Intrinsic vs extrinsic semiconductors: Worked examples

This page is the drill hall for Intrinsic vs extrinsic semiconductors. The parent note built the ideas; here we make sure no case can surprise you — every sign of doping, every degenerate input, every limit, plus a real-world word problem and an exam trap.

Before we compute anything, let us agree on the symbols so nothing appears unearned.

Definition The symbols we will reuse

$n$ — the number of free electrons per cubic centimetre ( $cm^{- 3}$ ). "How many wanderers can carry current."
$p$ — the number of holes per $cm^{- 3}$ . "How many empty seats drift around."
$n_{i}$ — the intrinsic carrier concentration: in a pure crystal $n = p = n_{i}$ . It is fixed once temperature is fixed.
$N_{D}$ — donor concentration: group-V atoms sprinkled in, each donating one electron.
$N_{A}$ — acceptor concentration: group-III atoms, each creating one hole.
$N_{c}, N_{v}$ — the effective density of states in the conduction band and valence band respectively ( $cm^{- 3}$ ). Loosely: "how many seats are available to sit in near the edge of each band." They come from the density-of-states counting done in Band gap and energy bands; we treat them here as known material numbers.
$k$ — the Boltzmann constant, $k = 8.617 \times 1 0^{- 5} eV/K$ . It converts a temperature $T$ (in kelvin) into a thermal energy $k T$ . "How much random jiggling energy each degree of temperature buys."
$T$ — absolute temperature in kelvin (K). Then $k T$ is the thermal energy that breaks bonds; at $300 K$ , $k T = 0.02585 eV$ .

The one law tying these together is the Law of Mass Action, $n p = n_{i}^{2}$ , true at equilibrium for any doping. See Fermi level and Fermi-Dirac statistics for where it comes from.

The scenario matrix

Every problem this topic throws is one of these cells. The examples below are tagged with the cell they cover.

Cell	Situation	What decides the method
A	Pure crystal	$n = p = n_{i}$ directly
B	n-type, $N_{D} ≫ n_{i}$	$n \approx N_{D}$ , then $p = n_{i}^{2} / N_{D}$
C	p-type, $N_{A} ≫ n_{i}$	$p \approx N_{A}$ , then $n = n_{i}^{2} / N_{A}$
D	p-type doping comparable to $n_{i}$ ( $N_{A} \sim n_{i}$ )	shortcut fails → neutrality quadratic for $p$
D′	n-type doping comparable to $n_{i}$ ( $N_{D} \sim n_{i}$ )	shortcut fails → neutrality quadratic for $n$
E	Zero / degenerate input ( $N_{A} = 0$ , or $T \to 0$ )	limiting behaviour
F	Both donors and acceptors present (compensation), net $≫ n_{i}$	net doping $
F′	Compensation with net $	N_D-N_A
G	Temperature limit — device "goes intrinsic"	solve $n_{i} (T) = N_{D}$
H	Real-world word problem + exam twist	translate words → the cells above

Unless stated, use silicon at $300 K$ with $n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ and band gap $E_{g} = 1.12 eV$ , with $k T = 0.02585 eV$ at $300 K$ (see Band gap and energy bands).

The whole matrix collapses to one branching question — carry this picture in your head:

Figure — Intrinsic vs extrinsic semiconductors

Example 1 — Cell A: the pure crystal (baseline)

Example 2 — Cell B: strong n-type

Worked example n-type, both carriers

Silicon is doped with phosphorus (group V), $N_{D} = 5 \times 1 0^{16} cm^{- 3}$ . Find $n$ and $p$ .

Forecast: by roughly what factor will electrons beat holes?

Check the shortcut condition $N_{D} ≫ n_{i}$ : $5 \times 1 0^{16} ≫ 1.5 \times 1 0^{10}$ — true by a factor of millions. Why this step? The shortcut $n \approx N_{D}$ is only legal when doping dwarfs intrinsic pairs.
Set majority carrier $n \approx N_{D} = 5 \times 1 0^{16} cm^{- 3}$ . Why? Each donor releases one nearly-free electron (Silicon crystal structure and covalent bonding), and thermal pairs are negligible beside them.
Get the minority carrier from mass action $p = n_{i}^{2} / n$ . Why? $n p = n_{i}^{2}$ always holds; we know $n$ , so $p$ is forced. $p = \frac{( 1.5 \times 1 0 ^{10} ) ^{2}}{5 \times 1 0 ^{16}} = \frac{2.25 \times 1 0 ^{20}}{5 \times 1 0 ^{16}} = 4.5 \times 1 0^{3} cm^{- 3}$

Verify: $n p = (5 \times 1 0^{16}) (4.5 \times 1 0^{3}) = 2.25 \times 1 0^{20} = n_{i}^{2}$ . ✓ Electrons beat holes by $\sim 1 0^{13}$ — decisively n-type.

Example 3 — Cell C: strong p-type

Example 4 — Cell D: p-type doping comparable to $n_{i}$ (the shortcut trap)

This is the cell where the careless student loses marks. When doping is not hugely larger than $n_{i}$ , we cannot say $p \approx N_{A}$ . We must enforce two facts at once.

Combined with $n p = n_{i}^{2}$ , this gives one equation in one unknown. Substituting $n = n_{i}^{2} / p$ — why substitute? because we want a single equation in the single unknown $p$ , and mass action lets us trade the unknown $n$ for $n_{i}^{2} / p$ : $p = \frac{n _{i}^{2}}{p} + N_{A} ⟹ p^{2} - N_{A} p - n_{i}^{2} = 0$ Multiplying through by $p$ turned it into a quadratic — why a quadratic? because two competing conditions (neutrality and mass action) generically meet at the roots of a degree-2 equation; there is no linear shortcut when the two terms are comparable. Solving with the quadratic formula: $p = \frac{N _{A}}{2} + (\frac{N _{A}}{2})^{2} + n_{i}^{2}$

We take the + root — why? because a concentration cannot be negative, and the $-$ root would give $p < 0$ .

The geometry of why the shortcut fails is in the figure below.

Worked example Weakly p-type sample

$N_{A} = 5 \times 1 0^{9} cm^{- 3}$ , $n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ . Find $p$ and $n$ . Is it "clearly p-type"?

Forecast: will $p \approx N_{A} = 5 \times 1 0^{9}$ ? (It won't — watch.)

Spot the trap: $N_{A} < n_{i}$ . Why this step? The shortcut needs $N_{A} ≫ n_{i}$ ; here doping is smaller than intrinsic, so we go to the quadratic.
Compute the root. $\frac{N _{A}}{2} = 2.5 \times 1 0^{9}$ . Then $(\frac{N _{A}}{2})^{2} + n_{i}^{2} = (2.5 \times 1 0^{9})^{2} + (1.5 \times 1 0^{10})^{2} = 6.25 \times 1 0^{18} + 2.25 \times 1 0^{20} = 2.3125 \times 1 0^{20}$ $2.3125 \times 1 0^{20} = 1.52069 \times 1 0^{10}$ $p = 2.5 \times 1 0^{9} + 1.52069 \times 1 0^{10} = 1.77069 \times 1 0^{10} cm^{- 3}$ Why the square root? It is the solution of the quadratic — the only way to satisfy neutrality and mass action simultaneously.
Electrons from mass action: $n = n_{i}^{2} / p = 2.25 \times 1 0^{20} /1.77069 \times 1 0^{10} = 1.27069 \times 1 0^{10} cm^{- 3}$ .

Verify: Neutrality $p - n = 1.77069 \times 1 0^{10} - 1.27069 \times 1 0^{10} = 5.0 \times 1 0^{9} = N_{A}$ . ✓ Mass action $n p = 2.25 \times 1 0^{20} = n_{i}^{2}$ . ✓ Note $p$ only slightly exceeds $n$ ( $1.77$ vs $1.27$ ) — the sample is barely p-type, not the clean $p = N_{A}$ the shortcut would wrongly predict.

Example 5 — Cell D′: n-type doping comparable to $n_{i}$ (mirror trap)

The same trap exists on the n-side, and the algebra mirrors Example 4. Now donors are positive fixed ions, so charge neutrality for an n-type-only sample reads: $n = p + N_{D}$ "electrons = holes + the fixed positive donor ions." Substituting $p = n_{i}^{2} / n$ — why substitute? to reduce everything to one unknown $n$ , exactly as before: $n^{2} - N_{D} n - n_{i}^{2} = 0 ⟹ n = \frac{N _{D}}{2} + (\frac{N _{D}}{2})^{2} + n_{i}^{2}$ We keep the + root — why? a negative electron concentration is physically impossible.

Example 6 — Cell E: zero and degenerate inputs

To probe the temperature limit here (and again in Cell G), we need the full temperature model of $n_{i}$ .

Worked example Turning off the doping and the temperature

(a) Put $N_{A} = 0$ into the quadratic of Example 4. (b) What happens to $n_{i}$ as $T \to 0$ ?

Forecast: should the quadratic collapse back to the pure-crystal answer?

Set $N_{A} = 0$ in $p = \frac{N _{A}}{2} + (\frac{N _{A}}{2})^{2} + n_{i}^{2}$ . Why? Zero doping must reproduce Cell A, or our formula is wrong. $p = 0 + 0 + n_{i}^{2} = n_{i}$ and then $n = n_{i}^{2} / p = n_{i}$ . So $n = p = n_{i}$ — exactly the intrinsic case. The general formula contains the pure crystal as a limit. ✓
Temperature limit $n_{i} (T) = N_{c} N_{v} e^{- E_{g} /2 k T}$ . As $T \to 0$ , the exponent $- E_{g} /2 k T \to - \infty$ , so $e^{- \infty} \to 0$ . Why? With no thermal energy, no bond can break — a semiconductor at absolute zero is a perfect insulator (Temperature dependence of resistance in semiconductors). (This is also where dopants "freeze out" — the full-ionisation assumption fails alongside.)

Verify: At $N_{A} = 0$ the closed form gives $p = n_{i}^{2} = n_{i} = 1.5 \times 1 0^{10}$ , matching Example 1. ✓ The $T \to 0$ limit gives $n_{i} \to 0$ : no carriers. ✓ (No sign ambiguity ever arises: we always chose the $+$ root, guaranteeing $p > 0$ .)

Example 7 — Cell F: compensation (donors AND acceptors together)

Real wafers sometimes contain both impurity types. The dopants partly cancel — only the net count matters.

Example 8 — Cell F′: compensation with net doping near $n_{i}$

What if donors and acceptors nearly cancel, leaving a net count comparable to $n_{i}$ ? The shortcut fails again — but the fix is elegant: replace $N_{D}$ (or $N_{A}$ ) in the earlier quadratic by the net doping.

Worked example Nearly-cancelling dopants

$N_{D} = 3.1 \times 1 0^{10}$ , $N_{A} = 3.0 \times 1 0^{10} cm^{- 3}$ . Find the type, $n$ , and $p$ .

Forecast: with net doping only $1 \times 1 0^{9}$ (below $n_{i}$ !), will $n \approx Δ$ ?

Net doping $Δ = 3.1 \times 1 0^{10} - 3.0 \times 1 0^{10} = 1.0 \times 1 0^{9} cm^{- 3} > 0$ → n-type, but barely. Why this step? Only the net excess acts as effective doping; here it is smaller than $n_{i}$ , flagging the trap.
Use the quadratic with $Δ$ . $\frac{Δ}{2} = 5 \times 1 0^{8}$ . Then $(\frac{Δ}{2})^{2} + n_{i}^{2} = (5 \times 1 0^{8})^{2} + (1.5 \times 1 0^{10})^{2} = 2.5 \times 1 0^{17} + 2.25 \times 1 0^{20} = 2.2525 \times 1 0^{20}$ $2.2525 \times 1 0^{20} = 1.50083 \times 1 0^{10}$ $n = 5 \times 1 0^{8} + 1.50083 \times 1 0^{10} = 1.55083 \times 1 0^{10} cm^{- 3}$ Why the quadratic? Net doping $\sim n_{i}$ , so both terms of neutrality matter — no linear shortcut.
Holes $p = n_{i}^{2} / n = 2.25 \times 1 0^{20} /1.55083 \times 1 0^{10} = 1.45083 \times 1 0^{10} cm^{- 3}$ .

Verify: Neutrality $n - p = 1.55083 \times 1 0^{10} - 1.45083 \times 1 0^{10} = 1.0 \times 1 0^{9} = Δ$ . ✓ Mass action $n p = 2.25 \times 1 0^{20} = n_{i}^{2}$ . ✓ The sample is only a whisker n-type — a shortcut answer of $n \approx 1 0^{9}$ would be off by 15×.

Example 9 — Cell G: the temperature ceiling

Worked example When does an n-type device go intrinsic?

An n-type Si device is doped $N_{D} = 1 \times 1 0^{15} cm^{- 3}$ . It loses its designed behaviour once $n_{i} (T)$ climbs to equal $N_{D}$ . Using the model $n_{i} (T) = 3.9 \times 1 0^{16} T^{3/2} e^{- E_{g} /2 k T} cm^{- 3}$ (from Example 6) with $E_{g} = 1.12 eV$ and $k = 8.617 \times 1 0^{- 5} eV/K$ , is the device still safely extrinsic at $T = 450 K$ ?

Forecast: at $450 K$ , do you expect $n_{i}$ to be well below $1 0^{15}$ , or dangerously close?

Compute the exponent $- E_{g} / (2 k T)$ at $T = 450$ : $2 k T = 2 (8.617 \times 1 0^{- 5}) (450) = 0.077553 eV$ ; exponent $= - 1.12/0.077553 = - 14.4419$ . Why? The Boltzmann factor governs how many bonds break; the exponent is the whole story.
Evaluate $n_{i} (450) = 3.9 \times 1 0^{16} \times 45 0^{1.5} \times e^{- 14.4419}$ . Here $45 0^{1.5} = 9545.94$ and $e^{- 14.4419} = 5.336 \times 1 0^{- 7}$ . $n_{i} (450) = 3.9 \times 1 0^{16} \times 9545.94 \times 5.336 \times 1 0^{- 7} \approx 1.986 \times 1 0^{14} cm^{- 3}$
Compare with $N_{D} = 1 0^{15}$ . Why? Extrinsic behaviour survives while $n_{i} ≪ N_{D}$ . Here $n_{i} \approx 2 \times 1 0^{14}$ is only ~5× below $N_{D}$ — the safety margin is thinning.

Verify: $n_{i} (450) / N_{D} \approx 1.986 \times 1 0^{14} /1 0^{15} = 0.199$ , i.e. intrinsic carriers are already ~20% of the doping. Still extrinsic but marginal — a real chip would be de-rated below this. ✓

Example 10 — Cell H: a real-world word problem with an exam twist

Worked example Sizing a diode's lightly-doped side (and the twist)

A PN junction diode has an n-side doped $N_{D} = 1 \times 1 0^{16} cm^{- 3}$ . The reverse saturation current depends on the minority-hole concentration on that side. (a) Find $p$ . (b) By what factor does $p$ change if the doping is doubled? (c) Exam twist: a lab technician instead lightly dopes a fresh wafer to $N_{D} = 1 \times 1 0^{10} cm^{- 3}$ "to be gentle." Is the shortcut $n \approx N_{D}$ still valid? What is the true $n$ ?

Forecast: (b) doubling $N_{D}$ — does minority $p$ halve, double, or stay put? (c) is "gentle" doping safe to shortcut?

Translate words to a cell. "n-side, heavily doped, want minority holes" = Cell B. Why? Word problems are disguised matrix cells; naming the cell picks the tool.
(a) Minority holes $p = n_{i}^{2} / N_{D} = 2.25 \times 1 0^{20} /1 \times 1 0^{16} = 2.25 \times 1 0^{4} cm^{- 3}$ . Why? Mass action fixes minority carriers; these are what leak across a reverse-biased junction.
(b) Double the doping to $N_{D}^{'} = 2 \times 1 0^{16}$ : $p^{'} = 2.25 \times 1 0^{20} /2 \times 1 0^{16} = 1.125 \times 1 0^{4} cm^{- 3}$ . Why? $p \propto 1/ N_{D}$ , so doubling doping halves minority holes — and thus roughly halves that leakage current.
(c) The twist. Now $N_{D} = 1 \times 1 0^{10} < n_{i} = 1.5 \times 1 0^{10}$ : this is Cell D′, not B! Why this step? The shortcut needs $N_{D} ≫ n_{i}$ , which fails here. Use the quadratic (from Example 5, which used exactly these numbers): $n = 2.08114 \times 1 0^{10} cm^{- 3}$ , more than double the naive $1 0^{10}$ . The "gentle" wafer is only weakly n-type.

Verify: (a)/(b) $p / p^{'} = 2.25 \times 1 0^{4} /1.125 \times 1 0^{4} = 2$ exactly — inverse relationship holds. ✓ Units: $cm^{- 6} / cm^{- 3} = cm^{- 3}$ . ✓ (c) $n = 2.08114 \times 1 0^{10}$ vs naive $1 \times 1 0^{10}$ : shortcut error factor $\approx 2.08$ . ✓ (Recall from Carrier mobility and drift current that fewer minority carriers means a cleaner diode.)

Recall check

Recall When is the shortcut

$p \approx N_{A}$ illegal, and what replaces it? When $N_{A}$ is not $≫ n_{i}$ (comparable or smaller). Replace it with the charge-neutrality quadratic $p^{2} - N_{A} p - n_{i}^{2} = 0$ , taking $p = \frac{N _{A}}{2} + (\frac{N _{A}}{2})^{2} + n_{i}^{2}$ .

Recall Under compensation, what plays the role of "the doping" in every formula?

The net excess $Δ = N_{D} - N_{A}$ . If $∣Δ∣ ≫ n_{i}$ use the shortcut on $Δ$ ; if $∣Δ∣ \sim n_{i}$ use the quadratic with $Δ$ .

Recall Why is full dopant ionisation assumed, and when does it fail?

Because the donor/acceptor level sits ≈ $0.05 eV$ from its band, far inside $k T$ 's reach at $300 K$ . It fails at very low $T$ (freeze-out) or extremely heavy doping.

Flashcards

What replaces

p \approx N_{A}

when

N_{A} \sim n_{i}

The charge-neutrality quadratic

p = \frac{N _{A}}{2} + (\frac{N _{A}}{2})^{2} + n_{i}^{2}

What replaces

n \approx N_{D}

when

N_{D} \sim n_{i}

n = \frac{N _{D}}{2} + (\frac{N _{D}}{2})^{2} + n_{i}^{2}

Under compensation with

N_{D} > N_{A}

, the majority electron count is?

n \approx N_{D} - N_{A}

(if the net exceeds

n_{i}

Under compensation with net

∣ N_{D} - N_{A} ∣ \sim n_{i}

, what do you do?

Use the neutrality quadratic with

Δ = N_{D} - N_{A}

in place of the doping.

T \to 0

, what happens to

n_{i}

n_{i} \to 0

— no bonds break, perfect insulator (and dopants freeze out).

Doubling

N_{D}

does what to minority holes

p

Halves them, since

p = n_{i}^{2} / N_{D}

Setting

N_{A} = 0

in the neutrality quadratic gives?

p = n_{i}

, recovering the intrinsic case.

Why is full dopant ionisation assumed at 300 K?

The dopant level is ≈

0.05 eV

from its band, well within thermal reach

k T = 0.02585 eV

What do

N_{c}

and

N_{v}

mean?

The effective density of states (available seats) near the conduction- and valence-band edges;

n_{i} = N_{c} N_{v} e^{- E_{g} /2 k T}