Carrier concentration equations (n, p, ni): Exercises

Definition The toolbox you'll reuse (every symbol earned)

Before any problem, here is every quantity, in plain words:

$n$ = number of mobile electrons per cubic centimetre in the conduction band. Think "how many people are up in the balcony."
$p$ = number of holes (empty electron seats) per cubic centimetre in the valence band. "Empty chairs on the main floor."
$n_{i}$ = the intrinsic carrier count — how many of each you get in a pure crystal at a given temperature.
$N_{c}$ , $N_{v}$ = the effective number of seats at the conduction edge and valence edge (as if the whole band squeezed into one level).
$E_{c}$ = energy of the conduction-band floor; $E_{v}$ = energy of the valence-band ceiling; $E_{g} = E_{c} - E_{v}$ = the gap (the height of the jump).
$E_{F}$ = the Fermi level — the "occupancy dial." Where it sits tells you how full each band is.
$E_{i}$ = the Fermi level of a pure crystal (near midgap).
$k T$ = thermal energy. At 300 K, $k T = 0.0259 eV$ — memorise this number.

The five equations you will lean on: $n = N_{c} e^{- (E_{c} - E_{F}) / k T}, p = N_{v} e^{- (E_{F} - E_{v}) / k T}$ $n p = n_{i}^{2}, n_{i} = N_{c} N_{v} e^{- E_{g} /2 k T}$ $n = n_{i} e^{(E_{F} - E_{i}) / k T}, p = n_{i} e^{(E_{i} - E_{F}) / k T}$

The picture below anchors what "measure down from $E_{c}$ " and "measure up from $E_{v}$ " mean — refer back to it whenever a problem asks where $E_{F}$ sits.

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

L1 · Recognition

Problem 1.1

Which formula gives the product $n p$ that is independent of doping, and what does it equal?

Recall Solution

WHAT we want: the doping-independent product. WHY it exists: in $n = N_{c} e^{- (E_{c} - E_{F}) / k T}$ and $p = N_{v} e^{- (E_{F} - E_{v}) / k T}$ , multiplying makes the $E_{F}$ terms cancel (one has $+ E_{F}$ , the other $- E_{F}$ ). $n p = N_{c} N_{v} e^{- (E_{c} - E_{v}) / k T} = N_{c} N_{v} e^{- E_{g} / k T} = n_{i}^{2}$ Answer: the mass-action law, $n p = n_{i}^{2}$ . It holds for any doping because $E_{F}$ dropped out.

Problem 1.2

At 300 K, state the numeric value of $k T$ in eV, and convert a Fermi-level shift of one decade of carriers into eV.

Recall Solution

$k T = 0.0259 eV$ at 300 K. A "decade" means $\times 10$ in $n$ . From $n = N_{c} e^{- (E_{c} - E_{F}) / k T}$ , multiplying $n$ by 10 needs the exponent to grow by $ln 10 = 2.303$ , i.e. $E_{c} - E_{F}$ drops by $k T ln 10 = 0.0259 \times 2.303 = 0.0596 eV$ . Answer: $k T = 0.0259$ eV; one decade $\approx 0.0596$ eV $\approx 60$ meV.

L2 · Application

Problem 2.1

Compute $n_{i}$ for silicon at 300 K. Given: $N_{c} = 2.8 \times 1 0^{19} cm^{- 3}$ , $N_{v} = 1.04 \times 1 0^{19} cm^{- 3}$ , $E_{g} = 1.12 eV$ , $k T = 0.0259 eV$ .

Recall Solution

Step 1 — geometric mean of the seat counts. WHY: intrinsic means $n = p$ , which forces $n_{i} = N_{c} N_{v}$ . $N_{c} N_{v} = 2.8 \times 1 0^{19} \cdot 1.04 \times 1 0^{19} = 1.706 \times 1 0^{19} cm^{- 3}$ Step 2 — the affordability factor. WHY: $e^{- E_{g} /2 k T}$ is the probability a floor electron can afford the jump. $\frac{E _{g}}{2 k T} = \frac{1.12}{2 \times 0.0259} = 21.62, e^{- 21.62} = 4.07 \times 1 0^{- 10}$ Step 3 — multiply. $n_{i} = 1.706 \times 1 0^{19} \times 4.07 \times 1 0^{- 10} \approx 6.9 \times 1 0^{9} cm^{- 3}$ Answer: $n_{i} \approx 7 \times 1 0^{9} cm^{- 3}$ (textbooks quote $\sim 1 \times 1 0^{10}$ ; the gap is rounding in $m^{*}$ / $E_{g}$ ).

Problem 2.2

An n-type sample has $n = 1 0^{16} cm^{- 3}$ . Find the minority-hole concentration $p$ , using $n_{i} = 1 0^{10} cm^{- 3}$ .

Recall Solution

WHY mass-action: $n p = n_{i}^{2}$ holds no matter the doping, so once you know $n$ , you know $p$ . $p = \frac{n _{i}^{2}}{n} = \frac{( 1 0 ^{10} ) ^{2}}{1 0 ^{16}} = \frac{1 0 ^{20}}{1 0 ^{16}} = 1 0^{4} cm^{- 3}$ Answer: $p = 1 0^{4} cm^{- 3}$ . Notice holes are twelve orders of magnitude below electrons — that's what "minority carrier" means.

L3 · Analysis

Problem 3.1

For the n-type sample above ( $n = 1 0^{16}$ , $N_{c} = 2.8 \times 1 0^{19}$ ), find how far $E_{F}$ sits below $E_{c}$ at 300 K.

Recall Solution

WHY this formula: invert $n = N_{c} e^{- (E_{c} - E_{F}) / k T}$ to isolate the distance $E_{c} - E_{F}$ . Fewer electrons than seats ( $n < N_{c}$ ) means $E_{F}$ must sit below the edge — see the figure. $E_{c} - E_{F} = k T ln \frac{N _{c}}{n} = 0.0259 ln \frac{2.8 \times 1 0 ^{19}}{1 0 ^{16}}$ $= 0.0259 ln (2800) = 0.0259 \times 7.937 = 0.206 eV$ Answer: $E_{F}$ lies $\approx 0.206 eV$ below $E_{c}$ . Add more donors → $n$ rises → the ratio $N_{c} / n$ shrinks → $E_{F}$ climbs toward $E_{c}$ .

Problem 3.2

Locate the intrinsic level $E_{i}$ relative to midgap for silicon. Use $N_{c} = 2.8 \times 1 0^{19}$ , $N_{v} = 1.04 \times 1 0^{19}$ , $k T = 0.0259$ eV.

Recall Solution

WHY there's an offset: setting $n = p$ gives $E_{i} = \frac{E _{c} + E _{v}}{2} + \frac{k T}{2} ln \frac{N _{v}}{N _{c}}$ . The extra term appears because $N_{c} \neq = N_{v}$ (the two bands don't have equal seat counts). $\frac{k T}{2} ln \frac{N _{v}}{N _{c}} = \frac{0.0259}{2} ln \frac{1.04 \times 1 0 ^{19}}{2.8 \times 1 0 ^{19}} = 0.01295 ln (0.3714)$ $= 0.01295 \times (- 0.9905) = - 0.0128 eV$ Answer: $E_{i}$ sits $\approx 12.8 meV$ below midgap. Tiny — so " $E_{i} \approx$ midgap" is a fine rule of thumb.

L4 · Synthesis

Problem 4.1

A silicon sample is doped with acceptors $N_{A} = 5 \times 1 0^{17} cm^{- 3}$ (all ionized), plus a smaller donor background $N_{D} = 1 \times 1 0^{17} cm^{- 3}$ . Using $n_{i} = 1 0^{10}$ , find $p$ and $n$ at 300 K.

Recall Solution

Step 1 — net doping via charge neutrality. WHY: donors give electrons, acceptors take them; only the net count matters. Net acceptors: $N_{A} - N_{D} = 5 \times 1 0^{17} - 1 \times 1 0^{17} = 4 \times 1 0^{17} cm^{- 3}$ This is a p-type sample (acceptors win). Step 2 — majority holes. WHY: since $N_{A} - N_{D} ≫ n_{i}$ , essentially $p \approx N_{A} - N_{D}$ . $p \approx 4 \times 1 0^{17} cm^{- 3}$ Step 3 — minority electrons via mass-action. $n = \frac{n _{i}^{2}}{p} = \frac{( 1 0 ^{10} ) ^{2}}{4 \times 1 0 ^{17}} = \frac{1 0 ^{20}}{4 \times 1 0 ^{17}} = 250 cm^{- 3}$ Answer: $p \approx 4 \times 1 0^{17} cm^{- 3}$ , $n \approx 250 cm^{- 3}$ .

Problem 4.2

Same sample. How far is $E_{F}$ from the intrinsic level $E_{i}$ ? Is it above or below?

Recall Solution

WHY the $E_{i}$ form: $p = n_{i} e^{(E_{i} - E_{F}) / k T}$ measures the Fermi level relative to midgap directly. Solve for $E_{i} - E_{F}$ : $E_{i} - E_{F} = k T ln \frac{p}{n _{i}} = 0.0259 ln \frac{4 \times 1 0 ^{17}}{1 0 ^{10}}$ $= 0.0259 ln (4 \times 1 0^{7}) = 0.0259 \times 17.505 = 0.453 eV$ Since $E_{i} - E_{F} > 0$ , we have $E_{F}$ below $E_{i}$ — as expected for p-type (Fermi level drops toward the valence band). See the amber marker in the figure. Answer: $E_{F}$ lies $\approx 0.453 eV$ below $E_{i}$ .

L5 · Mastery

Problem 5.1

At what temperature does silicon's intrinsic concentration $n_{i}$ equal $1 0^{16} cm^{- 3}$ ? Treat $N_{c} N_{v}$ as roughly constant at its 300 K value $N_{c} N_{v} = 1.71 \times 1 0^{19}$ , and use $E_{g} = 1.12$ eV, $k = 8.617 \times 1 0^{- 5} eV/K$ . (This is the temperature where a $1 0^{16}$ device "goes intrinsic" and stops working.)

Recall Solution

WHY this matters: when $n_{i}$ climbs up to your doping level, doping no longer dominates — the transistor loses control. We solve $n_{i} (T) = 1 0^{16}$ . Step 1 — isolate the exponent. $\frac{n _{i}}{N _{c} N _{v}} = \frac{1 0 ^{16}}{1.71 \times 1 0 ^{19}} = 5.85 \times 1 0^{- 4} = e^{- E_{g} /2 k T}$ Step 2 — take logs. $- \frac{E _{g}}{2 k T} = ln (5.85 \times 1 0^{- 4}) = - 7.443$ Step 3 — solve for $T$ . $T = \frac{E _{g}}{2 k \times 7.443} = \frac{1.12}{2 \times 8.617 \times 1 0 ^{- 5} \times 7.443} = \frac{1.12}{1.283 \times 1 0 ^{- 3}} \approx 873 K$ Answer: $T \approx 8.7 \times 1 0^{2} K$ ( $\approx 60 0^{\circ}$ C). (Ignoring the mild $T^{3/2}$ growth of $N_{c}, N_{v}$ ; including it lowers this slightly — a good self-check for the truly ambitious.)

Problem 5.2 (defend a subtlety)

A student doped silicon to $n = 2 \times 1 0^{19} cm^{- 3} \approx N_{c}$ and used $n = N_{c} e^{- (E_{c} - E_{F}) / k T}$ to place $E_{F}$ . Compute the "distance $E_{c} - E_{F}$ " this gives, then explain in one sentence why the number should not be trusted.

Recall Solution

Step 1 — plug in anyway. $E_{c} - E_{F} = k T ln \frac{N _{c}}{n} = 0.0259 ln \frac{2.8 \times 1 0 ^{19}}{2 \times 1 0 ^{19}} = 0.0259 ln (1.4) = 0.0259 \times 0.3365$ $= 0.00871 eV \approx 8.7 meV$ Step 2 — why distrust it. The Boltzmann approximation $f \approx e^{- (E - E_{F}) / k T}$ requires $E_{c} - E_{F} ≳ 3 k T \approx 78 meV$ . Here the result ( $9$ meV) is inside that forbidden zone — $E_{F}$ is essentially at the band edge, the "+1" in Fermi–Dirac matters, and occupancy saturates at 1. The crystal is degenerate; you must use the Fermi–Dirac integral $F_{1/2}$ , not this formula. Answer: formula gives $\approx 8.7 meV$ , but it's invalid — the sample is degenerate ( $E_{c} - E_{F} < 3 k T$ ), so Boltzmann over-counts and the true $E_{F}$ may even lie inside the band.

Q: Why does the mass-action product

n p

not depend on doping?

Because

n \propto e^{+ E_{F} / k T}

and

p \propto e^{- E_{F} / k T}

, so multiplying cancels

E_{F}

, leaving only

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

Q: One-step route to minority carriers?

p = n_{i}^{2} / n

(or

n = n_{i}^{2} / p

), valid at any doping.

Q: The validity check for the Boltzmann carrier formulas?

E_{c} - E_{F} ≳ 3 k T

(n-type) or

E_{F} - E_{v} ≳ 3 k T

(p-type); else the sample is degenerate.

Q: In a compensated sample, what sets the majority carrier?

The net doping

∣ N_{A} - N_{D} ∣

, not

N_{A}

N_{D}

alone.