The breakdown in a reverse-biased p–n junction diode is more likely to occur due to (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
Correct option is A) When it comes to he breakdown in a reverse biased PN junction diode, it will probably happen only because of the accumulation of the higher charge at the biased region and large velocity of the minority charge if the doping concentration is small. This is the main cause the breakdown
