Concept:An internal shunt resistance \( R_{\text{sh}} \) acts in parallel with the PN junction, creating a leakage path that drains photo-generated current and reduces both \( V_{\text{oc}} \) and the Fill Factor.
Formula:$$I = I_L - I_0 \left(e^{\frac{q(V + I R_s)}{k T}} - 1\right) - \frac{V + I R_s}{R_{\text{sh}}}$$
Solution:- Ideally, shunt resistance is infinite (\( R_{\text{sh}} \to \infty \)).
- When \( R_{\text{sh}} \) is low, alternate leakage paths allow photo-generated carriers to bypass the external load.
- This reduces \( V_{\text{oc}} \), flattens the I-V curve slope near short-circuit conditions, and lowers the Fill Factor.
Why other options are incorrect:- Option B: Shunt leakage reduces power output rather than increasing it.
- Option C: Shunt paths increase overall leakage current.
- Option D: Solar cells are broad-area energy converters, not coherent lasers.
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