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There has been many posts with similar questions and great andwers, but I've not found an answer to my question, so either I am missing something or the simplified standard explanation doesn't cover it?

Im thinking purely in terms of wandering electrons in the conduction band of the n-side and the conductor that is connected to the diode, and "wandering holes"/electrons wandering through holes in the valence band of the p-side.

My question
I'm wondering why/how the electrons of a diode in forward bias can go from the valence band to the conduction band at the p-side of a diode where it's connected to a conductor and why this doesn't happen at the p-side to n-side connection in a diode when it's in reverse bias?

Attempt to explain my question:
In reverse bias, there's no free holes on the p-side and no free electrons on the n-side, so no electricity can flow. But since the p-side is then negatively charged, the electrons in the now filled holes of the p-side would go into the conduction band and go to the n-side if they could. This doesn't happen, but isn't it exactly what happens at the p-side to conductor connection of a diode in forward bias?

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and "wandering holes"/electrons wandering through holes in the valence band of the p-side.

Well don't do that. In the valence band only concern yourself with holes. Trying to reason through everything based on the underlying electrons will likely only create more confusion.

But since the p-side is then negatively charged, the electrons in the now filled holes of the p-side would go into the conduction band and go to the n-side if they could.

You do have some charge carriers in the depletion region, contrary to what basic models would tell you. They are thermally generated and it is a relatively slow process. But it does occur and is one of the reasons that reverse bias current in a diode is not zero.

This doesn't happen, but isn't it exactly what happens at the p-side to conductor connection of a diode in forward bias?

At the metal/semiconductor junction the generation and recombination rates are greatly increased. Unless you are concerned with the behavior of a Schottky diode you can typically assume generation and recombination are instantaneous at metal/semiconductor junctions. At these junctions electrons and holes are generated or recombined as needed in order to conduct electrons into or out of the metal.

At an n-type semiconductor and metal junction your majority carriers are already electrons, so you don't need much carrier reorganization. However, at a p-type semiconductor and metal junction, you either recombine a hole approaching the metal with an electron approaching the semiconductor, or you generate an electron/hole pair and they leave in opposite directions (electron into the metal and hole into the semiconductor)

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  • $\begingroup$ Thank you for your answer. This pretty much covers my questions. It seems I wouldn't have struggled with these questions if we had covered band diagrams, but the class I'm taking is just a very basic electronics course, hence the unsatisfying model. Thanks again. $\endgroup$
    – klingeron
    Dec 3, 2022 at 19:24
  • $\begingroup$ Does this also mean that if you had the same amount of generation and recombination at the p-n junction as there is at the metal/semiconductor junction, the diode would conduct electricity in both biases? $\endgroup$
    – klingeron
    Dec 3, 2022 at 19:26

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