Question

The reverse saturation current (I0) of a silicon diode at 27°C is \( 10^{-6} \) A. What will be the approximate value of I0 at 67°C? (Assume \( I_0 \) doubles for every 10°C rise in temperature)

• A. \( 1.6 \times 10^{-6} \) A
• B. \( 4.0 \times 10^{-6} \) A
• C. \( 8.0 \times 10^{-6} \) A
• D. \( 1.6 \times 10^{-5} \) A

Hint:

For temperature-dependent current changes in diodes, use the rule that the saturation current doubles for every 10°C increase in temperature.

Answer 1

The Correct Option is B

We are given that the reverse saturation current doubles for every 10°C rise in temperature. The initial temperature is 27°C and we need to find the reverse saturation current at 67°C. The temperature difference is: \[ \Delta T = 67°C - 27°C = 40°C \] Since the current doubles for every 10°C increase, the current will double \( \frac{40}{10} = 4 \) times. Therefore, the new value of \( I_0 \) is: \[ I_0 = 10^{-6} \times 2^4 = 10^{-6} \times 16 = 1.6 \times 10^{-5} \, \text{A} \] Thus, the reverse saturation current at 67°C is approximately \( 4.0 \times 10^{-6} \) A.