Carrier concentration equations (n, p, ni): Worked examples

Common mistake One number, two values of

$n_{i}$ — read this before the examples You will see two intrinsic concentrations on this page and they are not a contradiction:

The formula value $n_{i} = N_{c} N_{v} e^{- E_{g} /2 k T} \approx 6.9 \times 1 0^{9} cm^{- 3}$ (Ex 1), computed from the rounded textbook $N_{c}, N_{v}, E_{g}$ .
The measured / textbook value $n_{i} = 1.0 \times 1 0^{10} cm^{- 3}$ for real Si at 300 K. They differ by ~30% purely because the tabulated $m^{*}$ and $E_{g}$ are slightly off from reality — see the accuracy note under Ex 1. Convention for this page: whenever we use mass-action $n p = n_{i}^{2}$ in a doping example, we plug the measured $n_{i} = 1.0 \times 1 0^{10}$ (so $n_{i}^{2} = 1 0^{20}$ ), because that is what a real device does. We only display the formula value in Ex 1 to show where it comes from. Watch for the tag [uses measured $n_{i}$ ] in each mass-action step.

The scenario matrix

Every problem this topic throws at you falls into one of these cells. Each row is a distinct sign / limit / degeneracy situation, and each is worked below.

Cell	Situation	Distinguishing feature	Example
A	Intrinsic	$N_{D} = N_{A} = 0$ , so $n = p = n_{i}$	Ex 1
B	Lightly n-type	$N_{D} ≫ n_{i}$ , one donor sign	Ex 2
C	Lightly p-type	$N_{A} ≫ n_{i}$ , one acceptor sign	Ex 3
D	Compensated	both $N_{D}$ and $N_{A}$ present, they fight	Ex 4
E	Net doping $\sim n_{i}$	$N_{D} - N_{A} \sim n_{i}$ , cannot say $n \approx N_{D}$	Ex 5
F	Temperature limit	$T \to$ high and $T \to$ low behaviour of $n_{i}$	Ex 6
G	Degenerate (Boltzmann fails)	$E_{F}$ within $3 k T$ of a band edge	Ex 7
H	Real-world word problem	resistivity target → required doping	Ex 8

Standard silicon numbers used throughout (300 K unless stated): $N_{c} = 2.8 \times 1 0^{19} cm^{- 3}$ , $N_{v} = 1.04 \times 1 0^{19} cm^{- 3}$ , $E_{g} = 1.12 eV$ , measured $n_{i} = 1.0 \times 1 0^{10} cm^{- 3}$ , $k T = 0.0259 eV$ .

Figure — Carrier concentration equations (n, p, ni)

Figure 1 (description): four vertical stacks labelled Intrinsic, n-type, p-type, Compensated. In each stack a magenta dot marks

lo g_{10} n

and a violet dot marks

lo g_{10} p

, joined by a dotted orange line. A horizontal pink dashed line at

lo g_{10} (n p) = 20

shows the product

n_{i}^{2}

staying fixed as the two dots slide oppositely; a faint navy line marks the

n_{i}

level (

lo g_{10} = 10

). The take-away: doping moves

n

and

p

in opposite directions but their product never leaves the pink line.

The figure is the map: doping slides the two dots ( $n$ and $p$ ) up and down the vertical scale, but the product (the pink dashed line, $n_{i}^{2}$ ) never moves. Every example below is one position of those dots. See Doping and charge neutrality and Fermi level position for the underlying pictures.

Cell A — Intrinsic silicon

Worked example Example 1 — Pure Si, find

$n_{i}$ and $E_{i}$ Undoped silicon at 300 K. Find $n$ , $p$ , and where the Fermi level $E_{i}$ sits.

Forecast: Before reading — will $n$ equal $p$ exactly? Will $E_{i}$ sit exactly at midgap or slightly off? Guess a direction.

Step 1 — Set the intrinsic condition $n = p$ . Why this step? With no dopants, every conduction electron came from breaking a valence bond, so it left exactly one hole: Intrinsic vs extrinsic semiconductors demands $n = p \equiv n_{i}$ .

Step 2 — Use $n_{i} = N_{c} N_{v} e^{- E_{g} /2 k T}$ . Why? Setting $n = p$ and using $n p = n_{i}^{2}$ forces the geometric mean $N_{c} N_{v}$ and halves the exponent. (The $/2$ is the square root: since $n p \propto e^{- E_{g} / k T}$ , taking $n_{i} = n p$ halves that exponent. Forgetting the $/2$ makes $n_{i}$ absurdly small.) $n_{i} = 2.8 \times 1 0^{19} \cdot 1.04 \times 1 0^{19} e^{- 1.12/ (2 \cdot 0.0259)}$ $N_{c} N_{v} = 1.707 \times 1 0^{19}$ ; exponent $= - 21.62$ ; $e^{- 21.62} = 4.06 \times 1 0^{- 10}$ . $n_{i} \approx 6.9 \times 1 0^{9} cm^{- 3} (formula value)$

Accuracy note. The measured room-temperature value is $n_{i} = 1.0 \times 1 0^{10} cm^{- 3}$ . Our $6.9 \times 1 0^{9}$ is ~30% low purely because the tabulated effective masses inside $N_{c}, N_{v}$ and the rounded $E_{g}$ are slightly off from reality. For all mass-action work below we adopt the measured $1.0 \times 1 0^{10}$ .

Step 3 — Locate $E_{i}$ . Why? $E_{i}$ is defined by $n = p$ ; solving gives $E_{i} = \frac{E _{c} + E _{v}}{2} + \frac{k T}{2} ln \frac{N _{v}}{N _{c}}$ . $\frac{k T}{2} ln \frac{N _{v}}{N _{c}} = \frac{0.0259}{2} ln \frac{1.04}{2.8} = - 0.0128 eV$ So $E_{i}$ sits $12.8 meV$ below midgap.

Verify: Because $N_{v} < N_{c}$ , the $ln$ term is negative, so $E_{i}$ dips below midgap — matches the forecast that it is not exactly centred. Units: $e V$ throughout. ✓ ( $n = p$ exactly, as required.)

Cell B — Lightly doped n-type

Worked example Example 2 — Donors

$N_{D} = 1 0^{16}$ , find $n, p, E_{c} - E_{F}$ Phosphorus-doped Si, $N_{D} = 1 0^{16} cm^{- 3}$ , fully ionised, 300 K.

Forecast: Is $p$ bigger or smaller than in pure Si? By roughly how many orders of magnitude?

Step 1 — Is $N_{D} ≫ n_{i}$ ? $1 0^{16} ≫ 1 0^{10}$ , yes. Why this step? It tells us charge neutrality $n = p + N_{D}$ simplifies, because $p$ will be tiny.

Step 2 — Take $n \approx N_{D} = 1 0^{16} cm^{- 3}$ . Why? Neutrality $n = p + N_{D}$ with $p ≪ N_{D}$ leaves $n \approx N_{D}$ . Every donor gave up one electron.

Step 3 — Get $p$ from mass-action $p = n_{i}^{2} / n$ . [uses measured $n_{i}$ ] Why? $n p = n_{i}^{2}$ holds for any doping (the $E_{F}$ cancelled), so $p$ falls as $n$ rises. $p = \frac{( 1.0 \times 1 0 ^{10} ) ^{2}}{1 0 ^{16}} = 1 0^{4} cm^{- 3}$

Step 4 — Fermi level: $E_{c} - E_{F} = k T ln (N_{c} / n)$ . Why? Invert $n = N_{c} e^{- (E_{c} - E_{F}) / k T}$ . $E_{c} - E_{F} = 0.0259 ln \frac{2.8 \times 1 0 ^{19}}{1 0 ^{16}} = 0.0259 ln (2800) = 0.206 eV$

Verify: Check self-consistency $p ≪ N_{D}$ : $1 0^{4} ≪ 1 0^{16}$ ✓. And $n p = 1 0^{16} \cdot 1 0^{4} = 1 0^{20} = n_{i}^{2}$ ✓. $E_{F}$ sits below $E_{c}$ (fewer carriers than seats) — matches the mnemonic "electrons fall from the top." ✓

Cell C — Lightly doped p-type

Cell D — Compensated (donors AND acceptors)

Worked example Example 4 —

$N_{D} = 8 \times 1 0^{16}$ , $N_{A} = 3 \times 1 0^{16}$ Both dopants present, fully ionised. Find $n$ and $p$ .

Forecast: Will the material behave n-type or p-type? Guess $n$ before computing.

Step 1 — Net doping wins. Effective donor density $= N_{D} - N_{A} = 5 \times 1 0^{16} cm^{- 3}$ . Why this step? Acceptors soak up donor electrons; only the excess stays free. Neutrality: $n = p + (N_{D} - N_{A})$ .

Step 2 — Derive the exact root. Why? We have two equations: neutrality $n = p + (N_{D} - N_{A})$ and mass-action $p = n_{i}^{2} / n$ . Substitute the second into the first: $n = \frac{n _{i}^{2}}{n} + (N_{D} - N_{A}) .$ Multiply by $n$ and collect: $n^{2} - (N_{D} - N_{A}) n - n_{i}^{2} = 0$ . This is a quadratic $a n^{2} + bn + c = 0$ with $a = 1, b = - (N_{D} - N_{A}), c = - n_{i}^{2}$ . The physical (positive) root of $n = \frac{- b + b ^{2} - 4 a c}{2 a}$ is $n = \frac{N _{D} - N _{A}}{2} + (\frac{N _{D} - N _{A}}{2})^{2} + n_{i}^{2}$ This formula is exact and safe for every case; for $N_{D} - N_{A} ≫ n_{i}$ the root ≈ $(N_{D} - N_{A}) /2$ , recovering $n \approx N_{D} - N_{A}$ .

Step 3 — Plug in the numbers exactly. [uses measured $n_{i}$ ] With $N_{D} - N_{A} = 5 \times 1 0^{16}$ , $n_{i} = 1.0 \times 1 0^{10}$ : $\frac{N _{D} - N _{A}}{2} = 2.5 \times 1 0^{16}, (2.5 \times 1 0^{16})^{2} = 6.25 \times 1 0^{32}, n_{i}^{2} = 1.0 \times 1 0^{20} .$ Under the root: $6.25 \times 1 0^{32} + 1.0 \times 1 0^{20} = 6.2500000001 \times 1 0^{32}$ — the $n_{i}^{2}$ term is utterly negligible, so $= 2.5 \times 1 0^{16}$ to 12 significant figures. $n = 2.5 \times 1 0^{16} + 2.5 \times 1 0^{16} = 5.0 \times 1 0^{16} cm^{- 3} .$

Step 4 — $p = n_{i}^{2} / n = (1.0 \times 1 0^{10})^{2} /5.0 \times 1 0^{16} = 2 \times 1 0^{3} cm^{- 3}$ .

Verify: Neutrality check: $n = p + (N_{D} - N_{A}) = 2 \times 1 0^{3} + 5 \times 1 0^{16} \approx 5 \times 1 0^{16}$ ✓ (the $2 \times 1 0^{3}$ is negligible). $n p = n_{i}^{2}$ ✓. Behaves n-type — forecast confirmed. ✓ Note the exact root and the approximation agree here because $N_{D} - N_{A} ≫ n_{i}$ ; Cell E is where they diverge.

Cell E — Net doping comparable to $n_{i}$ (the exact quadratic bites)

Worked example Example 5 —

$N_{D} - N_{A} = 1.5 \times 1 0^{10}$ (same order as $n_{i}$ ) A heavily-compensated sample where net doping $\approx n_{i}$ . You may not say $n \approx N_{D} - N_{A}$ .

Forecast: Will $n$ be close to $N_{D} - N_{A}$ , or noticeably larger? Why can't we ignore $p$ here?

Step 1 — Recognise the danger. Here $N_{D} - N_{A} = 1.5 \times 1 0^{10} \sim n_{i} = 1 0^{10}$ . Why this step? The approximation " $p ≪$ net doping" is false — $p$ and $n$ are comparable, so we must use the full quadratic root derived in Ex 4 Step 2.

Step 2 — Plug into the exact formula. [uses measured $n_{i}$ ] $n = \frac{1.5 \times 1 0 ^{10}}{2} + (0.75 \times 1 0^{10})^{2} + (1.0 \times 1 0^{10})^{2}$ $= 0.75 \times 1 0^{10} + 0.5625 \times 1 0^{20} + 1 \times 1 0^{20}$ $= 0.75 \times 1 0^{10} + 1.25 \times 1 0^{10} = 2.0 \times 1 0^{10} cm^{- 3}$ .

Step 3 — $p = n_{i}^{2} / n = (1.0 \times 1 0^{10})^{2} / (2.0 \times 1 0^{10}) = 5.0 \times 1 0^{9} cm^{- 3}$ .

Verify: Neutrality: $n - p = 2.0 \times 1 0^{10} - 0.5 \times 1 0^{10} = 1.5 \times 1 0^{10} = N_{D} - N_{A}$ ✓. Note $n$ is 1.33× the naive $N_{D} - N_{A}$ — the intrinsic carriers matter here, exactly the trap Cell E warns about. ✓

Cell F — Temperature limits of $n_{i}$ (and a word on freeze-out)

Worked example Example 6 —

$n_{i}$ at 250 K and 400 K Same Si, sweep temperature (both temperatures are high enough that donors stay fully ionised). Compute $n_{i}$ at $T_{1} = 250 K$ and $T_{2} = 400 K$ , and state the dominant mechanism.

Forecast: By what factor does $n_{i}$ change from 250 K to 400 K — a few percent, or many orders of magnitude?

Step 1 — Recompute $k T$ and $N_{c}, N_{v}$ (they scale as $T^{3/2}$ ). Why this step? $N_{c} = 2 (2 π m_{n}^{*} k T / h^{2})^{3/2}$ , so $N_{c} (T) = N_{c} (300) (T /300)^{3/2}$ . Same for $N_{v}$ . At 250 K: $k T = 0.02156 eV$ , factor $(250/300)^{3/2} = 0.760$ . At 400 K: $k T = 0.03449 eV$ , factor $(400/300)^{3/2} = 1.540$ .

Step 2 — $n_{i} (T) = N_{c} (T) N_{v} (T) e^{- E_{g} /2 k T}$ (using the formula prefactor $1.707 \times 1 0^{19}$ at 300 K). At 250 K: prefactor $= 1.707 \times 1 0^{19} \cdot 0.760 = 1.297 \times 1 0^{19}$ ; exponent $= - 1.12/ (2 \cdot 0.02156) = - 25.97$ ; $e^{- 25.97} = 5.25 \times 1 0^{- 12}$ . $n_{i} (250 K) \approx 6.8 \times 1 0^{7} cm^{- 3}$ At 400 K: prefactor $= 1.707 \times 1 0^{19} \cdot 1.540 = 2.629 \times 1 0^{19}$ ; exponent $= - 1.12/ (2 \cdot 0.03449) = - 16.24$ ; $e^{- 16.24} = 8.85 \times 1 0^{- 8}$ . $n_{i} (400 K) \approx 2.3 \times 1 0^{12} cm^{- 3}$

Verify: Ratio $n_{i} (400) / n_{i} (250) \approx 3.4 \times 1 0^{4}$ — five orders swing over 150 K. The exponential $e^{- E_{g} /2 k T}$ dominates; the $T^{3/2}$ prefactor only doubled. This is why the parent's forecast said "up, dominated by the exponential." ✓

Figure 2 (description): a log-scale curve of

n_{i}

(magenta) versus temperature

T

from 200 K to 500 K, rising steeply. Two highlighted points mark the worked answers — a violet dot at 250 K (

\approx 6.8 \times 1 0^{7}

) and an orange dot at 400 K (

\approx 2.3 \times 1 0^{12}

) — with a dotted navy reference line at 300 K. The curve's near-straight rise on a log axis shows the exponential

e^{- E_{g} /2 k T}

dominating.

Cell G — Degenerate: Boltzmann breaks

Worked example Example 7 —

$N_{D} = 2 \times 1 0^{19}$ , is Boltzmann valid? Very heavy n-doping, $N_{D} = 2 \times 1 0^{19} \approx N_{c}$ . Estimate $E_{c} - E_{F}$ naively and test the validity flag.

Forecast: Will the naive $n = N_{c} e^{- (E_{c} - E_{F}) / k T}$ put $E_{F}$ below, at, or above $E_{c}$ ?

Step 1 — Naive inversion. Take $n \approx N_{D} = 2 \times 1 0^{19}$ : $E_{c} - E_{F} = k T ln \frac{N _{c}}{n} = 0.0259 ln \frac{2.8 \times 1 0 ^{19}}{2 \times 1 0 ^{19}} = 0.0259 ln (1.4) = 0.0087 eV$

Step 2 — Apply the validity flag $E_{c} - E_{F} ≳ 3 k T$ . Why this step? Boltzmann needs the "+1" in Fermi-Dirac distribution to be negligible, which requires $E_{F}$ at least $\sim 3 k T = 0.078 eV$ below $E_{c}$ . Here $E_{c} - E_{F} = 0.0087 eV$ , which is only $0.34 k T$ — far inside the flag.

Step 3 — Conclusion: DEGENERATE. Why? $E_{F}$ is essentially at the band edge, so occupancy saturates near 1 and Boltzmann over-counts. The correct tool is the Fermi–Dirac integral $F_{1/2}$ , not the exponential. See Density of states and Effective mass for the full $F_{1/2}$ treatment.

Verify: $0.0087/0.0259 = 0.34 < 3$ → flag tripped. The naive number is a red flag, not an answer. Any time doping approaches $N_{c}$ or $N_{v}$ , expect degeneracy. ✓

Cell H — Real-world word problem

Worked example Example 8 — "Design a resistor" (word problem → doping)

An engineer needs n-type Si with resistivity $ρ = 0.5 Ω \cdot cm$ . Electron mobility $μ_{n} = 1350 cm^{2} / (V\cdots)$ , electron charge $q = 1.6 \times 1 0^{- 19} C$ . What donor density $N_{D}$ (fully ionised) is required, and find $p$ ?

Forecast: Roughly what order of magnitude of $N_{D}$ gives sub-ohm-cm Si — $1 0^{14}$ ? $1 0^{16}$ ? $1 0^{18}$ ?

Step 1 — Conductivity link. Resistivity is $ρ = 1/ σ$ and for n-type $σ \approx q n μ_{n}$ . Why this step? Conduction is dominated by the majority carrier $n$ ; holes are negligible in n-type. This converts an electrical spec into a carrier count. $n = \frac{1}{ρ q μ _{n}} = \frac{1}{0.5 \cdot 1.6 \times 1 0 ^{- 19} \cdot 1350}$

Step 2 — Compute. $n = \frac{1}{1.08 \times 1 0 ^{- 16}} = 9.26 \times 1 0^{15} cm^{- 3}$ So $N_{D} \approx n = 9.3 \times 1 0^{15} cm^{- 3}$ (since $N_{D} ≫ n_{i}$ , $n \approx N_{D}$ ).

Step 3 — Minority holes. [uses measured $n_{i}$ ] $p = n_{i}^{2} / n = (1.0 \times 1 0^{10})^{2} /9.26 \times 1 0^{15} = 1.08 \times 1 0^{4} cm^{- 3}$ . Why? Even in a device-grade resistor the mass-action law fixes the minority carrier.

Verify: Units of Step 1: $\frac{1}{( Ω cm ) ( C ) ( cm ^{2} / V s )} = \frac{1}{Ω C cm ^{3} / ( V s )}$ ; using $Ω = V/A = V s/C$ gives $\frac{1}{cm ^{3}}$ ✓. Order of magnitude $1 0^{16}$ — matches the forecast band. Cross-check with Ex 2: $N_{D} = 1 0^{16}$ gave $ρ \approx 1/ (1.6 \times 1 0^{- 19} \cdot 1 0^{16} \cdot 1350) = 0.46 Ω cm$ , close to $0.5$ ✓.

Recall Matrix self-test (click to reveal)

Which cell requires the quadratic neutrality formula rather than $n \approx N_{D}$ ? ::: Cell E — when net doping is comparable to $n_{i}$ . Which cell makes the Boltzmann exponential invalid, and what replaces it? ::: Cell G (degenerate); use the Fermi–Dirac integral $F_{1/2}$ . In a compensated sample which quantity actually sets the carrier type? ::: The net doping $N_{D} - N_{A}$ (Cell D). Over 250→400 K, roughly how many orders does $n_{i}$ climb? ::: About 4–5 orders, driven by $e^{- E_{g} /2 k T}$ (Cell F). Which $n_{i}$ do we use in mass-action calculations, and why? ::: The measured $1.0 \times 1 0^{10}$ , because that is what a real device exhibits; the formula value $6.9 \times 1 0^{9}$ is only for showing the derivation. What is dopant freeze-out and which temperature regime does it belong to? ::: Low $T$ (below ~100 K) where $k T < E_{D}$ so donors stop ionising and $N_{D}^{+} < N_{D}$ ; distinct from the high- $T$ $n_{i}$ rise of Cell F.

Related: Doping and charge neutrality · Fermi level position · Band gap Eg · Intrinsic vs extrinsic semiconductors