Concept:The reverse saturation current approximately doubles for every \( 10^\circ\text{C} \) increase in operating temperature.
Formula:$$I_0(T_2) = I_0(T_1) \times 2^{\frac{\Delta T}{10}}$$
Solution:- Calculate temperature change: \( \Delta T = 55^\circ\text{C} - 25^\circ\text{C} = 30^\circ\text{C} \).
- Calculate number of \( 10^\circ\text{C} \) doublings: \( n = \frac{30}{10} = 3 \).
- Calculate leakage current:
- $$I_0(55^\circ\text{C}) = 10\text{ nA} \times 2^3 = 10\text{ nA} \times 8 = 80\text{ nA}$$
Why other options are incorrect:- Option A: \( 30\text{ nA} \) assumes a linear increase (\( 10 + 30 \)) rather than exponential doubling.
- Option C: \( 40\text{ nA} \) corresponds to two doublings (\( 20^\circ\text{C} \) rise to \( 45^\circ\text{C} \)).
- Option D: \( 160\text{ nA} \) corresponds to four doublings (\( 40^\circ\text{C} \) rise to \( 65^\circ\text{C} \)).
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