Physics behind depletion region? I was reading the PN junction. During the formation of PN junction diffusion process takes place in which the free electrons cross the boundary of the P-N and joined with the holes in the P-type to form the negative ion and holes on the N side cause positive ions. So, the P-side has negative ions and N-side has positive that constitutes the depletion region
From doping we know there are other free electrons in the N-type and holes in the P-side. What I am not able to understand that why not the electron can join the N-side positive ions to make it neutral atom and why not on P side electrons from negative ions drift further away from the boundary to join with other holes. Why they always stay near the boundary (depletion region). 
Please keep the things simple without heavy duty equations.
 A: In a general sense both of your questions ask "Why is the arrangement of free and bound charges in a PN junction in equilibrium different from $X$?", where $X$ is one of the alternative arrangements you suggest. The simple answer to these questions is that such alternative arrangements would necessarily put the system out of equilibrium and the arrangement is therefore unstable. A system's tendency towards equilibrium will act to undo the modification you suggest. This is essentially a restatement of Le Chatlier's principle. It is not a very illuminating answer but one that I find is helpful to keep in mind when analyzing problems of equilibrium.
We can move closer to the physics by asking, "Why is the equilibrium arrangement of free and bound charges in a PN junction more stable than some other arrangement?" 
In particular, you suspect that the PN junction might achieve greater stability if electrons recombine with donor sites and holes recombine with acceptor sites, since such processes lead to a reduction in energy. The flaw in this reasoning is that the binding between an electron and a donor ion (or a hole and an acceptor ion) is so weak that, at normal temperatures, the rate at which thermal agitation will ionize a donor-electron (or acceptor-hole) pair is much higher than the rate at which an electron happens to get trapped by a donor ion (same for holes and acceptors). Therefore, in  order for the overall rate of formation to balance the overall rate of ionization (i.e. to meet the condition of thermal equilibrium), we would require a relatively low concentration of donor-electron (hole-acceptor) pairs and relatively high concentration of electrons and ionized donors. (This is basically a restatement of the law of mass action.)
Put another way, if some small region of the junction happened to have a relatively large concentration of donor-electron (hole-acceptor) pairs, then this region would have a lower temperature compared to its surroundings. Heat would then flow into the section, and this heat would go into ionizing the donor-electron (hole-acceptor) pairs until the concentration equaled that of its surroundings. (This is another restatement of Le Chatlier's principle, tailored specifically to the case doped semiconductors).
So the answer to your questions depends very much on the non-zero temperature of the junction.
