Question

In $P-N$ junction under the condition of reverse bias the electric field is max at,

• A. the edge of the depletion region on the $p$ side
• B. the edge of the depletion region on the $n$ side
• C. $p-n$ junction
• D. the centre of depletion region on the $n$ side

Answer 1

The correct option is C $p-n$ junction

Under the condition when a p-n junction is reverse biased, the applied reverse voltage is increased, the electric field across the p-n junction increases because in the depletion region across the p-n junction, there are equal two opposite charges stored. As a result of high electron field across the junction under reverse biased condition, the breakdown of charge carriers takes place. Therefore, current increases rapidly. This is because $ϵ=-\frac{dV}{dr}$, when both p and n regions are heavily doped, a depletion layer is very thin. So, electron field is maximum at the junction.