Minority vs majority carriers: Worked examples — Hardware deep dive

Quick symbol reminder, so line one is readable by anyone:

The scenario matrix

Every carrier problem is one of these cells. The examples below are labelled with the cell they hit.

Cell	Situation	What decides the answer	Approximation OK?
A	n-type, $N_{D} ≫ n_{i}$	majority $\approx N_{D}$	yes, use $n \approx N_{D}$
B	p-type, $N_{A} ≫ n_{i}$	majority $\approx N_{A}$	yes, use $p \approx N_{A}$
C	undoped ( $N_{D} = N_{A} = 0$ )	intrinsic: $n = p = n_{i}$	exact by symmetry
D	light doping, $N_{D} \sim n_{i}$	must solve the quadratic	no — approximation fails
E	compensation, both dopants, $N_{A} \neq = N_{D}$	net dopant $	N_A-N_D
F	exact compensation, $N_{A} = N_{D}$	net $= 0$ ⇒ behaves intrinsic	exact: $n = p = n_{i}$
G	temperature change (limiting behaviour)	$n_{i} (T)$ rises fast ⇒ everything shifts	quadratic near onset
H	real-world / exam twist	translate words → the two equations	depends

The master pair of equations, valid every time (thermal equilibrium, full ionisation):

The green hyperbola above is the single most important picture on this page: the point $(n, p)$ always lands on that curve. Doping just slides you along it — push $n$ right and $p$ must drop, because the product is locked at $n_{i}^{2}$ .

Worked examples

Cell A — heavily doped n-type

Worked example

A. Silicon, $n_{i} = 1.0 \times 1 0^{10} cm^{- 3}$ , doped $N_{D} = 1.0 \times 1 0^{17} cm^{- 3}$ . Find majority and minority concentrations, and the ratio.

Forecast: guess the majority (round number?) and guess how many orders of magnitude separate majority from minority before reading on.

Check the regime: is $N_{D} ≫ n_{i}$ ? Here $1 0^{17} ≫ 1 0^{10}$ — yes, by seven orders. Why this step? The clean approximation $n \approx N_{D}$ is only licensed when donors vastly outnumber the intrinsic trickle; otherwise you'd need the quadratic (Cell D).
Majority electrons: $n \approx N_{D} = 1.0 \times 1 0^{17} cm^{- 3}$ . Why this step? In neutrality $n = p + N_{D}$ , the hole term $p$ is minuscule, so $n \approx N_{D}$ .
Minority holes: $p = \frac{n _{i}^{2}}{n} = \frac{( 1 0 ^{10} ) ^{2}}{1 0 ^{17}} = 1 0^{3} cm^{- 3}$ . Why this step? Mass action pins the product; once $n$ is known, $p$ is forced.
Ratio majority/minority $= 1 0^{17} /1 0^{3} = 1 0^{14}$ .

Verify: product $n \cdot p = 1 0^{17} \times 1 0^{3} = 1 0^{20} = (1 0^{10})^{2} = n_{i}^{2}$ . ✓ Units: (cm⁻³)(cm⁻³) = cm⁻⁶, matches $n_{i}^{2}$ . ✓

Cell B — heavily doped p-type

Cell C — undoped (the symmetric baseline)

Cell D — light doping (approximation FAILS)

Worked example

D. Silicon, $n_{i} = 1.0 \times 1 0^{10}$ , doped only $N_{D} = 5.0 \times 1 0^{9} cm^{- 3}$ (n-type).

Forecast: the naive answer would say $n \approx N_{D} = 5 \times 1 0^{9}$ . Is that too high, too low, or right?

Regime check: is $N_{D} ≫ n_{i}$ ? No — here $N_{D} = 5 \times 1 0^{9}$ is smaller than $n_{i} = 1 0^{10}$ . Do not use $n \approx N_{D}$ . Why this step? When doping is comparable to (or below) the intrinsic level, the hole term in neutrality is not negligible; ignoring it gives a wrong number.
Solve exactly. From neutrality $n = p + N_{D}$ and mass action $p = n_{i}^{2} / n$ : $n = \frac{n _{i}^{2}}{n} + N_{D} \Rightarrow n^{2} - N_{D} n - n_{i}^{2} = 0.$ Why this step? Substituting one equation into the other collapses two unknowns to one quadratic in $n$ .
Positive root (concentrations can't be negative): $n = \frac{N _{D}}{2} + (\frac{N _{D}}{2})^{2} + n_{i}^{2} .$ Plug in: $\frac{N _{D}}{2} = 2.5 \times 1 0^{9}$ , and $(2.5 \times 1 0^{9})^{2} + (1 0^{10})^{2} = 6.25 \times 1 0^{18} + 1 0^{20} = 1.0625 \times 1 0^{20} = 1.0308 \times 1 0^{10}$ .
So $n = 2.5 \times 1 0^{9} + 1.0308 \times 1 0^{10} = 1.2808 \times 1 0^{10} cm^{- 3}$ , and $p = n_{i}^{2} / n = 1 0^{20} /1.2808 \times 1 0^{10} = 7.808 \times 1 0^{9} cm^{- 3}$ .

Verify: $n - p = 1.2808 \times 1 0^{10} - 7.808 \times 1 0^{9} = 5.0 \times 1 0^{9} = N_{D}$ ✓ (neutrality). Note $n$ is only $\approx 28%$ above $n_{i}$ — the naive $5 \times 1 0^{9}$ would have been badly wrong (too low, and even the wrong side of $n_{i}$ ).

Cell E — compensation, net acceptor

Cell F — exact compensation (a degenerate input)

Cell G — temperature limiting behaviour

Worked example

G. Take the Cell A sample ( $N_{D} = 1 0^{17} cm^{- 3}$ ) and heat it until $n_{i}$ rises to $5 \times 1 0^{16} cm^{- 3}$ (small-band-gap material or high $T$ ).

Forecast: at low $T$ , $n \approx N_{D}$ . As $n_{i}$ climbs toward $N_{D}$ , does the material stay clearly n-type or start looking intrinsic?

Now $n_{i} = 5 \times 1 0^{16}$ is comparable to $N_{D} = 1 0^{17}$ — approximation region D applies (thermally, we've entered the intrinsic-onset regime). Why this step? $n_{i}$ grows steeply with temperature; once it is not $≪ N_{D}$ , thermal pairs rival the dopant electrons and we must solve exactly.
Quadratic: $n = \frac{N _{D}}{2} + (\frac{N _{D}}{2})^{2} + n_{i}^{2}$ . $\frac{N _{D}}{2} = 5 \times 1 0^{16}$ , $(\frac{N _{D}}{2})^{2} = 2.5 \times 1 0^{33}$ , $n_{i}^{2} = 2.5 \times 1 0^{33}$ , sum $= 5 \times 1 0^{33}$ , $= 7.071 \times 1 0^{16}$ .
$n = 5 \times 1 0^{16} + 7.071 \times 1 0^{16} = 1.2071 \times 1 0^{17} cm^{- 3}$ ; $p = n_{i}^{2} / n = 2.5 \times 1 0^{33} /1.2071 \times 1 0^{17} = 2.071 \times 1 0^{16} cm^{- 3}$ .

Verify: $n - p = 1.2071 \times 1 0^{17} - 2.071 \times 1 0^{16} = 1.000 \times 1 0^{17} = N_{D}$ ✓. Limiting insight: the minority $p$ has ballooned from $1 0^{3}$ (Cell A) to $2 \times 1 0^{16}$ — the majority/minority ratio collapsed from $1 0^{14}$ to under $6$ . Heat enough and every doped semiconductor drifts toward intrinsic behaviour.

Cell H — exam-style word twist

Worked example

H. "A silicon resistor ( $n_{i} = 1 0^{10}$ ) is measured to have $p = 4.0 \times 1 0^{4} cm^{- 3}$ at 300 K. Is it n-type or p-type, and what is the doping?"

Forecast: you're given the minority-looking small number. Which carrier is it, and how do you back out the dopant?

Compare $p$ with $n_{i}$ : $p = 4 \times 1 0^{4} ≪ n_{i} = 1 0^{10}$ . Why this step? A carrier concentration below $n_{i}$ can only be the minority carrier (holes here). So holes are scarce ⇒ electrons are majority ⇒ the sample is n-type.
Get the majority from mass action: $n = n_{i}^{2} / p = 1 0^{20} / (4 \times 1 0^{4}) = 2.5 \times 1 0^{15} cm^{- 3}$ . Why this step? The product is fixed; knowing the minority gives the majority directly.
Doping from neutrality: $N_{D} = n - p = 2.5 \times 1 0^{15} - 4 \times 1 0^{4} \approx 2.5 \times 1 0^{15} cm^{- 3}$ .

Verify: $n p = 2.5 \times 1 0^{15} \times 4 \times 1 0^{4} = 1 0^{20} = n_{i}^{2}$ ✓. And $p < n_{i} < n$ , consistent with n-type ✓.

Reading the whole matrix at once

Recall

Why does the minority explode when temperature rises? ::: $n_{i}$ grows steeply with $T$ , and minority $= n_{i}^{2} / majority$ — the numerator grows while the majority stays $\approx$ dopant, so the minority shoots up toward intrinsic. A sample shows a carrier concentration below $n_{i}$ . Which carrier is it? ::: The minority carrier — nothing can be pushed below $n_{i}$ except the scarce type. Two wafers, same dopants but one has $N_{A} = N_{D}$ . Which behaves intrinsic? ::: The exactly-compensated one ( $N_{A} = N_{D}$ ): net dopant zero ⇒ $n = p = n_{i}$ .

Connections

Parent: Minority vs majority carriers
Law of mass action — the product constraint used in every cell.
Charge neutrality condition — the difference constraint $n - p = N_{D} - N_{A}$ .
Intrinsic carrier concentration $n_i$ — the yardstick every " $≫$ " is measured against, and what rises with temperature in Cell G.
Doping — donors and acceptors — where $N_{D}$ , $N_{A}$ come from.
Fermi level position vs doping — the same shift viewed in energy.
Diffusion and drift currents and pn junction diode — where these minority numbers become device currents.