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My textbook ("Electronic Principles" by Malvino and Bates) seems to suggest the free electrons in conduction band move left and reach the positive terminal of the battery. Does this mean electrons in valence band are not allowed to move left(hole moving right) and enter the positive terminal of the battery?

Similarly, it says the electrons from the negative terminal of the battery enter from the right directly into the valence band holes. Does this mean the electrons from the battery are not allowed to enter into the conduction band?

In summary: Electrons in an intrinsic semiconductor always leave from the conduction band and always enter from the valence band. This doesn't feel right. What am I missing?

EDIT: Saying electrons in the valence band are not free, so the battery cannot pull them doesn't make sense because of the following scenario: If a p-type semiconductor is connected across a battery, the battery's positive terminal has no problem attracting the electrons from valence band.

extract from text book

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If it's intrinsic, there's as many electrons in the conduction band as holes in the valence band by definition, because each electron leaves a hole. The electrons in the conduction band are bound to their atoms by (relatively weak) electrostatic forces, and can move fairly freely.

Electrons in the valence band are bound more tightly, and largely move only from one atom to an adjacent one that contains a hole.

Bottom line is, all the electrons are moving to the left--some in long unfettered paths, and some from one atom to the next. Also, in the valence band, it's kind of tag-team...an individual electron moves to an adjacent atom leaving a hole, then another electron from the other side fills that hole. It's just as easy to model it as a hole propagating than a tag-team of electrons.

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  • \$\begingroup\$ Thank you @Cristbol In pure crystal, I see that the conduction band free electrons move toward the positive side of battery and the valence band free holes move toward the negative side of the battery. But in a p-type doped crystal, battery has to pull the electrons from valence band because the number of conduction band free electrons is almost 0. How can the battery pull the valence band electrons here if they are bound tightly to the atoms as you said? \$\endgroup\$
    – across
    Feb 28, 2020 at 17:31
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    \$\begingroup\$ @beccabecca if it is p-doped you placed traps near the top of the valence band. That will make electron from the valence band jump into the traps just with a bit of thermal energy, thus freeing holes in the valence band. These holes will flow in the direction of the E field, while the bulk of the electrons in valence band can be thought of flowing the opposite way. There are nice pictures of the 'liquid in a pipe' analogy on Mueller and Kamins book. I could not find the picture online, tho. \$\endgroup\$ Feb 28, 2020 at 18:46
  • \$\begingroup\$ It works the same way, but the current in the valence band is much higher than the current in the conduction band. You don't need equal quantities of conduction electrons and valence holes...that's why doping works, it gives you an excess of one or the other. \$\endgroup\$ Feb 28, 2020 at 19:24
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Saying electrons in the valence band are not free, so the battery cannot pull them doesn't make sense

The valence band is localized around an individual atom or molecule, so nothing in the valence band can be moved without first being removed from the valence band (which takes energy from the battery, thus contributing to resistance).

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  • \$\begingroup\$ user, I maybe wrong, but if we replaced the intrinsic crystal with a p-type doped crystal, then the battery moves the electrons directly from valence band? \$\endgroup\$
    – across
    Feb 28, 2020 at 17:40
  • \$\begingroup\$ My understanding is that in a p-type doped crystal there is "nothing" in the conduction band. Since the hole concentration is very high compared to the free electrons, almost all the thermally generated free electrons have already fallen into the holes and vanished.. \$\endgroup\$
    – across
    Feb 28, 2020 at 17:43
  • \$\begingroup\$ I feel that forces the battery to pull the electrons directly from the valence band... \$\endgroup\$
    – across
    Feb 28, 2020 at 17:44
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    \$\begingroup\$ Doping adds additional energy levels above the valence band, reducing the energy required to move charges. See: hyperphysics.phy-astr.gsu.edu/hbase/Solids/dsem.html#c1 \$\endgroup\$ Feb 28, 2020 at 18:31
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In an intrinsic semiconductor like pure silicon crystal, the valence band is entirely occupied at 0deg K, thus can't contribute to current flow, just as an entirely full bottle of water can't slosh about inside. At room temp, there are a relative few free electrons in the conduction band and holes in the valence band, though not enough to be a good conductor.

I think Cristobol's comment "Electrons in the valence band are bound more tightly, and largely move only from one atom to an adjacent one that contains a hole." is just wrong. It's my understanding that existing holes move as freely within the valence band as existing electrons do in the conduction band, which just means electrons are moving freely in the opposite direction. And both are technically electron current, although hole current is more like a line of people (electrons) moving forward in a queue in the direction opposite to the hole current. There just aren't many free electrons and holes in pure silicon. The addition of P and N type dopant atoms greatly increase the population of free electrons in the conduction band and free holes in the valence band to serve as carriers of current.

But beccaboo, doesn't a high concentration of holes in the valence band imply electrons being free to move in the opposite direction of hole current (see above).

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