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My question (I hope) is a simple one. Are there two components of diffusion current (valence band AND conduction band) contributing to the generation of the PN junction voltage? If so, do those components stay exclusively within their respective bands across the junction and throughout?

Two components

Or is it like this:

Jump in band gap

Or are BOTH pictured paths involved? Can anyone refer me to a source that DOES go into that sort of detail in terms of current in both bands?

Also, at the interface between the diode and the conductors attached to them, do valence and conduction bands both participate in electron current to and from their respective conductors, since the valence and conduction bands of conductors overlap? Or is this pictured the ONLY transport to and from the semiconductors?

Conductor Interfaces

All the sources I've seen start by mentioning that N-doped silicon creates extra electrons in the conduction band in the N-type, and P-doped silicon creates extra holes in the valence band in the P-type. Then they make no further mention of bands when discussing diffusion and drift current and treat both types as black boxes. But perhaps they do in the language of crossing the junction and filling holes. Here are two examples:

https://www.youtube.com/watch?v=7ukDKVHnac4&feature=youtu.be&t=158

More accurate picture of diffusion current (amended):

https://www.youtube.com/watch?v=btOIDQeMrMg&feature=youtu.be&list=PLfYdTiQCV_p711DywXAh53wL3xI7S55lg&t=192

Diffusion Current

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  • \$\begingroup\$ (1) You say that "energetic photon can induce electron-hole generation". Actually the first video on transistor says that at room temperature, in other words, heat/thermal energy can also do that. (2) The second video on p-n junction / diode does say that photons, ie light energy can also do that (induce electron-hole pairs. In other words, light/photon is not the only way, The other way is heat/thermal energy. Please let me know if you agree with me, or comment on my statement, before I suggest where to move on from here to answer your question.. Cheers. \$\endgroup\$ – tlfong01 Jul 26 at 2:55
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    \$\begingroup\$ The gap between valence and conduction bands in pure silicon is about 1.1eV. Between donor and conduction bands and acceptor and valence bands in P and N doped silicon is about 50meV, a factor of about 22. There IS a small amount of thermal generation (ionization) at room temp producing electron and hole carriers, but it's small (minority carriers). The addition of dopants vastly increase the concentration of charge carriers in both bands (majority carriers) and reduce the minority carriers. Thermal generation is not the primary source of charge carriers in either band in the PN junction. \$\endgroup\$ – Ted Jackson Jul 26 at 3:54
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    \$\begingroup\$ The addition of an N-type dopant increases the number of conduction band electrons without a corresponding increase in hole concentration in the valence band (and even decreases it with more electrons available to thermally neutralize those holes through recombination). Similarly, a P-type dopant increases the number of valence band holes without a corresponding increase in electron concentration in the conduction band (and even decreases it with more holes available to thermally neutralize those electrons through recombination). \$\endgroup\$ – Ted Jackson Jul 26 at 4:11
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    \$\begingroup\$ But I was wrong in saying that a photon is the ONLY bridge between valence and conduction bands. I've removed that part from my question. So, thanks for your contribution. I'm learning, myself. I'm hoping an expert will come along who knows the physics well. An electron in the conduction band of the N-type could cross the junction and form a net negative region in the P-type's conduction band or it could fall down to the P-type's valence band and annihilate a hole and still create a net negative region. And vice-versa. Hence the confusion. But the vids say 'fill a hole', so... \$\endgroup\$ – Ted Jackson Jul 26 at 4:26
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    \$\begingroup\$ @TedJackson Note that electrons and holes from dopants are thermally generated. If the temperature is too low you lose them. This is called "freeze out". So I disagree that thermal generation is not the primary source of charge carriers in semiconductors in equilibrium. \$\endgroup\$ – Matt Jul 27 at 2:20
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Are there two components of diffusion current (valence band AND conduction band) contributing to the generation of the PN junction voltage?

Yes

If so, do those components stay exclusively within their respective bands across the junction and throughout?

Electrons and holes that diffuse initially stay within their respective bands. Electrons in the conduction band and holes in the valence band. After enough time in the "wrong" band they will recombine with a carrier of the opposite type.

at the interface between the diode and the conductors attached to them, do valence and conduction bands both participate in electron current to and from their respective conductors since the valence and conduction bands of conductors overlap? Or is this pictured the ONLY transport to and from the semiconductors?

Electrons contribute to current in the conduction band and holes contribute to current in the valence band. Metals only have the conduction band available for use and can therefore only have electron current. At these metal-semiconductor junctions you therefore have electrons contributing to the current.

I don't really understand your diagrams, so I can't be sure about if they are right or not. I suggest drawing (or learning about) standard band diagrams instead. But, that being said, it looks like your first drawing depicts diffusion current as it initially occurs, and your second appears to depict recombination that will eventually occur. So, if that is what you intended, then the answer is: you get both.

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  • \$\begingroup\$ It appears that majority carriers in the N-type's conduction band cross the junction still in the conduction band and then quickly drop down to the P-type's valence band. And presumably majority holes in the P-type do the same in the valence band (floating up in the rising bubbles and falling drops analogy). youtu.be/… I've added a third picture to my original question that seems to capture it. Although not mentioned specifically in the video, I'm assuming the red portion contributes too. Correct me if I'm wrong. \$\endgroup\$ – Ted Jackson Jul 27 at 14:10
  • \$\begingroup\$ What I find counter-intuitive is this: The high concentration of majority carrier electrons in the N-Type conduction band are the driving force behind diffusion (hence the same name as liquid diffusion). In terms of reverse bias, the battery's positive terminal attached to the N-type reduces the concentration of those electrons. Yet it also widens the depletion region. Reducing the concentration that drives diffusion, yet causing more of the effect of diffusion seems counterintuitive. youtube.com/… \$\endgroup\$ – Ted Jackson Jul 27 at 14:34
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    \$\begingroup\$ @TedJackson Reverse biasing a diode does not change the electron or hole majority carrier concentrations in the neutral region. It does however increase the barrier height they must overcome. This reduces the current. That video is either very confusing with its animation or outright wrong. \$\endgroup\$ – Matt Jul 27 at 20:40
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    \$\begingroup\$ @TedJackson The neutral region is the semiconductor outside the depletion region. It is charge neutral. My comment that it was confusing or wrong was in response to watching it. \$\endgroup\$ – Matt Jul 28 at 0:05
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    \$\begingroup\$ @TedJackson Yes, the free electrons (and holes) will move but the number of them doesnt get reduced the way the animation shows. \$\endgroup\$ – Matt Jul 28 at 12:31

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