JEE Mains · Physics · STD 12 - Semicondutor electronics
- JEE Mains 2026
- Medium
- MCQ
In a semiconductor p-n diode, the doping concentrations on p-side and n-side are \(10^{15}\text{ atoms/cm}^3\) and \(10^{18}\text{ atoms/cm}^3\), respectively. Which one of the following statements is true ?
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) The depletion region width is more on p-side compared to that in n-side
Step-by-step Solution
Detailed explanation
In a p-n junction diode, the total charge on both sides of the depletion region must be equal to maintain charge neutrality.
Let \(x_p\) and \(x_n\) be the widths of the depletion region on the p-side and n-side, respectively. Let \(N_A\) and \(N_D\) be the doping concentrations on the p-side and n-side.
The charge neutrality condition is given by:
\(N_A x_p = N_D x_n\)
Given \(N_A = 10^{15} \text{ atoms/cm}^3\) and \(N_D = 10^{18} \text{ atoms/cm}^3\).
Substituting the values, we get:
\(10^{15} x_p = 10^{18} x_n\)
\(x_p = 10^3 x_n\)
Since \(x_p > x_n\), the depletion region penetrates more into the lightly doped side. Therefore, the depletion region width is more on the p-side compared to that in the n-side.
Answer: The depletion region width is more on p-side compared to that in n-side
Let \(x_p\) and \(x_n\) be the widths of the depletion region on the p-side and n-side, respectively. Let \(N_A\) and \(N_D\) be the doping concentrations on the p-side and n-side.
The charge neutrality condition is given by:
\(N_A x_p = N_D x_n\)
Given \(N_A = 10^{15} \text{ atoms/cm}^3\) and \(N_D = 10^{18} \text{ atoms/cm}^3\).
Substituting the values, we get:
\(10^{15} x_p = 10^{18} x_n\)
\(x_p = 10^3 x_n\)
Since \(x_p > x_n\), the depletion region penetrates more into the lightly doped side. Therefore, the depletion region width is more on the p-side compared to that in the n-side.
Answer: The depletion region width is more on p-side compared to that in n-side
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