Concept:In heavily doped diodes, the depletion layer is extremely narrow (\(< 10\text{ nm}\)). Even a moderate reverse voltage produces an electric field exceeding \(10^6\text{ V/m}\), which pulls valence electrons directly across the narrow bandgap into the conduction band via field ionization (Zener breakdown).
Formula:$$E = \frac{V_r}{W} > 10^6\text{ V/m}$$
Solution:- Heavy doping produces a very thin depletion width \(W\).
- A strong electric field \(E = V / W\) ruptures covalent bonds directly, allowing valence electrons to tunnel into the conduction band.
- This field-emission process is known as Zener breakdown (typically occurring at \(V_z < 6\text{ V}\)).
Why other options are incorrect:- Option A: Forward diffusion is irrelevant under strong reverse bias.
- Option C: Breakdown is a solid-state junction phenomenon, not lead-wire melting.
- Option D: Donor ions are fixed in the semiconductor lattice and do not recombine with external contact atoms.
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