N-type doping with donor atoms (phosphorus, arsenic): Worked examples

This page is the "no surprise left" drill for N-type doping with donor atoms. We build a scenario matrix — every kind of question this topic can ask — then work an example for each cell. If you meet a case in an exam, it lives in one of these rows.

Before we start, one habit: every symbol we use is defined the first time it appears. If the parent note used it, we re-anchor it here.

Definition The core symbols (introduced now because Examples 1–6 all need them)

$N_{D}$ — donor concentration: how many donor atoms (phosphorus, arsenic...) we put in, per cubic centimetre. Think "how many spare-hand people we sneaked into the dance."
$n_{i}$ — intrinsic carrier density: how many electrons pure silicon frees on its own from heat alone ( $\approx 1.5 \times 1 0^{10} cm^{- 3}$ at room temperature).
$n$ — electron density: total mobile electrons per cm³.
$p$ — hole density: mobile "missing-electron" spots per cm³. A hole acts like a positive carrier.
$q$ — the electron charge magnitude, $1.6 \times 1 0^{- 19}$ coulombs.
$μ_{n}, μ_{p}$ — mobilities: how fast electrons / holes drift per unit electric field ( $cm^{2} / V\cdots$ ). Bigger = they move more easily.
$σ$ — conductivity: how well the material carries current, units $(Ω cm)^{- 1}$ .

(The extra symbols $k_{B}$ , $m^{*}$ , $ε_{r}$ , and $N_{A}$ are defined at the exact example that first needs them, so you meet each one in context.)

The scenario matrix

We are not solving quadrants of an angle here, but N-type doping has its own "every case" grid: the size of $N_{D}$ relative to $n_{i}$ , the identity of the carriers, and the boundary/limiting extremes. Here is every cell.

#	Cell class	What makes it special	Example
A	Standard heavy doping $N_{D} ≫ n_{i}$	The everyday case: $n \approx N_{D}$	Ex 1
B	Comparable doping $N_{D} \sim n_{i}$	Approximation fails, need full quadratic	Ex 2
C	Zero-doping boundary $N_{D} = 0$	Formula must reduce to intrinsic, $n = p = n_{i}$	Ex 3
D	Minority-carrier / mass-action	Find the tiny $p$ from a known $n$	Ex 4
E	Conductivity jump	Turn densities into $σ$ ; drop minority term	Ex 5
F	Limiting behaviour: very high $N_{D}$	Mobility droop, sub-linear $σ$	Ex 6
G-hi	High- $T$ limit: intrinsic crossover	$n_{i}$ grows past $N_{D}$ , " $n \approx N_{D}$ " breaks	Ex 7a
G-lo	Low- $T$ limit: freeze-out	Donors recapture electrons, $n < N_{D}$	Ex 7b
H	Binding-energy sanity check	Is the donor really ionised? Hydrogen model	Ex 8
I	Real-world word problem	Choose $N_{D}$ to hit a target resistance	Ex 9
J	Exam twist: compensation	Donors and acceptors together	Ex 10

Each example says which cell it fills. (Note: "zero-doping boundary" is a limiting case, not "degenerate doping" — in semiconductor language "degenerate" means the opposite extreme, very heavy doping, which shows up in Cell F.)

Example 1 — Cell A: standard heavy doping

Worked example Phosphorus at

$N_{D} = 1 \times 1 0^{16} cm^{- 3}$ Find the electron density $n$ . Given $n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ .

Forecast: Before reading on — will $n$ be closer to $N_{D}$ , to $n_{i}$ , or halfway? Guess.

Compare $N_{D}$ to $n_{i}$ . $N_{D} = 1 0^{16}$ , $n_{i} = 1.5 \times 1 0^{10}$ , so $N_{D} / n_{i} \approx 6.7 \times 1 0^{5}$ . Why this step? The whole choice of formula depends on this ratio. When $N_{D} ≫ n_{i}$ , donor electrons drown out thermal ones.
Apply the approximation. Because $N_{D} ≫ n_{i}$ , neutrality $n = p + N_{D}$ has $p$ negligible, so $n \approx N_{D} = 1 \times 1 0^{16} cm^{- 3}$ . Why this step? We drop $p$ only because we checked it is tiny — never drop a term you haven't justified.

Answer: $n \approx 1 \times 1 0^{16} cm^{- 3}$ .

Verify: Plug into the exact solution (Ex 2's formula) and you get $1 0^{16}$ to 6 significant figures — the approximation error is $\sim n_{i}^{2} / N_{D} \approx 2 \times 1 0^{4}$ , utterly negligible. Units: cm⁻³ ✓.

Example 2 — Cell B: comparable doping (approximation fails)

Here $N_{D}$ is not huge compared to $n_{i}$ , so " $n \approx N_{D}$ " would lie. We need the exact solution.

Deriving the exact $n$ (why a quadratic?): Combine the two laws. From neutrality $p = n - N_{D}$ . Substitute into mass-action $n p = n_{i}^{2}$ : $n (n - N_{D}) = n_{i}^{2} \Rightarrow n^{2} - N_{D} n - n_{i}^{2} = 0.$ Why a quadratic? We have two unknowns ( $n, p$ ) and two equations; eliminating $p$ leaves one equation, and because $n p$ multiplies the unknowns it is second-degree. Solving with the quadratic formula and keeping the positive root (a density can't be negative): $n = \frac{N _{D}}{2} + (\frac{N _{D}}{2})^{2} + n_{i}^{2}$

Worked example

$N_{D} = 2 \times 1 0^{10} cm^{- 3}$ (same order as $n_{i}$ ) $n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ .

Forecast: Will $n$ equal $N_{D} = 2 \times 1 0^{10}$ ? (If you say yes, this example is the trap.)

Check the ratio. $N_{D} / n_{i} = 1.33$ — not $≫ 1$ . So the shortcut is banned. Why this step? Same first move as Ex 1, but now it tells us to use the exact formula.
Plug into the boxed exact formula. $\frac{N _{D}}{2} = 1.0 \times 1 0^{10}$ , and $(\frac{N _{D}}{2})^{2} + n_{i}^{2} = 1 0^{20} + 2.25 \times 1 0^{20} = 3.25 \times 1 0^{20}$ . $3.25 \times 1 0^{20} = 1.803 \times 1 0^{10}$ . Why this step? Both terms in the neutrality equation matter now, so we keep the full square root.
Add: $n = 1.0 \times 1 0^{10} + 1.803 \times 1 0^{10} = 2.803 \times 1 0^{10} cm^{- 3}$ . Why this step? The two positive pieces of the quadratic root — the "half of $N_{D}$ " drift and the thermal square-root — combine to give the true electron count; adding them is the final assembly of the formula.

Answer: $n \approx 2.80 \times 1 0^{10} cm^{- 3}$ — noticeably more than $N_{D}$ , because thermal electrons still contribute here.

Verify: $p = n - N_{D} = 0.803 \times 1 0^{10}$ ; then $n p = (2.803) (0.803) \times 1 0^{20} = 2.25 \times 1 0^{20} = n_{i}^{2}$ ✓. Mass-action law restored.

Example 3 — Cell C: the zero-doping boundary

Example 4 — Cell D: minority carriers via mass-action

Example 5 — Cell E: the conductivity jump (with figure)

The figure below is the whole argument in one glance: two bars, one per carrier type, showing each carrier's contribution to conductivity (the product $n μ_{n}$ vs $p μ_{p}$ ) on a logarithmic height scale.

Figure — N-type doping with donor atoms (phosphorus, arsenic)

Worked example Conductivity of the Ex 1 sample

$n = 1 0^{16}$ , $p = 2.25 \times 1 0^{4} cm^{- 3}$ (from mass-action), $μ_{n} = 1350$ , $μ_{p} = 480 cm^{2} / V\cdots$ , $q = 1.6 \times 1 0^{- 19} C$ .

Forecast: Will the hole term change the answer's first two digits?

Write full conductivity. $σ = q (n μ_{n} + p μ_{p})$ . Why this step? Both carriers conduct in parallel; we start honest with both, then justify dropping one.
Compare the two terms. $n μ_{n} = 1 0^{16} \times 1350 = 1.35 \times 1 0^{19}$ ; $p μ_{p} = 2.25 \times 1 0^{4} \times 480 = 1.08 \times 1 0^{7}$ . The hole term is $\sim 1 0^{12}$ times smaller. Why this step? This is exactly the bar comparison in the figure — only once the teal bar is proven negligible may we drop it.
Compute. $σ \approx q n μ_{n} = (1.6 \times 1 0^{- 19}) (1 0^{16}) (1350) = 2.16 (Ω cm)^{- 1}$ . Why this step? One dominant term left; multiply through.

Answer: $σ \approx 2.16 (Ω cm)^{- 1}$ .

Verify: Intrinsic $σ_{i} \approx q n_{i} (μ_{n} + μ_{p}) = 1.6 \times 1 0^{- 19} \times 1.5 \times 1 0^{10} \times 1830 \approx 4.4 \times 1 0^{- 6}$ . Ratio $2.16/4.4 \times 1 0^{- 6} \approx 4.9 \times 1 0^{5}$ — about the "500,000×" the parent quoted ✓. See Conductivity and carrier mobility.

Example 6 — Cell F: limiting behaviour, very high $N_{D}$

Worked example Does 100× more doping give 100× more

$σ$ ? Sample A: $N_{D} = 1 0^{16}$ , $μ_{n} = 1350$ . Sample B: $N_{D} = 1 0^{18}$ , but heavy doping scatters electrons so $μ_{n}$ drops to $\approx 300 cm^{2} / V\cdots$ .

Forecast: Guess the conductivity ratio $σ_{B} / σ_{A}$ . Is it 100?

Sample A. $σ_{A} = q N_{D} μ_{n} = 1.6 \times 1 0^{- 19} \times 1 0^{16} \times 1350 = 2.16 (Ω cm)^{- 1}$ . Why this step? Reuse the majority-only formula (justified as in Ex 5).
Sample B. $σ_{B} = 1.6 \times 1 0^{- 19} \times 1 0^{18} \times 300 = 48 (Ω cm)^{- 1}$ . Why this step? Same formula, but we must use the reduced mobility — that's the whole lesson. (At this extreme, doping is called degenerate — carriers so dense the material behaves almost metallic.)
Take the ratio. $σ_{B} / σ_{A} = 48/2.16 \approx 22.2$ , not 100. Why this step? Compare to the naive expectation to see the sub-linearity.

Answer: $σ$ rose only $\sim 22 \times$ for a $100 \times$ carrier increase — mobility droop from ionised-impurity scattering ate the rest.

Verify: If mobility had stayed 1350, ratio would be exactly 100. Since $300/1350 = 0.222$ , the true ratio $= 100 \times 0.222 = 22.2$ ✓. Matches the parent's "sub-linear" note.

Example 7 — Cell G: temperature limits (both edges)

First we need a temperature-to-energy converter.

At very low temperature the donor electrons don't have enough thermal energy to jump into the conduction band — they "freeze out" back onto the donor, so $n < N_{D}$ . At very high temperature so many bonds break that $n_{i}$ grows past $N_{D}$ and the crystal behaves intrinsic again. Both edges break " $n \approx N_{D}$ ". We work each one numerically.

The figure sweeps temperature (plotted as $1000/ T$ , so hot is left, cold is right) and shows the electron density riding a flat "extrinsic plateau" in the middle, then bending up on the hot side (intrinsic) and drooping on the cold side (freeze-out).

Example 7a — Cell G-hi: high-temperature intrinsic crossover

Worked example Heated until

$n_{i}$ climbs to $N_{D}$ A sample with $N_{D} = 1 0^{15} cm^{- 3}$ is heated until $n_{i}$ rises to $1 0^{15} cm^{- 3}$ . Is " $n \approx N_{D}$ " still valid?

Forecast: At $n_{i} = N_{D}$ , is $n$ still $\approx N_{D}$ , or noticeably larger?

Check the ratio. $N_{D} / n_{i} = 1$ — we are exactly in Cell B territory, not Cell A. Why this step? The validity test is always $N_{D}$ vs $n_{i}$ ; temperature only enters through $n_{i}$ .
Use the exact formula. $n = \frac{N _{D}}{2} + (N_{D} /2)^{2} + n_{i}^{2} = 0.5 \times 1 0^{15} + 0.25 \times 1 0^{30} + 1 0^{30}$ . Why this step? With $n_{i}$ now comparable, thermal electrons contribute — same math as Ex 2.
Evaluate the root and add. $0.25 \times 1 0^{30} + 1.0 \times 1 0^{30} = 1.25 \times 1 0^{30} = 1.118 \times 1 0^{15}$ , so $n = 0.5 \times 1 0^{15} + 1.118 \times 1 0^{15} = 1.618 \times 1 0^{15} cm^{- 3}$ . Why this step? We finish the arithmetic — the "half $N_{D}$ " drift plus the thermal square-root assemble into the true $n$ .

Answer: $n = 1.62 \times 1 0^{15} cm^{- 3}$ , about $62%$ above $N_{D}$ . The doping shortcut has failed — we've entered the intrinsic crossover (orange curve bending up on the hot/left side of the figure).

Verify: $p = n - N_{D} = 0.618 \times 1 0^{15}$ ; $n p = 1.618 \times 0.618 \times 1 0^{30} = 1.0 \times 1 0^{30} = n_{i}^{2}$ ✓. The Fermi level slides back toward mid-gap here.

Example 7b — Cell G-lo: low-temperature freeze-out

At low $T$ , not every donor is ionised. A standard statistical-mechanics result (analogous to the intrinsic case, but with the donor level instead of the full gap) gives the freed-electron density in the freeze-out regime as $n \approx N_{C} N_{D} exp (- \frac{E _{D}}{2 k _{B} T}),$ where $N_{C}$ is the conduction-band "effective density of states" (the number of available seats), $\approx 2.8 \times 1 0^{19} cm^{- 3}$ for silicon, and $E_{D} \approx 26 meV$ is the donor ionisation energy (built in Example 8). This is the plum dashed limb in the figure.

Worked example Cool a

$N_{D} = 1 0^{15}$ sample to $T = 20 K$ Use $E_{D} = 26 meV$ , $N_{C} = 2.8 \times 1 0^{19} cm^{- 3}$ , $k_{B} = 8.617 \times 1 0^{- 5} eV/K$ . Roughly how many electrons remain free?

Forecast: At 20 K (very cold), is $n$ still $\approx 1 0^{15}$ , or dramatically less?

Find the thermal energy. $k_{B} T = 8.617 \times 1 0^{- 5} \times 20 = 1.72 \times 1 0^{- 3} eV = 1.72 meV$ . Why this step? Everything hinges on comparing $E_{D} = 26 meV$ to this. Here $E_{D} / k_{B} T \approx 15$ — energy is very scarce, so deep freeze-out is expected.
Form the exponent. $- \frac{E _{D}}{2 k _{B} T} = - \frac{0.026}{2 \times 0.00172} = - 7.56$ , so $e^{- 7.56} = 5.2 \times 1 0^{- 4}$ . Why this step? The exponential is the fraction of donors thermally lifted; with a big positive $E_{D} / k_{B} T$ it becomes small — this is the mathematical face of "freeze-out."
Multiply the prefactor. $N_{C} N_{D} = 2.8 \times 1 0^{19} \times 1 0^{15} = 2.8 \times 1 0^{34} = 1.673 \times 1 0^{17}$ . Then $n = 1.673 \times 1 0^{17} \times 5.2 \times 1 0^{- 4} = 8.7 \times 1 0^{13} cm^{- 3}$ . Why this step? Prefactor × exponential gives the freed count; here it lands well below $N_{D} = 1 0^{15}$ , confirming true freeze-out (the formula is valid precisely when $n ≪ N_{D}$ ).

Answer: $n \approx 8.7 \times 1 0^{13} cm^{- 3}$ — only about $9%$ of the donors are still ionised at 20 K. The rest have recaptured their electrons (plum limb diving on the cold/right side of the figure).

Verify: Sanity: $n / N_{D} = 8.7 \times 1 0^{13} /1 0^{15} = 0.087 < 1$ ✓ (freeze-out formula self-consistent). Warm back to room T and $E_{D} / k_{B} T \approx 1$ pushes the exponential to order 1, restoring $n \approx N_{D}$ (the plateau).

Example 8 — Cell H: is the donor really ionised?

Two crystal constants enter here, so we define them right where they're needed.

The parent note plugs these into a hydrogen-atom model to get the donor ionisation energy $E_{D} = 13.6 eV \times \frac{m ^{*} / m _{0}}{ε _{r}^{2}} = 13.6 eV \times \frac{0.26}{11. 7 ^{2}} .$ The $13.6 eV$ is the ordinary hydrogen binding energy; the two crystal factors crush it down to millielectron-volts.

Worked example Ionisation fraction from binding energy

Compute $E_{D}$ and compare it to room-temperature thermal energy $k_{B} T \approx 26 meV$ . Roughly how easily is the electron freed?

Forecast: Is $E_{D}$ big or small compared to the thermal "kick" available?

Recompute $E_{D}$ from the model. $13.6 \times 0.26/11. 7^{2} = 13.6 \times 0.26/136.89 = 0.0258 eV = 25.8 meV$ . Why this step? We rebuild the number from the defined constants ( $m^{*}, ε_{r}$ ) instead of trusting a memorised value.
Form the ratio $E_{D} / k_{B} T$ . $= 25.8/26 \approx 1$ . Why this step? Whether a state is occupied or empty is governed by how its energy compares to $k_{B} T$ — the natural energy of thermal jostling. A dimensionless ratio near or below 1 means "easily excited."
Interpret. Because $E_{D} ≲ k_{B} T$ , the thermal bath routinely supplies enough energy to lift the electron into the conduction band. Nearly every donor is ionised. Why this step? This justifies the $N_{D}^{+} \approx N_{D}$ assumption used in every example above.

Answer: $E_{D} / k_{B} T \approx 1$ ⇒ essentially full ionisation at room temperature. Our " $n \approx N_{D}$ " rests on this.

Verify: $13.6 \times 0.26/11. 7^{2} = 0.02583 eV = 25.8 meV$ ✓, matching the parent's derivation.

Example 9 — Cell I: real-world word problem

Worked example Design a resistor

You need a silicon bar with conductivity $σ = 10 (Ω cm)^{- 1}$ . Assume standard-doping mobility $μ_{n} = 1350 cm^{2} / V\cdots$ and the majority-only formula. What phosphorus concentration $N_{D}$ do you specify?

Forecast: Order of magnitude of $N_{D}$ — $1 0^{15}$ ? $1 0^{16}$ ? $1 0^{17}$ ?

Invert the conductivity formula. From $σ \approx q N_{D} μ_{n}$ , solve $N_{D} = \frac{σ}{q μ _{n}}$ . Why this step? We know the target output ( $σ$ ) and want the input ( $N_{D}$ ); algebra runs the equation backwards.
Plug numbers. $N_{D} = \frac{10}{( 1.6 \times 1 0 ^{- 19} ) ( 1350 )} = \frac{10}{2.16 \times 1 0 ^{- 16}} = 4.63 \times 1 0^{16} cm^{- 3}$ . Why this step? Direct substitution; units $\frac{( Ω cm ) ^{- 1}}{C \cdot cm ^{2} / V\cdots}$ reduce to cm⁻³ (since $C/V\cdots = A/V = Ω^{- 1}$ ).
Sanity-check the regime. $N_{D} = 4.6 \times 1 0^{16} ≫ n_{i}$ ✓ so the majority-only formula we used is valid. Why this step? Always re-confirm the approximation you assumed at the start — otherwise the answer is self-inconsistent.

Answer: Specify $N_{D} \approx 4.6 \times 1 0^{16} cm^{- 3}$ of phosphorus.

Verify: Forward-check: $σ = 1.6 \times 1 0^{- 19} \times 4.63 \times 1 0^{16} \times 1350 = 10.0 (Ω cm)^{- 1}$ ✓.

Example 10 — Cell J: exam twist, compensation

Real wafers may already hold acceptor atoms (from boron) when you add donors. This introduces one new symbol.

Worked example

$N_{D} = 5 \times 1 0^{16}$ , $N_{A} = 2 \times 1 0^{16} cm^{- 3}$ Find $n$ and $p$ . $n_{i} = 1.5 \times 1 0^{10}$ .

Forecast: Will $n$ equal $N_{D}$ , or the difference $N_{D} - N_{A}$ ?

Compute net doping. $N_{D} - N_{A} = 5 \times 1 0^{16} - 2 \times 1 0^{16} = 3 \times 1 0^{16} cm^{- 3}$ . Why this step? Acceptors cancel an equal number of donors before any electrons stay free; only the surplus survives.
Check it dwarfs $n_{i}$ . $3 \times 1 0^{16} ≫ 1.5 \times 1 0^{10}$ ✓, so $n \approx N_{D} - N_{A} = 3 \times 1 0^{16} cm^{- 3}$ . Why this step? Same $≫ n_{i}$ test as Cell A, but applied to the net value — the exam trap is forgetting to subtract $N_{A}$ .
Minority holes. $p = n_{i}^{2} / n = 2.25 \times 1 0^{20} /3 \times 1 0^{16} = 7.5 \times 1 0^{3} cm^{- 3}$ . Why this step? Mass-action still holds at fixed $T$ regardless of compensation.

Answer: $n \approx 3 \times 1 0^{16} cm^{- 3}$ , $p \approx 7.5 \times 1 0^{3} cm^{- 3}$ . Still N-type, but weaker than uncompensated $5 \times 1 0^{16}$ .

Verify: $n p = 3 \times 1 0^{16} \times 7.5 \times 1 0^{3} = 2.25 \times 1 0^{20} = n_{i}^{2}$ ✓. Ties into PN Junction formation where opposing dopants meet.

Recall Which cell does each trigger phrase belong to?

" $N_{D} ≫ n_{i}$ , find $n$ " ::: Cell A → $n \approx N_{D}$ (Ex 1) " $N_{D}$ same order as $n_{i}$ " ::: Cell B → exact quadratic (Ex 2) "Undoped / $N_{D} = 0$ " ::: Cell C → $n = p = n_{i}$ (Ex 3) "Find the tiny $p$ " ::: Cell D → mass-action $p = n_{i}^{2} / n$ (Ex 4) "How much did $σ$ jump" ::: Cell E → drop hole term (Ex 5) "100× doping, less than 100× $σ$ " ::: Cell F → mobility droop (Ex 6) "Heated until $n_{i}$ near $N_{D}$ " ::: Cell G-hi → intrinsic crossover (Ex 7a) "Cooled to freeze-out" ::: Cell G-lo → donors recapture electrons (Ex 7b) "Is the donor ionised" ::: Cell H → $E_{D} / k_{B} T \approx 1$ (Ex 8) "Design for a target $σ$ " ::: Cell I → invert $σ = q N_{D} μ_{n}$ (Ex 9) "Both donors and acceptors" ::: Cell J → $n \approx N_{D} - N_{A}$ (Ex 10)