Concept:Excessive forward current increases junction temperature (\( T_j \)) through \( I^2 R \) heating, which increases non-radiative Auger recombination, causes carrier overflow, and reduces internal quantum efficiency (thermal roll-off).
Formula:$$P_{\text{opt}}(T_j) = P_0 \exp\left(-\frac{T_j}{T_0}\right) \quad \text{where } T_j = T_a + I_F V_F \theta_{ja}$$
Solution:- High current increases internal power dissipation (\( P = I_F V_F \)), raising the junction temperature \( T_j \).
- Higher temperatures promote non-radiative recombination and increase carrier leakage out of the active region.
- This causes optical output power to saturate and drop, a phenomenon known as thermal roll-off.
Why other options are incorrect:- Option A: Heating degrades efficiency; it does not produce superconductivity.
- Option B: Thermal heating is an electronic/lattice effect and does not affect atomic nuclei.
- Option C: Increasing temperature slightly narrows the band gap; it does not expand it.
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