Geoffrey_Coram
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Jason - If the dopant density (which I think you are calling p(x)) is smaller on the n side, then this forces the depletion width to be larger on the n side. It's simple math: the depleted charge on either side must be equal (at equilibrium), and the depleted charge is the density times the volume. Assuming equal cross-sectional area, you're left with the depletion width.
As to why n doping is less than p doping, this is process-specific. One could build an (n)(p+) diode or an (n+)(p) diode, ie, with either side more highly doped than the other. I don't know if there's a reason that diodes would typically be doped stronger on the p side.
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