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

Definition Every symbol on this page (keep this open while you work)

$n$ = free electron concentration in the conduction band (number per cm³).
$p$ = free hole concentration (empty valence states) per cm³.
$n_{i}$ = intrinsic carrier concentration, the common value $n = p$ in a pure crystal — see Intrinsic carrier concentration $n_i$.
$N_{D}$ = donor concentration (atoms that each donate one electron); $N_{A}$ = acceptor concentration (each accepts one electron, creating a hole) — see Doping: donors and acceptors.
$N = N_{D} - N_{A}$ = net doping.
$N_{C}$ , $N_{V}$ = effective density of states at the conduction / valence band edge — the "number of available seats" near each band — see [[Effective density of states $N_{C}$ , $N_{V}$ ]].
$E_{g}$ = band gap, the energy an electron must gain to jump valence → conduction — see [[Band gap $E_{g}$ and its temperature dependence]].
$E_{C}$ , $E_{V}$ = the energies of the conduction-band bottom and valence-band top; $E_{F}$ = the Fermi level, the energy that marks the 50% filling probability — see Fermi level & Fermi–Dirac statistics.
$k$ = Boltzmann's constant, $T$ = absolute temperature (kelvin); $k T$ = the thermal energy scale, "how hard heat kicks."

The single rule under attack all through this page: $n p = n_{i}^{2} (thermal equilibrium, non-degenerate only)$

Every trap below pokes at one of the words in that parenthesis: thermal equilibrium, non-degenerate, or the meaning of the product.

Intuition Why a

product, not a sum — the collision picture Look at the figure above. Heat constantly kicks electrons up onto the stage (conduction band), creating a pair: one electron above, one hole below. That is generation, and it only needs heat and a filled seat below, so its rate is $G = G (T)$ — independent of how crowded things already are. To recombine, one electron and one hole must physically meet. The chance of a meeting per unit time is proportional to how many electrons times how many holes — like the chance two randomly walking dancers bump: double the electrons and each hole has twice the chance of a hit; double the holes too and you double again. So the recombination rate is $R = r n p,$ where $r$ (units: cm³/s) is the recombination coefficient — a fixed material constant that simply converts "how often a meeting leads to an annihilation." It is the proportionality factor turning the count of possible pairings $n p$ into an actual rate. At equilibrium generation balances recombination, $G (T) = r n p$ , so $n p = \frac{G ( T )}{r} = const (T),$ and evaluating in pure material ( $n = p = n_{i}$ ) names that constant $n_{i}^{2}$ . The and in "an electron and a hole must meet" is exactly why it is a product.

True or false — justify

Doubling the donor concentration

N_{D}

doubles the total number of free carriers

n + p

Roughly true for the total, because the majority electrons track

N_{D}

— but false if you meant both carriers rise: the minority

p = n_{i}^{2} / N_{D}

actually halves. The product is conserved, so one climbs only as the other collapses.

n p = n_{i}^{2}

can be used to find

n

and

p

separately in a doped crystal.

False on its own — it is one equation with two unknowns. You must pair it with charge neutrality

n + N_{A} = p + N_{D}

(see Charge neutrality condition) to pin both values.

In a heavily n-type sample there are literally no holes.

False — holes never reach zero.

p = n_{i}^{2} / n

is tiny but strictly positive; generation always keeps a trickle of holes present at equilibrium.

n p = n_{i}^{2}

holds inside a solar cell that is generating current in sunlight.

False — sunlight injects excess pairs, breaking equilibrium, so

n p > n_{i}^{2}

. You then use quasi-Fermi levels:

n p = n_{i}^{2} e^{(F_{n} - F_{p}) / k T}

The mass action law depends on where the Fermi level

E_{F}

sits.

False, and this is the whole point — multiplying

n

and

p

cancels

E_{F}

, leaving only

N_{C}

N_{V}

E_{g}

T

. Doping moves

E_{F}

but cannot move the product.

n_{i}

is a property of the individual sample, so two silicon wafers can have different

n_{i}

at the same temperature.

False —

n_{i}

depends only on the material and temperature, not on doping. Both wafers share the same

n_{i}

; only their

n

and

p

split differently.

If a semiconductor is intrinsic then

n = p

; if

n = p

then the semiconductor is intrinsic.

The first implication is true by definition of intrinsic. The second is false — a compensated sample with

N_{D} = N_{A}

is doped yet still has

n = p = n_{i}

. Equal carriers does not prove purity.

T = 0 K

the mass action law still gives a meaningful non-zero product.

False in practice — as

T \to 0

e^{- E_{g} / k T} \to 0

, so

n_{i}^{2} \to 0

. Both

n

and

p

vanish; the crystal becomes a perfect insulator with no free carriers.

Raising temperature raises

n_{i}

but leaves the product

n p

in a doped sample unchanged.

False —

n p = n_{i}^{2}

still holds, but

n_{i}^{2}

itself grows with

T

, so the product rises. Eventually

n_{i}

can exceed

N_{D}

and the sample goes "intrinsic" again (loses its n-type character).

Spot the error

n_{i}^{2} = N_{C} N_{V} e^{- E_{g} / k T}

, therefore

n_{i} = N_{C} N_{V} e^{- E_{g} / k T}

The exponent must also be square-rooted:

e^{- E_{g} / k T} = e^{- E_{g} /2 k T}

. Correct is

n_{i} = N_{C} N_{V} e^{- E_{g} /2 k T}

— square-root the whole thing, gap halves.

"Recombination rate

R \propto n + p

because you need electrons and holes."

Wrong operator — a recombination event needs an electron and a hole to meet simultaneously, a two-body encounter, so

R = r n p

. This product is exactly why the product

n p

is what gets pinned.

"Generation

G

depends on how many carriers are already present, so

G \propto n p

too."

No — thermal generation is an electron being kicked across the gap by heat, needing only a filled valence state (essentially always available). So

G = G (T)

depends on temperature and gap, not on existing carrier counts.

n = N_{C} e^{- (E_{F} - E_{C}) / k T}

Sign flipped in the exponent. Since

E_{C} > E_{F}

for non-degenerate material, the correct form is

n = N_{C} e^{- (E_{C} - E_{F}) / k T}

, a number less than

N_{C}

— see Fermi level & Fermi–Dirac statistics.

"In n-type material the neutrality condition is

n = N_{D}

exactly."

Only approximate. The exact statement is

n + N_{A} = p + N_{D}

; you get

n \approx N_{D}

only when

N_{D} ≫ n_{i}

and

N_{D} ≫ N_{A}

. Near intrinsic doping you need the full quadratic.

"Since

n p = n_{i}^{2}

is constant, adding donors doesn't change conductivity."

The product is constant, but conductivity depends on the carriers themselves (mostly the majority). Adding donors raises

n

dramatically, so conductivity rises — the pinned product doesn't freeze conductivity.

Why questions

Why is it the product

n p

that is conserved and not the sum

n + p

Because recombination is a two-body meeting: its rate is

R = r n p

(each hole's chance of being hit scales with how many electrons roam, and vice-versa), while generation

G (T)

is heat-only. Balancing

G = r n p

pins the product; no physical process forces the sum.

Why does the Fermi level cancel when we multiply

n

and

p

Write them out:

n = N_{C} e^{- (E_{C} - E_{F}) / k T}

carries

+ E_{F}

in the exponent, while

p = N_{V} e^{- (E_{F} - E_{V}) / k T}

carries

- E_{F}

. Multiplying adds the exponents,

(E_{F} - E_{C}) + (E_{V} - E_{F}) = E_{V} - E_{C} = - E_{g}

, so

E_{F}

vanishes and

n p = N_{C} N_{V} e^{- E_{g} / k T}

Why must we insist on "non-degenerate" doping, and what does non-degenerate mean quantitatively?

Non-degenerate means the Fermi level stays at least about $3 k T$ inside the gap, i.e.

E_{C} - E_{F} ≳ 3 k T

and

E_{F} - E_{V} ≳ 3 k T

. Only then is the Boltzmann exponential a valid stand-in for Fermi–Dirac. Heavy doping pushes

E_{F}

into a band, so you must use full Fermi–Dirac statistics and

n p = n_{i}^{2}

picks up corrections.

Why is

n_{i}

so ferociously temperature-sensitive?

It carries the factor

e^{- E_{g} /2 k T}

; a small change in

T

in the denominator of a large exponent produces a huge multiplicative change, so

n_{i}

can double every few kelvin — see Band gap $E_g$ and its temperature dependence.

Why does the minority carrier concentration drop when we dope, even though we're adding stuff to the crystal?

We add donors (electrons), not holes. The extra electrons give every wandering hole vastly more partners to recombine with, so holes are consumed faster, driving

p

down to keep

n p = n_{i}^{2}

Why does illumination make

n p > n_{i}^{2}

rather than

< n_{i}^{2}

Light creates excess electron–hole pairs on top of the equilibrium population, pushing both

n

and

p

above their equilibrium values, so their product exceeds

n_{i}^{2}

. Equilibrium is the minimum-product state.

Why can a doped semiconductor "become intrinsic" at high temperature?

When

T

rises enough that

n_{i} > N_{D}

, thermally generated carriers swamp the fixed dopant contribution, so

n \approx p \approx n_{i}

and the doping is drowned out.

Edge cases

By product symmetry, if

n = \frac{N}{2} + (N /2)^{2} + n_{i}^{2}

, what is the matching closed form for

p

Since

p = n_{i}^{2} / n

and also

p = n - N

, the symmetric root is

p = - \frac{N}{2} + (N /2)^{2} + n_{i}^{2}

— the same square root, just with the sign of

N /2

flipped. Their product is

n_{i}^{2}

automatically.

What does the general solution

n = \frac{N}{2} + (N /2)^{2} + n_{i}^{2}

give when net doping

N = N_{D} - N_{A} = 0

The

N /2

term vanishes, leaving

n = n_{i}^{2} = n_{i}

— perfectly compensated material behaves exactly like intrinsic material,

n = p = n_{i}

What does that same formula give as

N \to + \infty

(heavy n-type)?

The square root

\approx N /2

, so

n \approx N /2 + N /2 = N

, recovering

n \approx N_{D}

. The general formula smoothly contains the simple n-type limit.

For a heavily p-type sample, does

n p = n_{i}^{2}

still hold, and which carrier collapses?

Yes, it always holds at equilibrium. Now holes are the majority (

p \approx N_{A}

) and electrons collapse to

n = n_{i}^{2} / N_{A}

— the seesaw simply tips the other way.

What is

n p

in a perfectly pure (intrinsic) crystal?

Exactly

n_{i}^{2}

, since

n = p = n_{i}

there. Intrinsic material is the special case where the two seesaw arms are level.

As doping

N_{D} \to n_{i}

(comparable, not dominant), can you still write

n \approx N_{D}

No — the approximation needs

N_{D} ≫ n_{i}

. When they're comparable, intrinsic generation is non-negligible and you must use the quadratic

n^{2} - N n - n_{i}^{2} = 0

If a sample is compensated with

N_{D} = N_{A} = 1 0^{18}

(both large), is it a good conductor?

No — net doping

N = 0

gives

n = p = n_{i}

, so despite

1 0^{18}

dopants it conducts like intrinsic material. The donors' electrons fall into the acceptors' empty states.

Recall One-line summary of every trap

Every trap here is really one of three: (1) confusing product with sum, (2) forgetting "thermal equilibrium" (light/current breaks it → quasi-Fermi levels), or (3) forgetting "non-degenerate" (heavy doping breaks Boltzmann). Keep those three guardrails and $n p = n_{i}^{2}$ never surprises you.

Connections

Mass action law (np = ni²) (index 2.1.7)
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

Question bank — Mass action law (np = ni²)

True or false — justify

Spot the error

Why questions

Edge cases

Connections

Concept Map