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Confused: Doping and Depletion Zone

  1. Apr 5, 2010 #1
    Background: Math & Physics undergrad

    Reason for asking:
    So I was trying to understand semiconductor diodes and realized it all hinged on the PN junction which in turn is worthy of attention because of the effect created by the depletion zone. As I continued on studying up on this I began understanding why the depletion zone is important but then realized I didn't understand the mechanism behind the spontaneous formation of the depletion zone because I don't understand doping.

    Let me describe what I believe happens in doping and the PN junction, and you guys correct me:
    In doping we take a semiconductor, and add impurities to the semiconductor so that it has excess charge in the form of extra electrons or extra holes (is this where I'm wrong?). When we have P-type semiconductor (extra holes) and an N-type semiconductor (extra electrons) adjacent to each other we'll have a region we call the PN junction at their point of contact. Here is where the depletion zone forms spontaneously as holes and electrons diffuse across the artificial boundary (really just electrons right?). I get a little lose here because according to my sources, positive ions will form on the N side and negative ions on the P side due to diffusion, but here i thought when you recombine electrons and holes you have neutralization and so I would expect neutral atoms in this region (and worse this neutral region should spread across the whole material till all of it becomes locally electrically neutral). Ignoring this caveat, we will have an E-field due to the separation of ions going across the junction and this is what gives rise to the properties of a diode.

    Please correct me wherever I'm wrong, and throw in equations or further information if you find it interesting or relevant.
     
  2. jcsd
  3. Apr 6, 2010 #2
    Wow, I would've written the same question about two weeks ago.

    Background: CS major, also not Physics/EE, but interested in the field

    Regarding the depletion zone & why it doesn't spread indefinitely. (Basically, why the electric field gets set up when APPARENTLY we are neutralizing charge carriers)

    Let's imagine we're dealing with Silicon. We use Phosphorus (+1 elec) as the Group-V N dopant and Aluminum as the Group-III P dopant.

    When the free electron migrates off of the phosphorus atom, and wanders across the junction to join with the Aluminum atom (giving both 8 electrons, instead of 9 & 7) the charge is not "neutralized": the electron is in a comfortable-but-wrong location now. Each dopant atom is now an ion. The Phosphorus has 9 protons, but only 8 electrons. Likewise, the Aluminum has 7 protons, and an extra electron.

    Even though the electron (via lattice forces) is happier to be where it is, filling a rule-of-8 slot, it has actually moved, acquiring a voltage potential in the process.

    My vocabulary is probably making all the "real" physicists here cringe, but I'm pretty sure that's where the field comes from. The field, in turn, explains why the depletion zone doesn't spread indefinitely.

    Hope this helps,
    -Jeff Evarts
     
  4. Apr 6, 2010 #3
    Thanks for the reply Jeff.

    I think I'm starting to get pushed in the right direction now. So it seems that once the depletion zone is set up to a certain width, there is a field there of enough strength to lower the likelihood of electron diffusion to the other side to zero.

    Now I think that I my problem is that I am misunderstanding what is going on during doping. When we dope the material, we are introducing foreign atoms artificially to the semiconductor, this much I know. But what does doping do to the electrical properties of the material? Does introduce an excess of charge? And in what sense? Is it because the dopant is an ion and so when you throw it into the mix you end up with a material with net charge? I don't believe this is the case as it would imply charge neutralization across the junction as opposed to ion formation. Any input?
     
  5. Apr 6, 2010 #4
    OK, in *grievously* simplistic terms, undoped semiconductors have enormous resistance because all their valence electrons are full (usually), their conduction band is empty, and there is a substantial (but not insurmountable) energy gap that needs to be crossed to get an electron free to roam. In practical terms, it's MOSTLY an insulator.

    Doping causes this to stop being true: The dopant atoms either add extra electrons (which can bop freely around the N- and depletion- regions) or extra holes (which can blink from place to place within the P- and depletion- regions).

    Now obviously, if there's no junction, the depletion region doesn't exist, so in the monolithic-singly-doped-crystal case, you have (say) one extra electron for every phosphorus atom. These extra electrons are all in the conduction band, so they're free to roam from one side of the crystal to the other (it's all N-, baby) and conduct electricity. Likewise, in an all-P environment, the hole that starts out on the aluminum atom can wander freely all over the crystal and IT can conduct electricity.

    In the junction case, the charge carriers (electrons or holes) can move about their RESPECTIVE regions, and also throughout the depletion region, where they can meet one another. The forward/reverse current thing becomes important now. Both the P and the N areas have a "junction" side and a "wire" side.

    If you try to supply electrons to the P-wire side of the crystal, and holes to the N-wire side, this is called "reverse bias". Imagine what will happen: Holes near the wire-edge of the P crystal (far from the depletion zone) will head for the wire, get filled, and vanish. Likewise, the electrons "near" the wire-edge of the N crystal will fill holes in the wire (actually just join the other electrons in the wire's conduction band, but that's not important right now) and likewise, they leave the picture. Once that little bit of (capacitive) current flows, nothing happens: the current stops.

    If, on the other hand, you try to supply electrons to the N side and holes to the P side, the reverse occurs: the electrons in the N side will flee (similar charges repel) the wire side of the crystal and enter the depletion zone. Likewise, the holes in the P side will also head for the depletion zone. Once they (both) get there, they annihilate one another (virtually), and vanish. Lather, rinse, repeat. More electrons and holes migrate into the crystal, head for the depletion zone, and vanish. Over and over: current flows. This is a forward-bias current, which flows perfectly well through the PN junction.

    Clear as mud?
    -Jeff Evarts
     
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