JEE Main 2021 · 24 February · evening · Single correct · +4 / −1
Zener breakdown occurs in a p - n junction having p and n both :
- (a) heavily doped and have wide depletion layer.
- (b) lightly doped and have narrow depletion layer.
- (c) heavily doped and have narrow depletion layer.
- (d) lightly doped and have wide depletion layer.
Answer and solution
Answer (c) heavily doped and have narrow depletion layer.
Solution
Physics sir · 4 steps
Zener breakdown needs a huge field, and heavy doping with a thin depletion layer is what gives it.
- Doping sets width. Heavy doping on both sides packs many fixed ions close to the junction. Only a thin slice of them has to be uncovered to build the barrier, so the depletion layer is narrow, typically below 1 μm.
- Field across it. In reverse bias, the electric field is roughly the voltage divided by the depletion width. A narrow layer therefore gives a very strong field even at a few volts.
E ≈ V/d, e.g. (5 V)/(10⁻⁷ m) = 5 × 10⁷ V m⁻¹
- Field emission. A field this strong pulls valence electrons straight out of their covalent bonds. The reverse current rises sharply at a fixed voltage, and that is Zener breakdown.
- So. Both sides are heavily doped and the depletion layer is narrow, which is option (c).
The idea to keep. Heavy doping means a thin depletion layer, a strong field and Zener breakdown at low voltage. Light doping means a wide layer and avalanche breakdown at high voltage.
Answer from the NTA final answer key. Solution by PrepWiser.