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

Step 1 — Draw the stage: two energy bands and a gap

WHAT. Before any formula, we draw the world the electrons live in. Energy runs upward on the vertical axis. Two horizontal shelves are marked:

$E_{C}$ — the bottom edge of the conduction band, the lowest energy a free electron can have.
$E_{V}$ — the top edge of the valence band, the highest energy a bound electron sits at.

Between them is a forbidden zone of width $E_{g} = E_{C} - E_{V}$ , the band gap (see [[Band gap $E_{g}$ and its temperature dependence]]).

WHY. Every quantity later ( $n$ , $p$ , the exponentials) is a distance measured on this vertical energy axis. If we don't fix the axis first, the exponents have no meaning.

PICTURE. Look at the amber gap in the figure. An electron on the lower shelf ( $E_{V}$ ) must be lifted a height $E_{g}$ to reach the upper shelf ( $E_{C}$ ). That "lift height" is the whole story.

Step 2 — The heat budget: where the Fermi level sits

WHAT. We add one more horizontal line, the Fermi level $E_{F}$ . Roughly, $E_{F}$ is the energy at which a state is 50% likely to be occupied — the "water line" of the electron sea (full treatment in Fermi level & Fermi–Dirac statistics).

WHY. The number of electrons that make it up to $E_{C}$ depends on how far $E_{C}$ is above this water line. The larger the distance $E_{C} - E_{F}$ , the rarer a filled state up there. So we must mark $E_{F}$ before we can count anyone.

PICTURE. The dashed cyan water line sits inside the gap. Two labelled gaps appear:

the distance $E_{C} - E_{F}$ (how far electrons must climb above the water line),
the distance $E_{F} - E_{V}$ (how far below the water line the holes hang).

Notice they share $E_{F}$ — that shared line is the pivot everything will hinge on.

Step 3 — Count the electrons: the conduction-band population

WHAT. The number of free electrons per cm³ is $n = how many seats near E_{C} N_{C} \times chance a seat that high is filled e^{- (E_{C} - E_{F}) / k T}$

Reading it left to right:

$N_{C}$ — the effective density of states at the conduction edge: think of it as the count of available seats bunched near $E_{C}$ (see Effective density of states $N_C$, $N_V$).
$E_{C} - E_{F}$ — the climb height from Step 2.
$k T$ — the thermal energy scale: $k$ is Boltzmann's constant, $T$ the absolute temperature. It sets how big a climb heat can pay for.
$e^{- (E_{C} - E_{F}) / k T}$ — the Boltzmann factor: a fraction between 0 and 1 saying "what share of the climb-height can thermal energy afford?"

WHY this tool — the exponential? Because the chance of a thermal kick of size $Δ E$ dies off exponentially with $Δ E / k T$ . Not linearly, not as a power — exponentially. That's the defining behaviour of random thermal energy, and it's why doubling temperature barely changes the picture but doubling the gap changes it enormously.

PICTURE. The blue shaded blob above $E_{C}$ is the electron population. Its area is $n$ . Watch it shrink as the arrow $E_{C} - E_{F}$ grows — bigger climb, tinier blob.

Step 4 — Count the holes: the valence-band population

WHAT. A hole is an empty valence seat — a missing electron. Its population is the mirror image: $p = seats near E_{V} N_{V} \times chance a seat that low is empty e^{- (E_{F} - E_{V}) / k T}$

$N_{V}$ — effective density of states at the valence edge (seats bunched near $E_{V}$ ).
$E_{F} - E_{V}$ — the drop from the water line down to the valence shelf, from Step 2.
$e^{- (E_{F} - E_{V}) / k T}$ — Boltzmann factor for emptiness this far below $E_{F}$ .

WHY. Same physics as Step 3, reflected. Electrons are counted above the water line; holes are counted below it. The signs in the two exponents therefore come out opposite: electrons carry $+ E_{F}$ , holes carry $- E_{F}$ . Remember that opposite sign — it is the entire trick of Step 5.

PICTURE. The amber blob below $E_{V}$ is the hole population; its area is $p$ . It's the up-down mirror of the blue blob.

Step 5 — Multiply, and watch the water line vanish

WHAT. Multiply the two counts. Exponents add: $n p = N_{C} N_{V} e^{- (E_{C} - E_{F}) / k T} e^{- (E_{F} - E_{V}) / k T}$ In the combined exponent, $+ E_{F}$ from the electron term and $- E_{F}$ from the hole term cancel: $- (E_{C} - E_{F}) - (E_{F} - E_{V}) = - (E_{C} - E_{V}) = - E_{g}$ $n p = N_{C} N_{V} e^{- E_{g} / k T}$

Term by term in the result:

$N_{C} N_{V}$ — pure material seat-counts, no doping in sight.
$e^{- E_{g} / k T}$ — depends only on the fixed gap height and temperature.
$E_{F}$ — gone.

WHY this matters. Doping is the act of sliding the water line $E_{F}$ up (donors) or down (acceptors), see Doping: donors and acceptors. Step 5 shows $E_{F}$ cancels — so no amount of sliding changes the product $n p$ . That is the mass action law.

PICTURE. The two exponential arrows point in opposite directions from $E_{F}$ ; when laid head-to-tail they span exactly $E_{g}$ , and the $E_{F}$ line drops out of the sum. The residual span is the amber gap arrow — doping-independent.

Step 6 — Name the constant: evaluate it in pure crystal

WHAT. In a pure (intrinsic) crystal there are no dopants, so every electron came from breaking one bond, leaving one hole: $n = p$ . Call this common value $n_{i}$ , the intrinsic carrier concentration (see [[Intrinsic carrier concentration $n_{i}$ ]]). Then $n p = n_{i} \cdot n_{i} = n_{i}^{2}$ , and comparing with Step 5: $n_{i}^{2} = N_{C} N_{V} e^{- E_{g} / k T} ⟹ n p = n_{i}^{2}$ Taking the square root (the whole right side, so the exponent halves): $n_{i} = N_{C} N_{V} e^{- E_{g} /2 k T}$

WHY. The constant on the right of Step 5 was some number depending only on the material. Step 6 gives that number a physical name by measuring it in the one situation where $n$ and $p$ are equal.

PICTURE. The pure crystal panel: blue and amber blobs are the same size ( $n = p = n_{i}$ ), and $E_{F}$ sits mid-gap. The product of the two equal areas is $n_{i}^{2}$ .

Step 7 — The seesaw: what doping does to $n$ and $p$

WHAT. Since the product is pinned, $n$ and $p$ trade off: $p = \frac{n _{i}^{2}}{n}$ Dope n-type with $N_{D}$ donors ( $N_{D} ≫ n_{i}$ ): charge neutrality (see Charge neutrality condition) gives $n \approx N_{D}$ , so $p \approx \frac{n _{i}^{2}}{N _{D}}$

For silicon at room temperature ( $n_{i} = 1 0^{10} cm^{- 3}$ ), doping $N_{D} = 1 0^{16}$ :

$n \approx 1 0^{16}$ /cm³ (majority, tracks doping),
$p = \frac{( 1 0 ^{10} ) ^{2}}{1 0 ^{16}} = \frac{1 0 ^{20}}{1 0 ^{16}} = 1 0^{4}$ /cm³ (minority, collapses),
product $n p = 1 0^{16} \cdot 1 0^{4} = 1 0^{20} = (1 0^{10})^{2}$ . ✅

WHY. This is the payoff picture: raising one blob's area forces the other's area down by the same factor, because their product is a constant.

PICTURE. A seesaw. Left seat $n$ rises, right seat $p$ falls; a rigid bar labelled " $n p = n_{i}^{2}$ " pins the pivot. The areas trade, the product holds.

Step 8 — Every regime at once (the degenerate cases)

WHAT. To cover all doping levels — including the awkward middle where doping $\approx n_{i}$ — substitute $p = n_{i}^{2} / n$ into neutrality $n - p = N$ (net doping $N = N_{D} - N_{A}$ ): $n^{2} - N n - n_{i}^{2} = 0 ⟹ n = \frac{N}{2} + (\frac{N}{2})^{2} + n_{i}^{2}$

Check the extreme cases so no reader is ever stranded:

Heavy n-type $N ≫ n_{i}$ : root $\to N /2$ , so $n \to N$ . Majority = doping. ✓
Undoped $N = 0$ : $n = n_{i}^{2} = n_{i}$ . Pure crystal recovered. ✓
Heavy p-type $N < 0$ (i.e. $N_{A} ≫ N_{D}$ ): the $+$ root gives a small positive $n$ , and $p = n_{i}^{2} / n$ becomes large ≈ $∣ N ∣$ . The seesaw simply tips the other way. ✓
Zero temperature limit $T \to 0$ : $e^{- E_{g} /2 k T} \to 0$ , so $n_{i} \to 0$ and both minority populations vanish — carriers freeze out.

WHY. One quadratic, four physical corners. The single formula is continuous across all of them, which is exactly what "covers every case" means.

PICTURE. A curve of $n$ versus net doping $N$ (log axes for the amber tails, linear near zero). It hugs $n = ∣ N ∣$ far out on both sides and bottoms out gently at $n = n_{i}$ when $N = 0$ — the smooth floor set by the intrinsic value.

The one-picture summary

Everything on one canvas: the two bands with $E_{F}$ between them, the electron arrow up (carrying $+ E_{F}$ ) and the hole arrow down (carrying $- E_{F}$ ) meeting head-to-tail to span $E_{g}$ , the cancelled $E_{F}$ crossed out, and the seesaw at the side showing $n$ and $p$ trading while the bar $n p = n_{i}^{2}$ stays level.

Recall Feynman retelling of the whole walkthrough

We drew two shelves — a high one (conduction, $E_{C}$ ) and a low one (valence, $E_{V}$ ) — with a forbidden gap $E_{g}$ between them. A dashed "water line" $E_{F}$ sits in the gap; how high above it the top shelf is decides how few electrons climb up there, and how far below it the bottom shelf is decides how many empty spots (holes) sit down there. Both counts fall off exponentially with distance from the water line — that's just how heat spreads energy. Here's the trick: electrons count the distance going up from the line and holes count the distance going down, so when you multiply the two counts, the water line's position cancels out. What's left only knows the fixed gap $E_{g}$ and the temperature. Since doping only slides the water line, the product $n p$ can't feel it — push electrons up and holes drop to keep $n p$ nailed to the same number, $n_{i}^{2}$ . We named that number by looking at pure crystal (where $n = p = n_{i}$ ), then wrote one quadratic that handles pure, n-type, p-type and frozen-out crystals in a single stroke.

Recall Quick self-test

Why does $E_{F}$ cancel when you multiply $n$ and $p$ ? ::: Electrons carry $e^{+ E_{F} / k T}$ (measured up from $E_{F}$ ), holes carry $e^{- E_{F} / k T}$ (measured down); the product's exponents add and $E_{F}$ subtracts to zero, leaving $e^{- E_{g} / k T}$ . In the summary picture, what does the amber gap arrow represent? ::: The surviving $E_{g}$ span after $E_{F}$ cancels — the only energy left in the product $n p = N_{C} N_{V} e^{- E_{g} / k T}$ . Set net doping $N = 0$ in Step 8's formula — what is $n$ ? ::: $n = n_{i}^{2} = n_{i}$ , the pure-crystal value.

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