Build an abrupt pn junction, apply a voltage, and follow how charge, field, potential and the energy bands respond, together with the current and the junction capacitance.
The depletion region is almost entirely on the lightly doped side: with ND ≫ NA the n side barely depletes (a one-sided junction, as for a MOSFET source or drain against its body). The charge areas on the two sides are equal, so the tall thin n-side block matches the short wide p-side block.
Reverse bias raises the barrier ψc = φbi + VR, widens the depletion region as √ψc, raises the peak field and lowers C′j. In the band diagram the quasi-Fermi levels separate by qVA while staying flat across the depletion region.
Forward bias lowers the barrier and the current grows by a decade every φt ln10 ≈ 60 mV, while C′j rises. Close to φbi the depletion approximation fails (the widget floors ψc at 0.05 V).
Try this
Set NA = 10¹⁷ and ND = 10¹⁹ cm⁻³ at VA = 0. Read φbi, d1 and d2. Which side holds almost all the depletion width?
Sweep VA from 0 to −5 V. By what factor do d2 and C′j change, and does that match the √ dependence on ψc?
Make NA = ND = 10¹⁷ cm⁻³. Compare the field and charge plots with the one-sided case.
Switch the I–V plot to linear and find the voltage at which the current becomes visible on the scale of the plot (the “knee”). Is it the same on the log plot?
Set VA = +0.5 V and raise T from 300 to 400 K. φt grows, which alone would lower eV/φt, yet the current rises. Why? (Watch I₀ ∝ ni².)