In order for electrons to leave their lower shells to higher shells, they need energy i.e voltage. Electrons also move from high concentration to lower concentration.

If we take a single piece of N-type (which is not joined with P-type piece) and connect one side of N-type with negative terminal of a battery and another side with a positive terminal of a battery, there will be a current. And if we connect the terminals again but oppositely, there will also be a current going in opposite direction.

If we take a single piece of P-type (which is not joined with N-type piece) and connect one side of P-type with negative terminal of a battery and another side with a positive terminal of a battery, there will be a current. And if we connect the terminals again but oppositely, there will also be a current going in opposite direction.

If we connect these two pieces together, high concentration of electrons from N type would naturally migrate into low concentration of electrons on P type, resulting in negative charge depletion layer on P type, and a positive charge depletion layer on N side, which will prevent further migration.

My question, why can’t the N electrons that got combined in P side to form a PN junction move further into P side? Is it because there’s no enough energy for that (i.e no voltage?) to excite electrons? If the answer is the depletion layer has become depleted of free carriers to do that, please continue reading below.

If the PN junction is physically there in forward bias(if it doesn’t disappear), why dont electrons that got pushed from N type to P type widen the PN junction as they also get recombined as mentioned in the above paragraph?

If the PN junction disappears in forward bias, why can’t current flow in reverse bias? As the electrons can move both sides in single pieces of either N type or P type. So if the electrons started to flow from P type, they should also flow to N type in the same direction?

 

It's not just the movement of electrons, it's the generation of adding (- + -)and subtracting ( - + +) electric fields that cause the electrons to migrate or not across the PN junction boundary.

https://www.pveducation.org/pvcdrom/pn-junctions/formation-of-a-pn-junction

P-n junctions are formed by joining n-type and p-type semiconductor materials, as shown below. Since the n-type region has a high electron concentration and the p-type a high hole concentration, electrons diffuse from the n-type side to the p-type side. Similarly, holes flow by diffusion from the p-type side to the n-type side. If the electrons and holes were not charged, this diffusion process would continue until the concentration of electrons and holes on the two sides were the same, as happens if two gasses come into contact with each other. However, in a p-n junction, when the electrons and holes move to the other side of the junction, they leave behind exposed charges on dopant atom sites, which are fixed in the crystal lattice and are unable to move. On the n-type side, positive ion cores are exposed. On the p-type side, negative ion cores are exposed. An electric field E forms between the positive ion cores in the n-type material and negative ion cores in the p-type material. This region is called the "depletion region" since the electric field quickly sweeps free carriers out, hence the region is depleted of free carriers. A "built-in" potential Vbi is formed at the junction due to E. The animation below shows the formation of the E at the junction between n and p-type material.

So, we have the "built-in" electric field E and a applied electric field (from the external circuit) that can be in the same direction or the opposite direction.


Forward Bias
If a positive voltage is applied to the p-type side and a negative voltage to the n-type side, current can flow (depending upon the magnitude of the applied voltage). This configuration is called "Forward Biased" (see Figure 5). At the p-n junction, the "built-in" electric field and the applied electric field are in opposite directions. When these two fields add, the resultant field at the junction is smaller in magnitude than the magnitude of the original "built-in" electric field. This results in a thinner, less resistive depletion region. If the applied voltage is large enough, the depletion region's resistance becomes negligible. In silicon, this occurs at about 0.6 volts forward bias. From 0 to 0.6 volts, there is still considerable resistance due to the depletion region. Above 0.6 volts, the depletion region's resistance is very small and current flows virtually unimpeded.





https://www.unsw.edu.au/content/dam...23-08-UNSW_Understanding_the_p-n_Junction.pdf