I have read in my textbook that if a PN junction is in Zener breakdown and if we now reduce the external voltage, the PN junction is not damaged and returns back to its initial state. However, the same is not applicable in avalanche breakdown, where the damage is permanent.

I tried to analyse this from the mechanism of each breakdown, but couldn't arrive at a possible explanation.

  • $\begingroup$ With appropriate current limiting, avalanche does not have to be destructive. $\endgroup$
    – Jon Custer
    Dec 24, 2018 at 16:36

1 Answer 1


During Avalanche breakdown carriers(electrons) are developed due to the collision of free electrons (generated by reverse biased voltage) with adjacent atoms(diffusion current).

While in Zener breakdown, high doping cause strong electric field at the junction. Breaking of covalent bonds of lattice atoms and the generation of carriers happens due to this electric field. It does not involve collision of carriers with the lattice atoms(drift current).

The collisions happening in avalanche breakdown dissippiate thermal energy and may cause damage of normal diodes. Whereas zener doesn't have much thermal dissippiation. Thus after reaching zener breakdown voltage zener remains undamaged, can be used again.

  • $\begingroup$ Ig zener breakdown doesn't involve breaking of covalent bonds. Wikipedia states [in zener breakdown] strong electric fields enable tunneling of electrons across the depletion region... [While] avalanche breakdown involves electrons being accelerated by the electric field, to energies sufficient for freeing electron-hole pairs via collisions with bound electrons. $\endgroup$
    – Shub
    Oct 16, 2023 at 13:30

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