p-n junction: depletion region and barrier potential
At a p-n junction, electrons and holes diffuse and recombine, leaving fixed ions. This depletion region creates a barrier potential: about 0.7 V (Si) or 0.3 V (Ge).
How the depletion region and barrier form
- 1One crystal is doped with acceptors on one side (p) and donors on the other (n). The p-side has many holes; the n-side has many free electrons.
- 2Because of this difference in concentration, electrons diffuse into the p-side and holes into the n-side.
- 3There they recombine. Each recombination removes one free electron and one hole.
- 4Electrons leaving the n-side leave fixed positive donor ions. Holes leaving the p-side leave fixed negative acceptor ions.
- 5This narrow layer of fixed ions has no mobile carriers. It is the depletion region (space-charge region), about 0.5 μm thick.
- 6★The ions set up an electric field from n to p. The potential difference across the layer is the barrier potential. It stops further diffusion.
★ key step
The chain of causes at a p-n junction
How to draw the junction and the barrier
- Draw a rectangle. Label the left half p and the right half n.
- Shade a narrow band in the middle and label it depletion region.
- In it, draw circled − signs on the p-side and circled + signs on the n-side for the fixed ions.
- Show small open circles (holes) in the rest of the p-side and dots (electrons) in the rest of the n-side.
- Draw an arrow labelled E from the n-side to the p-side, and mark the barrier potential, n-side positive.
- Beside it, draw the diode symbol: a triangle (anode A, p-side) pointing to a bar (cathode K, n-side).
Identifying a real diode
- A ring or band printed near one end marks the cathode (n-side).
- Some diodes use colour: a red mark for the anode and a blue mark for the cathode. Others have the symbol painted on.
- Codes: the first letter gives the material (A germanium, B silicon). The second gives the use (Y rectifier, Z Zener, A detection).
- So BY127 is a silicon rectifier diode and BZ148 is a silicon Zener diode.
- Always use a diode within its rated current and voltage, or it will be damaged.
Worked example: the field inside the depletion region
A silicon p-n junction has a barrier potential of 0.7 V across a depletion region 0.5 μm thick. Estimate the electric field in the region, taking it as uniform.
- Formula: E = V/d.
- d = 0.5 μm = 0.5 × 10⁻⁶ m = 5 × 10⁻⁷ m.
- E = 0.7 V ÷ 5 × 10⁻⁷ m = 1.4 × 10⁶ V m⁻¹.
- The voltage is small, but the layer is so thin that the field is very strong.
About 1.4 × 10⁶ V m⁻¹
Every LED bulb, mobile charger and rooftop solar panel contains p-n junctions. In a solar panel, the field of the depletion region pulls apart the electrons and holes made by sunlight, and this drives the current.
2 marks: “When p-type and n-type regions meet in one crystal, a thin layer at the junction loses its free carriers. What is this layer called, and how does it form?” Electrons diffuse from n to p and holes from p to n and recombine. Fixed + ions are left on the n-side and − ions on the p-side. The layer is emptied (depleted) of mobile carriers, hence the name. Simple labelled diagram.
1 marks: “What is the barrier potential of a silicon p-n junction?” About 0.7 V (germanium about 0.3 V).
2 marks: “Which charges make up the barrier potential of a p-n junction, and which side of an unbiased junction is positive?” Fixed (immobile) donor and acceptor ions near the junction. The n-side is positive and the p-side negative.
Diffuse, recombine, deplete, block: carriers cross, cancel out, leave fixed ions, and the ions' field blocks further crossing.