plz counter my inadequate knowledge of semiconductor physics

[zwol]

Suppose you have a semiconducting wire which has been doped in a gradient, so that one end of the wire is strongly n-type, the other end is strongly p-type, and the middle is neutral. This wire is suspended in air, and is at the same temperature throughout.

Is there a voltage between the ends of the wire?

If not, why not?

If so, what happens if you connect the wire to an electrical load?

Comments

[rthecuban.livejournal.com]

...a certain Ms. E.A. Williams asked that I answer:

Most of the time (assuming an ideal system) you wouldn't have a voltage across it. Remember Ohm's Law - voltage is proportional to resistance times current - in an open circuit there's no current, and with no applied voltage (battery, power supply, etc.), there shouldn't be a voltage across the wire. Semiconductors generally don't behave like batteries because they don't have anywhere to store the electrons - although there is a little capacitance that builds up when this sort of structure is connected to power, the charge will dissapate after it's been disconnected (the depletion regions collapse).

Hmm ... put another way - if you measure across a normal wire, it'll read zero. If you then convince all the electons in the wire to move left, and all the holes to move right ... it'll still read zero, unless you're measuring at some point other than the ends.

There's two other things that might come up here, though - it is possible that if you measured across the wire with a voltmeter, that you would get a small voltage in response, but it would probably keep shifting. This has more to do with the way that voltmeters take measurements than there being any real voltage across the wire. (...more or less, the voltmeter will sorta charge the wire.)

Also, in a non-ideal system, you could run into contact potential - see the article below. ...these are pretty small in most cases, though.

A couple of articles for more reading:

http://en.wikipedia.org/wiki/Voltmeter
http://en.wikipedia.org/wiki/PIN_diode
http://en.wikipedia.org/wiki/Volta_potential

[elisaana.livejournal.com]

(By "E.A.", he means me.) :)

Good luck!

[zwol.livejournal.com]

aha, the mystery demystified!

[zwol.livejournal.com]

Hm. I think either I've set up the thought experiment wrong or described it in a way that made you latch onto irrelevant details.

The intent of the thought experiment is to mess around with the thermoelectric effect (http://en.wikipedia.org/wiki/Thermoelectric_effect), because I can't quite convince myself it's impossible to build Maxwell's demon (http://en.wikipedia.org/wiki/Maxwell%27s_demon) out of it. The notion is that a composition gradient along a wire might induce a thermoelectric voltage difference even if both ends of the wire are at the same temperature. (I think it has to be a gradient, not a junction, for this to have any chance of working.) Semiconductor doping was just a wild guess at the right sort of composition gradient, since I know that affects electron/hole density.

[rthecuban.livejournal.com]

Ah! ...sorry, have no idea what your background is or what you were really asking. : )

...so, sorta a reverse Thomson effect?

[zwol.livejournal.com]

very sort of, but yes. Possibly a copper-nickel gradient is what I actually wanted.

[brooksmoses]

I don't know from doped semiconductors, but I've seen this in similar sorts of things, and I suspect essentially the same trick applies.

The trick, fundamentally, is that you have to connect the wire to the electrical load. And those connections are joints across materials with dissimilar electron affinities. And those junction effects exactly counteract the effects of the composition gradient along the wire, in this sort of fundamental "if you integrate around a closed curve you have to get back to zero" way.

Meanwhile, the thermoelectric effect works, if I'm understanding it correctly, because electrons carry different amounts of thermal energy in the two materials. Thus, the heat flow along the wires pushes the electrons differently, creating voltage gradients that set up a loop current. But that's fundamentally a dynamic effect, driven by the heat flow along the wire in the form of electron motion; it's not a static effect at all.

[rthecuban.livejournal.com]

::agrees::