Difficulty: Easy
Correct Answer: A decrease in the material's resistance (increase in conductivity)
Explanation:
Introduction / Context:
Photoconductors are light-sensitive resistive devices. When photons are absorbed, additional electron–hole pairs are generated, which changes the carrier concentration and therefore the conductivity of the material. With ohmic contacts and no built-in junction, the device operates in photoconductive mode rather than photovoltaic mode.
Given Data / Assumptions:
Concept / Approach:
Conductivity sigma of a semiconductor is sigma = q * (n * mu_n + p * mu_p). If mu_n = mu_p and photo-generation increases carrier densities n and p by Delta n and Delta p, conductivity increases approximately in proportion to the excess carriers. In a photoconductor, the measurable effect is a reduction in resistance R = L / (sigma * A). There is generally no light-induced open-circuit voltage because ohmic contacts do not create a built-in electric field; instead, a bias is applied and the photocurrent is read as an increase over the dark current.
Step-by-Step Solution:
Verification / Alternative check:
In practice, I–V curves of photoconductors under different light levels are straight lines with increasing slope as illumination increases, confirming lower resistance and no significant open-circuit photovoltage for purely ohmic contacts.
Why Other Options Are Wrong:
Option (a) and (e): open-circuit voltage generation is characteristic of photovoltaic junctions (p–n, Schottky), not purely ohmic contacts. Option (b): at zero bias an ohmic photoconductor does not sustain a notable short-circuit current. Option (d): illumination does not increase resistance; it decreases it.
Common Pitfalls:
Confusing photoconductive devices with photovoltaic cells, or assuming any illuminated semiconductor creates a DC voltage without a junction.
Final Answer:
A decrease in the material's resistance (increase in conductivity)
Discussion & Comments