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

Constants used throughout (memorise these three):

Intrinsic density of silicon: $n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$
Elementary charge: $q = 1.6 \times 1 0^{- 19} C$
Thermal energy at room temperature: $k_{B} T \approx 0.026 eV = 26 meV$

Definition Every symbol used on this page (nothing assumed)

$n$ — number of free electrons per cm³ (the movers that carry current in N-type).
$p$ — number of holes per cm³ (empty bond-slots that act like positive movers).
$N_{D}$ — number of donor atoms per cm³ we deliberately added (e.g. phosphorus).
$N_{D}^{+}$ — a fixed, positive donor ion: the donor atom after it has released its 5th electron. It is locked into the lattice and cannot move, so it carries no current, but its $+ 1$ charge cancels the freed electron's $- 1$ , keeping the crystal neutral. "Full ionisation" means every donor has become $N_{D}^{+}$ , so $N_{D}^{+} \approx N_{D}$ .
$μ_{n}, μ_{p}$ — mobilities: how fast electrons / holes drift per unit electric field ( $cm^{2} / Vs$ ).
$σ$ — conductivity: how well the material conducts.
$E_{D}$ — donor ionisation energy: the tiny energy needed to free the 5th electron.
$m_{0}$ — the free-electron mass (the mass of an electron in vacuum, $9.11 \times 1 0^{- 31} kg$ ). It is the reference against which we measure the effective mass.
$m^{*}$ — the effective mass: how heavy an electron behaves when moving through the silicon lattice rather than empty space. We write it as a fraction of $m_{0}$ , e.g. $m^{*} \approx 0.26 m_{0}$ .
$ε_{r}$ — the relative dielectric constant of silicon ( $\approx 11.7$ ): a number saying how strongly the crystal screens (weakens) electric attraction compared to vacuum.

Level 1 — Recognition

Can you recall and identify the core facts?

L1.1

Which of these elements can act as a donor in silicon: boron (group III), phosphorus (group V), arsenic (group V), silicon (group IV)? For each, say why or why not.

Recall Solution

A donor must have one more valence electron than silicon's 4 — i.e. group V (5 valence electrons).

Phosphorus (V): ✅ donor — 4 electrons bond, the 5th is spare.
Arsenic (V): ✅ donor — same reason.
Boron (III): ❌ — only 3 valence electrons, it is short one, so it accepts an electron (an acceptor, see P-type doping with acceptor atoms (boron)).
Silicon (IV): ❌ — exactly 4 electrons, no spare, no shortage. It is the host, not a dopant.

L1.2

In N-type silicon, name the majority carrier and the minority carrier. What immobile object is left behind when a donor releases its electron?

Recall Solution

Majority carrier: electrons (they are abundant).
Minority carrier: holes (very few).
Left behind: a fixed positive donor ion $N_{D}^{+}$ (defined in the symbol box above). It cannot move — it is locked into the lattice — so it does not carry current, but it keeps the crystal neutral.

Level 2 — Application

Plug the numbers into the formulas from the parent note.

L2.1

Silicon is doped with phosphorus at $N_{D} = 5 \times 1 0^{16} cm^{- 3}$ . Assume full ionisation. Find the electron density $n$ and hole density $p$ .

Recall Solution

Step 1 — electrons. Full ionisation means every donor gives one electron. Since $N_{D} = 5 \times 1 0^{16} ≫ n_{i} = 1.5 \times 1 0^{10}$ , the donor electrons swamp the thermal ones: $n \approx N_{D} = 5 \times 1 0^{16} cm^{- 3} .$ Why? The neutrality equation $n = p + N_{D}$ with $p$ tiny gives $n \approx N_{D}$ .

Step 2 — holes. Use the mass-action law $n p = n_{i}^{2}$ (holds at fixed temperature): $p = \frac{n _{i}^{2}}{n} = \frac{( 1.5 \times 1 0 ^{10} ) ^{2}}{5 \times 1 0 ^{16}} = \frac{2.25 \times 1 0 ^{20}}{5 \times 1 0 ^{16}} = 4.5 \times 1 0^{3} cm^{- 3} .$ Electrons outnumber holes by more than $1 0^{13}$ — hence "majority" vs "minority".

L2.2

Compute the conductivity of the sample in L2.1. Use $μ_{n} = 1350 cm^{2} / Vs$ and $μ_{p} = 480 cm^{2} / Vs$ .

Recall Solution

The full formula is $σ = q (n μ_{n} + p μ_{p})$ . Because $p ≪ n$ , the hole term is negligible: $σ \approx q n μ_{n} = (1.6 \times 1 0^{- 19}) (5 \times 1 0^{16}) (1350) .$ Multiply: $1.6 \times 1 0^{- 19} \times 5 \times 1 0^{16} = 8 \times 1 0^{- 3}$ , then $\times 1350 = 10.8$ . $σ \approx 10.8 (Ω cm)^{- 1} = 10.8 S/cm .$ Check the dropped term: $q p μ_{p} = (1.6 \times 1 0^{- 19}) (4.5 \times 1 0^{3}) (480) \approx 3.5 \times 1 0^{- 13}$ — utterly negligible next to $10.8$ . See Conductivity and carrier mobility.

Level 3 — Analysis

Now reason about relationships, not just plug-ins.

L3.1

Estimate the donor ionisation energy $E_{D}$ for a hydrogen-like donor in silicon using $E_{D} = 13.6 eV \times \frac{m ^{*} / m _{0}}{ε _{r}^{2}},$ with $m^{*} / m_{0} = 0.26$ and $ε_{r} = 11.7$ . Then explain, using $k_{B} T$ , why "full ionisation at room temperature" is a good assumption.

Intuition Why a donor behaves like a hydrogen atom (read this before solving)

Look at the figure. A phosphorus donor, after four of its electrons join the bonds, is a +1 core with one electron loosely circling it — that is exactly the picture of a hydrogen atom (one proton, one electron). So we reuse hydrogen's binding-energy formula, but with two edits for living inside silicon:

The pull is weakened because silicon's atoms screen the charge — divide by the dielectric constant $ε_{r} \approx 11.7$ (and it appears squared in energy).
The electron is lighter to move through the lattice — use the effective mass $m^{*} \approx 0.26 m_{0}$ (recall $m_{0}$ is the free-electron mass, defined in the symbol box).

These two edits shrink hydrogen's $13.6 eV$ down to a few tens of meV — a whisper of energy. That is the whole reason the electron is essentially free at room temperature.

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

Figure (s02): Left — a phosphorus donor drawn as a mini hydrogen atom: a red $+ 1$ core with one blue electron (the loosely bound 5th electron) circling on a faint orbit, annotated with the two silicon edits $ε_{r} \approx 11.7$ and $m^{*} \approx 0.26 m_{0}$ . Right — the energy-band diagram: the blue conduction band $E_{C}$ at top, the yellow valence band $E_{V}$ at bottom, the full gap $\approx 1.12$ eV marked, and a green dashed donor level $E_{D}$ sitting just below $E_{C}$ with a short red arrow ( $\approx 26$ meV) showing the tiny nudge needed to free the electron.

Recall Solution

Plug in: $E_{D} = 13.6 \times \frac{0.26}{11. 7 ^{2}} = 13.6 \times \frac{0.26}{136.89} = 13.6 \times 1.899 \times 1 0^{- 3} \approx 0.0258 eV \approx 26 meV .$ Interpretation. Thermal energy at room temperature is $k_{B} T \approx 26 meV$ . Since $E_{D} \approx k_{B} T$ , a typical thermal kick is already enough to knock the 5th electron loose. So essentially every donor is ionised — "full ionisation" is justified. (Real P in Si is $\approx 45$ meV, still comparable to $k_{B} T$ , so the conclusion survives.)

L3.2

A student doubles $N_{D}$ from $1 \times 1 0^{16}$ to $2 \times 1 0^{16} cm^{- 3}$ . By what factor does $n$ change? By what factor does $p$ change? Explain the opposite directions.

Recall Solution

Electrons: $n \approx N_{D}$ , so $n$ doubles (factor $2$ ).
Holes: $p = n_{i}^{2} / N_{D}$ , so doubling $N_{D}$ halves $p$ (factor $0.5$ ). Why opposite? The product $n p = n_{i}^{2}$ is pinned by temperature alone (the seesaw figure shows this). If you push $n$ up, the mass-action law forces $p$ down by the same factor to keep the product constant. More free electrons means holes recombine faster, thinning them out. See Mass-action law and carrier concentrations.

Figure (s03): A balance seesaw. A green triangular pivot is labelled "product $n p = n_{i}^{2}$ (fixed by temperature)". The white plank tilts so the right end (a large blue ball marked $n$ , electrons) is UP and the left end (a smaller yellow ball marked $p$ , holes) is DOWN, with arrows showing $n$ pushed up and $p$ pushed down. The caption reads: push $n$ up → $p$ slides down, product stays level.

Level 4 — Synthesis

Combine several ideas, or work a formula backwards.

L4.1

You measure a phosphorus-doped silicon sample and find conductivity $σ = 3.2 (Ω cm)^{- 1}$ . Assuming N-type with $μ_{n} = 1350 cm^{2} / Vs$ and the hole term negligible, find the doping level $N_{D}$ . Then find $p$ .

Recall Solution

Step 1 — invert conductivity for $n$ . From $σ \approx q n μ_{n}$ : $n = \frac{σ}{q μ _{n}} = \frac{3.2}{( 1.6 \times 1 0 ^{- 19} ) ( 1350 )} = \frac{3.2}{2.16 \times 1 0 ^{- 16}} \approx 1.48 \times 1 0^{16} cm^{- 3} .$ Step 2 — donors. With full ionisation, $N_{D} \approx n \approx 1.48 \times 1 0^{16} cm^{- 3}$ . Step 3 — holes. $p = n_{i}^{2} / n = (1.5 \times 1 0^{10})^{2} / (1.48 \times 1 0^{16}) \approx 1.52 \times 1 0^{4} cm^{- 3}$ . Sanity check: $p ≪ n$ confirms our decision to drop the hole term in Step 1 was self-consistent.

L4.2

The graph below plots $lo g_{10} n$ and $lo g_{10} p$ versus $lo g_{10} N_{D}$ for N-type silicon. Predict the slopes of the two lines before looking, then confirm from the figure. Where do the lines meet, and what does that crossing point physically mean?

Figure (s01): A log-log plot. Horizontal axis is $lo g_{10} N_{D}$ ; vertical axis is $lo g_{10}$ of carrier density. A blue line for $lo g_{10} n$ rises with slope $+ 1$ (it tracks $N_{D}$ one-for-one). A yellow line for $lo g_{10} p$ falls with slope $- 1$ . Two green dashed guide lines mark the level $n_{i}$ and the doping $N_{D} = n_{i}$ . A red dot marks where blue and yellow cross ( $n = p = n_{i}$ ), labelled "intrinsic limit". To the right of the dot is annotated "extrinsic (useful doping)".

Recall Solution

$n \approx N_{D}$ ⟹ on a log-log plot, $lo g n = lo g N_{D}$ : a line of slope $+ 1$ (the blue line).
$p = n_{i}^{2} / N_{D}$ ⟹ $lo g p = 2 lo g n_{i} - lo g N_{D}$ : a line of slope $- 1$ (the yellow line).
They cross where $n = p$ , i.e. where $N_{D} = n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ . At that crossing the material stops behaving "extrinsic" — the added donors no longer swamp the thermal carriers, and it looks nearly intrinsic again (see Intrinsic vs Extrinsic Semiconductors). For any useful doping ( $N_{D} ≫ n_{i}$ ) you are far to the right of the crossing.

Level 5 — Mastery

Push to limiting cases and edge conditions the formulas usually hide.

L5.1 (Low doping / intrinsic crossover)

Silicon is very lightly doped: $N_{D} = 3 \times 1 0^{10} cm^{- 3}$ , comparable to $n_{i} = 1.5 \times 1 0^{10}$ . The approximation $n \approx N_{D}$ fails here. Solve for the exact $n$ and $p$ using both $n p = n_{i}^{2}$ and $n = p + N_{D}$ .

Recall Solution

We cannot approximate because $N_{D}$ is not $≫ n_{i}$ . Combine the two exact equations. Substitute $p = n_{i}^{2} / n$ into $n = p + N_{D}$ : $n = \frac{n _{i}^{2}}{n} + N_{D} \Rightarrow n^{2} - N_{D} n - n_{i}^{2} = 0.$ This is the general quadratic that always holds. Solve with the quadratic formula: $n = \frac{N _{D} \pm N _{D}^{2} + 4 n _{i}^{2}}{2} .$ Why keep only the $+$ root? The discriminant $N_{D}^{2} + 4 n_{i}^{2}$ is larger than $N_{D}$ , so the $-$ root gives a negative $n$ . A carrier density can never be negative (you cannot have fewer than zero electrons), so that root is unphysical and we discard it. Only the $+$ root survives. Plug numbers ( $N_{D} = 3 \times 1 0^{10}$ , $n_{i} = 1.5 \times 1 0^{10}$ ):

$N_{D}^{2} = 9 \times 1 0^{20}$ , $4 n_{i}^{2} = 4 (2.25 \times 1 0^{20}) = 9 \times 1 0^{20}$ .
$9 \times 1 0^{20} + 9 \times 1 0^{20} = 1.8 \times 1 0^{21} \approx 4.243 \times 1 0^{10}$ .
$n = (3 \times 1 0^{10} + 4.243 \times 1 0^{10}) /2 \approx 3.62 \times 1 0^{10} cm^{- 3}$ .
$p = n_{i}^{2} / n = 2.25 \times 1 0^{20} /3.62 \times 1 0^{10} \approx 6.22 \times 1 0^{9} cm^{- 3}$ .

Note $n$ is not simply $N_{D}$ — the naive answer $3 \times 1 0^{10}$ is off by ~20%, and $p$ is not negligible here. This is exactly the crossover region from L4.2's figure.

L5.2 (Zero doping — degenerate check)

Set $N_{D} = 0$ in the exact quadratic $n^{2} - N_{D} n - n_{i}^{2} = 0$ from L5.1. Solve for $n$ and $p$ , and show the result correctly reproduces intrinsic silicon (where electrons and holes are equal).

Recall Solution

Put $N_{D} = 0$ into the quadratic: $n^{2} - (0) n - n_{i}^{2} = 0 \Rightarrow n^{2} = n_{i}^{2} \Rightarrow n = \pm n_{i} .$ A density cannot be negative, so we keep the positive root: $n = n_{i} = 1.5 \times 1 0^{10} cm^{- 3} .$ Then from the mass-action law: $p = \frac{n _{i}^{2}}{n} = \frac{n _{i}^{2}}{n _{i}} = n_{i} = 1.5 \times 1 0^{10} cm^{- 3} .$ So $n = p = n_{i}$ — exactly intrinsic silicon, where electrons and holes are equal because none were added by doping. The general formula degrades gracefully to the pure-crystal limit, which is our sign that it was set up correctly (see Intrinsic vs Extrinsic Semiconductors).

L5.3 (Heavy doping — the mobility catch)

Two samples: (A) $N_{D} = 1 \times 1 0^{16}$ with $μ_{n} = 1350$ ; (B) $N_{D} = 1 \times 1 0^{19}$ with $μ_{n} = 100$ (mobility has dropped from impurity scattering). Compute $σ$ for each. Did conductivity rise proportionally to $N_{D}$ ? Explain.

Figure (s04): A semi-log plot of electron mobility $μ_{n}$ (vertical axis) versus doping $N_{D}$ (horizontal, log scale). A green curve stays high and nearly flat at low doping, then plunges as $N_{D}$ climbs past $\sim 1 0^{17}$ due to ionised-impurity scattering. A blue dot marks sample A ( $N_{D} = 1 0^{16}$ , $μ_{n} \approx 1350$ ) on the flat part; a red dot marks sample B ( $N_{D} = 1 0^{19}$ , $μ_{n} \approx 100$ ) far down the plunge.

Recall Solution

Sample A: $σ_{A} = q N_{D} μ_{n} = (1.6 \times 1 0^{- 19}) (1 0^{16}) (1350) = 2.16 (Ω cm)^{- 1}$ .
Sample B: $σ_{B} = (1.6 \times 1 0^{- 19}) (1 0^{19}) (100) = 160 (Ω cm)^{- 1}$ .
$N_{D}$ rose by a factor $1000$ , but $σ$ rose only by $160/2.16 \approx 74$ — sub-linear. Why? $σ = q n μ_{n}$ : raising $N_{D}$ raises $n$ , but the extra ionised donors also scatter the electrons, dragging $μ_{n}$ down (the falling curve in the figure). The two effects fight, so conductivity grows more slowly than doping. This is the real-world ceiling the parent note warns about.

Active recall — final sweep

Recall Quick self-quiz (hide answers)

Which group makes a donor? ::: Group V (5 valence electrons), e.g. P, As. What is $N_{D}^{+}$ ? ::: A fixed, immobile positive donor ion — the donor after it released its electron. What is $m_{0}$ versus $m^{*}$ ? ::: $m_{0}$ is the free-electron mass in vacuum; $m^{*}$ is how heavy the electron behaves inside the lattice. Exact equation linking n, p, N_D (no approximations)? ::: $n^{2} - N_{D} n - n_{i}^{2} = 0$ , from $n p = n_{i}^{2}$ and $n = p + N_{D}$ . Why keep only the $+$ root of that quadratic? ::: The $-$ root is negative, and a carrier density cannot be negative. If N_D doubles, what happens to p? ::: It halves (mass-action law pins $n p = n_{i}^{2}$ ). Why is σ sub-linear at heavy doping? ::: Mobility $μ_{n}$ falls due to ionised-impurity scattering. What does the n=p crossing at $N_{D} = n_{i}$ mean? ::: Material behaves intrinsic; doping no longer dominates. $(Ω cm)^{- 1}$ equals what other unit? ::: $S/cm$ (siemens per cm), since $S = Ω^{- 1}$ .