Although I've seen the typical no-bias, reverse-bias and forward-bias band diagrams associated with the PN junction and seen the overlapping valence and conduction bands of a typical metallic conductor, I can't find any example that include all four sandwiched together (conductor-Ntype-Ptype-conductor). I'm curious about the physics of electron flow between the conductors' bands and those of the doped semiconductors in between. Does that flow include not only their conduction bands, but their valence bands as well?


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If there is no bias, then there is no current flowing in the conductors on the sides of the PN junction. In this case, there is only a drift and diffusion current in the space charge region that cancels each other.

If there is a bias, then there is a net current flowing through the whole structure. In the metallic conductors, the current is carried by the free electrons near the Fermi level. In the N-doped semiconductor, the current is mainly carried by the electrons in the conduction band. In the P-type semiconductor, the current is mainly carried by the holes in the valance band.

  • $\begingroup$ Thanks. I agree. So, you're saying that both valence and conduction bands of the doped semiconductors are able to transmit electrons to and from the conductors' overlapped valence and conduction bands. And the exception to 'mainly' is minority carrier holes in the N-type valence band and minority carrier electrons in the P-type conduction band produced by thermal ionization? Does a barrier exist also between the conductors and their respective semiconductors as well? Can you refer me to a conductor-P-N-conductor band diagram? Thanks. $\endgroup$ Jul 28, 2020 at 12:43
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    $\begingroup$ That is correct what you are saying. Depending on the position of the fermilevel in the metal and semiconductor there can be a barrier at the interface. Here you can find a band diagram of the metal-semiconductor contact: eng.libretexts.org/Bookshelves/Materials_Science/… $\endgroup$
    – Frederic
    Jul 28, 2020 at 13:26
  • $\begingroup$ Thanks. I assume that's a typo at Eq. 3. and should be X = Eo - Ec $\endgroup$ Jul 28, 2020 at 22:29

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