Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 361 of 494
When the forward bias voltage across a silicon \( \text{P-N} \) junction exceeds \( 0.7\text{ V} \), the forward current increases rapidly because:
A
The semiconductor transitions into a permanent superconducting state
B
The internal barrier potential is overcome and dynamic resistance becomes very small
C
Minority carriers outnumber majority carriers across both semiconductor regions
D
Covalent bonds throughout the neutral regions break simultaneously
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: The internal barrier potential is overcome and dynamic resistance becomes very small
Concept:

When the applied forward voltage exceeds the built-in potential barrier (\( V_{\text{applied}} > V_B \)), the opposition to majority carrier diffusion is largely removed, and current is limited mainly by the small bulk resistance.

Formula:

$$I_f = I_0 \left( e^{\frac{e V_f}{\eta k_B T}} - 1 \right)$$

Solution:

  • Below \( 0.7\text{ V} \), the built-in potential barrier prevents most majority carriers from crossing the junction.


  • Once \( V_f \ge 0.7\text{ V} \), the potential barrier is eliminated.


  • The dynamic resistance of the depletion layer becomes negligible, and forward current increases exponentially with small voltage increments.


Why other options are incorrect:

  • Option A: Silicon behaves as an extrinsic semiconductor with finite resistance, not a superconductor.
  • Option C: Forward conduction is driven by majority carriers injected across the junction.
  • Option D: Mass covalent bond rupture is characteristic of high-voltage reverse breakdown, not normal forward conduction.

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