Q157 / 180 · COMEDK 2023 (Evening Shift) · Physics
The reverse current in the semiconductor diode changes from $20 \mu \mathrm{A}$ to $40 \mu \mathrm{A}$ when the reverse potential is changed from $10 \mathrm{~V}$ to $15 \mathrm{~V}$, then the reverse resistance of the junction diode will be :
- Correct answerA$250 \mathrm{~k} \Omega$
- B$250 \Omega$
- C$400 \Omega$
- D$400 \mathrm{~k} \Omega$
Solution
The reverse resistance $R$ of a junction diode is defined as the ratio of the change in reverse potential $\Delta V$ to the change in reverse current $\Delta I$.
Given:
Initial potential $V_1 = 10 \text{ V}$
Final potential $V_2 = 15 \text{ V}$
Initial current $I_1 = 20 \mu \text{A} = 20 \times 10^{-6} \text{ A}$
Final current $I_2 = 40 \mu \text{A} = 40 \times 10^{-6} \text{ A}$
Change in potential $\Delta V = V_2 - V_1 = 15 \text{ V} - 10 \text{ V} = 5 \text{ V}$
Change in current $\Delta I = I_2 - I_1 = 40 \mu \text{A} - 20 \mu \text{A} = 20 \mu \text{A} = 20 \times 10^{-6} \text{ A}$
The reverse resistance $R$ is calculated as:
$R = \dfrac{\Delta V}{\Delta I} = \dfrac{5 \text{ V}}{20 \times 10^{-6} \text{ A}}$
$R = \dfrac{5}{20} \times 10^{6} \Omega = 0.25 \times 10^{6} \Omega = 250 \times 10^{3} \Omega = 250 \text{ k}\Omega$
Answer: $250 \mathrm{~k} \Omega$
Given:
Initial potential $V_1 = 10 \text{ V}$
Final potential $V_2 = 15 \text{ V}$
Initial current $I_1 = 20 \mu \text{A} = 20 \times 10^{-6} \text{ A}$
Final current $I_2 = 40 \mu \text{A} = 40 \times 10^{-6} \text{ A}$
Change in potential $\Delta V = V_2 - V_1 = 15 \text{ V} - 10 \text{ V} = 5 \text{ V}$
Change in current $\Delta I = I_2 - I_1 = 40 \mu \text{A} - 20 \mu \text{A} = 20 \mu \text{A} = 20 \times 10^{-6} \text{ A}$
The reverse resistance $R$ is calculated as:
$R = \dfrac{\Delta V}{\Delta I} = \dfrac{5 \text{ V}}{20 \times 10^{-6} \text{ A}}$
$R = \dfrac{5}{20} \times 10^{6} \Omega = 0.25 \times 10^{6} \Omega = 250 \times 10^{3} \Omega = 250 \text{ k}\Omega$
Answer: $250 \mathrm{~k} \Omega$
Hence, the correct answer is option: 'A'.