Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 265 of 494
When a forward-conducting PN junction diode is rapidly switched into reverse bias, current does not drop to zero immediately because of:
A
Magnetic core hysteresis in the leads
B
Instantaneous dielectric breakdown of ambient air
C
Reverse recovery time (\( t_{rr} \)) required to sweep out and recombine stored excess minority carriers near the junction
D
Nuclear decay of dopant atoms
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: Reverse recovery time (\( t_{rr} \)) required to sweep out and recombine stored excess minority carriers near the junction
Concept:

During forward conduction, excess minority carriers are stored near the depletion layer edges. When reverse bias is suddenly applied, these stored charges produce a transient reverse current pulse until they are removed via drift and recombination over the reverse recovery time \( t_{rr} \).

Formula:

$$t_{rr} = t_s \text{ (storage time)} + t_t \text{ (transition time)}$$

Solution:

  • In forward bias, large numbers of minority carriers are injected across the junction and stored in the neutral regions.


  • When the applied voltage switches to reverse bias, these stored carriers flow backward as a substantial reverse current.


  • The diode recovers its blocking state only after these carriers are swept out or recombine, over the duration \( t_{rr} \).


Why other options are incorrect:

  • Option A: Packaging lead inductance is negligible compared to carrier storage effects at standard switching speeds.
  • Option B: Air dielectric breakdown does not occur at normal operating voltages.
  • Option D: Dopant atoms are chemically stable and do not undergo nuclear decay.

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