I'm just a bit confused here about the specifics of the depletion zone when a p-n junction is established. Is it the same as the potential barrier? If not, when does the depletion zone form? Or is it like the potential barrier prevents charges from moving across the junction which then results in a depletion zone?
The depletion zone is the area within which charge carriers have all recombined with atoms (charge carriers have been
depleted) - This indeed sets up a potential difference which eventually stops more charge carriers from moving and depleting! When this happens, we call the balance equilibrium
so I suppose it is the depletion region that sets up the potential barrier
I also read in my textbook that after the potential difference is established across a p-n junction (after charges drift across junction to opposite ends) that the semiconductor becomes electrically charged. Is this right? Because for a substance to become charged, shouldn't charges be removed/added, not moved?
Before the N and P type semiconductors are brought together, they aren't charged. P and N type semiconductors might have additional electrons/holes, but the charge is still neutral (the protons in the atoms balance everything out). Now, when we bring the two together, we've got negative and positive charges moving around. They were neutral to begin - So once everything has moved,
their must now be regions of charge. Specifically, the electrons that drift to the P-type semiconductor make it negatively charged. The holes that drift to the N-type semiconductor make it positively charged.
Further, could I get an explanation of the equilibrium business that occurs just after the p-n junction is formed as i've read two different things. I've read that some electrons move back to the n-type and the holes to the p-type and this will continue to happen until the depletion region no longer has any charges passing through. I've also read that charges will diffuse across the p-n junction until the number of electrons (which are now on the p-type’s boundary) have accumulated a large enough electrical charge to repel/prevent any more charge carriers from cross over the junction. Which is correct and why?
I'd say the second is correct more than the first, because electrons move to the P-type, not the N-type. Does a better job explaining it. But it sounds like both are trying to say the same thing
Finally is the following note correct? If a voltage is applied to the p-n junction, it will act as a diode, allowing current to flow from p --> n
Are we talking about an external voltage and when there is forward bias?
Yep - The junction itself
is a diode, and yes, applying an external voltage to forward bias it will allow current to flow!