Question

A p-n junction diode is reverse biased with a voltage of 8 V. If the resistance of the diode is \( 4 \times 10^7 \, \Omega \), then the reverse saturation current is

• A. \( 32 \, \mu A \)
• B. \( 2 \, \mu A \)
• C. \( 0.2 \, \mu A \)
• D. \( 0.5 \, \mu A \)

Hint:

For calculating current in resistive components, use Ohm’s law. Make sure to convert units correctly when working with small values such as microamperes.

Answer 1

The Correct Option is C

The reverse saturation current in a diode can be calculated using Ohm's Law: \[ I = \frac{V}{R} \] where \( I \) is the current, \( V \) is the applied reverse bias voltage, and \( R \) is the resistance of the diode. Given: \[ V = 8 \, \text{V}, \quad R = 4 \times 10^7 \, \Omega \] Substitute the values into the equation: \[ I = \frac{8}{4 \times 10^7} = 0.2 \times 10^{-6} \, \text{A} = 0.2 \, \mu A \] Thus, the reverse saturation current is \( 0.2 \, \mu A \). Therefore, the correct answer is \( \boxed{0.2 \, \mu A} \).

Answer 2

Step 1: Understand the given data.
- Reverse bias voltage, \( V = 8 \, \text{V} \)
- Resistance of the diode in reverse bias, \( R = 4 \times 10^7 \, \Omega \)

Step 2: Use Ohm's Law to calculate the current.
\[ I = \frac{V}{R} = \frac{8}{4 \times 10^7} = \frac{8}{40,000,000} = 2 \times 10^{-7} \, \text{A} \]

Step 3: Convert the result to microamperes.
\[ 2 \times 10^{-7} \, \text{A} = 0.2 \, \mu \text{A} \]

Step 4: Conclusion.
The reverse saturation current is \( \boxed{0.2 \, \mu \text{A}} \).