- In a PN junction diode, as far my understanding, when we apply a forward bias, the free electrons from n type material will cross the junction and then it will again recombine with the holes of p type material. And hence the p type material will become more negative. So in this situation if any remaining free ectron from n type material wants to cross the p type material to reach the positive terminal of the supply then it needs more energy comparing to that was required at the beginning when there were only few electrons(only in the depletion region)??
- In some places I saw that all the electrons will not recombine during crossing the p type material, but why so?
The electrons are in the valence band in the "P region", simply because they have lost too much energy overcoming the barrier potential to remain in the conduction band. Since unlike charges attract, the positive side of the bias-voltage source attracts the valence electrons toward the left end of the "P region". The holes in the "P region" provide the medium or “pathway” for these valence electrons to move through the "P region".
The valence electrons move from one hole to the next toward the left. The holes, which are the majority carriers in the p region, effectively (not actually) move to the right toward the junction. This effective flow of holes is the hole current. You can also view the hole current as being created by the flow of valence electrons through the p region, with the holes providing the only means for these electrons to flow. As the electrons flow out of the p region through the external connection (conductor) and to the positive side of the bias-voltage source, they leave holes behind in the p region; at the same time, these electrons become conduction electrons in the metal conductor. Recall that the conduction band in a conductor overlaps the valence band so that it takes much less energy for an electron to be a free electron in a conductor than in a semiconductor and that metallic conductors do not have holes in their structure. There is a continuous availability of holes effectively moving toward the P-N junction to combine with the continuous stream of electrons as they come across the junction into the "P region".
reference: Electronic Devices -Conventional Current Version- Ninth Edition - Thomas L. Floyd
For your first question; The diffusion currents of minority carriers (electrons in the P side, and holes in the N side) won't affect the overall polarity of each side of the junction very much. Electrons diffusing to the P-side, for example, can be replaced by the negative terminal of the battery and holes diffusing to the N-side can be replaced by the positive terminal of the battery. Thus the battery maintains the voltage across it.
As for your second; If the physical length of the diode is too short, the probability that an electron or hole will recombine is very low. The recombination of a given electron-hole pair is essentially random but particular diode structures can affect the probability. This is actually a major issue in the design of LEDs as radiative recombination is necessary for their operation. Typically, more complicated structures are used to ensure recombination rates are high such as Quantum Wells and Heterojunctions.