- #1
skweiler
- 7
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I am studying to become and electrical engineer and am currently taking Electronics I. I am having trouble understanding the change in the depletion region that results from a bias. I understand that at the time of manufacture the free electrons from the n-type semiconductor are attracted to the vacancies in the p-type semiconductor and move until the electric field produced by the newly created ions prohibits further majority carriers from crossing the junction. What I don't understand is exactly which mechanism causes the depletion region to shrink when a diode is forward biased (+ on the p-side, - on the n-side) or vice versa when reversed biased? Is the depletion region defined as the width of the region of ions surrounding the p-n junction or the gap created when a potential "pushes" or "pulls" the majority carriers towards or away from the junction? If it is defined by the ions how does the applied potential fill (or empty) more holes and thus create more (or less) ions? How do the ions and their electrons surrounding the junction react to the applied potential?