P-type doping with acceptor atoms (boron): Worked examples

Before we start, let us re-anchor every symbol so no notation is used before it is earned.

Definition The symbols, in plain words

$N_{A}$ = number of acceptor (boron) atoms per cubic centimetre we deliberately added. Units: $cm^{- 3}$ (things counted inside one cubic centimetre).
$N_{D}$ = number of donor (phosphorus) atoms per cm³. A donor is the mirror of an acceptor — it gives away a spare electron (see N-type doping with donor atoms (phosphorus)). For pure p-type there are none, so $N_{D} = 0$ ; it only matters in the compensation case (Cell H).
$n_{i}$ = intrinsic carrier concentration: how many free electrons (and equally, holes) pure undoped silicon makes on its own from heat alone. In silicon at room temperature $n_{i} \approx 1.5 \times 1 0^{10} cm^{- 3}$ . See Intrinsic Semiconductors.
$p$ = how many free holes per cm³ actually exist after doping.
$n$ = how many free electrons per cm³ actually exist after doping.
$q$ = the charge of one electron in magnitude, $q = 1.6 \times 1 0^{- 19}$ coulombs (C). We will only need $q$ once, in Cell G, where it converts a carrier count into an actual electric current — until then every equation is pure counting and $q$ does not appear.
$μ_{p}$ = hole mobility, units $cm^{2} / (V\cdots)$ — how fast (drift speed, cm/s) a hole moves per unit of electric push (field, V/cm). Bigger $μ_{p}$ = faster holes. In silicon $μ_{p} \approx 480 cm^{2} / (V\cdots)$ .
$μ_{n}$ = electron mobility, same units $cm^{2} / (V\cdots)$ . In silicon electrons are actually faster than holes ( $μ_{n} \approx 1350 cm^{2} / (V\cdots)$ vs $μ_{p} \approx 480$ ), which we will need in Cell G when we argue why minority electrons still carry negligible current.

Dimensional check for later: in $σ = q p μ_{p}$ the units multiply as $C \times cm^{- 3} \times cm^{2} / (V\cdots) = C / (V\cdots\cdotcm) = A / (V\cdotcm) = (Ω \cdotcm)^{- 1}$ , using $C/s = A$ and $A/V = Ω^{- 1}$ . So conductivity comes out in $(Ω \cdotcm)^{- 1}$ automatically — good to keep in mind before we ever plug numbers.

The two laws we lean on the whole way (both from the parent note):

The scenario matrix

Every doping problem is really one of these cells. The whole point of a "deep dive" is to fill every box, not just the easy one.

Cell	Regime	What controls $p$ ?	Which formula?	Example
A	Heavy doping $N_{A} ≫ n_{i}$	the dopant	$p \approx N_{A}$	Ex 1
B	Comparable $N_{A} \sim n_{i}$	dopant and heat	full quadratic	Ex 2
C	Zero doping $N_{A} = 0$	pure heat	$p = n_{i}$	Ex 3
D	Trace doping $N_{A} ≪ n_{i}$	mostly heat, tiny tilt	full quadratic + series expansion	Ex 4
E	Temperature raised (hot)	$n_{i}$ grows, may overtake $N_{A}$	recompute $n_{i}$ , then quadratic	Ex 5
F	Temperature lowered (freeze-out)	only some acceptors ionized	$N_{A}^{-} < N_{A}$	Ex 6
G	Real device: resistance/current	conductivity from $p$	$σ = q p μ_{p}$	Ex 7
H	Compensation, net p-type ( $N_{A} > N_{D}$ )	$N_{A} - N_{D}$	net-doping neutrality	Ex 8
I	Compensation, net n-type ( $N_{D} > N_{A}$ )	$N_{D} - N_{A}$	net-doping neutrality	Ex 9
J	Extreme doping (degenerate)	statistics break down	formulas above invalid	note in Cell J

The figure below is the map we will keep returning to. Figure 1 plots, on log–log axes, the hole count $p$ (purple) and electron count $n$ (coral) as we sweep $N_{A}$ (horizontal axis, $cm^{- 3}$ ) from far below $n_{i}$ to far above it; both vertical and horizontal axes are carrier concentrations in $cm^{- 3}$ . The green dashed line marks the $n_{i}$ level, and the shaded bands + text labels tell you exactly which cell and which example each region belongs to.

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

Figure 1 — Carrier concentrations $p$ and $n$ (both cm⁻³, vertical) versus acceptor doping $N_{A}$ (cm⁻³, horizontal), log–log. Mint band = Cells C/D (below $n_{i}$ ); butter band = Cell B bend ( $N_{A} \sim n_{i}$ ); lavender band = Cell A ( $N_{A} ≫ n_{i}$ ).

Cell A — Heavy doping (the everyday case)

Worked example Example 1 — Cell A

Silicon doped with $N_{A} = 1 0^{16} cm^{- 3}$ boron, $n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ . Find $p$ and $n$ . (This is the far-right end of Figure 1, where the purple line sits flat on $N_{A}$ .)

Forecast: Will $p$ be close to $1 0^{16}$ , or noticeably less? Guess before reading.

Compare sizes: $N_{A} / n_{i} = 1 0^{16} / (1.5 \times 1 0^{10}) \approx 6.7 \times 1 0^{5}$ . Why this step? The whole decision "shortcut vs quadratic" rests on whether $N_{A} ≫ n_{i}$ . Here it is 600,000× bigger — deep in Cell A.
So the $N_{A}^{2} + 4 n_{i}^{2}$ term is essentially $N_{A}^{2} = N_{A}$ , giving $p \approx N_{A} = 1 0^{16} cm^{- 3}$ . Why this step? $4 n_{i}^{2}$ is $\sim 1 0^{- 11}$ of $N_{A}^{2}$ — invisible under the square root.
$n = n_{i}^{2} / p = (1.5 \times 1 0^{10})^{2} /1 0^{16} = 2.25 \times 1 0^{20} /1 0^{16} = 2.25 \times 1 0^{4} cm^{- 3}$ . Why this step? Once $p$ is fixed, mass-action pins $n$ automatically.

Verify: Plug into neutrality: $n + N_{A} = 2.25 \times 1 0^{4} + 1 0^{16} \approx 1 0^{16} = p$ . ✓ The electron term is a rounding speck. Units: $cm^{- 3}$ throughout. ✓

Cell B — Comparable $N_{A} \sim n_{i}$ (shortcut breaks!)

Worked example Example 2 — Cell B

$N_{A} = 2 \times 1 0^{10} cm^{- 3}$ , so $N_{A}$ is barely above $n_{i} = 1.5 \times 1 0^{10}$ . Find $p$ and $n$ . (This is the bend in Figure 1, where purple and coral start to separate.)

Forecast: The shortcut says $p = 2 \times 1 0^{10}$ . Do you trust it here? By how much might it be off?

Do not drop $n_{i}$ . Compute the discriminant: $N_{A}^{2} + 4 n_{i}^{2} = (2 \times 1 0^{10})^{2} + 4 (1.5 \times 1 0^{10})^{2} = 4 \times 1 0^{20} + 9 \times 1 0^{20} = 1.3 \times 1 0^{21}$ . Why this step? $4 n_{i}^{2}$ is now bigger than $N_{A}^{2}$ — the term we would have thrown away dominates, so we keep it.
$1.3 \times 1 0^{21} = 3.606 \times 1 0^{10}$ . Why this step? The exact formula needs the square root of that discriminant; taking it turns the "sum of squares" back into a quantity with the same units ( $cm^{- 3}$ ) as $N_{A}$ , so it can be added to $N_{A}$ in the next step.
$p = (2 \times 1 0^{10} + 3.606 \times 1 0^{10}) /2 = 2.803 \times 1 0^{10} cm^{- 3}$ . Why this step? This is literally the quadratic-formula answer $p = \frac{1}{2} (N_{A} + \dots)$ — we average $N_{A}$ with the root we just found to get the one physical (positive) hole count.
$n = n_{i}^{2} / p = 2.25 \times 1 0^{20} /2.803 \times 1 0^{10} = 8.03 \times 1 0^{9} cm^{- 3}$ . Why this step? Mass-action fixes $n$ once $p$ is known.

Verify: neutrality check $n + N_{A} = 8.03 \times 1 0^{9} + 2 \times 1 0^{10} = 2.803 \times 1 0^{10} = p$ . ✓ The shortcut $p \approx N_{A} = 2 \times 1 0^{10}$ would be 28% too low — real lesson of Cell B.

Cell C — Zero doping (degenerate input)

Cell D — Trace doping $N_{A} ≪ n_{i}$ (limiting behaviour + series expansion)

Before crunching numbers, let us do the promised analytical expansion so we understand the shape of the answer when doping is tiny.

Worked example Example 4 — Cell D

$N_{A} = 3 \times 1 0^{9} cm^{- 3}$ , five times below $n_{i} = 1.5 \times 1 0^{10}$ . Find $p$ and $n$ both ways (exact and the expansion above), and show the material is barely p-type. (Figure 1: just right of the far-left merge — purple is a hair above coral.)

Forecast: Is $p$ closer to $N_{A}$ (tiny) or to $n_{i}$ (big)? Which carrier wins?

Exact route — discriminant: $N_{A}^{2} + 4 n_{i}^{2} = (3 \times 1 0^{9})^{2} + 4 (1.5 \times 1 0^{10})^{2} = 9 \times 1 0^{18} + 9 \times 1 0^{20} = 9.09 \times 1 0^{20}$ . Why this step? $N_{A}^{2}$ is 1% of $4 n_{i}^{2}$ — the crystal's own heat dominates, so we must keep the $n_{i}$ term.
$9.09 \times 1 0^{20} = 3.015 \times 1 0^{10}$ , so $p = (3 \times 1 0^{9} + 3.015 \times 1 0^{10}) /2 = 1.657 \times 1 0^{10} cm^{- 3}$ . Why this step? Direct quadratic-formula answer $p = \frac{1}{2} (N_{A} + \dots)$ .
Expansion route — plug into $p \approx n_{i} + N_{A} /2 + N_{A}^{2} / (8 n_{i})$ : $= 1.5 \times 1 0^{10} + 1.5 \times 1 0^{9} + (9 \times 1 0^{18}) / (1.2 \times 1 0^{11}) = 1.5 \times 1 0^{10} + 0.15 \times 1 0^{10} + 7.5 \times 1 0^{7} = 1.6575 \times 1 0^{10} cm^{- 3}$ . Why this step? This shows the series and the exact formula agree to 4 figures — confirming the expansion is trustworthy in Cell D and giving physical insight (the $N_{A} /2$ tilt).
$n = n_{i}^{2} / p = 2.25 \times 1 0^{20} /1.657 \times 1 0^{10} = 1.357 \times 1 0^{10} cm^{- 3}$ . Why this step? Mass-action.

Verify: $n + N_{A} = 1.357 \times 1 0^{10} + 3 \times 1 0^{9} = 1.657 \times 1 0^{10} = p$ . ✓ Exact and expansion match to $< 0.1%$ . ✓ Notice $p / n = 1.22$ — holes only slightly outnumber electrons. As $N_{A} \to 0$ , $p / n \to 1$ (back to intrinsic, Cell C). The doping merely tilts the balance.

Cell E — Heat it up ( $n_{i}$ grows)

The intrinsic concentration is not a constant — it rises steeply with temperature because heat rips more bonds. If you heat p-type silicon enough, $n_{i}$ can catch up to $N_{A}$ , dragging you from Cell A back toward Cell B. This is why power devices lose their doping character when they overheat. Figure 2 plots $n_{i}$ climbing (coral) toward the fixed $N_{A}$ line (purple) against temperature — the yellow crossing marker is exactly the temperature used in Example 5.

Figure 2 — Temperature $T$ (K, horizontal) vs concentration (cm⁻³, vertical log axis). Coral = $n_{i}$ rising with heat; lavender dashed = fixed $N_{A} = 1 0^{15} cm^{- 3}$ ; butter dot = the Example 5 operating point where $n_{i}$ has grown to half of $N_{A}$ .

Worked example Example 5 — Cell E

A wafer with $N_{A} = 1 0^{15} cm^{- 3}$ is heated until $n_{i}$ rises to $5 \times 1 0^{14} cm^{- 3}$ . Find $p$ and $n$ , and compare to the room-temperature shortcut $p = N_{A}$ .

Forecast: At this temperature $n_{i}$ is half of $N_{A}$ . Will $p$ still be near $N_{A}$ ?

Discriminant: $N_{A}^{2} + 4 n_{i}^{2} = (1 0^{15})^{2} + 4 (5 \times 1 0^{14})^{2} = 1 \times 1 0^{30} + 1 \times 1 0^{30} = 2 \times 1 0^{30}$ . Why this step? Now $4 n_{i}^{2}$ equals $N_{A}^{2}$ — we are firmly in the "must use quadratic" zone (Cell B territory, reached by heating instead of by light doping).
$2 \times 1 0^{30} = 1.4142 \times 1 0^{15}$ . Why this step? Take the root of the discriminant so it carries units $cm^{- 3}$ and can be added to $N_{A}$ .
$p = (1 0^{15} + 1.4142 \times 1 0^{15}) /2 = 1.2071 \times 1 0^{15} cm^{- 3}$ . Why this step? Quadratic-formula answer $p = \frac{1}{2} (N_{A} + \dots)$ .
$n = n_{i}^{2} / p = 2.5 \times 1 0^{29} /1.2071 \times 1 0^{15} = 2.071 \times 1 0^{14} cm^{- 3}$ . Why this step? Mass-action fixes $n$ from the $p$ we just computed; note $n$ is now only ~6× below $p$ , showing the material is drifting toward intrinsic.

Verify: $n + N_{A} = 2.071 \times 1 0^{14} + 1 0^{15} = 1.2071 \times 1 0^{15} = p$ . ✓ The shortcut $p = N_{A} = 1 0^{15}$ is now 21% low. Heat pushes any doped crystal toward intrinsic.

Cell F — Cool it down (freeze-out)

At room temperature boron's acceptor level sits just above the valence band, so essentially all acceptors have grabbed an electron and are ionized: $N_{A}^{-} \approx N_{A}$ . Cool the crystal and there isn't enough thermal energy to complete that hop — some boron atoms sit un-ionized, holding onto their empty bond without releasing a hole. Then $N_{A}^{-} < N_{A}$ and $p < N_{A}$ . This is why the parent note warns " $p = N_{A}$ breaks at low T."

Worked example Example 6 — Cell F

$N_{A} = 1 0^{16} cm^{- 3}$ , but at a cold temperature only $40%$ of acceptors are ionized, and $n_{i}$ has dropped to a negligible $1 \times 1 0^{6} cm^{- 3}$ . Find $p$ and $n$ .

Forecast: Will $p$ be $1 0^{16}$ , or the fraction $0.4 \times 1 0^{16}$ , or something else?

Ionized acceptors: $N_{A}^{-} = 0.40 \times 1 0^{16} = 4 \times 1 0^{15} cm^{- 3}$ . Why this step? Only ionized boron has released a hole; neutral boron contributes none. Charge neutrality uses $N_{A}^{-}$ , not $N_{A}$ .
Since $n_{i} = 1 0^{6} ≪ N_{A}^{-}$ , we are deep in Cell A relative to the ionized count: $p \approx N_{A}^{-} = 4 \times 1 0^{15} cm^{- 3}$ . Why this step? The quadratic with $N_{A}^{-}$ still applies; $n_{i}$ is utterly tiny so the shortcut is fine.
$n = n_{i}^{2} / p = (1 0^{6})^{2} / (4 \times 1 0^{15}) = 1 0^{12} / (4 \times 1 0^{15}) = 2.5 \times 1 0^{- 4} cm^{- 3}$ . Why this step? Mass-action fixes $n$ from $p$ ; the answer being far below 1 per cm³ just tells us electrons are essentially absent at this cold temperature.

Verify: $p = 4 \times 1 0^{15}$ , well below the full $N_{A} = 1 0^{16}$ — freeze-out has hidden 60% of the holes. As temperature rises back to room level, $N_{A}^{-} \to N_{A}$ and $p$ climbs back to $1 0^{16}$ . ✓

Cell G — Real device: resistance & drift current

This is the cell where the charge $q$ (defined in the symbols block) finally does work: it turns a carrier count ( $cm^{- 3}$ ) into an electric current. Recall the dimensional check from the symbols block — $σ = q p μ_{p}$ comes out in $(Ω \cdotcm)^{- 1}$ .

Worked example Example 7 — Cell G

A bar of p-type silicon: length $L = 2 mm = 0.2 cm$ , cross-section $A = 0.01 cm^{2}$ , doping $N_{A} = 1 0^{16} cm^{- 3}$ (so $p \approx N_{A}$ from Cell A), hole mobility $μ_{p} = 480 cm^{2} / (V\cdots)$ , $q = 1.6 \times 1 0^{- 19} C$ . Find the conductivity $σ$ and the resistance $R$ .

Forecast: Which carrier carries the current — holes or the minority electrons? Guess the order of magnitude of $R$ .

Conductivity from holes: $σ = q p μ_{p} = (1.6 \times 1 0^{- 19}) (1 0^{16}) (480) = 0.768 (Ω \cdotcm)^{- 1}$ . Why this step? Current in p-type is holes drifting (see Drift and Diffusion Currents). Minority electrons ( $n \approx 2.25 \times 1 0^{4}$ ) would add $q n μ_{n} = (1.6 \times 1 0^{- 19}) (2.25 \times 1 0^{4}) (1350) \sim 5 \times 1 0^{- 12}$ — about $1 0^{12}$ × smaller, so we drop it even though $μ_{n} > μ_{p}$ .
Resistivity $ρ = 1/ σ = 1/0.768 = 1.302 Ω \cdotcm$ . Why this step? Resistivity is just the reciprocal of conductivity.
Resistance $R = ρ L / A = 1.302 \times 0.2/0.01 = 26.04 Ω$ . Why this step? Standard geometry law: longer bar = more resistance, fatter bar = less.

Verify: Units: $(Ω cm) \cdot (cm) / (cm^{2}) = Ω$ . ✓ Recompute $σ$ backward: $q p μ_{p} = 1.6 \times 1 0^{- 19} \cdot 1 0^{16} \cdot 480 = 0.768$ . ✓

Cell H — Compensation, still net p-type

Sometimes a wafer already has some phosphorus donors (N-type doping with donor atoms (phosphorus)) when we add boron. The donors' free electrons cancel an equal number of holes. Only the excess dopant survives. Neutrality is the full $p + N_{D} = n + N_{A}$ , so the effective doping is $N_{A} - N_{D}$ .

Worked example Example 8 — Cell H

A crystal has $N_{A} = 8 \times 1 0^{15} cm^{- 3}$ boron and $N_{D} = 3 \times 1 0^{15} cm^{- 3}$ phosphorus, $n_{i} = 1.5 \times 1 0^{10}$ . Is it p-type or n-type? Find $p$ and $n$ .

Forecast: Which dopant wins? Will the leftover be big enough for the shortcut?

Net acceptor doping: $N_{A} - N_{D} = 8 \times 1 0^{15} - 3 \times 1 0^{15} = 5 \times 1 0^{15} cm^{- 3}$ (positive → still p-type). Why this step? Each donor electron fills one acceptor hole; only the surplus of holes remains mobile.
Since $5 \times 1 0^{15} ≫ n_{i}$ , Cell A shortcut applies to the net: $p \approx N_{A} - N_{D} = 5 \times 1 0^{15} cm^{- 3}$ . Why this step? The net doping plays exactly the role $N_{A}$ did before.
$n = n_{i}^{2} / p = (1.5 \times 1 0^{10})^{2} / (5 \times 1 0^{15}) = 2.25 \times 1 0^{20} / (5 \times 1 0^{15}) = 4.5 \times 1 0^{4} cm^{- 3}$ . Why this step? Mass-action fixes $n$ from $p$ ; because $p$ is large, $n$ is heavily suppressed — holes are still the majority.

Verify: neutrality $p + N_{D} = n + N_{A} \Rightarrow 5 \times 1 0^{15} + 3 \times 1 0^{15} = 8 \times 1 0^{15}$ , and $n + N_{A} = 4.5 \times 1 0^{4} + 8 \times 1 0^{15} \approx 8 \times 1 0^{15}$ . ✓ Both sides $= 8 \times 1 0^{15}$ .

Cell I — Compensation flips it to n-type

If the donors outnumber the acceptors, the surplus is now electrons, and the crystal is n-type despite containing boron. The same neutrality equation handles it — you just find the sign of $N_{A} - N_{D}$ flips, so you solve for $n$ (majority) instead of $p$ .

Worked example Example 9 — Cell I

A crystal has $N_{A} = 2 \times 1 0^{15} cm^{- 3}$ boron and $N_{D} = 7 \times 1 0^{15} cm^{- 3}$ phosphorus, $n_{i} = 1.5 \times 1 0^{10}$ . Find the majority carrier, $n$ and $p$ .

Forecast: With more donors than acceptors, which carrier now dominates — and by roughly how much?

Net donor doping: $N_{D} - N_{A} = 7 \times 1 0^{15} - 2 \times 1 0^{15} = 5 \times 1 0^{15} cm^{- 3}$ (positive donor surplus → n-type). Why this step? Each acceptor hole is filled by a donor electron; the leftover $5 \times 1 0^{15}$ electrons are the mobile majority.
Since $5 \times 1 0^{15} ≫ n_{i}$ , the n-type shortcut applies: $n \approx N_{D} - N_{A} = 5 \times 1 0^{15} cm^{- 3}$ . Why this step? This is the mirror of Cell A with electrons as majority — see N-type doping with donor atoms (phosphorus).
$p = n_{i}^{2} / n = (1.5 \times 1 0^{10})^{2} / (5 \times 1 0^{15}) = 2.25 \times 1 0^{20} / (5 \times 1 0^{15}) = 4.5 \times 1 0^{4} cm^{- 3}$ . Why this step? Mass-action fixes the minority holes from the majority $n$ .

Verify: neutrality $p + N_{D} = n + N_{A} \Rightarrow 4.5 \times 1 0^{4} + 7 \times 1 0^{15} \approx 7 \times 1 0^{15}$ and $n + N_{A} = 5 \times 1 0^{15} + 2 \times 1 0^{15} = 7 \times 1 0^{15}$ . ✓ Both sides $= 7 \times 1 0^{15}$ . Adding boron did not make it p-type — the donor surplus wins.

Cell J — Extreme doping (degenerate): the formulas break

Everything above uses the non-degenerate mass-action law $n p = n_{i}^{2}$ . That law quietly assumes carriers are dilute enough to obey Boltzmann statistics. When doping gets very high, that assumption fails and we need a quantitative flag for when.

Worked example Example 10 — Cell J (recognise, don't misuse)

A wafer is doped $N_{A} = 5 \times 1 0^{19} cm^{- 3}$ . Someone plugs it into $p \approx N_{A}$ then $n = n_{i}^{2} / p$ . Why is this wrong, and what should be used instead?

Forecast: Is $p$ really $5 \times 1 0^{19}$ , larger, or smaller than that?

Test the criterion: $N_{A} / N_{V} = 5 \times 1 0^{19} /1.04 \times 1 0^{19} \approx 4.8 > 0.1$ . Why this step? This is the single check that decides "legal vs illegal formula." Here $N_{A}$ is nearly 5× the band's capacity — deeply degenerate.
So $n p = n_{i}^{2}$ is invalid; you cannot get $p$ from the simple formula. Instead $p$ must be found from the Fermi–Dirac integral $p = N_{V} F_{1/2} (\frac{E _{V} - E _{F}}{k _{B} T})$ where $F_{1/2}$ is the Fermi–Dirac integral of order $\frac{1}{2}$ (it replaces the simple exponential of Boltzmann statistics). Not every acceptor even releases a free hole, so $p < N_{A}$ here. Why this step? At these densities the holes fill a finite number of valence-band seats; Fermi–Dirac counting (from Mass-Action Law and Carrier Statistics and Fermi Level in Doped Semiconductors) is required.

Verify (sanity, not a plug-in): the warning sign $N_{A} > 0.1 N_{V}$ is what flags degeneracy. Numerically $5 \times 1 0^{19} > 0.1 \times 1.04 \times 1 0^{19} = 1.04 \times 1 0^{18}$ . ✓ So the honest answer is "degenerate — non-degenerate formulas do not apply; use $p = N_{V} F_{1/2} (\dots)$ ."

Recall Which cell am I in? (decision flow)

Answer these in order. First: is $N_{D}$ present? ::: If yes, replace $N_{A}$ with $N_{A} - N_{D}$ (Cell H if positive → p-type, Cell I if negative → n-type). Next: is the net doping near $0.1 N_{V} \approx 1 0^{18}$ – $1 0^{19} cm^{- 3}$ or above? ::: Then it is degenerate (Cell J) — the $n p = n_{i}^{2}$ formulas do not apply; use $p = N_{V} F_{1/2} (\dots)$ . Then: how does the net doping compare to $n_{i}$ at this temperature? ::: $≫ n_{i}$ → shortcut $p \approx N_{A}$ (Cell A). Comparable or below → full quadratic or its series expansion (Cells B, C, D). Cold crystal? ::: Use ionized $N_{A}^{-}$ , not full $N_{A}$ (Cell F). Hot crystal? ::: Recompute the larger $n_{i}$ first, then decide (Cell E).

Recall

When is the shortcut

p \approx N_{A}

safe?

Only when

N_{A} ≫ n_{i}

and all acceptors ionized and below degeneracy (

N_{A} ≪ N_{V}

) — Cell A.

In Example 2 (

N_{A} = 2 \times 1 0^{10}

) why is

p = 2.8 \times 1 0^{10}

not

2 \times 1 0^{10}

Because

n_{i} \sim N_{A}

, so

4 n_{i}^{2}

dominates the discriminant and cannot be dropped, pushing

p

about 28% above the naive

N_{A}

What does the exact formula give when

N_{A} = 0

p = n = n_{i}

— it reduces to the intrinsic result.

Series expansion of

p

for

N_{A} ≪ n_{i}

p \approx n_{i} + N_{A} /2 + N_{A}^{2} / (8 n_{i})

— leading intrinsic term, then a half-doping tilt.

Why does heating p-type silicon reduce its "p-ness"?

n_{i}

rises steeply, catches up to

N_{A}

, and the carrier ratio

p / n

falls toward 1.

What is freeze-out?

At low temperature only a fraction of acceptors ionize, so

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

and

p < N_{A}

For a compensated crystal, what sets the majority carrier?

The net dopant

N_{A} - N_{D}

: positive → p-type (Cell H), negative → n-type (Cell I).

Quantitative test for when

n p = n_{i}^{2}

breaks down?

Degeneracy when

p ≳ 0.1 N_{V}

(i.e.

N_{A} ≳ 1 0^{19} cm^{- 3}

in Si); then use the Fermi–Dirac integral

p = N_{V} F_{1/2} ((E_{V} - E_{F}) / k_{B} T)

(Cell J).

Conductivity of p-type silicon formula, units, and why holes only?

σ = q p μ_{p}

(Ω \cdotcm)^{- 1}

; minority electrons contribute

q n μ_{n}

which is ~

1 0^{12}

× smaller even though

μ_{n} > μ_{p}

Connections

Parent: P-type doping (Hinglish) — the base theory these examples exercise.
Intrinsic Semiconductors — the $N_{A} = 0$ / Cell C limit.
N-type doping with donor atoms (phosphorus) — supplies $N_{D}$ in the compensation cases (H and I).
Mass-Action Law and Carrier Statistics — the $n p = n_{i}^{2}$ we use in every cell, and its Fermi–Dirac replacement in Cell J.
Fermi Level in Doped Semiconductors — how these carrier counts move $E_{F}$ , and why degeneracy (Cell J) breaks the formulas.
Drift and Diffusion Currents — the physics behind Cell G's conductivity.
PN Junction Formation — where p-type and n-type finally meet.

Worked examples — P-type doping with acceptor atoms (boron)