Concept:A solar cell generates power (acting as a source rather than a load), which corresponds to the fourth quadrant of the standard diode I-V characteristic curve (positive terminal voltage, negative output current).
Formula:$$P = V \cdot I < 0 \quad (\text{Power delivered to external load in 4th quadrant})$$
$$I(V) = I_0 \left( e^{\frac{q V}{k T}} - 1 \right) - I_L$$
Solution:- When illuminated, the photo-generated current \( I_L \) shifts the diode's I-V curve downward by \( I_L \).
- In the fourth quadrant, the terminal voltage is positive (\( 0 < V < V_{\text{oc}} \)) while current flows out of the positive terminal (\( I < 0 \)), delivering power (\( P = V \cdot I < 0 \)) to the external load.
Why other options are incorrect:- Option A: In the first quadrant, a forward-biased diode consumes electrical power.
- Option B: The second quadrant is not accessible in standard passive or solar junction configurations.
- Option D: In the third quadrant, a reverse-biased photodiode consumes power from an external supply.
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