Question:

In a pn junction, dopant concentration on p-side is higher than n-side. Which statements are correct when the junction is unbiased?

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More doping → narrower depletion width; less doping → wider depletion width.
Updated On: Dec 12, 2025
  • The width of the depletion layer is larger on the n-side.
  • Fermi energy is higher on the p-side.
  • Negative charge per unit area on p-side equals positive charge per unit area on n-side.
  • Built-in potential depends on dopant concentration.
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The Correct Option is A, C, D

Solution and Explanation

Step 1: Depletion widths.
Depletion width on a side is inversely proportional to doping. p-side is more doped → depletion region on n-side is wider. So (A) true.

Step 2: Charge neutrality.
Total negative charge in n-side depletion region = total positive charge in p-side region. Thus (C) true.

Step 3: Built-in potential.
\[ V_{bi} = \frac{kT}{e} \ln\left(\frac{N_A N_D}{n_i^2}\right) \] Depends on doping ⇒ (D) true.

Step 4: Fermi level.
At equilibrium, Fermi levels align → (B) is false.

Step 5: Conclusion.
Correct: (A), (C), (D).

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