Bonds in the crystal
A silicon atom has four valence electrons. In the crystal each atom sits among four neighbours and shares one electron with each of them. Each shared pair is a covalent bond. So every atom is held by four bonds, and every valence electron is locked in a bond. This is the full valence band of lesson 1, seen as bonds between atoms.
When thermal energy breaks a bond, an electron is set free. It now wanders through the crystal: it is in the conduction band. The bond it left is short of one electron. This empty place in a bond is called a hole. The atom next to it is left with a net charge \(+e\), because it has lost a negative electron.
A hole moves the other way
A hole can move without any free electron. An electron from a neighbouring bond can hop into the empty place. That fills the old hole and leaves a new hole where the electron came from. In an electric field the bound electrons hop against the field, one after another. The hole moves step by step along the field, like a positive charge.
So a semiconductor has two kinds of carrier. Free electrons move in the conduction band, against the field. Holes move in the valence band, along the field. Both give a conventional current in the same direction, along the field. The total current is the sum of the electron current and the hole current, \(I = I_e + I_h\).
An intrinsic semiconductor is a pure semiconductor, with no impurity atoms. Its free electrons and holes are made only in pairs by thermal energy. So their numbers per unit volume are equal: \(n_e = n_h = n_i\), the intrinsic carrier concentration.
In pure silicon at 300 K, \(n_i = 1.5 \times 10^{16}\,\text{m}^{-3}\). Pairs are made all the time by heat. A free electron that meets a hole falls back into the bond, and both disappear; this is recombination. At a steady temperature the pairs are made and lost at the same rate, and \(n_i\) stays fixed.
A field points from left to right in a piece of pure silicon. Which way do the holes move, and which way is the current that they carry?
Doping with a donor
Pure silicon conducts too poorly to be useful. So a tiny number of impurity atoms is added on purpose, often a few per million or fewer. This is called doping. Suppose a phosphorus atom takes the place of a silicon atom. Phosphorus has five valence electrons. Four of them form bonds with the four neighbours. The fifth has no bond to join.
The fifth electron is held only loosely, in a wide orbit around the phosphorus ion. About 0.05 eV frees it, only about twice the thermal energy \(kT = 0.026\,\text{eV}\). So at room temperature almost every phosphorus atom has given up its fifth electron to the conduction band. The phosphorus atom is called a donor. It stays fixed in the crystal as a positive ion.
Silicon doped with donors has many more free electrons than holes. Electrons are the majority carriers, and holes the minority carriers. It is called n-type silicon, n for the negative charge of its majority carriers. Arsenic and antimony also have five valence electrons and work the same way.
Doping with an acceptor
Now put a boron atom in place of a silicon atom. Boron has only three valence electrons. It can complete three bonds, and the fourth is short of an electron: a hole. An electron from a nearby bond easily moves in to complete it, and the hole moves away into the crystal. The boron atom, holding the extra electron, becomes a fixed negative ion. It is called an acceptor.
Silicon doped with acceptors has holes as majority carriers and electrons as minority carriers. It is p-type, p for the positive charge of its majority carriers. Aluminium and indium also have three valence electrons and act as acceptors.
A doped crystal is still electrically neutral. In n-type silicon, every extra free electron came from a donor that is now a positive ion. The charges balance exactly. In general the positive charges, donor ions and holes, equal the negative ones, acceptor ions and free electrons: \(N_D + n_h = N_A + n_e\). Here \(N_D\) and \(N_A\) are the numbers of donor and acceptor atoms per unit volume.
Silicon is doped with indium, which has three valence electrons. What kind of semiconductor results, and what charge does the crystal carry?
Which of these are true of n-type silicon?
More electrons, fewer holes
Add donors, and the free electrons increase. With more electrons around, each hole meets an electron sooner and recombines sooner. So the holes become fewer. The balance is exact. Recombination goes at a rate proportional to \(n_e n_h\), and pairs are made at a rate set by the temperature alone. At a steady temperature the product \(n_e n_h\) is therefore fixed, whatever the doping.
In a semiconductor in thermal equilibrium, the product of the free-electron and hole concentrations is fixed: \(n_e n_h = n_i^2\). Here \(n_i\) is the intrinsic concentration at that temperature. It holds for pure and for doped crystals alike.
Silicon has \(5 \times 10^{28}\) atoms per cubic metre. One atom in a million is replaced by arsenic. Find the concentrations of free electrons and holes at 300 K.
\(N_D = 5 \times 10^{28} / 10^{6} = 5 \times 10^{22}\,\text{m}^{-3}\). Each arsenic atom gives one free electron.
\(n_e \approx N_D\), and \(n_h = n_i^2 / n_e\).
\(n_e = 5 \times 10^{22}\,\text{m}^{-3}\). Then \(n_h = (1.5 \times 10^{16})^2 / 5 \times 10^{22} = 2.25 \times 10^{32} / 5 \times 10^{22} = 4.5 \times 10^{9}\,\text{m}^{-3}\).
The electrons have risen from \(1.5 \times 10^{16}\) to \(5 \times 10^{22}\), and the holes have fallen to \(4.5 \times 10^{9}\). There are now about \(10^{13}\) electrons for every hole.
How much better it conducts
The electrons and holes drift in a field with speeds proportional to the field. The speed per unit field is the mobility \(\mu\). In silicon \(\mu_e = 0.135\,\text{m}^2/\text{V s}\) for electrons and \(\mu_h = 0.048\,\text{m}^2/\text{V s}\) for holes. The conductivity \(\sigma\), the inverse of the resistivity \(\rho\), adds the two contributions.
For pure silicon, \(\sigma = 1.6 \times 10^{-19} \times 1.5 \times 10^{16} \times (0.135 + 0.048) = 4.4 \times 10^{-4}\,\text{S/m}\). Its resistivity is \(2.3 \times 10^{3}\,\Omega\) m. Now dope it with \(2 \times 10^{21}\) donors per cubic metre, one for 25 million atoms. The hole term is negligible, and \(\sigma = 1.6 \times 10^{-19} \times 2 \times 10^{21} \times 0.135 = 43\,\text{S/m}\). A trace of phosphorus has raised the conductivity about \(10^5\) times.
A crystal may hold both donors and acceptors. Then they cancel one another first: the donors' electrons fill the acceptors' vacancies. Only the excess decides the type. With \(N_D = 3 \times 10^{22}\) and \(N_A = 10^{22}\,\text{m}^{-3}\), neutrality gives \(n_e \approx N_D - N_A = 2 \times 10^{22}\,\text{m}^{-3}\). The mass action law then gives \(n_h = 2.25 \times 10^{32} / 2 \times 10^{22} = 1.1 \times 10^{10}\,\text{m}^{-3}\). Chip makers use this to turn an n-type layer into a p-type one by adding more acceptors.
Silicon at 300 K is doped with \(4 \times 10^{21}\) acceptor atoms per cubic metre. How many free electrons are there per cubic metre? Give the answer in units of \(10^{10}\,\text{m}^{-3}\).
Adding donors to silicon gives it more electrons. Why does it also leave it with fewer holes than pure silicon has?
Heat makes electron-hole pairs at a rate set by the temperature alone, while recombination goes at a rate proportional to the product of the electron and hole concentrations. With many more electrons, a hole finds an electron and recombines much sooner. Balance is reached with far fewer holes, so that the product stays \(n_i^2\).
The silicon for chips is refined until only about one atom in a thousand million is not silicon. That is 99.9999999 % pure, called "nine nines". The purity is needed because the doping itself is so small. A stray impurity at one part in a million would swamp a doping of one part in a hundred million. After refining, the chip maker adds exactly the donors and acceptors it wants, where it wants them.
- Thinking n-type silicon is negatively charged.n-type names the majority carriers. Each free electron is balanced by a fixed positive donor ion, so the crystal is neutral.
- Using \(n_e + n_h = 2n_i\) for a doped crystal.The product is fixed, not the sum: \(n_e n_h = n_i^2\). The minority carriers fall far below \(n_i\).
- Saying holes move because free electrons fill them.A hole moves when a bound electron from a neighbouring bond hops in. No free electron is needed.