Physics Electronics MDCAT 2000
PMDC Verified Question 494 of 494
In a standard current-voltage (\( \text{I-V} \)) characteristic plot of a \( \text{P-N} \) junction diode, the forward-bias conduction region is located in:

+V-V+I-IV_knee
A
The first quadrant (positive voltage and positive current)
B
The second quadrant (negative voltage and positive current)
C
The third quadrant (negative voltage and negative current)
D
The fourth quadrant (positive voltage and negative current)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: The first quadrant (positive voltage and positive current)
Concept:

Under forward bias, a positive voltage applied to the anode results in a positive forward current from anode to cathode.

Formula:

$$V_f > 0, \quad I_f > 0 \implies (V_f, I_f) \in \text{Quadrant I}$$

Solution:

  • The horizontal axis represents applied voltage (\( +V \) to the right, \( -V \) to the left).


  • The vertical axis represents diode current (\( +I \) upward, \( -I \) downward).


  • Because forward bias corresponds to positive voltage (\( +V \)) and positive current (\( +I \)), the forward conduction curve lies in the first quadrant.


Why other options are incorrect:

  • Option B: The second quadrant (\( -V, +I \)) would represent power generation, which does not occur in passive diodes.
  • Option C: The third quadrant (\( -V, -I \)) represents reverse-bias leakage and breakdown.
  • Option D: The fourth quadrant (\( +V, -I \)) is non-physical for passive semiconductor diodes.

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