P-type doping with acceptor atoms (boron): Question bank

Definition Symbol glossary — read this first

Every symbol used on this page, in plain words, so the page stands on its own:

$n$ — the electron concentration (free electrons per cm³).
$p$ — the hole concentration (mobile "missing-electron" spots per cm³).
$n_{i}$ — the intrinsic carrier concentration: the value of both $n$ and $p$ in pure silicon, set only by temperature and band gap ( $n_{i} \approx 1.5 \times 1 0^{10} cm^{- 3}$ for Si at room temperature).
$N_{A}$ — the total number of acceptor (boron) atoms you put in, per cm³.
$N_{A}^{-}$ — the number of those acceptors that are actually ionized (have grabbed an electron and become $B^{-}$ ). Only these have released a hole. At room temperature $N_{A}^{-} \approx N_{A}$ ; when cold, $N_{A}^{-} < N_{A}$ .
$N_{D}, N_{D}^{+}$ — the same two quantities for donor (phosphorus) atoms in n-type material.
$q$ — the magnitude of the electron charge, $1.6 \times 1 0^{- 19} C$ .
$μ_{p}, μ_{n}$ — the mobility of holes and of electrons: how fast each drifts per unit electric field ( $cm^{2} / Vs$ ). In silicon $μ_{p} \approx 480$ , $μ_{n} \approx 1350$ .
$E_{F}$ — the Fermi level: the energy at which a state has a 50% chance of being occupied. Its position between the valence and conduction band tells you whether holes or electrons dominate.
$E_{V}, E_{C}$ — the top of the valence band and the bottom of the conduction band.

Figure — P-type doping with acceptor atoms (boron)

Two short derivations you will lean on below

The traps repeatedly invoke two facts. Rather than send you off-page, here is the why in miniature.

True or false — justify

A p-type silicon block, sitting on your desk, carries a net positive charge.

False. Every mobile positive hole is matched by a fixed negative acceptor ion (

B^{-}

, counted as

N_{A}^{-}

), so the total charge is zero — only the mobile population is positive.

Boron is called an acceptor because it donates a hole to the crystal.

False naming logic. It is called acceptor because it accepts an electron from a neighbouring bond; the hole is what's left behind, not what boron "gives out."

In p-type silicon, electrons no longer exist.

False. Electrons are still there as minority carriers at concentration

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

— tiny, but never zero. The mass-action law

n p = n_{i}^{2}

forbids

n = 0

for any finite

p

At room temperature, essentially all boron atoms are ionized.

True. Boron's acceptor level sits just above the valence band, so thermal energy (

k T

) easily promotes valence electrons into it — hence

N_{A}^{-} \approx N_{A}

The product

n p

increases when you add more boron.

False.

n p = n_{i}^{2}

depends only on temperature and band gap (it is

G / r

), not on doping. Adding boron raises

p

and lowers

n

by exactly the same factor.

A hole is a real particle that boron injects into the crystal.

False. A hole is the absence of an electron in a bond — an emergent quasiparticle that behaves like a positive charge, not a physical object boron carries in.

Doping silicon with boron makes it a better conductor than intrinsic silicon.

True. Intrinsic Si has only

\sim n_{i}

carriers; boron supplies

\sim N_{A} ≫ n_{i}

mobile holes, raising conductivity

σ = q (n μ_{n} + p μ_{p}) \approx q p μ_{p}

by many orders of magnitude.

If you heat p-type silicon high enough, it eventually behaves like intrinsic silicon.

True. At very high

T

, thermally generated pairs make

n_{i}

grow until

n_{i} ≳ N_{A}

; then

n \approx p

again and the dopant no longer dominates.

Spot the error

"Boron has one extra proton, and that proton is the positive hole."

The hole is a missing electron in a bond, never a proton. Boron simply has 3 valence electrons instead of 4; the shortfall of one bond-electron is the hole.

"Charge neutrality is

p = n + N_{A}

, using the total boron count."

Strictly it is

p = n + N_{A}^{-}

: only the ionized acceptors contribute a fixed negative charge. Using

N_{A}

is a shortcut valid only when

N_{A}^{-} \approx N_{A}

(room temperature); during freeze-out

N_{A}^{-} < N_{A}

and the distinction matters.

"Charge neutrality gives

p = n + N_{A}

, so if

N_{A} = 0

then

p = 0

Wrong conclusion. With

N_{A} = 0

you get intrinsic silicon:

p = n = n_{i}

, not zero. The neutrality equation still holds because

p = n

satisfies it.

"Since holes are the majority, current in p-type is carried only by holes and electrons carry nothing."

Electrons do carry a current, it is just negligible: from

σ = q n μ_{n} + q p μ_{p}

, the electron term is about

1 0^{- 12}

of the hole term (see the derivation above). "Negligible" is not "zero."

"Approximation

p \approx N_{A}

is always valid because

N_{A}

is huge."

Only valid when

N_{A} ≫ n_{i}

. For lightly doped Si (

N_{A}

near

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

) you must use

p = \frac{1}{2} (N_{A} + N_{A}^{2} + 4 n_{i}^{2})

, else you get ~30% error.

"When boron ionizes it becomes

B^{+}

because it lost something."

It becomes

B^{-}

. Boron gains an electron to complete its fourth bond, so it acquires a net negative charge; the released mobile hole carries the

+ 1

"The acceptor level lies just below the conduction band."

It lies just above the valence band (

E_{V}

). That closeness to

E_{V}

is exactly why a valence electron needs only a tiny energy to jump up and create a free hole.

"Adding boron pushes the Fermi level up toward the conduction band."

In p-type material

E_{F}

shifts down toward the valence band

E_{V}

, because holes (empty states near the top of the valence band) dominate. See Fermi Level in Doped Semiconductors and the band figure above.

Why questions

Why does an "empty" bond let charge flow, when nothing positive was added?

A neighbouring electron slides in to fill the empty bond, moving the emptiness elsewhere. A travelling absence-of-electron is mathematically identical to a positive charge moving the opposite way — that is conduction.

Why is

n p = n_{i}^{2}

independent of how much boron you add?

Pair generation rate

G

depends only on temperature/band gap; recombination rate is

r n p

. Equilibrium

G = r n p

forces

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

, a constant. Doping only redistributes carriers between

n

and

p

. (Detail in Mass-Action Law and Carrier Statistics.)

Why is the crystal neutral even though we deliberately added positive carriers?

For every free hole (+1) created, boron kept an extra electron and became a fixed ion (

N_{A}^{-}

, charge −1). The bookkeeping is exact: mobile

+ 1

+ fixed

- 1 = 0

Why do we take only the positive root of

p^{2} - N_{A} p - n_{i}^{2} = 0

A concentration cannot be negative. The negative root gives

p < 0

, which is physically meaningless, so it is discarded.

Why does n-type doping do the opposite to p-type?

A Group V donor (phosphorus) has one extra valence electron, releasing a mobile electron instead of a hole — making electrons the majority. It is the mirror image, see N-type doping with donor atoms (phosphorus).

Why must holes and electrons both be tracked in a pn junction even though one is a minority on each side?

At the junction the minority carriers are precisely what diffuse across and drive the diffusion current; ignoring them would erase the entire physics of the junction.

Why does the conductivity formula use hole mobility

μ_{p}

, not electron mobility?

Because holes carry essentially all the current in p-type. Starting from

σ = q n μ_{n} + q p μ_{p}

, the ratio

\frac{n}{p} \frac{μ _{n}}{μ _{p}} \sim 1 0^{- 12}

, so only

q p μ_{p}

survives.

Edge cases

What is

p

when

N_{A} = 0

(undoped silicon)?

The formula collapses to

p = \frac{1}{2} (0 + 4 n_{i}^{2}) = n_{i}

, correctly reproducing the intrinsic case where

n = p = n_{i}

. See Intrinsic Semiconductors.

What happens to

p

at very low temperature ("freeze-out"), quantitatively?

Thermal energy can no longer ionize every acceptor, so

N_{A}^{-} < N_{A}

. Charge neutrality becomes

p = n + N_{A}^{-}

, so the quadratic changes to

p^{2} - N_{A}^{-} p - n_{i}^{2} = 0

with the reduced coefficient

N_{A}^{-}

. Its positive root is smaller, hence

p < N_{A}

; the clean

p \approx N_{A}

result fails.

What if

N_{A}

is exactly equal to

n_{i}

Then

N_{A}^{2} + 4 n_{i}^{2} = 5 n_{i} \approx 2.24 n_{i}

, giving

p \approx 1.62 n_{i}

— the simple

p \approx N_{A}

would be off by ~38%, so the full quadratic is mandatory here.

What happens if you dope with both boron and phosphorus in equal amounts (

N_{A} = N_{D}

They compensate exactly: in

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

the ionized-dopant terms cancel, leaving

p = n = n_{i}

— the material behaves intrinsic despite being heavily doped.

In the limit of extremely heavy doping (degenerate p-type), does

n p = n_{i}^{2}

still hold, and what is the criterion?

No. The threshold is roughly when

E_{F}

moves to within about

3 k T

of the valence band edge

E_{V}

(i.e.

E_{V} - E_{F} ≲ 3 k T

), which happens near

N_{A} \sim 1 0^{19} cm^{- 3}

. Then the Boltzmann approximation fails, the density of states must be integrated with full Fermi–Dirac occupancy, and

n p = n_{i}^{2}

no longer applies.

What is the minority electron concentration when

N_{A} = n_{i}

exactly?

Since

n p = n_{i}^{2}

and

p \approx 1.62 n_{i}

, we get

n = n_{i}^{2} / p \approx 0.62 n_{i}

— comparable to

p

, so "majority/minority" language barely applies at this weak doping.

Connections

Intrinsic Semiconductors — the $N_{A} \to 0$ limit these traps keep returning to.
N-type doping with donor atoms (phosphorus) — the mirror-image traps.
Mass-Action Law and Carrier Statistics — the source of the "why is $n p$ fixed" answers.
Fermi Level in Doped Semiconductors — where the level-direction trap is resolved.
PN Junction Formation — why minority carriers still matter.
Drift and Diffusion Currents — conductivity and carrier-motion traps.