Conductivity, carriers and how temperature changes them

Conductivity is the number of carriers times how easily each one moves

Intuition

Think of a corridor in a train station. How many people get through depends on two things: how many people there are, and how easily each one walks. Conductivity is the same. More carriers help. Carriers that are rarely bumped help too.

Conductivity, $\sigma$ (in S cm⁻¹), is the inverse of resistivity, $\rho$.

$\sigma = \frac{1}{\rho}$

Small changes in structure or composition can drastically change the band structure and so the conductivity. The conductivities of materials span more than 25 orders of magnitude. The slide’s chart puts metals near the top (about $10^4$ to $10^6$ S cm⁻¹), semiconductors in the middle (Ge, Si) and insulators at the bottom (MgO, diamond, polythene, glass, about $10^{-12}$ to $10^{-20}$). The chart did not extract in full detail.

Resistivity depends on the carrier concentration, $n$, and the mobility, $\mu$.

$\sigma = n e \mu$

$e$ is the electronic charge. Only electrons that can easily be excited from occupied states into empty states count towards $n$. With no light or voltage, the push is thermal, of order $kT$, about 0.03 eV at room temperature. So only electrons within a few $kT$ of the Fermi level, $E_F$, can conduct. In general $E_F$ must cut a band for a large $n$.

Mobility falls whenever something knocks the carrier off course

Mobility is how freely a carrier travels between collisions.

$\mu = \frac{e\tau}{m^*}$

$\tau$ is the relaxation time between scattering events. $m^*$ is the effective mass. It depends on the band width, which depends on orbital overlap. Two things scatter carriers and reduce $\mu$. A defect or impurity: $\tau$ is inversely proportional to the amount of impurity. Lattice vibrations (phonons): $\tau$ is inversely proportional to $T$.

Putting the pieces together gives the Drude model.

$\sigma = \frac{n e^2 \tau}{m^*}$

Exam detail

Learn the four equations as a chain: $\sigma = 1/\rho$, $\sigma = n e \mu$, $\mu = e\tau/m^*$ and Drude $\sigma = ne^2\tau/m^*$. Know which symbol is which, and which of $n$ or $\tau$ moves with $T$.

Metals lose conductivity when hot, semiconductors gain it

Compare

Metal

$n$ barely changes with $T$. $\tau$ falls as $T$ rises (more phonons). Conductivity decreases as $T$ increases. $\sigma$ against $1/T$, or $\rho$ against $T$, is essentially linear.

Semiconductor

$n$ rises as $T$ rises because more electrons cross the gap $E_g$. $\tau$ also falls, but the exponential in $n$ wins. Conductivity increases with $T$. $\ln\sigma$ against $1/T$ is essentially linear.

For semiconductors the carrier density follows

$n \propto \exp\left\{-\frac{E_g}{2kT}\right\}$

Because the lecture classifies by temperature dependence rather than size alone, the pattern on slide 18 is this. A superconductor jumps to zero resistivity at $T_c$. A metal’s conductivity falls gently with $T$. A semiconductor’s and insulator’s conductivity rise from zero, the semiconductor’s more steeply.

Try it: worked example

A material's conductivity falls steadily as it warms from 100 K to 300 K. Is it a metal or a semiconductor, and why?

Show answer and steps

AnswerA metal. Carrier concentration barely changes, while the relaxation time falls as phonon scattering grows.

1. Conductivity is $\sigma = n e^2 \tau / m^*$.
2. If $\sigma$ falls while $T$ rises, the factor that falls must outweigh any rise in $n$.
3. In a metal $n$ changes very slowly with $T$, and $\tau$ is inversely proportional to $T$, so $\sigma$ falls.
4. In a semiconductor, $n \propto \exp\{-E_g/2kT\}$ rises faster than $\tau$ falls, so $\sigma$ would rise. That does not match, so it is a metal.

Pitfall

Do not classify a material by the size of its conductivity alone. Lecture 9 says to use the temperature dependence. A semiconductor can have a conductivity close to a poor metal at one temperature.

Optional aside

The lecture’s slide 17 plots a semiconductor rising steeply and an insulator rising more slowly from the origin. The exact curve shapes are in the figure, not in the text, so only the trend is given here.

Recall

Why does heating help a semiconductor but hurt a metal?

Heating adds many carriers across a band gap in a semiconductor, and that beats the extra scattering. A metal has no gap to cross, so heating only adds scattering.