Final answer:
In semiconductors, increasing the doping density introduces more charges, leading to a stronger electric field across the depletion region in a p-n junction.
Step-by-step explanation:
The question relates to the behavior of semiconductors and specifically how the depletion width in a p-n junction changes with doping density. When semiconductor materials, such as silicon, are doed with impurities, the electrical properties are altered.
Doping refers to the addition of donor or acceptor atoms to create n-type or p-type semiconductors, respectively. In a p-n junction, a depletion region is formed where p-type and n-type materials contact each other.
This region is void of free charge carriers because they have recombined, creating a fixed region of positive and negative ions which results in an electric field.
In the context of this question, as the doping density increases, more donor or acceptor atoms are introduced, which increases the number of charges on each side of the junction.
This increase in charge density leads to a stronger electric field across the depletion region. Because the field is stronger, it can support the same potential difference with a smaller physical separation of charges, thus decreasing the maximum depletion width.
It is similar to how bringing two oppositely charged plates closer in a capacitor increases the capacitance for the same voltage.