Question

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?

• A. Widths of depletion region on either side of the interface are equal
• B. The depletion region width is more on p-side compared to that in n-side
• C. The depletion region width is more on n-side compared to that in p-side
• D. No depletion region forms because of unequal doping concentrations

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In a p-n junction, the depletion region must satisfy the condition of charge neutrality. This means the total ionized charge on the p-side must equal the total ionized charge on the n-side.
: Key Formula or Approach:
Charge neutrality condition: \( q \cdot N_A \cdot x_p = q \cdot N_D \cdot x_n \).
Where \( N_A, N_D \) are acceptor/donor concentrations and \( x_p, x_n \) are widths of depletion layers on each side.
Step 2: Detailed Explanation:
Given:
\( N_A (\text{p-side}) = 10^{15} \text{ cm}^{-3} \).
\( N_D (\text{n-side}) = 10^{18} \text{ cm}^{-3} \).
From neutrality: \( N_A \cdot x_p = N_D \cdot x_n \).
\[ \frac{x_p}{x_n} = \frac{N_D}{N_A} = \frac{10^{18}}{10^{15}} = 1000 \].
Since \( x_p = 1000 \cdot x_n \), the depletion region extends much further into the p-side.
This is a general principle: the depletion region is always wider on the more lightly doped side.
Step 3: Final Answer:
The depletion region width is more on the p-side compared to that in the n-side.