Question

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

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

Answer 1

Answer: (A), (D)

large velocity of the minority charge carriers if the doping concentration is small.


During reverse biasing, ionization of the atoms takes place. Since, the polarity of the voltage applied is negative, the minority charge carries are accelerated.

When the doping concentration is small, these minority charge carriers collide with other atoms releasing new electrons and thus producing more charge carriers, which causes breakdown.

When the doping concentration is large, there are a large number of ions in the depletion region, which creates a strong electric field, which causes breakdown.

$\text{Final Answer :}\left(a\right),\left(d\right)$