The p-n Junction & Diodes
A one-way valve for current.
Lesson at a glanceShow the key points
- Depletion layer: fixed ions, field from n to p, potential barrier.
- Forward bias lowers the barrier (large current); reverse bias leaves µA drift current.
- Dynamic resistance ΔV/Δi.
- Photodiode, LED; Zener breakdown holds voltage constant.
Join p-type and n-type silicon in one crystal and you get a one-way valve for current. That simple junction is a diode, and with light or a carefully chosen breakdown it becomes a light sensor, an LED or a voltage regulator.
Making the junction
A p-n junction is a single crystal doped with donors on one side and acceptors on the other. One way to make it is to dope a whole wafer n-type, then diffuse a stronger dose of acceptors in from one face to turn that layer p-type. Metal contacts are added to both sides.
The depletion layer
Imagine the junction just formed. The p side is full of holes, the n side full of conduction electrons. Each kind spreads from where it is crowded to where it is scarce: holes diffuse into the n side, electrons into the p side, and near the junction they meet and cancel. The heavy ions cannot move, so they are left behind uncovered: negative acceptor ions on the p side, positive donor ions on the n side.
These fixed charges create a field pointing from the n side to the p side. It pushes any hole near the junction back to the p side and any electron back to the n side, so a thin region around the junction is left with no mobile carriers at all: the depletion layer. The n side ends up at a higher potential than the p side: a potential barrier of a few tenths of a volt.
Two currents that balance
Only the few holes and electrons energetic enough to climb the barrier diffuse across. They make a current from the p side to the n side.
Heat keeps creating electron–hole pairs everywhere. A pair born in the depletion layer is pulled apart at once by the field: the electron to the n side, the hole to the p side. That makes a small current from the n side to the p side.
With no battery connected the two are equal, and no net charge crosses the junction.
Forward and reverse bias
Forward bias: the battery’s positive terminal goes to the p side. That raises the p side’s potential, lowering the barrier and narrowing the depletion layer. Many more carriers can diffuse across, while the drift current (set by how fast pairs are made) barely changes. The net current from p to n grows very rapidly with voltage, not in proportion to it: a junction does not obey Ohm’s law.
Reverse bias: the positive terminal goes to the n side. The barrier grows and the depletion layer widens, so diffusion almost stops. What remains is the small drift current, typically microamperes, flowing from n to p. The junction now behaves like a very large resistance.
Dynamic resistance
Because the i–V graph is curved, a single “resistance” doesn’t describe it. Instead we use the response to a small change:
From a diode’s characteristic: the current is 1.0 mA at 0.50 V and 2.0 mA at 0.55 V; further up, 20 mA at 0.70 V and 30 mA at 0.72 V. Find the dynamic resistance in each region.
A junction has a barrier of 0.70 V across a depletion layer 1.0 μm wide. (a) Estimate the field in the layer. (b) An electron approaches from the n side at 6.0 × 10⁵ m/s. How fast is it moving when it reaches the p side?
Points A and B are joined by two parallel branches. Branch 1 has an ideal diode D₁ pointing from A to B in series with 15 Ω. Branch 2 has an ideal diode D₂ pointing from B to A in series with 30 Ω. A 3 V battery is connected between A and B. Find the current when the positive terminal is at (a) A, (b) B.
Diodes that use light
Photodiode: if light with photon energy hc/λ at least equal to the gap falls on the junction, it breaks bonds and creates extra electron–hole pairs. In reverse bias, where the current is just the drift of such pairs, the current rises in proportion to the light intensity: a light sensor.
Light-emitting diode (LED): in forward bias, conduction electrons fall into holes near the junction, and the energy released can come out as a photon. Silicon and germanium give infrared; compounds such as gallium arsenide, gallium phosphide and indium phosphide give visible light (380–780 nm). LEDs are now everywhere, from indicator lamps to room lighting.
A photodiode’s reverse current rises only for light of wavelength 870 nm or shorter. Assuming the light acts by creating electron–hole pairs, find the band gap.
Breakdown and the Zener diode
Increase a reverse bias far enough and, at a particular breakdown voltage, the reverse current suddenly shoots up. Two mechanisms can cause it:
- Avalanche breakdown: minority carriers accelerated by the field gain enough energy to knock valence electrons free; each new pair is accelerated in turn.
- Zener breakdown: in a very thin, heavily doped junction the field is strong enough to tear electrons out of their bonds directly.
Diodes built to work in breakdown (avalanche or Zener diodes) have a useful property: once broken down, the voltage across them stays almost constant whatever the current. Placed in reverse across a load, with a resistor in series, one keeps the load voltage steady even when the supply or the load changes.
A 6.0 V Zener diode is fed from a 10 V supply through a 200 Ω resistor, with a 1.0 kΩ load across the diode. Find the currents in the resistor, the load and the diode.
- 1Diffusion across a p-n junction leaves fixed ions, a field from n to p, a barrier, and a carrier-free depletion layer.
- 2Unbiased: diffusion current (p → n) = drift current (n → p).
- 3Forward bias lowers the barrier: current grows rapidly. Reverse bias raises it: only µA drift current.
- 4Dynamic resistance ΔV/Δi = 1/slope of the i–V curve.
- 5Photodiode: light makes pairs (reverse current ∝ intensity). LED: recombination emits photons.
- 6Zener/avalanche breakdown holds the voltage nearly constant: used for regulation.
Check yourself
Keep exploring
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