Concept:Charge neutrality requires that total uncompensated negative charge on the P-side equals total uncompensated positive charge on the N-side: \( q N_A x_p = q N_D x_n \).
Formula:$$\frac{x_n}{x_p} = \frac{N_A}{N_D} \implies x_n \gg x_p \quad (\text{when } N_A \gg N_D)$$
Solution:- Because \( N_A \gg N_D \), fewer atomic layers on the heavily doped P-side are needed to provide the balancing space charge.
- The lightly doped N-side must deplete across a much larger distance \( x_n \) to uncover an equal amount of charge.
- Consequently, the depletion layer extends predominantly into the lightly doped N-region (\( W \approx x_n \)).
Why other options are incorrect:- Option A: Equal penetration occurs only for symmetrical junctions where \( N_A = N_D \).
- Option B: Depletion depth is inversely related to doping concentration.
- Option C: Depletion layers are confined within the semiconductor and do not extend into ohmic metal contacts.
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