First of all, it is wrong(misleading) to refer P type as positively charge and N type as negatively charged, both P type and N type are neutral in nature, however it is right to say that P type contains free charge carriers in form of holes and N type contains carriers in form of electrons.
Secondly, a depletion region/layer is already in picture from the beginning i.e. while fabricating P-N Junction, due to abrupt change in concentration of electrons/holes in two types of materials, electrons from N type material and holes from P type material diffuses
to P type and N type materials respectively. This leads to formation of depletion region/layer which contains ions (Positive and negative ions), not electrons or holes. These ions are generally immobile in nature. And in this way, region nearby p-n interface loose its neutrality and becomes charged.Since space charges in depletion region leads to an electric field which opposes further movement of electrons and holes due to process of diffusion, P-N junction reach to a state of equilibrium.
Next thing is, again, applying a positive current is somehow misleading, we apply positive voltage to P material and negative voltage (zero voltage) to N material, and when battery connected this way, its Forward Bias/Biasing.With a battery connected this way, the holes in the P-type region and the electrons in the N-type region are pushed toward the junction. This reduces the width of the depletion zone. The positive charge applied to the P-type material repels the holes, while the negative charge applied to the N-type material repels the electrons. As electrons and holes are pushed toward the junction, the distance between them decreases.Only majority carriers (electrons in N-type material or holes in P-type) can flow through a semiconductor for a macroscopic length. With this in mind, consider the flow of electrons across the junction. The forward bias causes a force on the electrons pushing them from the N side toward the P side. With forward bias, the depletion region is narrow enough that electrons can cross the junction and inject into the P-type material. However, they do not continue to flow through the P-type material indefinitely, because it is energetically favorable for them to recombine with holes. Although the electrons penetrate only a short distance into the P-type material, the electric current continues uninterrupted, because holes (the majority carriers) begin to flow in the opposite direction. The total current (the sum of the electron and hole currents) is constant in space, because any variation would cause charge buildup over time
Therefore, the current flow through the diode involves electrons flowing through the N-type region toward the junction, holes flowing through the P-type region in the opposite direction toward the junction, and the two species of carriers constantly recombining in the vicinity of the junction. The electrons and holes travel in opposite directions, but they also have opposite charges, so the overall current is in the same direction on both sides of the diode, as required.
Same analogy can be obtained/derived for Reverse Bias situation as well.
I think i answered most of the questions of yours, rest you can answer by yourself.
Though, i will also suggest you to go through some standard book (Streetman and Banerjee is good) to understand concepts fully, once you understand them, there will be no doubt in future as well, but its really difficult to understand P-N junction or physics concepts through a 1/2 hour video.
A hole is not the result of a charge-neutral atom losing an electron. A hole is created when an "acceptor" atom is located in a silicon crystal but that atom does not have as many electrons available for bonding as do the silicon atoms. Silicon atoms bond by sharing a pair of electrons, each atom contributes one electron to the bond. The acceptor atom leaves one bond unfilled, and it's that unfilled bond that constitutes the hole. Note that although this hole exists the atoms are completely charge-neutral. It's easy for a wandering electron to get stuck in the hole, and when that happens the acceptor atom actually has one more electron than it normally would...thus it has become a negative ion. The captured electron came from somewhere...some atom that was also previously charge-neutral...so that atom has become a positive ion. Since we have an immobile negative ion and an immobile positive ion, an electric field exists between them.
As holes are filled with wayward electrons the e-field increases in strength until it prevents any more movement of electrons. At this point the depletion region has been created. This region is depleted of free (mobile) charge carriers but the impurity (non-silicon) atoms are ionized.
I've mentioned silicon but the same thing can be done with some other materials, such as germanium and gallium-arsenide.
Best Answer
Light DOES destroy the static (or built-in) potential of the PV cell. That's how PV cells work! (At least, at the macro level, that's how the internal voltage-drops behave.)
During darkness, we might find 1.0V static potential between the n-type and p-type parts of the solar cell. This potential is really there, it's the energy-hill appearing in the depletion zone. But it can only be detected by non-contact electrometers (where no metal probes make contact, so no unwanted metal/silicon junctions are formed.)
Besides the junction-potential, we'd also find another potential where the metal terminal connects to the n-type, and a third where the other terminal connects to the p-type. These two "ohmic" or "non-rectifying" built-in potentials are much like static potential of any thermocouple ...though quite a bit higher volts. Roughly 0.5V each. They sum to the same value as the PN junction potential. That way, if the solar-cell contacts are shorted, there will be zero current, even though in darkness the junction exhibits an energy-hill of one entire volt!
And so, when light strikes the junction, carriers flood the depletion zone, and the staticf junction-potential is "shorted out." But this doesn't affect the two 0.5V steps located where metal contacts touch silicon.
As a result, the potential measured across the terminals has an upper limit: roughly 1.0V. PV output-voltage doesn't rise continuously as light-intensity rises. That one-volt limit was the same value as static built-in potential of the PN junction during darkness. That potential is now missing, so it will now appear at the output terminals of the PV cell.
Very strange, eh? Solar cells are actually driven by the sum of the two potential-steps found at their ohmic contacts. It almost seems like perpetual motion! But the electromotive force at the metal junctions can only pump some charges when photons are injecting energy into the PN junction, and therefore "promoting" valence electrons into the conduction band, without them having to be pushed there by an external power supply. The depletion zone is shorted out by the sudden new population of carriers there. The diode becomes forward-biased. Yet the energy needed to do this ...is exactly the energy being injected by the ohmic contacts, as carriers "fall down" the contact-potentials found at the metal-semiconductor bonds.
"Contact potentials" are weird stuff!
And now you can get an idea about Peltier module function. A thermoelectric module works because p-type and n-type blocks are welded onto little copper straps, with no diodes being formed. Yet the potential-hills are really there. They all sum to zero around the circuit ...unless half of the metal-semiconductor contacts are heated up, and the other half cooled down. As with solar cells, the heated contacts become "shorted out" because of injected energy at the micro-level: phonon absorption, which knocks the carriers uphill over the built-in junction potentials. Heh, a Peltier thermogenerator is a bunch of PHONO-voltaic cells in series! (As in: lattice thermal energy, junctions "illuminated" by phonons not photons.)
Confusing regarding measuring barrier potential of a pn junction using a voltmeter