If a reverse-biased voltage is applied across a p-n junction, the depletion layer width is

Analyze the question: It asks what happens to the depletion layer of a p-n junction when a reverse-bias voltage is applied.

Concept:

• A p-n junction has a depletion layer, a region near the junction with no free charge carriers. Electrons from the n-side diffuse across and recombine with holes on the p-side. This leaves behind fixed positive ions on the n-side and fixed negative ions on the p-side.
• Reverse bias means the positive terminal of the battery is connected to the n-side and the negative terminal to the p-side.

Solution:

1. With reverse bias, the positive terminal pulls free electrons in the n-region away from the junction.
2. The negative terminal pulls holes in the p-region away from the junction.
3. More fixed ions are uncovered near the junction, so the region without free carriers grows wider.
4. The barrier potential increases, so almost no majority-carrier current flows. Only a tiny leakage current from minority carriers flows.

In forward bias the opposite happens: the depletion layer becomes narrower, and current flows easily.

Answer: The depletion layer width is increased, which is option A.

Tip: Remember "Reverse = Wider, Forward = Fine (narrow)". In reverse bias the layer widens and acts like an insulator. In forward bias it narrows and conducts.