Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 133 of 494
What causes Zener breakdown in a reverse-biased PN junction diode?
A
Thermal runaway from high forward current
B
Collisions of high-velocity carriers in lightly doped junctions
C
Direct rupture of covalent bonds by an intense internal electric field in heavily doped junctions
D
Recombination of majority carriers across a wide barrier
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: Direct rupture of covalent bonds by an intense internal electric field in heavily doped junctions
Concept:

Zener breakdown occurs in heavily doped diodes with narrow depletion regions (\( < 10\text{ nm} \)), where a reverse voltage produces a very strong electric field (\( > 10^6\text{ V/m} \)) that pulls electrons directly out of covalent bonds.

Formula:

$$E = \frac{V_R}{W} \ge 10^6\text{ V/m} \quad (\text{Field Emission / Direct Band Tunneling})$$

Solution:

  • Heavy doping produces an extremely thin depletion layer.


  • Applying a modest reverse voltage (typically \( < 6\text{ V} \)) creates an electric field strong enough to pull valence electrons into the conduction band via quantum tunneling.


  • This produces a sharp, non-destructive increase in reverse current.


Why other options are incorrect:

  • Option A: Thermal runaway is an overheating failure mode, not the quantum mechanism of Zener breakdown.
  • Option B: Impact ionization by high-velocity carriers in lightly doped junctions defines Avalanche breakdown (typically \( > 6\text{ V} \)).
  • Option D: Recombination decreases during reverse bias because carriers are separated rather than combined.

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