The reverse saturation current of the Ge transistor is 2 μA at room temperature of 25°C and increases by a factor of 2 for each temperature increase of 10°C. Find the reverse saturation current of the transistor at a temperature of 75°C.
Question 4.24: The reverse saturation current of the Ge transistor is 2 μA ...
Given I_{o1} = 2 \mu A at T_{1} = 25°C, T_{2} = 75°C
Therefore, the reverse saturation current of the transistor at T_{2} = 75°C is
I_{o2}=I_{o1}\times 2^{(T_{2}-T_{1} )/10}=2\times 10^{-6} \times 2^{\left(\frac{75-25}{10} \right) }
= 2 × 10^{ – 6} × 2^{5} = 64 \mu A
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