Q.Explain $P^+\,n^-$ junction under thermal equilibrium.

The rectifying junction of a power diode is between the heavily doped p⁺ anode and the lightly doped n⁻ drift region; at thermal equilibrium diffusion of carriers sets up a depletion region and a built-in potential that stops further net flow.

[!ANSWER]

Under thermal equilibrium the $P^+ n^-$ junction forms a depletion region that extends deep into the lightly doped n⁻ side, and a built-in barrier $V_{bi}$ balances drift and diffusion so the net current is zero.

[!TLDR]

At equilibrium the $P^+ n^-$ junction develops a wide depletion layer (mainly into the n⁻ side) and a built-in potential that gives zero net current.

In a power diode the main rectifying junction lies between the heavily doped p⁺ anode layer and the very lightly doped n⁻ drift region. Under thermal equilibrium (no external bias applied) majority holes from the p⁺ side diffuse into the n⁻ region and majority electrons from the n⁻ side diffuse into the p⁺ region. This diffusion leaves behind immobile ionised acceptor ions on the p⁺ side and ionised donor ions on the n⁻ side, forming a depletion (space-charge) region around the junction.

Because the n⁻ region is only lightly doped, it contains very few donor atoms, so the depletion region must extend deep into the n⁻ drift region to expose enough fixed charge to balance the charge on the narrow, heavily doped p⁺ side. The exposed charges set up a built-in potential barrier $V_{bi}$ that opposes further diffusion. At equilibrium the diffusion current is exactly balanced by the drift current produced by this barrier field, so the net current across the junction is zero.

[!ANSWER]

At thermal equilibrium the $P^+ n^-$ junction has a depletion layer spreading mainly into the lightly doped n⁻ drift region and a built-in barrier potential; drift and diffusion currents cancel, giving zero net current, and this wide n⁻ depletion is what later lets the diode block a large reverse voltage.