Intrinsic vs extrinsic semiconductors: Question bank

This page has no new formulas — it only stress-tests the ones you already met: the pair $n = p = n_{i}$ (pure crystal), the Law of Mass Action $n p = n_{i}^{2}$ , and the doping shortcuts $n \approx N_{D}$ , $p \approx N_{A}$ . Every reveal explains why.

The three-region temperature map (keep this picture open)

Before the traps, anchor the one curve that ties this whole topic together — electron concentration in a doped sample versus temperature. It has three regions, and half the edge-case traps below are really just "which region am I in?"

Figure — Intrinsic vs extrinsic semiconductors

True or false — justify

TRUE / FALSE — At

T = 0

K an intrinsic semiconductor is a perfect insulator.

True — every valence electron is locked in a covalent bond, so no thermal energy is available to break bonds and there are zero free carriers. See Silicon crystal structure and covalent bonding.

TRUE / FALSE — In an n-type crystal there are literally no holes.

False — holes survive as minority carriers at

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

; they are few but they drive diode and transistor action.

TRUE / FALSE — Doping adds net electric charge to the crystal.

False — each donor also brings an extra proton (and each acceptor one fewer), so the fixed ionised dopant exactly balances the mobile carrier; the crystal stays neutral.

TRUE / FALSE — The Law of Mass Action

n p = n_{i}^{2}

only holds for pure (intrinsic) material.

False — it holds at thermal equilibrium for any doping level; that universality is precisely what lets us compute minority carriers.

TRUE / FALSE — Raising temperature increases

n_{i}

but leaves the majority carrier of a heavily-doped n-type sample roughly fixed at

N_{D}

True in the extrinsic region — while

N_{D} ≫ n_{i} (T)

the electrons stay pinned near

N_{D}

(the plateau); only when

n_{i} (T)

catches up does the sample "go intrinsic."

TRUE / FALSE — A larger band gap

E_{g}

makes a material a better conductor.

False —

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

, so a bigger gap means exponentially fewer thermal carriers; diamond (

E_{g} \approx 5.5

eV) is an insulator.

TRUE / FALSE — Doubling the doping always doubles the useful conductivity.

False — heavier doping raises impurity scattering, which lowers mobility; there is an optimum, not "more is always better."

TRUE / FALSE — In a pure crystal, breaking one bond creates one electron and one hole simultaneously.

True — a freed electron is the vacated bond, so pair generation is the only intrinsic mechanism, forcing

n = p = n_{i}

TRUE / FALSE — A donor electron needs the full band-gap energy

E_{g}

to become free.

False — the extra fifth electron is bound only weakly (≈0.05 eV), far less than

E_{g} \approx 1.12

eV, so it ionises easily at room temperature. See Band gap and energy bands.

Spot the error

ERROR? "Sample:

N_{D} = 1 0^{16}

n_{i} = 1.5 \times 1 0^{10}

, so

p = N_{D} - n_{i}

Wrong subtraction — minority carriers come from mass action,

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

, not

N_{D} - n_{i}

. Carriers of opposite type multiply to

n_{i}^{2}

; they do not simply cancel.

ERROR? "

N_{A} = 5 \times 1 0^{9}

n_{i} = 1.5 \times 1 0^{10}

, therefore

p \approx N_{A} = 5 \times 1 0^{9}

The shortcut

p \approx N_{A}

requires

N_{A} ≫ n_{i}

; here

N_{A} < n_{i}

, so intrinsic pairs dominate and you must solve charge neutrality

p = N_{A} + n

with

n p = n_{i}^{2}

ERROR? "n-type means electrons were added, so the crystal now has negative charge."

The donor nucleus supplies a matching positive charge; only mobile carrier count changed, net charge is zero.

ERROR? "Since

n p = n_{i}^{2}

is fixed, adding donors cannot change

n

Adding donors raises

n

toward

N_{D}

; the product stays

n_{i}^{2}

, so

p

falls to compensate. The constraint fixes the product, not each factor.

ERROR? "Germanium leaks more current than silicon because germanium is doped and silicon is pure."

Doping is irrelevant here — Ge leaks because its smaller gap (

E_{g} \approx 0.66

eV vs 1.12 eV) gives a far larger intrinsic

n_{i}

ERROR? "A group-V dopant creates one electron and one hole, just like breaking a bond."

No — a donor supplies an electron without a hole; only intrinsic bond-breaking makes pairs. That asymmetry is what makes it extrinsic.

ERROR? "In n-type, since electrons are majority, the Fermi level sits at mid-gap like in intrinsic material."

Doping shifts the Fermi level toward the conduction band in n-type (toward the valence band in p-type); mid-gap is the intrinsic case.

Why questions

WHY do we multiply

n

and

p

instead of adding them when deriving mass action?

Because each carrier count already carries

E_{F}

in an exponent — electrons as

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

(occupancy factor

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

above

E_{F}

) and holes as

p = N_{v} e^{- (E_{F} - E_{v}) / k T}

— so the product

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

makes the

+ E_{F}

and

- E_{F}

terms in the exponents add to zero, leaving

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

; adding would keep

E_{F}

and tell you nothing universal.

WHY is a hole treated as a real positive charge carrier and not just "absence of electron"?

Because a neighbouring electron hops in to fill the vacancy, the empty spot drifts through the crystal and responds to fields exactly like a mobile charge of

+ e

— tracking one empty seat beats tracking every electron.

WHY does an intrinsic device become unreliable at high temperature but a doped one stays stable over a range?

n_{i}

rises steeply as

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

; a doped device holds its designed

N_{D}

carriers only while

N_{D} ≫ n_{i} (T)

(on the plateau), so there is a temperature ceiling where it reverts to intrinsic behaviour. See Temperature dependence of resistance in semiconductors.

WHY does

E_{F}

cancelling matter physically, not just algebraically?

It means the equilibrium product

n p

is a material property (set by

E_{g}

and

T

) independent of how you doped — so once you know one carrier type you instantly know the other.

WHY does resistance of an intrinsic semiconductor fall with rising temperature, opposite to a metal?

More heat breaks more bonds, exponentially multiplying carriers; that carrier surge overwhelms the mild mobility drop, so conductivity rises. In a metal, carrier count is fixed and scattering dominates. See Temperature dependence of resistance in semiconductors.

WHY can we use

n \approx N_{D}

as a shortcut at all?

When

N_{D} ≫ n_{i}

, the thermal pairs are negligible next to donated electrons and nearly every donor is ionised, so the electron count is set almost entirely by the doping.

WHY does the same crystal need both a donor's proton and its electron accounted for?

The proton is fixed and ionised (immobile positive charge); the electron is mobile. Charge neutrality holds because the fixed positive dopant ions balance the extra mobile electrons.

Edge cases

EDGE — What happens when

N_{D} = n_{i}

exactly?

Neither doping nor thermal pairs dominates; you must solve the full neutrality quadratic

n = N_{D} + p

n p = n_{i}^{2}

, giving

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

— the shortcut

n \approx N_{D}

is invalid.

EDGE — What is the carrier type of a crystal doped equally with

N_{D} = N_{A}

The dopants compensate: donors fill acceptor holes, so the net doping is zero and the material behaves intrinsic (

n = p = n_{i}

) despite containing impurities.

EDGE — As

T \to 0

in an n-type sample, what happens to the free electrons?

They "freeze out" — thermal energy can no longer ionise even the weakly-bound donors, so electrons fall back onto the donor atoms and conduction collapses (the freeze-out region, just below the extrinsic plateau on the map above).

EDGE — As

T

rises very high in any doped sample, what does

n

approach?

n_{i} (T)

eventually exceeds

N_{D}

, so

n \to p \to n_{i} (T)

and both doping types converge to intrinsic behaviour — the device loses its designed asymmetry.

EDGE — Can

n_{i}

ever exceed the doping and still leave the sample usefully extrinsic?

No — once

n_{i} ≳ N_{D}

the majority/minority distinction blurs and mass-action shortcuts break; usefulness requires the doping to dwarf

n_{i}

at the operating temperature.

EDGE — Is a hypothetical semiconductor with

E_{g} = 0

intrinsic, extrinsic, or neither?

Neither in the usual sense — with no gap the conduction and valence bands overlap, giving metal-like conduction at all temperatures rather than thermally-activated carriers.

Recall One-line summary of the traps

Product not sum ( $n p = n_{i}^{2}$ ); the shortcut $n \approx N_{D}$ (and likewise $p \approx N_{A}$ ) needs the doping to greatly exceed $n_{i}$ ; doping keeps the crystal neutral; big gap ⇒ few carriers; and every doped device eventually "goes intrinsic" if you heat it enough.