P-type doping with acceptor atoms (boron): Exercises

This page is a self-testing ladder. Each problem is stated cleanly, then a hidden [!recall]- solution unfolds every step: WHAT we do, WHY we do it, and WHAT the numbers mean. Work each one on paper first, then reveal.

Everything here builds on the parent note. If a symbol feels unfamiliar, here is the entire toolkit before we use it:

Definition The symbols you need (defined before use)

$N_{A}$ = number of acceptor (boron) atoms packed into each cubic centimetre. Units $cm^{- 3}$ . Think "dopant crowd size."
$N_{A}^{-}$ = number of those acceptors that have actually grabbed an electron and become fixed negative ions. At room temperature $N_{A}^{-} \approx N_{A}$ (fully ionized).
$N_{D}$ = number of donor (phosphorus) atoms per $cm^{- 3}$ — the opposite dopant, which each give one free electron. In pure p-type silicon $N_{D} = 0$ ; it only appears when the crystal is also doped with donors (a "compensated" sample). Defined here so Level 4 can use it.
$p$ = number of free holes per $cm^{- 3}$ (the mobile "+" carriers).
$n$ = number of free electrons per $cm^{- 3}$ (the minority carriers here).
$n_{i}$ = intrinsic carrier density of pure silicon. In Si, $n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ at room temperature. This is the count when $n = p$ with no doping.
$q = 1.6 \times 1 0^{- 19} C$ = the charge on one electron/hole.
$μ_{p}, μ_{n}$ = mobility = how fast holes/electrons drift per unit electric field. Units $cm^{2} / (V \cdot s)$ . Bigger $μ$ = easier to push.
$σ$ = conductivity = how well the material conducts. Units $(Ω \cdot cm)^{- 1}$ .
$k T / q$ at room temperature $\approx 0.0259 V$ (the "thermal voltage" $V_{T}$ ). This is energy-per-charge from heat jiggling.

LEVEL 1 — Recognition

(Can you spot which carrier is which and read the definitions?)

Exercise 1.1

A silicon sample is doped with boron. State (a) the majority carrier, (b) the minority carrier, and (c) the sign of the fixed dopant ion left behind.

Recall Solution 1.1

WHAT: Boron has 3 valence electrons, one fewer than Si's 4, so it makes holes. WHY: Each boron eventually accepts an electron from a neighbouring bond, releasing a mobile hole and becoming stuck as a negative ion.

(a) Majority carrier = holes (positive, mobile).
(b) Minority carrier = electrons.
(c) The boron ion is == $B^{-}$ , a fixed negative charge==. The crystal stays neutral: mobile $+ 1$ hole balanced by fixed $- 1$ ion.

Exercise 1.2

Pure (intrinsic) silicon has $n = p = n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ . After doping with $N_{A} = 1 0^{16} cm^{- 3}$ , which of $n$ or $p$ goes up and which goes down relative to $n_{i}$ ?

Recall Solution 1.2

WHAT: Holes are injected, so $p$ rises far above $n_{i}$ . WHY: $p \approx N_{A} = 1 0^{16} ≫ 1.5 \times 1 0^{10} = n_{i}$ , so $p$ increases by ~6 orders of magnitude. Mass-action fixes the product: $n = n_{i}^{2} / p$ . Since $p$ shot up, $n$ must drop below $n_{i}$ .

$p ↑$ (majority), $n ↓$ (minority). Their product stays $n_{i}^{2}$ .

LEVEL 2 — Application

(Plug into the master equations.)

Exercise 2.1

Silicon with $N_{A} = 5 \times 1 0^{16} cm^{- 3}$ , $n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ . Find $p$ and $n$ .

Recall Solution 2.1

WHAT: Check whether $N_{A} ≫ n_{i}$ . Here $5 \times 1 0^{16} ≫ 1.5 \times 1 0^{10}$ (a factor $\sim 3 \times 1 0^{6}$ ), so the simple limit applies. WHY: the $N_{A}^{2} + 4 n_{i}^{2} \approx N_{A}$ , so $p \approx N_{A}$ .

$p \approx N_{A} = 5 \times 1 0^{16} cm^{- 3}$ .
$n = n_{i}^{2} / p = (1.5 \times 1 0^{10})^{2} / (5 \times 1 0^{16}) = 2.25 \times 1 0^{20} /5 \times 1 0^{16} = 4.5 \times 1 0^{3} cm^{- 3}$ . Holes outnumber electrons by $\sim 1 0^{13}$ — overwhelmingly p-type.

Exercise 2.2

A p-type sample has hole density $p = 2 \times 1 0^{17} cm^{- 3}$ , hole mobility $μ_{p} = 400 cm^{2} / (V \cdot s)$ . Compute the conductivity $σ$ and resistivity $ρ = 1/ σ$ .

Recall Solution 2.2

WHAT: Use $σ \approx q p μ_{p}$ (electrons negligible since $n ≪ p$ ). WHY: conductivity = charge × carrier count × how mobile each carrier is. $σ = (1.6 \times 1 0^{- 19}) (2 \times 1 0^{17}) (400) = 12.8 (Ω \cdot cm)^{- 1}$ WHAT it means: resistivity $ρ = 1/ σ = 1/12.8 = 0.078 Ω \cdot cm$ — a decent conductor, far better than intrinsic Si.

Exercise 2.3

You want a p-type resistor with $σ = 5 (Ω \cdot cm)^{- 1}$ using boron, $μ_{p} = 450 cm^{2} / (V \cdot s)$ . What boron concentration $N_{A}$ do you need? (Assume $p \approx N_{A}$ .)

Recall Solution 2.3

WHAT: Invert $σ = q N_{A} μ_{p}$ to solve for $N_{A}$ . WHY: we know the target conductivity and the mobility, so the only unknown is the dopant crowd size. $N_{A} = \frac{σ}{q μ _{p}} = \frac{5}{( 1.6 \times 1 0 ^{- 19} ) ( 450 )} = \frac{5}{7.2 \times 1 0 ^{- 17}} = 6.94 \times 1 0^{16} cm^{- 3}$ Check $N_{A} ≫ n_{i}$ ? $6.94 \times 1 0^{16} ≫ 1.5 \times 1 0^{10}$ ✓, so $p \approx N_{A}$ was valid.

LEVEL 3 — Analysis

(When does the approximation break? Handle BOTH the near-intrinsic case AND the very-high, degenerate case.)

Exercise 3.1

$N_{A} = 3 \times 1 0^{10} cm^{- 3}$ (comparable to $n_{i} = 1.5 \times 1 0^{10}$ ). Find $p$ and $n$ using the full quadratic. By what percent does the crude $p \approx N_{A}$ underestimate $p$ ?

Recall Solution 3.1

WHAT: $N_{A}$ is only $2 n_{i}$ , so we must NOT drop $n_{i}$ . Use the full root. WHY: the $4 n_{i}^{2}$ term inside the square root is now the same order as $N_{A}^{2}$ . $N_{A}^{2} + 4 n_{i}^{2} = (3 \times 1 0^{10})^{2} + 4 (1.5 \times 1 0^{10})^{2}$ $= 9 \times 1 0^{20} + 9 \times 1 0^{20} = 1.8 \times 1 0^{21} = 4.243 \times 1 0^{10}$ $p = \frac{3 \times 1 0 ^{10} + 4.243 \times 1 0 ^{10}}{2} = 3.62 \times 1 0^{10} cm^{- 3}$ $n = \frac{n _{i}^{2}}{p} = \frac{2.25 \times 1 0 ^{20}}{3.62 \times 1 0 ^{10}} = 6.22 \times 1 0^{9} cm^{- 3}$ Error of crude estimate: crude gives $p \approx N_{A} = 3 \times 1 0^{10}$ . $underestimate = \frac{3.62 - 3.0}{3.62} \times 100% = 17.2%$ Lesson: near intrinsic, the crude formula is off by ~17%. Use the quadratic.

Exercise 3.2

Verify charge neutrality for your Exercise 3.1 answer: does $p = n + N_{A}$ hold?

Recall Solution 3.2

WHAT: Add electrons plus acceptor ions and check it equals holes. WHY: neutrality is the physical constraint the quadratic was built from — a good self-check. $n + N_{A} = 6.22 \times 1 0^{9} + 3 \times 1 0^{10} = 0.622 \times 1 0^{10} + 3 \times 1 0^{10} = 3.62 \times 1 0^{10}$ This equals $p = 3.62 \times 1 0^{10}$ ✓. The books balance — our root is consistent.

Exercise 3.3 — the OTHER breakdown: degenerate doping

Now push the opposite extreme: $N_{A} = 2 \times 1 0^{20} cm^{- 3}$ (very heavy boron). The formula $p = \frac{1}{2} (N_{A} + N_{A}^{2} + 4 n_{i}^{2})$ still returns $p \approx N_{A}$ . Explain why this answer is no longer trustworthy, i.e. which hidden assumption behind ALL of our master equations has quietly failed.

Recall Solution 3.3

WHAT: Numerically the quadratic gives $p \approx N_{A} = 2 \times 1 0^{20}$ — nothing looks wrong. But the physics behind the equations has broken. WHY it fails: the mass-action law $n p = n_{i}^{2}$ and the simple formula $p = n_{i} e^{(E_{i} - E_{F}) / k T}$ both assume non-degenerate (Boltzmann) statistics — the carriers are so dilute that they never "bump into each other" for states. That holds only when the Fermi level $E_{F}$ stays a few $k T$ inside the band gap. At $N_{A} \sim 1 0^{20}$ the holes are so crowded that $E_{F}$ pushes into the valence band itself. Then:

The exponential Boltzmann approximation must be replaced by the full Fermi–Dirac statistics.
Not every boron atom stays ionized in the simple way; the band edges shift ("band-gap narrowing").
$n p = n_{i}^{2}$ is no longer exactly valid. WHAT it means: for degenerate material, $p \approx N_{A}$ is still roughly true as a count of holes, but the derived quantities ( $n$ , $E_{i} - E_{F}$ from the Boltzmann formula) are wrong. This is the ceiling of our toolbox — see Mass-Action Law and Carrier Statistics for the Fermi–Dirac fix. Rule of thumb: trust our formulas only for roughly $n_{i} ≪ N_{A} ≲ 1 0^{19} cm^{- 3}$ .

LEVEL 4 — Synthesis

(Combine doping, temperature effects, and compensation.)

Exercise 4.1 — Compensated silicon

A crystal is doped with BOTH boron $N_{A} = 8 \times 1 0^{16} cm^{- 3}$ and phosphorus $N_{D} = 3 \times 1 0^{16} cm^{- 3}$ . Is it p-type or n-type? Find the majority carrier density (assume all dopants ionized, $N_{A}, N_{D} ≫ n_{i}$ ).

Recall Solution 4.1

WHAT: Donors and acceptors partly cancel. Net acceptor density = $N_{A} - N_{D}$ . WHY: each donor electron (recall $N_{D}$ = donor/phosphorus count from the toolkit) fills one acceptor hole. Only the excess dopant sets the carrier type. See N-type doping with donor atoms (phosphorus) for the donor side. $N_{A} - N_{D} = 8 \times 1 0^{16} - 3 \times 1 0^{16} = 5 \times 1 0^{16} cm^{- 3}$ Acceptors win → p-type. With net doping $≫ n_{i}$ : $p \approx N_{A} - N_{D} = 5 \times 1 0^{16} cm^{- 3}$ $n = n_{i}^{2} / p = 2.25 \times 1 0^{20} /5 \times 1 0^{16} = 4.5 \times 1 0^{3} cm^{- 3}$ WHAT it means: even though $8 \times 1 0^{16}$ boron atoms are present, $3 \times 1 0^{16}$ of their holes are "eaten" by phosphorus electrons. Only $5 \times 1 0^{16}$ holes survive.

Exercise 4.2 — Minority carrier and diffusion current setup

For the compensated sample above ( $p = 5 \times 1 0^{16}$ , $n = 4.5 \times 1 0^{3} cm^{- 3}$ ), which carrier's diffusion current matters for a nearby pn junction, and why? Compute the ratio $p / n$ .

Recall Solution 4.2

WHAT: In a diode, the current is limited by injection of minority carriers. Here minorities are electrons. WHY: majority carriers are abundant; the bottleneck is the scarce minority carriers crossing the junction (see Drift and Diffusion Currents). $\frac{p}{n} = \frac{5 \times 1 0 ^{16}}{4.5 \times 1 0 ^{3}} = 1.11 \times 1 0^{13}$ WHAT it means: holes outnumber electrons by ~ $1 0^{13}$ . The tiny electron population is what governs the reverse/minority-injection behaviour of a junction made from this material.

LEVEL 5 — Mastery

(Temperature freeze-out, the Fermi level, and a physical design.)

Exercise 5.1 — Fermi level position

For $N_{A} = 1 0^{16} cm^{- 3}$ , the hole density is $p \approx N_{A}$ . Using $p = n_{i} exp (\frac{E _{i} - E _{F}}{k T})$ with $V_{T} = k T / q = 0.0259 V$ , find how far below the intrinsic level $E_{i}$ the Fermi level $E_{F}$ sits (in eV). (Recall from the toolkit: $E_{i}$ = midgap reference, $E_{F}$ = the electron "sea level.")

Recall Solution 5.1

WHAT: Solve $p = n_{i} e^{(E_{i} - E_{F}) / k T}$ for the energy gap $E_{i} - E_{F}$ . WHY: in p-type, more holes ⟺ $E_{F}$ moves down toward the valence band. This is exactly the shift described in Fermi Level in Doped Semiconductors. We invert the exponential with a natural log. $\frac{E _{i} - E _{F}}{k T} = ln \frac{p}{n _{i}} = ln \frac{1 0 ^{16}}{1.5 \times 1 0 ^{10}} = ln (6.667 \times 1 0^{5}) = 13.41$ $E_{i} - E_{F} = 13.41 \times k T = 13.41 \times 0.0259 eV = 0.347 eV$ WHAT it means: the Fermi level drops 0.35 eV below midgap — clearly on the valence-band side, confirming p-type.

The band diagram below draws exactly this: the black $E_{i}$ line at midgap and the red $E_{F}$ line pushed down by $0.347 eV$ . Notice how close $E_{F}$ now sits to the valence band $E_{v}$ — that "downward slide" is what "p-type" means on an energy diagram.

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

Exercise 5.2 — Freeze-out reasoning

At very low temperature ( $T = 30 K$ ) the boron acceptors are only 20% ionized. If $N_{A} = 1 0^{16} cm^{- 3}$ , what is the actual hole density $p$ (ignore $n_{i}$ , which is essentially zero at 30 K)? Contrast with room temperature.

Recall Solution 5.2

WHAT: Only ionized acceptors release holes. Ionized fraction = 20% = 0.20. WHY: at low $T$ there isn't enough thermal energy ( $k T$ is tiny) for valence electrons to jump up into boron's acceptor level. Un-ionized boron holds no free hole — this is freeze-out. $N_{A}^{-} = 0.20 \times 1 0^{16} = 2 \times 1 0^{15} cm^{- 3}$ With $n_{i} \approx 0$ and neutrality $p \approx N_{A}^{-}$ : $p \approx 2 \times 1 0^{15} cm^{- 3}$ Contrast: at room T, $p \approx N_{A} = 1 0^{16}$ . At 30 K only a fifth of that, $2 \times 1 0^{15}$ . The dopant is present but electrically dormant — the material is 5× less conductive.

The curve below traces free-hole density $p$ as temperature climbs: read off the red 30 K point (only 20% ionized, $p = 2 \times 1 0^{15}$ ) and watch it rise to the room-temperature plateau where $p \approx N_{A} = 1 0^{16}$ . The flat top is the "fully ionized" regime our earlier exercises assumed.

Recall One-line summary of the whole ladder

Compute $p$ from $\frac{1}{2} (N_{A} + N_{A}^{2} + 4 n_{i}^{2})$ , drop to $p \approx N_{A}$ only when $n_{i} ≪ N_{A} ≲ 1 0^{19}$ , subtract donors when compensated ( $p \approx N_{A} - N_{D}$ ), multiply by the ionized fraction under freeze-out, then get $n = n_{i}^{2} / p$ , $σ = q p μ_{p}$ , and $E_{i} - E_{F} = k T ln (p / n_{i})$ .

Connections

Parent: P-type doping — the theory these drills exercise.
Intrinsic Semiconductors — the $n_{i}$ baseline used throughout.
N-type doping with donor atoms (phosphorus) — needed for the compensation problem (4.1).
Mass-Action Law and Carrier Statistics — source of $n p = n_{i}^{2}$ and the Fermi–Dirac fix for degeneracy (3.3).
Fermi Level in Doped Semiconductors — used in Exercise 5.1.
Drift and Diffusion Currents — conductivity and minority-current problems.
PN Junction Formation — why minority carriers (4.2) matter.