Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 370 of 494
What occurs when the reverse bias voltage across a standard silicon \( \text{P-N} \) junction diode exceeds its Peak Inverse Voltage (\( \text{PIV} \)) rating without current-limiting resistance?
A
The depletion layer narrows, initiating forward conduction
B
The diode stabilizes the reverse voltage by decreasing its leakage current to zero
C
Avalanche breakdown occurs, causing excessive reverse current and potential thermal destruction
D
The built-in potential barrier permanently inverts its polarity
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: Avalanche breakdown occurs, causing excessive reverse current and potential thermal destruction
Concept:

Exceeding the reverse breakdown voltage provides minority carriers with sufficient kinetic energy to break covalent bonds via impact ionization, triggering carrier multiplication.

Formula:

$$P_{\text{dissipated}} = V_{\text{breakdown}} \times I_{\text{reverse}} > P_{\text{max}}$$

Solution:

  • High electric fields accelerate minority carriers to high velocities.


  • Collisions with lattice atoms create additional electron-hole pairs (avalanche multiplication).


  • This causes reverse current to increase sharply. Without a series resistor to limit current, excessive power dissipation (\( P = V I \)) can cause thermal runaway and destroy the device.


Why other options are incorrect:

  • Option A: Breakdown increases reverse carrier multiplication under high field conditions; it does not narrow the depletion layer into forward bias.
  • Option B: Reverse current increases rapidly rather than dropping to zero.
  • Option D: The space-charge polarity remains unchanged; breakdown is an electrical conduction phenomenon.

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