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I looked at "Why does a p-type conduction band have higher energy then n-type conduction band in a p-n junction?" and am still confused.

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From the image above we can see that the conduction band on the p-type part of the material is at a higher energy. Why is that?

In my head, the p-type region is electron deficient because we doped it with an element from group III. This means it should have lower energy right?

Applying the logic, the n-type region has extra electrons because we doped it with an element from group V, so shouldn't the conduction band be at a higher energy?

I then thought that maybe for electrons in the p-type region, we'd need more energy to excite the electron into the conduction band. But this energy would still be $$E_c - E_v$$ which is the same in both regions...

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  • \$\begingroup\$ A lot of simplifications out there. But a good model is hard to master. You are discussing solid-state physics and it's complicated because atoms are thrust into very close proximity to each other and the electrons interfere with each other, quite a lot. If you are seriously into this stuff, you might read this. It covers silicon structures as well as germanium and others. Keep in mind that a "band" is really a band -- a wide span of wave-function probabilities. \$\endgroup\$
    – jonk
    Oct 26, 2020 at 3:20
  • \$\begingroup\$ There's a simplified image here, towards the end there. It gives a cartoon image, so to speak, of various materials such as intrinsic and doped semiconductors, semi-metals, metals, etc. You can see that the N-type, for example, as the conduction band (it's a band -- never forget that) is quite near the Fermi-level. So there will be significant probabilities of thermally agitated electrons kicked up into the conduction band. Just look it over and see if it "clicks" in your mind. If not, dig into the long document I linked above. \$\endgroup\$
    – jonk
    Oct 26, 2020 at 3:24

2 Answers 2

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Your thinking is on the right track, but it misses a few small details. One way to visualize the process you're interested in is via the Fermi level -- this however is in some ways a post hoc rationalization of the end-result, rather than an explanation from first principles.

Notice the dashed line for the Fermi level, marked \$E_f\$ on the band diagram. It is flat because the structure is in thermodynamic equilibrium, and represents (among other interpretations) the energy level of a hypothetical electron level that is 50% likely to be occupied.

The consequence of doping is that the bands move relative to the Fermi level -- in a P-type semiconductor, the valence band is likely to have openings and the conduction band is likely to be near-empty, putting the Fermi level close to the valence band; in an N-type semiconductor the valence band is full and states are likely to be occupied in the conduction band, thus putting the Fermi level close to the conduction band. However, once the materials are connected and brought into equilibrium, the conduction and valence bands themselves must be aligned to maintain the Fermi level as flat.

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  • \$\begingroup\$ Why do the conduction and valence bands themselves have to move? Does this mean that the electrons at those energies somehow gain more energy when the two types of materials are brought together? (Just so I'm not going crazy... we want the Fermi level to be flat because we're at equilibrium right? No reason besides that???) \$\endgroup\$
    – BigBear
    Oct 26, 2020 at 0:43
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    \$\begingroup\$ @BigBear the electrons at those energies don't gain more energy -- whilst we are in a material and not talking about stuff like thermionic emission, we can deal with relative rather than absolute energy. On your PN junction, the P type material has the conduction band at a higher energy level, far above the fermi level, and so very few electrons will live in that band while the proximity of the valence band to Ef allows holes to develop; in the N type, the cond. band is close to the Fermi level and hence it's filled with thermally excited electrons. \$\endgroup\$
    – nanofarad
    Oct 26, 2020 at 0:46
  • \$\begingroup\$ OH!!! Wait are you saying that the p-region's E_c isn't actually at a higher energy, but we draw it higher because we need to place the Fermi level appropriately? \$\endgroup\$
    – BigBear
    Oct 26, 2020 at 0:49
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    \$\begingroup\$ @BigBear It's not really higher in isolation, since the blocks of material might just have a DC voltage between them for whatever reason. When they're brought in equilibrium, they do have a relative difference in energies. \$\endgroup\$
    – nanofarad
    Oct 26, 2020 at 0:53
  • \$\begingroup\$ But why do p-type materials inherently have higher conduction band energy then? I'm sorry for all the questions!! \$\endgroup\$
    – BigBear
    Oct 26, 2020 at 1:03
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For the material to be p-type there must be a substantial number of holes in the valence band.

That means the occupation probability at the upper edge of the valence band can't be too high.

That means the upper edge of the valence band must not be very far below the Fermi level. (In degenerately doped material the valence band edge will even be above the Fermi level)

So, by this logic when we dope the material with acceptors, we move the Fermi level down toward the valence band edge, not up toward the conduction band.

Conversely, when we dope the material with donors, the Fermi level moves up toward the conduction band edge so that there will be a substantial occupation probability in the conduction band and the number of conduction band electrons will be greater than in intrinsic material.

Yes but when we compare E_c of the p-type region with E_c of the the n-type region, we see that the p-type's E_c is much higher than the n-type's E_c. Why is that?

Because when we place the two materials in intimate contact and let them reach thermal equilibrium with each other, the Fermi levels must equalize.

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  • \$\begingroup\$ Yes but when we compare E_c of the p-type region with E_c of the the n-type region, we see that the p-type's E_c is much higher than the n-type's E_c(same for E_v). Why is that? (Sorry if I'm just not understanding your answer) \$\endgroup\$
    – BigBear
    Oct 26, 2020 at 0:46
  • \$\begingroup\$ The Fermi levels must equalize, but where do the energy come needed to lift the bands come from? \$\endgroup\$
    – BigBear
    Oct 26, 2020 at 0:58
  • \$\begingroup\$ @BigBear, from whatever process you use to inject the acceptors and donors (in the real world) or from the effort of pushing the two sides of the junction together (in the imaginary theoretical world where you make the p and n material seperately and then join them). \$\endgroup\$
    – The Photon
    Oct 26, 2020 at 2:14

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