Temperature Dependence of Resistivity

Equivalent forms

R = R_0[1+\alpha\,\Delta T]

\alpha = \frac{1}{\rho_0}\frac{d\rho}{dT}

A single coefficient α captures the whole electron-phonon story over everyday temperatures — the messy quantum scattering hides inside one slope.

Symbol	Name	SI unit	Dimension	Range
$ρ$	Resistivityoutput Resistivity at temperature T	Ω·m	M·L³·T⁻³·I⁻²	0 – 0.000001
$ρ₀$	Reference resistivity Resistivity at reference temperature T₀	Ω·m	M·L³·T⁻³·I⁻²	1e-8 – 1e-7
$α$	Temperature coefficient Fractional resistivity change per degree	K⁻¹	Θ⁻¹	0 – 0.006
$T$	Temperature Conductor temperature	°C	Θ	-50 – 300

Unit systems

Symbol	SI	CGS	Imperial
$ρ$	Ω·m	s	Ω·m
$ρ₀$	Ω·m	s	Ω·m
$α$	K⁻¹	K⁻¹	°F⁻¹
$T$	°C	°C	°F

Where it holds

Accurate for metals over roughly

-100^{\circ}C

to +

600^{\circ}C

where the linear term dominates. Fails near 0 K (residual resistivity), near melting, and for semiconductors/insulators where resistivity is exponential in 1/T.

Dimensional analysis

$\alpha (T-T_{0})$ is dimensionless $(K^{-1} \times K)$ , $so \rho$ keeps the units $of \rho _{0} = \Omega \cdot m \checkmark$

Discovery

Humphry Davy · 1821

Davy first noted that a metal's conducting power falls as it heats. The linear coefficient was later systematized by Matthiessen, whose rule splits resistivity into a temperature-dependent lattice part and a fixed impurity part.

Try this

A light-bulb filament is 10× more resistive glowing than cold. Why does heating a metal choke its current?

Copper has ρ₀ = 1.68×10⁻⁸ Ω·m at 20°C and α = 0.0039 /°C. Find its resistivity at 120°C.

Research status: stable

Real-world applications

Platinum resistance thermometers (PT100/PT1000)
Inrush current limiting in incandescent and heater circuits
Self-regulating PTC heaters
Compensating sensor drift in precision electronics

Common misconceptions

Resistance is a fixed property of a wire — it depends strongly on temperature
All materials get more resistive when heated — semiconductors do the opposite
The relation is exactly linear — it's a first-order approximation that curves at extremes

Experimental verification

Four-point probe measurements of wire resistance versus a controlled oven temperature trace a straight line over hundreds of degrees; the slope yields

\alpha

, matching tabulated values within a percent. Platinum's clean linearity is the basis of the PT100 thermometer.

Derivation

Lattice (phonon) scattering rate rises roughly linearly with T above the Debye temperature, and resistivity is proportional to the scattering rate:

\rho \propto 1/\tau \propto T

.

Expanding about a reference

T_{0} to

first order gives

\rho = \rho _{0}[1 + \alpha (T - T_{0})]

with

\alpha = (1/\rho _{0})(d\rho /dT)

.

Limiting cases

T = T_{0}

⟶

\rho = \rho _{0}

At the reference temperature the correction term vanishes.

\alpha

< 0⟶

\rho

falls with TSemiconductors gain carriers as they heat, so resistivity drops.

T \to 0\,\mathrm{K}

⟶ linear model failsReal metals approach a residual resistivity floor set by impurities (Matthiessen).

What if…

What if

T_{0} is

chosen as

0^{\circ}C

instead of

20^{\circ}C

?

$\rho _{0} and \alpha$ must be re-quoted at that reference; the physical $\rho (T)$ curve is unchanged.

What if the filament reaches

2500^{\circ}C

?

Tungsten's resistance rises $\sim 15\times$ , which is why a cold bulb draws a large inrush current at switch-on.

What

if \alpha is

negative?

You have an NTC thermistor — resistance drops as it warms, useful for inrush limiting and temperature sensing.

1

Copper at 120°C

Given ·

ρ₀:: 1.68e-8
α:: 0.0039
T:: 120

Find ·

\rho

Steps

$\Delta T = 120 - 20 = 100^{\circ}C$
Bracket $= 1 + 0.0039\times 100 = 1.39$
$\rho = 1.68\times 10^{-8} \times 1.39 \approx 2.34\times 10^{-8} \Omega \cdot m (39$ % higher)

Result ·

\rho = 1.68\times 10^{-8}[1 + 0.0039(120-20)] = 1.68\times 10^{-8}(1.39) \approx 2.34\times 10^{-8} \Omega \cdot m

Ohm's Law→Drift Velocity→