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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