FCC Exam Question: 3-19C5

What happens to the conductivity of a photosensitive semiconductor junction when it is illuminated?

A. The junction resistance is unchanged.
B. The junction resistance decreases.
C. The junction resistance becomes temperature dependent.
D. The junction resistance increases
Correct Answer: B

Explanation: When a photosensitive semiconductor junction is illuminated, photons carrying light energy strike the material. This energy is absorbed by electrons, exciting them from the valence band to the conduction band. This process creates additional free charge carriers (both electrons and holes) within the semiconductor. An increase in the number of free charge carriers means that current can flow more easily through the material for a given applied voltage. According to Ohm's Law, if current increases for a fixed voltage, the resistance must decrease (R = V/I). This fundamental principle is used in devices like photodiodes and solar cells. * **A) The junction resistance is unchanged.** This is incorrect because the material is *photosensitive*, meaning its electrical properties are designed to change with light. * **C) The junction resistance becomes temperature dependent.** All semiconductors inherently exhibit some temperature dependence; illumination introduces a separate effect rather than solely *making* it temperature-dependent. * **D) The junction resistance increases.** This is incorrect, as an increase in charge carriers directly leads to a decrease in resistance.

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Includes Elements 1, 3, 6, 7R, 8, and 9.