Carrier concentration equations (n, p, ni): Question bank

Before we start, we must earn every symbol these questions lean on. Look at the band-diagram figure below while reading the definitions — each symbol is anchored to a place on that picture.

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

Definition The band-diagram vocabulary (every symbol used on this page)

Energy runs upward on the vertical axis. On the sketch:

$E_{c}$ = the conduction-band edge — the lowest energy a mobile electron can have. Everything above the $E_{c}$ line is the conduction band (the "balcony" with empty seats).
$E_{v}$ = the valence-band edge — the highest energy of the packed-full band. Everything below the $E_{v}$ line is the valence band (the "main floor").
$E_{g} = E_{c} - E_{v}$ = the band gap, the forbidden energy strip between the two edges. (See Band gap Eg.)
$E_{F}$ = the Fermi level, the energy where a state has a 50% chance of being filled — the "water line" of the electron sea. (See Fermi level position.)
$n$ = mobile electrons per unit volume in the conduction band; $p$ = holes (missing electrons) per unit volume in the valence band.
$n_{i}$ = the value both $n$ and $p$ take in a pure (intrinsic) crystal — a property of the material and temperature only.

Definition The three functions the "spot the error" items rely on

$g_{c} (E)$ and $g_{v} (E)$ = the density of states: how many available seats exist per unit energy per unit volume, in the conduction and valence bands respectively. Near an edge $g_{c} (E) \propto E - E_{c}$ — the count grows like a square root as you climb (more $k$ -states on a bigger sphere). See Density of states. In the figure this is the shaded curve that fattens away from each edge.
$f (E) = \frac{1}{1 + e ^{(E - E_{F}) / k T}}$ = the Fermi–Dirac occupancy: the probability that a seat at energy $E$ is filled. It equals $1$ far below $E_{F}$ , drops through $\frac{1}{2}$ at $E = E_{F}$ , and falls toward $0$ far above. See Fermi-Dirac distribution.
$1 - f (E)$ = the probability a seat is empty — this is what holes are counted with.
Combine them: $n = \int_{E_{c}}^{\infty} g_{c} (E) f (E) d E$ (seats × filled), and after the Boltzmann simplification $n = N_{c} e^{- (E_{c} - E_{F}) / k T}$ , $p = N_{v} e^{- (E_{F} - E_{v}) / k T}$ .

True or false — justify

True or false:

n_{i}

increases when you add more donor atoms.

False.

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

contains only

N_{c}, N_{v}, E_{g}, T

— doping never appears. Donors raise

n

and lower

p

but leave their product

n_{i}^{2}

untouched.

True or false: in a doped semiconductor

n p

is still equal to

n_{i}^{2}

True (in equilibrium). The derivation of

n p = N_{c} N_{v} e^{- E_{g} / k T}

cancels

E_{F}

entirely, so the product is fixed by material and temperature regardless of doping level.

True or false:

n p = n_{i}^{2}

holds even when a battery is driving current through the device.

False. Mass-action assumes thermal equilibrium; under bias or illumination the carrier populations are pushed out of equilibrium and

n p

can exceed (injection) or fall below (extraction)

n_{i}^{2}

True or false: doubling temperature roughly doubles

n_{i}

False — the change is far larger.

n_{i} \propto T^{3/2} e^{- E_{g} /2 k T}

; the exponential dominates, so a modest temperature rise multiplies

n_{i}

by orders of magnitude, not by two.

True or false:

N_{c}

is the actual number of states sitting exactly at the energy

E_{c}

False.

N_{c}

is an effective count — the whole spread-out conduction band collapsed into one imaginary level at

E_{c}

that would hold the same Boltzmann-weighted number of states. No real states live precisely at

E_{c}

in that quantity.

True or false: if

E_{F}

sits exactly at midgap, the semiconductor is necessarily intrinsic.

Mostly false.

E_{F}

at midgap gives

n \approx p

, but the true intrinsic level

E_{i}

is nudged off midgap by

\frac{k T}{2} ln (N_{v} / N_{c})

. Being intrinsic means

E_{F} = E_{i}

, which is near — but not exactly at — the geometric middle.

True or false: the Boltzmann approximation

f (E) \approx e^{- (E - E_{F}) / k T}

works for any doping.

False. It fails once

E_{F}

approaches within about

3 k T

of a band edge (heavy doping/degeneracy), where the "+1" in Fermi–Dirac matters and Boltzmann over-counts.

True or false: holes are positively-charged particles that physically exist inside the crystal.

Partly false — holes are the absence of an electron in the valence band. The bookkeeping treats that absence as a mobile positive charge with its own effective mass, but there is no literal positive particle sitting there.

True or false: because electrons and holes are created in pairs,

n

always equals

p

False. Only in an intrinsic crystal does

n = p = n_{i}

. Doping breaks the symmetry: a donor supplies an electron without making a hole, so

n > p

(n-type) or

p > n

(p-type). See Intrinsic vs extrinsic semiconductors.

Spot the error

"To get

n_{i}

, use

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

The exponent is wrong: it should be

E_{g} /2 k T

. Since

n_{i} = n p

and

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

, the square root halves the exponent. Using

E_{g}

makes

n_{i}

absurdly (squared) too small.

n = N_{c} e^{- (E_{F} - E_{c}) / k T}

Sign error. Because

E_{c} > E_{F}

in a non-degenerate n-type gap, the correct exponent measures down from

E_{c}

n = N_{c} e^{- (E_{c} - E_{F}) / k T}

. The written form would give

n > N_{c}

, which is impossible for a non-degenerate band.

"The integral for

p

uses

f (E)

, the probability a state is filled."

Wrong occupancy factor. A hole is an empty valence state, so

p

uses

1 - f (E)

, the probability the state is empty:

p = \int_{- \infty}^{E_{v}} g_{v} (E) [1 - f (E)] d E

"Since more states exist higher in the band,

g_{c} (E)

grows linearly with energy."

It grows as a square root,

g_{c} \propto E - E_{c}

, not linearly. The count comes from the surface of a

k

-sphere converted to energy, which produces the

E - E_{c}

dependence.

"The integration limit for

n

starts at

E_{F}

It starts at

E_{c}

, the conduction band edge. Electrons only conduct once they occupy states above

E_{c}

;

E_{F}

merely sets the occupancy, it is not a band boundary.

n p = n_{i}^{2}

lets me claim

n

and

p

are always small."

The product being fixed says nothing about the individual sizes. In heavy n-type doping

n

can be

1 0^{16}

while

p

drops to

1 0^{4}

— one is huge, the other tiny, product still

n_{i}^{2}

N_{c}

and

N_{v}

are equal because both bands are symmetric."

They differ because

N_{c} \propto (m_{n}^{*})^{3/2}

and

N_{v} \propto (m_{p}^{*})^{3/2}

, and the electron and hole effective masses are generally unequal (for Si,

N_{c} \approx 2.8 \times 1 0^{19}

N_{v} \approx 1.04 \times 1 0^{19}

^{- 3}

Why questions

Why do the

E_{F}

terms cancel when you multiply

n

and

p

Because

n \propto e^{- (E_{c} - E_{F}) / k T}

carries

+ E_{F}

and

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

carries

- E_{F}

; multiplying adds the exponents so

E_{F}

vanishes, leaving only the gap

E_{c} - E_{v} = E_{g}

Why does the intrinsic level

E_{i}

not sit exactly at midgap?

Because

n = p

requires balancing

N_{c}

against

N_{v}

, which are unequal. The correction

\frac{k T}{2} ln (N_{v} / N_{c})

pushes

E_{i}

slightly toward the band with the smaller effective density of states.

Why is

N_{c} N_{v}

(a geometric mean) the right prefactor for

n_{i}

, not the average?

Setting

n = p

n = N_{c} / N_{v} \dots

style algebra, the condition forces

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

; taking the square root gives

N_{c} N_{v}

. The geometric mean falls out of the product

n p

, not a sum.

Why does raising temperature increase

n_{i}

so dramatically?

More thermal energy lets more electrons afford the jump across

E_{g}

; the occupancy factor

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

rises steeply as

T

grows, swamping the mild

T^{3/2}

growth of

N_{c}, N_{v}

Why can we replace the full Fermi–Dirac function with a simple exponential for lightly-doped silicon?

For lightly-doped material

E_{F}

sits deep in the gap, so

E - E_{F} ≫ k T

for all conduction states; then

e^{(E - E_{F}) / k T} ≫ 1

and the "+1" in the denominator is negligible, leaving the Boltzmann tail.

Why do we integrate (states × occupancy) rather than just counting filled states directly?

We can't enumerate

1 0^{18}

electrons; instead we count available seats per energy via

g (E)

and multiply by the fraction occupied

f (E)

, then sum over energy. It converts an impossible count into a tractable integral.

Why does

E_{F}

move toward

E_{c}

as we add more donors?

More electrons fill more high-energy states, raising the "water line." Since

n = N_{c} e^{- (E_{c} - E_{F}) / k T}

, a larger

n

demands a smaller

E_{c} - E_{F}

, i.e.

E_{F}

climbs toward the conduction edge. See Doping and charge neutrality.

Edge cases

Edge case — what happens to

n

E_{F}

climbs above

E_{c}

(degenerate doping)?

The Boltzmann formula breaks: occupancy saturates near 1 and the "+1" in Fermi–Dirac matters, so

n

grows slower than

e^{(E_{F} - E_{c}) / k T}

would predict; one must use the Fermi–Dirac integral

F_{1/2}

Edge case — as

T \to 0

, what do

n

p

, and

n_{i}

approach?

All approach zero. With no thermal energy no electron can cross the gap, so

e^{- E_{g} /2 k T} \to 0

; the crystal becomes a perfect insulator with an empty conduction band.

Edge case — impurity freeze-out: in a doped crystal as

T \to 0

, is

n \approx N_{D}

still true?

No. At very low temperature the donor electrons no longer have enough thermal energy to ionize, so they stay bound to their donor atoms ("freeze-out") and

n

drops below

N_{D}

; full ionization (

n \approx N_{D}

) only holds over the middle "extrinsic" temperature range.

Edge case — a hypothetical zero-gap material (

E_{g} \to 0

): what does

n_{i}

become?

e^{- E_{g} /2 k T} \to e^{0} = 1

, so

n_{i} \to N_{c} N_{v}

— a huge intrinsic concentration. With no gap the material behaves like a semimetal, electrons flooding the conduction band even at low temperature.

Edge case — an n-type sample where

N_{D}

is comparable to

n_{i}

: is

n \approx N_{D}

still valid?

No. The clean

n \approx N_{D}

approximation assumes

N_{D} ≫ n_{i}

. When they are comparable you must solve charge neutrality

n = p + N_{D}

together with

n p = n_{i}^{2}

exactly, because thermally-generated carriers are no longer negligible.

Edge case — compensated material with equal donors and acceptors (

N_{D} = N_{A}

): what is

n

The extra electrons and holes cancel, so the sample behaves intrinsically:

n = p = n_{i}

. Charge neutrality reduces to

n = p

, recovering the intrinsic condition despite the crystal being doped.

Edge case — does

n p = n_{i}^{2}

survive if the two bands have very different effective masses?

Yes. Unequal masses change

N_{c}

N_{v}

(and shift

E_{i}

), but the product

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

still holds; the mass asymmetry is absorbed into

N_{c} N_{v}

Recall One-line summary of the traps (click to reveal)

$n_{i}$ is a material property (never doping); $n p = n_{i}^{2}$ needs equilibrium; Boltzmann needs non-degeneracy; signs measure down from $E_{c}$ , up from $E_{v}$ ; and every "obvious" symmetry ( $N_{c} = N_{v}$ , $E_{i}$ at midgap, $n = p$ ) is broken by unequal effective masses or doping.