Question

The reverse current in the semiconductor diode changes from 20 \(\mu\text{A}\) to 40 \(\mu\text{A}\) when the reverse potential is changed from 10 V to 15 V, then the reverse resistance of the junction diode will be:

• A. 250 \(\text{k}\Omega\)
• B. 400 \(\Omega\)
• C. 400 \(\text{k}\Omega\)
• D. 2500 \(\Omega\)

Hint:

To calculate reverse resistance, use the formula \( R = \frac{\Delta V}{\Delta I} \), where \( \Delta V \) is the change in reverse voltage and \( \Delta I \) is the change in reverse current.

Answer 1

The Correct Option is A

To find the reverse resistance of the diode, we can use Ohm's law, which is given by: \[ R = \frac{\Delta V}{\Delta I} \] where: - \( \Delta V \) is the change in voltage, - \( \Delta I \) is the change in current.
The reverse potential changes from 10 V to 15 V, so: \[ \Delta V = 15 \, \text{V} - 10 \, \text{V} = 5 \, \text{V} \] The reverse current changes from 20 \(\mu\text{A}\) to 40 \(\mu\text{A}\), so: \[ \Delta I = 40 \, \mu\text{A} - 20 \, \mu\text{A} = 20 \, \mu\text{A} = 20 \times 10^{-6} \, \text{A} \] Now substitute the values into the formula for resistance: \[ R = \frac{5 \, \text{V}}{20 \times 10^{-6} \, \text{A}} = \frac{5}{20 \times 10^{-6}} = 250 \times 10^{3} \, \Omega = 250 \, \text{k}\Omega \] Thus, the reverse resistance of the junction diode is \( 250 \, \text{k}\Omega \).