What happens when you touch an object with a positively charged object?
Ans: It gets positively charged.
Now, you have connected a semiconductor to a positive end of battery. What do you expect?
Ans: Yes, it gets positively charged.
Will the terminal pull
electrons out of the doped silicon,
or equivalently, inject holes into
it?
Yes, it will.
But the question is to what extent?
You figured it out correctly that undoped silicon is
unwilling to give away electrons,
or accept holes, because that
would rip gaps in the stable
crystalline bonding
I would guess that the p-type block
remains neutral, unlike a copper
block which would take on the
same charge as the terminal.
Think of a comb getting charged when rubbed with hair.
What do you observe?
Ans: You see that charge on comb is localised and do not spread out. The charge don't pass to your hand.
Similarly, the p semiconductor gets locally charged . The point in direct contact with battery gets charged and remaining remains uncharged.
Extent of positive charge and effective distance depends on strength and voltage of battery.
Now, when you connect it to negative potential
This will give you similar result but with little increased magnitude of negative charge on it as it has a lot of holes to be occupied.
I think to understand why holes are the preferred "thinking tool" for p-type semi-conductors one must understand the following concepts:
- In a crystalline materials there are so-called energy bands which define the possible relationships between an electron's energy and its (crystal) momentum "k".
That is, there is a function $E(k)$ which defines the set of possible energy states which may be present in a crystal.
In a semiconducting material there is a gap in the band structure such that there are energy levels which electrons may not possess (they are forbidden in a sense).
The energy-momentum relationship can be approximated as parabolas just above and below the energy gap (i.e. the band gap).
The parabola approximating the energy band just above the gap is pointing up / concave up. The parabola just below the gap is pointing down / concave down.
The average velocity of an electron can be expressed as:
$$v(k) = \frac{1}{\hbar}\frac{\partial E(k)}{\partial k}$$
- The current density associated with a single band is then calculated by adding up the contributions from all electrons.
$$ j = (-e)\int_{occupied}\frac{dk}{4\pi^3}v(k)$$
Here the flag occupied means to integrate over only those states which electrons occupy.
- In general, the relationship $E(k)$ is a complicated function and thus not very straightforward to manipulate mathematically. However, it is possible to show that the current density integral is mathematically equivalent to the following integral:
$$ j = (+e)\int_{unoccupied}\frac{dk}{4\pi^3}v(k)$$
The reason this simplification is important is that in the valence band (i.e. the band below the band gap) electrons fill almost all the states except those near the top of the band (i.e. the unoccupied states). However, as previously mentioned the function $E(k)$ can be represented by a downward facing parabola at the top of this band. Thus a previously intractable integral becomes "trivial" as one only needs to integrate a quadratic equation. This is a very key point: it is mathematically much simpler to consider that the current density is due to fictitious positively charged particles which 'occupy' the top of the valence band instead of considering the mathematical equivalent, but non-fictitious, fact that the electrons present in valance band occupy almost all the valence band states except those near very near to the top of the band.
Finally, to answer the posed question: why are the unoccupied states (i.e. holes) treated as the predominant charge carriers.
In a p-doped material, the doping atoms such as boron "accept" electrons present in the crystal. These electrons, generally speaking, are taken from the valence band thus leading to a state where more holes are present in the valance band than in the pure semiconducting material.
Also in this conception of semiconductors, it is incorrect to say that only electrons move. When an electron moves, if it moves to a previously unoccupied state (i.e. a hole), the hole must have moved as well. Actually, as @Matt indicated, the motion of electrons and holes under an electric field follow identical trajectories. This occurs because electrons in the upper part of the valence band move as if they had negative mass and thus in the opposite direction of an applied force.
So to conclude:
It is more mathematically convenient when calculating the current density to treat the valence band as being filled only with positively charged particles called "holes" which have energies near the top of the band. Though, in reality, the valence band is almost completely filled with electrons except near the top.
For p-doped semiconductors, atoms which accept electrons are added. The electrons picked up by these atoms are typically taken from the valance band and thus more "holes" are present than would be otherwise. Thus there are more holes in the valence band than electrons in the conduction band. Therefore upon the application of an electric field, the holes will be the dominant charge carriers.
Best Answer
No, Boron still bonds to 4 Si atoms. Since Boron has only 3 valence electrons, it can only form three covalent bonds with Si. This leaves one of the four silicon atoms with an unsatisfied bond. To compensate, a nearby electron (from an Si-Si bond) will "jump" to complete the B-Si bond leaving the Si-Si bond with a net positive charge (the hole). To fill that broken bond, another nearby electron will jump to fill in the vacancy etc... Thus the "hole" is highly mobile and can conduct when a voltage is applied. The Boron atom does become negatively charged, but that additional negative charge cannot conduct because it takes place in the bonding.