Question

A semiconductor \( pn \) junction at thermal equilibrium has the space charge density \( \rho(x) \) profile as shown in the figure. The figure that best depicts the variation of the electric field \( E \) with \( x \) is ( \( W \) denotes the width of the depletion layer): 

• A. A
• B. B
• C. C
• D. D

Hint:

In a \( pn \) junction, the electric field is obtained by integrating the charge density. A linear \( \rho(x) \) produces a linearly varying \( E(x) \) that changes sign at the metallurgical junction.

Answer 1

The Correct Option is A

Step 1: Recall the relation between charge density and electric field.
From Poisson’s equation, \[ \frac{dE}{dx} = \frac{\rho(x)}{\varepsilon}, \] where \( \rho(x) \) is the charge density and \( \varepsilon \) is the permittivity of the semiconductor.
Step 2: Analyze the given \( \rho(x) \) profile.
The charge density varies linearly with \( x \), being negative on the p-side and positive on the n-side. Therefore, the derivative of \( E \) with respect to \( x \) is linear, implying that \( E(x) \) must vary quadratically if \( \rho(x) \) were constant, but since \( \rho(x) \) is linear, \( E(x) \) will vary linearly.
Step 3: Determine the behavior of \( E(x) \).
- The slope of \( E(x) \) on the p-side is positive (since \( \rho(x) \) is negative). - The slope of \( E(x) \) on the n-side is negative (since \( \rho(x) \) is positive). At \( x = 0 \), the electric field passes through zero because the net electric field in equilibrium must balance across the junction.
Step 4: Graphical interpretation.
Thus, \( E(x) \) increases linearly from \( -W/2 \) to 0, and decreases linearly from 0 to \( W/2 \), forming an inverted V-shape centered at the origin.
Step 5: Final Answer.
Hence, the correct graph is the one shown in Option (B).