Thermal effects on conductivity: Question bank — Hardware deep dive

Before we lean on any symbol, let's earn every one of them in plain words.

Definition The five symbols this page uses

$n$ = number of charge carriers packed into one unit of volume (how many runners are in the room).
$q$ = the elementary charge, the fixed amount of electric charge on a single carrier. For an electron it is $q = 1.602 \times 1 0^{- 19}$ coulombs — a constant of nature, never changing with temperature.
$m$ = the effective mass of one carrier, how much "inertia" the electron carries as it is pushed by the field. Heavier carriers accelerate less for the same push.
$τ$ = the average time a carrier travels before smacking into a vibrating atom (its "memory" of the last collision).
$μ$ = mobility: how far a carrier drifts per unit of applied electric field (how freely each runner moves). From Newton's law it works out to $μ = q τ / m$ — bigger charge and longer free time help; heavier mass hurts.

Conductivity ties $n$ , $q$ and $μ$ together: $σ = n q μ$ . Resistivity is just its flip, $ρ = 1/ σ$ .

For the semiconductor items we also need two more constants:

The two curves those laws produce are worth staring at before the traps:

Figure — Thermal effects on conductivity

Look at the red metal line rising straight, and the mint semiconductor curve diving as heat spawns carriers — every trap below is really a question about which of these two shapes applies.

True or false — justify

A metal has a fixed sea of free electrons; a semiconductor starts nearly empty and gets carriers only when heat kicks electrons across the band gap. Keep that split in mind for each.

Heating a copper wire always increases its resistance (near room temperature).

True — in a metal

n

is essentially fixed, so hotter lattice means more scattering, smaller

τ

, lower

μ

, and

ρ \propto T

rises. The verdict is metal-specific, not universal.

Heating a pure silicon sample always increases its resistance.

False — thermally generated carriers make

n \propto e^{- E_{g} /2 k_{B} T}

explode, and that exponential growth dwarfs the mild mobility drop, so resistance falls.

In the formula

σ = n q μ

, only mobility can depend on temperature.

False —

n

and

μ

can both move (

q

is a fixed constant of nature). In metals

n

barely moves so

μ

dominates; in semiconductors

n

changes exponentially and wins.

The temperature coefficient

α

is positive for metals and negative for typical semiconductors.

True —

α = \frac{1}{R _{0}} \frac{d R}{d T}

; metals'

R

climbs with

T

(positive slope), while intrinsic semiconductors'

R

drops (negative slope).

Higher lattice temperature makes electrons drift faster, so mobility rises.

False — faster lattice motion means more vibrating scattering targets, which shortens

τ

and lowers

μ = q τ / m

. The carriers' drift is hurt, not helped.

The exponent for intrinsic carrier density is

e^{- E_{g} / k_{B} T}

(no factor of 2).

False — exciting one electron also leaves one hole, so pair creation splits

E_{g}

across two carriers, giving

e^{- E_{g} /2 k_{B} T}

A material with a very large band gap behaves more like an insulator at room temperature.

True — a large

E_{g}

makes

e^{- E_{g} /2 k_{B} T}

vanishingly small, so almost no carriers are excited and

n

(hence

σ

) is tiny.

Doubling the band gap halves the intrinsic conductivity.

False —

E_{g}

sits inside an exponent, so doubling it squares the tiny suppression factor, crushing

σ

far more than a mere halving. See Band Theory & Energy Gaps.

For a metal, resistivity is truly a straight line for all temperatures down to 0 K.

False — the linear law

ρ = ρ_{0} [1 + α (T - T_{0})]

is a near-room-temperature approximation; at very low

T

vibrations freeze out and

ρ

flattens toward a residual value set by impurities.

Spot the error

Each statement contains one flawed step. Name the flaw and correct it.

"In a metal, heat gives electrons more energy, so more of them become free to carry current, raising

σ

Error: assumes

n

grows. A metal's conduction band is already full of free electrons at any temperature, so

n

is fixed; heat only shortens

τ

and lowers

σ

"Because

μ = q τ / m

and heating adds energy,

τ

grows, so mobility grows."

Error:

τ

shrinks, not grows. Hotter lattice = larger vibration amplitude = more collisions per second = smaller mean free time

τ

, so

μ

falls. See Mobility and Drift Velocity.

"A semiconductor's conductivity rises with

T

because its mobility improves with heat."

Error: mobility actually decreases slightly with heat, just as in a metal. The rise comes from the exponential growth of carrier density

n

, which overwhelms the

μ

drop.

"Since

1/ τ \propto T

, resistivity

ρ \propto T^{2}

for a metal."

Error in the exponent:

ρ = m / (n q^{2} τ)

, and

ρ \propto 1/ τ \propto T

, a single power — linear, not quadratic.

"The factor of 2 in

e^{- E_{g} /2 k_{B} T}

appears because

k_{B}

must be doubled."

Error:

k_{B}

is untouched; the 2 comes from pair creation — one photon-or-phonon's worth of gap energy births both an electron and a hole, splitting

E_{g}

between them.

"Copper's resistance rose 20%, so its number of carriers must have dropped by roughly 20%."

Error: in a metal

n

is fixed. The 20% rise in

ρ

reflects a fall in

τ

(about 17%), i.e. a mobility change, not a carrier-count change.

"Adding donor dopants to silicon makes it behave like a metal, so its resistance now rises with temperature at all temperatures."

Error: only in the extrinsic (dopant-dominated) regime does

n

saturate and

μ

-limited behavior mimic a metal; once heat excites intrinsic pairs across the gap,

n

climbs again and resistance falls. See Doping and Carrier Concentration.

Why questions

Why is the master question "

n

μ

?" rather than just "does it heat up?"

Because

σ = n q μ

has two temperature-dependent factors that pull opposite ways; which one dominates decides whether resistance rises or falls, and that depends on the material, not on the heat alone.

Why does

τ

(mean time between collisions) shrink as temperature rises?

Higher

T

enlarges the atoms' vibration amplitude; since stored spring energy

\propto x^{2} \propto T

, the scattering target area grows linearly with

T

, so collisions happen more often and the average time between them drops.

Why does the carrier density in a semiconductor grow exponentially rather than linearly with

T

Because only carriers with thermal energy above

E_{g}

can cross the gap, and Boltzmann statistics make that high-energy fraction grow as

e^{- E_{g} /2 k_{B} T}

— a sharply accelerating curve, not a straight line.

Why is a metal's mobility drop with temperature "mild" compared to a semiconductor's carrier explosion?

Mobility falls only as a power law (

μ \propto 1/ T

), while

n

in a semiconductor grows exponentially; an exponential always outruns a power for large enough

T

Why does a semiconductor's resistivity–temperature curve look like a decaying exponential while a metal's looks like a straight line?

The metal's

ρ \propto T

from linear scattering growth; the semiconductor's

ρ \propto e^{+ E_{g} /2 k_{B} T}

from exponential carrier creation, which plots as a steeply falling curve as

T

rises (see the figure above).

Why can't we simply say "heat = more resistance" as a universal rule?

That rule only holds when

n

is constant (metals). Whenever heat creates carriers (semiconductors), the growing

n

can more than cancel the mobility loss, flipping the sign of the effect.

Edge cases

What happens to a metal's resistivity as

T \to 0

Lattice vibrations freeze out, so scattering from phonons vanishes and

ρ

approaches a nonzero residual resistivity set by fixed impurities and defects, not zero (unless it becomes a superconductor). See Superconductivity.

What happens to an intrinsic semiconductor's carrier density as

T \to 0

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

, so

n \to 0

: with no thermal energy, no electrons cross the gap and the material becomes essentially a perfect insulator. See Intrinsic vs Extrinsic Semiconductors.

At the exact instant

Δ T = 0

R = R_{0} [1 + α Δ T]

, what is

R

R = R_{0}

— the reference resistance. The linear law is anchored so that at the reference temperature the correction term is zero.

For a superconductor cooled below its critical temperature, what does the "resistance rises with heat" metal rule say — and does it hold?

It doesn't apply: below

T_{c}

resistance is exactly zero, a phase-transition effect outside the phonon-scattering picture; above

T_{c}

the ordinary metal rule resumes. See Superconductivity.

In a heavily doped semiconductor, is there a temperature range where it behaves almost like a metal (resistance rising with

T

Yes — in the extrinsic saturation regime

n

is pinned by the ionized dopants and roughly constant, so mobility loss dominates and resistance can rise with

T

until intrinsic generation takes over at high

T

What is the sign of

α

right at the crossover where a semiconductor transitions from extrinsic (n saturated) to intrinsic (n exploding) behavior?

α

passes through a change: it can be positive (metal-like) in the saturation plateau and swings negative once intrinsic carrier creation kicks in, so near the crossover

α \approx 0

momentarily.

E_{g} = 0

(bands overlap, a semimetal), what does the intrinsic exponent predict for carrier density?

e^{- 0} = 1

, a constant — no exponential suppression, so carriers are always plentiful and the material conducts even at low temperature, behaving metal-like.

Connections

1.3.08 Thermal effects on conductivity — parent topic
Free Electron Model of Metals
Band Theory & Energy Gaps
Intrinsic vs Extrinsic Semiconductors
Mobility and Drift Velocity
Doping and Carrier Concentration
Superconductivity