Concept:Reverse saturation current \( I_0 \) is caused by thermally generated minority carriers, and increases by approximately \( 7\% \) per \( ^\circ\text{C} \), causing it to roughly double for every \( 10^\circ\text{C} \) rise in temperature.
Formula:$$I_0(T_2) = I_0(T_1) \times 2^{\frac{T_2 - T_1}{10}}$$
Solution:- Thermal generation of electron-hole pairs increases exponentially with temperature.
- For both Silicon and Germanium diodes, a temperature rise of \( \Delta T = 10^\circ\text{C} \) approximately doubles the reverse saturation current: \( I_0(T + 10) \approx 2 \times I_0(T) \).
Why other options are incorrect:- Option A: Higher temperatures increase thermal carrier generation, so leakage current increases rather than decreases.
- Option C: A thousand-fold increase would require a much larger temperature rise (roughly \( 100^\circ\text{C} \)).
- Option D: Reverse saturation current is strongly temperature-dependent.
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