Question

Consider a pn junction. Why does the depletion layer become bigger when you reverse bias the junction?

Answer 1

@peter_pan
We know that under thermal equilibrium and without any applied voltage, in a PN junction the depletion region is made of fixed charges, negative in the P-side and positive in N-side; that's for the diffusion free electrons migrate from the N-side to the P-side while holes migrate from the P-side to the N-side. These charges tend to oppose the diffusion with an electric field barrier, creating an equilibrium and a depletion region (the one with fixed charges).
Under reverse bias (external applied voltage with negative on P-side terminal and positive on the N-side terminal) the electric field barrier is higher because more holes tend to migrate from the P-side to the N-side and more electrons tend to migrate from the N-side to the P-side, thus the depletion region is wider.
If we increase enough the reverse bias voltage the junction can reach the breakdown because the electric field becomes so strong to create an abrupt current peak that destroys the junction.

Answer 2

I'm not very sure about this: "the electric field barrier is higher because more holes tend to migrate from the P-side to the N-side and more electrons tend to migrate from the N-side to the P-side". This is the point I don't understand! In fact, if the electric barrier is higher the free carriers shouldn't migrate at all from p to n and viceversa! Instead, in forward bias more electrons and holes can move from one region to the other one.

Answer 3

@peter_pan
well, really first the external voltage cause the migration of free charge carriers, then fixed charges become more in number and extension (wider depletion) so the electric field (originated by fixed charges) grows up tending to balance the previous migration at a new equilibrium point related to the current reverse bias voltage level