The breakdown in a reverse biased p-n junction diode is more likely to occur due to

Options

A. large velocity of the minority charge carriers if the doping concentration is small.
B. large velocity of the minority charge carriers if the doping concentration is large.
C. strong electric field in a depletion region if the doping concentration is small.
D. strong electric field in the depletion region if the doping concentration is large.

Answer

A, D

Explanation

Reverse current obtained in reverse biased p-n junction is grateful to minority charge carriers. When amount of impurity (or doping level) is less, the minority charge (holes in N part and free electrons in P part) accelerate towards the junction. AT this time, they collide with the atoms in depletion layer, releasing more and more charge carriers subsequently, which also in turn, get accelerated continuing the process further. Rate of this process increases with the increase in reverse voltage. Finally at certain reverse voltage, breakdown takes place. Thus, option (A) is correct.
If doping level is kept high, then electric field in the depletion layer is strong from the beginning itself (because of high barrier potential), because of which, the charges freed in the depletion layer obtain tremendous high velocity, which in turn, release the charge carriers further by colliding with the atoms in the depletion layer. Rate of this process increases with the increase in reverse voltage. Finally, breakdown occurs at some reverse voltage. Thus, option (D) is also correct.