Diffusion current
When a p-n junction is formed, a diffusion phenomena causes electrons from the n-doped region to diffuse to the p-doped region. At the same time (even if it's an abstraction) holes diffuse from the p-type region to the n-type one. The atoms that lose a carrier (electron or hole) become ions, which means that instead of being neutral, they have a positive or negative net charge. This happens because the ideal equilibrium would have the same concentration of mobile carriers equal all over the region.
Ohmic current
However, this diffusion causes the growth of a region, populated by ions, called depletion region, because all atoms have lost their carrier. These ions, as we said, are electrically charged, and cause an electric field directed from the n-region to the p-region, pushing carriers in the opposite way than diffusion. Therefore an equilibrium is reached in which the current (movement of carriers) caused by diffusion is perfectly balanced by the current caused by the electric field (ohmic current).
Effect of biasing
Applying a potential to the junction causes a perturbation on this equilibrium, making one of the currents dominant on the other. Reverse biasing the junction causes the ohmic current to prevail, while forward biasing increases the diffusion current.
Now, the diffusion current is a much stronger phenomena, from which derives the exponential growth of the forward bias current with the bias voltage. Ohmic current, on the other side, is much weaker, and saturates quite soon (neglecting avalanche effect) because the width of the depletion region (which determines the resistivity) is proportional to the reverse bias voltage.
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.
Best Answer
Nope, for the most part, the positive ions in the n-side are the pos-charged dopant atoms, and those are locked into the crystal lattice.
It's confusing because there are actually four ions involved, not just two.
First, in the p-doped silicon, the neutral dopant atoms don't remain neutral. Instead, a "hole" is created by each dopant atom, and it moves away. This leaves the dopant atom with net negative charge. Yet the wandering "hole" is actually a positively-ionized silicon atom! At the same time, on average the entire hunk of p-type silicon has zero net-charge. After all, every positive-charged hole has a negative-charged dopant ion somewhere nearby. In other words, p-type silicon is actually made up of equal quantities of:
The n-type silicon is the opposite. The dopant atoms in n-type silicon will contribute wandering electrons. But when each electron initially leaves its dopant atom, that atom becomes a positive-charged ion. And, when the wandering electron is sitting upon some distant silicon atom, that atom temporarily becomes a negative silicon ion. So, n-type silicon is overall neutral, but is is composed of:
It gets worse!
Suppose that some holes wandered out of the p-side and invaded the n-type silicon? Thermal motion causes them to jump around randomly, and the random jumping can take them over into the n-side. They won't last long over there, but while they're briefly existing in the n-side, the holes are producing a region of positive net-charge! (After all, they no longer are near their negative-charged dopant atoms, which were all left behind in the p-type side.)
YES! Because actually a "hole" is a positive-charged silicon ion ...so if holes invade the n-side, electrically it's just as if some pos-charged silicon atoms were invading. Yet the atoms themselves don't have to move. Just their "ionization" is wandering around through the crystal. (Heh, but at the same time, the n-side is full of positive-charged dopant ions which cannot move. So, whenever n-type silicon is full of wandering holes, it actually contains two kinds of positive ions, but only one of them can move around.)
PS
Important question: does p-type silicon have a positive charge? Nope, since p-type silicon is full of non-movable negative dopant ions. Their quantity is exactly the same as the quantity of wandering, positive-charged holes. P-type silicon is a conductor of course, and that means we can give it a positive net-charge by hooking it to the positive terminal of a power supply.
And to make things even more interesting, if we connect p-type to n-type, some electrons will diffuse from n-type to p-type and become trapped in the depletion zone (because they encountered holes, and "fell in.") This causes the p-type side to become negative charged, and the n-type side to become equally positive. This is the "built-in potential" of semiconductor junctions, which is caused by the "built-in" trapping of mobile charges in the depletion-zone. Or in other words, a diode junction is also a spontaneously self-charged capacitor.