One crystal, two kinds of doping
A p-n junction is a single crystal of silicon that is p-type on one side and n-type on the other. It is not made by pressing a p-type piece against an n-type piece. Two separate pieces touch only at a few points, with gaps of air and dirt between them. Instead, one side of a single wafer is doped with acceptors and the other with donors. The boundary between them is the junction.
Diffusion across the junction
On the n side there are very many free electrons, and on the p side very few. Carriers wander at random, and wherever there are more of them, more wander away. So electrons spread from the n side into the p side. This spreading from a crowded region into an empty one is diffusion, like a drop of ink spreading through still water. In the same way holes diffuse from the p side into the n side.
An electron that crosses into the p side soon meets one of the many holes there, and the two recombine. A hole that crosses into the n side meets an electron and recombines too. So a thin region on both sides of the junction loses nearly all its free carriers.
What remains in that region are the fixed ions. On the n side, the donors that gave their electrons are left as positive ions. On the p side, the acceptors that took electrons are left as negative ions. This layer of fixed charge with no free carriers is the depletion layer. In a typical junction it is less than a micrometre thick.
Why are the fixed ions in the p side of the depletion layer negative?
A field and a step in potential
The fixed charges, positive on the n side and negative on the p side, set up an electric field across the depletion layer. It points from the positive ions to the negative ones, from n to p. A potential difference between two points is the work done per unit charge in carrying a charge from one to the other. The field makes such a difference: the n side stands at a higher potential than the p side.
This step in potential is the barrier potential \(V_0\). For silicon it is about \(0.7\,\text{V}\), and for germanium about \(0.3\,\text{V}\). It is called a barrier because of what it does to the majority carriers. An electron on the n side that tries to cross into p is pushed back by the field. To cross, it must climb an energy step of \(eV_0\), which is 0.7 eV for silicon.
The barrier potential \(V_0\) of a p-n junction is the step in potential across its depletion layer, the n side being higher. It is about \(0.7\,\text{V}\) for silicon and \(0.3\,\text{V}\) for germanium.
Fig 3.2 shows the barrier as a hill in energy. For an electron, a higher potential means a lower energy, because its charge is negative. So the bands on the n side lie lower, by \(eV_0\). The free electrons of the n side sit at the bottom of the hill. For a hole, energy is measured downwards, so the holes of the p side face the same hill from their side.
Two free electrons on the n side of a silicon junction head into the depletion layer, with kinetic energies of \(0.5\,\text{eV}\) and \(0.9\,\text{eV}\). Which one reaches the p side, and with how much kinetic energy? What would happen in a germanium junction?
(a) Crossing from n to p, an electron climbs an energy step \(eV_0\). For silicon that is \(0.7\,\text{eV}\).
(b) The \(0.5\,\text{eV}\) electron is turned back inside the layer. The \(0.9\,\text{eV}\) electron crosses and enters the p side with \(0.9 - 0.7 = 0.2\,\text{eV}\).
(c) In germanium the step is only \(0.3\,\text{eV}\). Both electrons cross, with \(0.2\,\text{eV}\) and \(0.6\,\text{eV}\).
Values used: barrier potential \(0.7\,\text{V}\) for silicon and \(0.3\,\text{V}\) for germanium.
A silicon junction has a barrier of \(0.7\,\text{V}\) across a depletion layer about \(0.5\,\mu\text{m}\) thick. Treat the field as uniform, so that it equals the potential difference divided by the thickness. Estimate the field, in units of \(10^{6}\,\text{V/m}\).
You cannot measure \(V_0\) by touching a voltmeter across a junction; it reads zero. Where the metal leads meet the silicon, two more contact potentials appear. Round the whole loop they cancel the barrier exactly. If they did not, an unlit junction at a steady temperature would drive a current for ever, a machine that makes energy from nothing.
In which direction does the field in the depletion layer point? Which way does it push a free electron that wanders in from the n side?
The barrier potential of a germanium junction is \(0.3\,\text{V}\). How much energy must an electron from the n side have to cross it? Give the answer in units of \(10^{-20}\,\text{J}\). Take \(e = 1.6 \times 10^{-19}\,\text{C}\).
Two currents that cancel
Now there are two flows across the junction. The first is the diffusion of majority carriers: holes from p to n and electrons from n to p. Both give a conventional current from p to n, the diffusion current. As the depletion layer grows, the barrier rises. Fewer majority carriers have the energy to climb it, and the diffusion current falls.
The second flow runs the other way. A few minority carriers wander into the depletion layer: electrons from the p side and holes from the n side. The field sweeps them straight across, electrons into n and holes into p. This motion under a field is drift. It gives a conventional current from n to p, the drift current.
In an unbiased p-n junction the depletion layer grows until the drift current of minority carriers equals the diffusion current of majority carriers. Then the net current across the junction is zero, and the width of the layer and the barrier potential stay fixed.
Why does the layer stop exactly there? Diffusion builds the barrier, and the barrier chokes the diffusion. The drift current depends on how many minority carriers are made nearby, not on the height of the barrier. So as the barrier rises, the diffusion current falls towards the steady drift current. When the two are equal, no more charge is uncovered, and the layer stops growing. Fig 3.1 shows the two arrows becoming equal.
Put the events in the forming of a p-n junction in order, first to last.
A p-n junction sits on the table with nothing connected to it. Which statement about the currents across the junction is correct?
At Bell Laboratories in 1940, Russell Ohl was testing rods of silicon as radio detectors. One rod behaved strangely. When a lamp shone on it, a voltage of about half a volt appeared between its ends. Ohl found that the rod had a crack, and that the impurities on the two sides of the crack were different. He called the two regions p and n, for the sign of the voltage each side took. He had found the p-n junction, and also the first silicon solar cell.
- Thinking the depletion layer is empty of charge because it is empty of carriers.It holds no free carriers, but it holds the fixed ions: negative on the p side and positive on the n side.
- Drawing the field in the depletion layer from p to n.A field points from positive charge to negative. The positive donor ions are on the n side, so the field points from n to p.
- Thinking the diffusion stops because the carriers run out.Diffusion stops growing the layer because the barrier turns back most carriers, and drift balances the rest.
- Expecting a voltmeter to read 0.7 V across a silicon junction.The contact potentials at the leads cancel the barrier round the loop, and the meter reads zero.