N_{D} - N_{A} = 0 - 5 \times 1 0^{15} = - 5 \times 1 0^{15}$ . Negative ⇒ p-type. Why this step? The sign of $N$ tells you which carrier wins — negative means holes.
Compare magnitude. $∣ N ∣ = 5 \times 1 0^{15} ≫ n_{i}$ . Cell B, shortcut valid. Why this step? Same $∣ N ∣ ≫ n_{i}$ test as Ex 1, just on the hole side.
Majority (holes). $p \approx ∣ N ∣ = 5 \times 1 0^{15} cm^{- 3}$ . Why this step? Neutrality $p - n = ∣ N ∣$ ; $n$ is tiny so $p \approx ∣ N ∣$ .
Minority (electrons). $n = \frac{n _{i}^{2}}{p} = \frac{1 0 ^{20}}{5 \times 1 0 ^{15}} = 2 \times 1 0^{4} cm^{- 3}$ . Why this step? Mass action again — product fixed, so $n$ is forced small.

Verify: $n p = (5 \times 1 0^{15}) (2 \times 1 0^{4}) = 1 0^{20} = n_{i}^{2}$ ✅. Note the mirror of Ex 1: roles of $n$ and $p$ swap when $N$ flips sign.

Example 3 — Cell C: zero net doping (intrinsic OR compensated)

Worked example Two ways to reach

$N = 0$ (a) Pure silicon, no dopants. (b) Silicon with $N_{D} = N_{A} = 3 \times 1 0^{15}$ (equal donors and acceptors). Find $n, p$ in each.

Forecast: are these two physically identical for carrier count? (Careful — mobility differs, but what about $n$ and $p$ ?)

Case (a): $N = 0$ . Neutrality $n - p = 0 \Rightarrow n = p$ . Why this step? No net charge from dopants means equal free charges.
Apply mass action. $n = p$ and $n p = n_{i}^{2} \Rightarrow n^{2} = n_{i}^{2} \Rightarrow n = p = n_{i} = 1 0^{10} cm^{- 3}$ . Why this step? Two equations, two unknowns; the symmetric solution is exactly $n_{i}$ .
Case (b): $N = N_{D} - N_{A} = 0$ . Same equation $n - p = 0$ , so again $n = p = n_{i} = 1 0^{10}$ . Why this step? Only the net doping enters neutrality; equal donors and acceptors cancel.

Verify: both give $n p = (1 0^{10})^{2} = n_{i}^{2}$ ✅. Carrier counts are identical; only scattering/mobility differ (that's a transport story, not a mass-action story). This is the $N \to 0$ limit of the quadratic — see Ex 4.

Example 4 — Cell D: lightly doped, $N$ comparable to $n_{i}$ (quadratic required)

Worked example When the shortcut breaks

A semiconductor with $n_{i} = 2 \times 1 0^{10}$ is doped $N_{D} = 3 \times 1 0^{10}$ , $N_{A} = 0$ . Find $n$ and $p$ exactly.

Forecast: if you naively used $n \approx N_{D}$ , how far off would you be?

Net doping. $N = 3 \times 1 0^{10}$ . Compare: $N / n_{i} = 1.5$ — not $≫ 1$ . Shortcut forbidden. Why this step? When $N$ is within a factor of a few of $n_{i}$ , the neglected term is not negligible.
Set up the quadratic. Substitute $p = n_{i}^{2} / n$ into $n - p = N$ : $n - \frac{n _{i}^{2}}{n} = N \Rightarrow n^{2} - N n - n_{i}^{2} = 0.$ Why this step? We cannot drop $p$ , so we keep both equations and eliminate $p$ .
Take the physical root (positive): $n = \frac{N}{2} + (\frac{N}{2})^{2} + n_{i}^{2} .$ Plug in: $\frac{N}{2} = 1.5 \times 1 0^{10}$ , $(N /2)^{2} = 2.25 \times 1 0^{20}$ , $n_{i}^{2} = 4 \times 1 0^{20}$ . $n = 1.5 \times 1 0^{10} + 6.25 \times 1 0^{20} = 1.5 \times 1 0^{10} + 2.5 \times 1 0^{10} = 4.0 \times 1 0^{10} cm^{- 3} .$ Why the $+$ root? Concentrations must be positive; the $-$ root is negative and unphysical.
Minority. $p = n_{i}^{2} / n = 4 \times 1 0^{20} /4 \times 1 0^{10} = 1.0 \times 1 0^{10} cm^{- 3}$ . Why this step? Mass action closes the problem.

Verify: $n - p = 4 \times 1 0^{10} - 1 \times 1 0^{10} = 3 \times 1 0^{10} = N$ ✅ (neutrality) and $n p = 4 \times 1 0^{20} = n_{i}^{2}$ ✅. The naive $n \approx N_{D} = 3 \times 1 0^{10}$ would be 33 % too low — this cell earns the quadratic.

Example 5 — Cell E: compensation (both dopants large, nearly equal)

Worked example Fighting dopants

Silicon: $N_{D} = 1.00 \times 1 0^{16}$ , $N_{A} = 9.90 \times 1 0^{15}$ . Find $n, p$ .

Forecast: with both dopings near $1 0^{16}$ , is $n$ near $1 0^{16}$ — or much smaller?

Net doping. $N = N_{D} - N_{A} = 1.00 \times 1 0^{16} - 0.990 \times 1 0^{16} = 1.0 \times 1 0^{14}$ . Why this step? Neutrality only cares about the difference; the donors and acceptors mostly cancel.
Compare. $N = 1 0^{14} ≫ n_{i} = 1 0^{10}$ still holds (four orders). So despite compensation, cell A shortcut is valid on the net value. Why this step? Compensation shrinks $N$ , but here it is still huge vs $n_{i}$ .
Carriers. $n \approx N = 1.0 \times 1 0^{14}$ , and $p = n_{i}^{2} / n = 1 0^{20} /1 0^{14} = 1.0 \times 1 0^{6} cm^{- 3}$ . Why this step? Standard heavy-doping shortcut, on the reduced $N$ .

Verify: $n p = 1 0^{14} \cdot 1 0^{6} = 1 0^{20} = n_{i}^{2}$ ✅. Lesson: compensation lowers the effective doping — $n$ dropped from $1 0^{16}$ (Ex 1) to $1 0^{14}$ purely because acceptors ate donors. This is why real crystals report net doping.

Example 6 — Cell F: non-equilibrium (illumination)

Worked example Shining light on it

A silicon sample sits in the dark with $n_{0} = 1 0^{16}$ , $p_{0} = 1 0^{4}$ (from Ex 1). Light injects $Δ n = Δ p = 2 \times 1 0^{14}$ excess pairs. Compute $n p$ and the quasi-Fermi splitting $F_{n} - F_{p}$ .

Forecast: will $n p$ still equal $n_{i}^{2}$ ? Bigger or smaller?

New carrier counts. $n = n_{0} + Δ n = 1 0^{16} + 2 \times 1 0^{14} \approx 1.02 \times 1 0^{16}$ ; $p = p_{0} + Δ p = 1 0^{4} + 2 \times 1 0^{14} \approx 2 \times 1 0^{14}$ . Why this step? Excess pairs add equally to both, but the tiny $p_{0}$ gets swamped.
Product. $n p = (1.02 \times 1 0^{16}) (2 \times 1 0^{14}) = 2.04 \times 1 0^{30}$ . Why this step? Compute directly — mass action no longer pins it.
Compare to $n_{i}^{2} = 1 0^{20}$ . $n p / n_{i}^{2} = 2.04 \times 1 0^{30} /1 0^{20} = 2.04 \times 1 0^{10} ≫ 1$ . So $n p ≫ n_{i}^{2}$ : the law is broken (as it must be out of equilibrium — see Quasi-Fermi levels & non-equilibrium carriers). Why this step? Injection means generation $>$ recombination locally, lifting the product.
Quasi-Fermi split. From $n p = n_{i}^{2} e^{(F_{n} - F_{p}) / k T}$ : $F_{n} - F_{p} = k T ln \frac{n p}{n _{i}^{2}} = 0.02585 ln (2.04 \times 1 0^{10}) = 0.02585 \times 23.74 \approx 0.614 eV .$ Why this step? The split measures how far from equilibrium we are; it's the log of the product's excess.

Verify: exponentiate back: $n_{i}^{2} e^{0.614/0.02585} = 1 0^{20} \cdot e^{23.74} = 1 0^{20} \cdot 2.04 \times 1 0^{10} = 2.04 \times 1 0^{30} = n p$ ✅.

Example 7 — Cell G: change the temperature

Worked example Heating silicon from 300 K to 400 K

At 300 K, $n_{i} = 1.0 \times 1 0^{10}$ . Using $n_{i}^{2} = N_{C} N_{V} e^{- E_{g} / k T}$ with $E_{g} = 1.12$ eV (treat $N_{C} N_{V}$ and $E_{g}$ as constant), find $n_{i}$ at 400 K, then $p$ in a sample with $N_{D} = 1 0^{16}$ at 400 K.

Forecast: $n_{i}$ grows steeply with $T$ — order-of-magnitude guess for the increase?

Ratio trick. Since $N_{C} N_{V}$ cancels in a ratio (ignoring its mild $T^{3}$ dependence for this estimate): $\frac{n _{i} ( T _{2} ) ^{2}}{n _{i} ( T _{1} ) ^{2}} = exp [- \frac{E _{g}}{k} (\frac{1}{T _{2}} - \frac{1}{T _{1}})] .$ Why this step? The prefactor drops out, isolating the dominant exponential — the reason $n_{i}$ is so $T$ -sensitive (see Band gap $E_g$ and its temperature dependence).
Plug numbers. $E_{g} / k = 1.12/8.617 \times 1 0^{- 5} = 1.300 \times 1 0^{4}$ K. Then $\frac{1}{400} - \frac{1}{300} = - 8.333 \times 1 0^{- 4} K^{- 1}$ . Exponent $= - 1.300 \times 1 0^{4} \times (- 8.333 \times 1 0^{- 4}) = + 10.83$ . So $\frac{n _{i} ( 400 ) ^{2}}{n _{i} ( 300 ) ^{2}} = e^{10.83} = 5.06 \times 1 0^{4}$ . Why this step? The exponent is large and positive ⇒ huge growth.
New $n_{i}$ . $n_{i} (400) = 1 0^{10} 5.06 \times 1 0^{4} = 1 0^{10} \times 225 \approx 2.25 \times 1 0^{12} cm^{- 3}$ . Why this step? Square-root because we scaled $n_{i}^{2}$ ; recall "square the concentration ⇒ halve the gap."
Minority hole at 400 K. Still $n \approx N_{D} = 1 0^{16}$ (since $n_{i} = 2.25 \times 1 0^{12} ≪ 1 0^{16}$ ), so $p = \frac{n _{i}^{2}}{N _{D}} = \frac{( 2.25 \times 1 0 ^{12} ) ^{2}}{1 0 ^{16}} = \frac{5.06 \times 1 0 ^{24}}{1 0 ^{16}} = 5.06 \times 1 0^{8} cm^{- 3} .$ Why this step? Mass action with the new, hotter $n_{i}$ .

Verify: $p$ jumped from $1 0^{4}$ (300 K, Ex 1) to $5.06 \times 1 0^{8}$ — a factor $\sim 5 \times 1 0^{4}$ , exactly the $n_{i}^{2}$ growth ✅. Heating floods the minority band even though majority barely moves.

Example 8 — Cell H: exam-style word twist

Worked example "Design a p-type wafer"

You must fabricate p-type silicon whose minority electron concentration is exactly $n = 1.0 \times 1 0^{5} cm^{- 3}$ at room temperature. What acceptor doping $N_{A}$ do you use (assume $N_{A} ≫ n_{i}$ , no donors)?

Forecast: invert the usual problem — you're given the minority and want the doping.

Identify the regime. Want $n$ tiny ( $1 0^{5} ≪ n_{i} = 1 0^{10}$ ) ⇒ heavily doped, cell B. Majority = holes $\approx N_{A}$ . Why this step? Small minority means large majority; picks cell B.
Use mass action backwards. $n = n_{i}^{2} / p = n_{i}^{2} / N_{A} \Rightarrow N_{A} = n_{i}^{2} / n$ . Why this step? We know the product and one factor, solve for the majority.
Compute. $N_{A} = \frac{( 1 0 ^{10} ) ^{2}}{1 0 ^{5}} = \frac{1 0 ^{20}}{1 0 ^{5}} = 1.0 \times 1 0^{15} cm^{- 3}$ . Why this step? Direct division; check $N_{A} = 1 0^{15} ≫ n_{i}$ ✅ so cell B assumption holds.

Verify: with $N_{A} = 1 0^{15}$ , $p \approx 1 0^{15}$ and $n = n_{i}^{2} / p = 1 0^{20} /1 0^{15} = 1 0^{5}$ ✅ — exactly the target minority. Self-consistent.

Recall Quick self-test across the matrix

Which cell is $N_{D} = 1 0^{17}$ , $N_{A} = 9.9 \times 1 0^{16}$ , $n_{i} = 1 0^{10}$ ? ::: Cell E (compensation): net $N = 1 0^{15} ≫ n_{i}$ , so it collapses to cell A on the net value; $n \approx 1 0^{15}$ . If $n p$ measured $= 3 n_{i}^{2}$ , what does it mean? ::: Non-equilibrium (cell F): excess carriers present, quasi-Fermi split $F_{n} - F_{p} = k T ln 3 > 0$ . Undoped and perfectly compensated silicon — same $n$ ? ::: Yes, both give $n = p = n_{i}$ (cell C); only mobility differs. When must you abandon $n \approx N$ and use the quadratic? ::: When $∣ N ∣$ is not $≫ n_{i}$ (cell D), i.e. within a small factor of $n_{i}$ .

Connections

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