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Connect the positive terminal of a battery to a piece of p-doped semiconductor, say, silicon doped with boron. Will the terminal pull electrons out of the doped silicon, or equivalently, inject holes into it?

The atomic structure of p-doped semiconductors certainly accepts electrons freely, but I feel like it should be stingier than an insulator when it comes to donating electrons. As an example of an insulator, undoped silicon is unwilling to give away electrons, or accept holes, because that would rip gaps in the stable crystalline bonding. P-type silicon should behave similarly here; thus, I would guess that the p-type block remains neutral, unlike a copper block which would take on the same charge as the terminal. However, if it were connected to a negative terminal, it would accept enough electrons to basically saturate it and give it a negative charge. Is this understanding correct? Or could the p-type somehow donate electrons/accept holes because of some kind of instability associated with doping?

EDIT: Admittedly, this is a rather petty question, and I doubt that anyone has ever bothered to check the answer in a published experiment. Still, knowing that the p-type stays neutral would confirm most people's (or at least my) model of why doped semiconductors conduct. Their conductivity possesses a kind of asymmetry, unlike that of copper, which gives and takes electrons equally freely.

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Your question actually breaks up into two sections: 1. what happens when we touch some lightly p-doped semiconductor against metal to either form a "cats-whisker" Schottky diode, or use heavy p-doped material to form a non-rectifying thermocouple-style junction? 2. What happens when touching either a diode or a thermocouple against a supply terminal which is positive with respect to ground? As for donating/accepting, these are basically symmetrical if the metal-P junction doesn't form a rectifier. – wbeaty Jul 5 '13 at 6:05

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.

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The answer to your main question is NO. Connecting only one side of a battery (positive or negative) does not alter the electrical state of the block (semiconductor, copper, etc.). It is only until the negative side of the battery is also applied, that something can happen. For an undoped block, since the resistance is large, a small current will flow. For a copper block, since the resistance is small, a large current will flow. For the doped block, the resistance is "medium," so a medium current will flow.

An example that might make this clear is the case of two plates separated by some distance (a capacitor), but even here, nothing happens when you connect only one side of the battery to one plate. It's only until the negative terminal is connected to the other plate that a potential difference exists between the plates causing some of the electrons on the plate attached to the positive terminal to be removed, causing the plate to be positively charged, (the opposite happens on the other plate). The plates continue to "charge" until the voltage across the plates equals the battery voltage and then the current flow stops.

Note: The battery terminal does not have a charge, what the battery has, is a voltage difference/potential between its terminals! This is why nothing happens, unless you use both terminals.

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This is false! I mean, for lab purposes sure, and no, the voltage difference on a battery doesn't completely specify everything, but the electric potential can be viewed as the result of a charge distribution, and so a change in potential means a change in the charge distribution, even if there's no obvious change to the naked eye. – NeuroFuzzy Apr 6 at 16:38

The P-type material already has a positive charge and thus a lack of free electrons so, no, it will not pull electrons from it making it more positively charged. You have to make a complete circuit in order to move electrons. Like Kirchoff said, if one electron leaves, then one has to take it's place. So if you take the negative terminal and connect it to the other side of the chunk o' P-type silicon then you'll have electrons moving. You will have also created a dead short... but that's ok. Everyone likes a little arcy sparky blowy upy.

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Actually, it is a common misconception that doped materials have charge. In fact, p-type semiconductor is neutral. The electron holes are only gaps in the bonding structure. Doping silicon with boron, for instance, means sprinkling a few atoms with 5 protons and 5 electrons (3 valence) among atoms with 14 protons and 14 electrons (4 valence). The difference in valence electrons creates unstable gaps in the otherwise diamond-like bonding pattern. Similarly for n-type. – Kent Apr 5 '13 at 20:10

protected by Qmechanic Apr 6 at 19:03

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