Why a Semiconductor Conducts Better as It Gets Hotter

Copper carries current easily, glass hardly at all, and silicon sits in between — yet silicon's conductivity can be raised enormously by heat, light or a trace of impurity. Energy bands explain all three behaviours and lead directly to the p-n junction inside every diode and chip.

This lesson covers energy bands, intrinsic and extrinsic semiconductors, and the p-n junction.

In a solid, electron energy levels merge into a valence band and a conduction band; conductors have overlapping or partly filled bands, insulators have a large gap between them, and semiconductors have a small gap of about 1 eV.

Why bands form. When many atoms come close, each atomic energy level splits into a huge number of closely spaced levels, forming continuous bands.

The three classes:

- Conductors — bands overlap, so electrons move freely even at low temperature
- Insulators — a wide forbidden gap, as in diamond with about 5.4 eV
- Semiconductors — a narrow gap, about 1.1 eV for silicon and 0.7 eV for germanium

Worked example 1 — thermal energy. At 300 K,

$$kT=\frac{1.38\times 10^{-23}\times 300}{1.6\times 10^{-19}}=0.026\text{ eV}$$

This is far below 1.1 eV, so only a small fraction of electrons cross silicon's gap, and more cross as temperature rises.

Worked example 2 — light and the gap. The longest wavelength that can lift an electron across silicon's gap is

$$\lambda =\frac{hc}{E_g}=\frac{6.63\times 10^{-34}\times 3.0\times 10^8}{1.1\times 1.6\times 10^{-19}}=1.13\times 10^{-6}\text{ m}$$

An everyday example. Rooftop solar panels use silicon because its band gap lets visible light and near-infrared free electrons.

The substance. Heating raises a semiconductor's conductivity but lowers a metal's — in silicon, many more carriers are freed, while in copper the same carriers simply collide more often.

An intrinsic semiconductor is pure, with equal numbers of free electrons and holes; an extrinsic one is doped, becoming n-type with pentavalent impurities that donate electrons, or p-type with trivalent impurities that create holes.

Intrinsic. Each electron freed into the conduction band leaves a hole behind, so $n_e=n_h=n_i$.

Doping:

- n-type — arsenic, phosphorus or antimony has five valence electrons; four bond with silicon and the fifth is nearly free. Electrons are the majority carriers
- p-type — boron, aluminium or indium has three valence electrons, leaving a missing bond, a hole. Holes are the majority carriers

The mass action law. At a given temperature, $n_e{n}_h=n_i^2$, so adding electrons suppresses holes.

Worked example. Silicon with $n_i=1.5\times 10^{16}$ m${}^{-3}$ is doped with $5.0\times 10^{22}$ m${}^{-3}$ arsenic atoms, so $n_e\approx 5.0\times 10^{22}$ m${}^{-3}$ and

$$n_h=\frac{n_i^2}{n_e}=\frac{(1.5\times 10^{16}{)}^2}{5.0\times 10^{22}}=4.5\times 10^9{\text{ m}}^{-3}$$

Electrons now outnumber holes by more than ten trillion to one.

An everyday example. Every chip in a mobile phone is built from regions of silicon doped p-type and n-type in precise patterns.

The substance. An n-type semiconductor is not negatively charged — each extra electron is balanced by a positive donor ion, so the crystal stays neutral.

When p-type and n-type regions meet, electrons and holes diffuse across and recombine, leaving a thin depletion region of fixed ions whose electric field sets up a potential barrier that stops further diffusion.

Step by step:

- Electrons diffuse from the n-side into the p-side, and holes from the p-side into the n-side
- They recombine near the junction, uncovering positive donor ions on the n-side and negative acceptor ions on the p-side
- These fixed charges create an electric field pointing from n to p, which pushes carriers back as drift current
- Equilibrium is reached when drift current equals diffusion current

Typical values. The barrier is about 0.7 V for silicon and 0.3 V for germanium, across a depletion region roughly a micrometre wide.

Worked example. A 0.70 V barrier across $1.0\times 10^{-6}$ m gives an average field of

$$E=\frac{V}{d}=\frac{0.70}{1.0\times 10^{-6}}=7.0\times 10^5{\text{ V m}}^{-1}$$

An electron needs $0.70\times 1.6\times 10^{-19}=1.1\times 10^{-19}$ J to climb this barrier.

An everyday example. The small LED indicator on a television contains a p-n junction, which glows when current is pushed across its barrier.

The substance. The depletion region has no free carriers but is not uncharged — it holds the fixed ions whose field creates the barrier.