Diode barrier potential

Discussion in 'General Electronics Chat' started by epsilonjon, Dec 24, 2011.

  1. epsilonjon

    Thread Starter Member

    Feb 15, 2011
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    1
    Hi, I'm learning about diodes and the formation of the PN junction, but there are a few things i'm unclear on.

    It is my understanding that the PN junction forms when electrons in the conduction band of the N region move across and combine with the holes in the valence band of the P region. This leaves positive ions on the N side and negative ions on the P side near to the junction, and this creates an electric field which is eventually strong enough to prevent any more electrons from moving across from N to P. The energy required (per unit charge) to move additional electrons is called the barrier potential. So we have the before and after energy diagrams as follows:

    [​IMG]

    According to the book i'm reading, the barrier potential is usually something like 0.7V, and they take this as the minimum forward bias required to establish a current. This will allow electrons in the conduction band of the N region to get over the barrier potential. Am I correct in thinking they then flow into the conduction band of the P region, and immediately combine with the holes in the valence band? This is where I start to get a bit confused.

    Firstly, this sounds a bit stupid, but why can't the electrons just go from conduction band in the N type to valence band in the P type? Why do they have to go from conduction band to conduction band? The lazy ones wouldn't have to bother getting over the barrier potential then :D

    Secondly, in the diagram above, where is the barrier potential? Is it the energy difference between the conduction bands in the two regions? That is what I was thinking it was for a few weeks, but then I thought about it some more and became convinced that this is incorrect. For example, if I forward bias the diode with a battery equal to the barrier potential I will give the electrons just enough energy to get up over this barrier and into the conduction band of the P region. They then immediately combine with holes in the valence band and lose a load of energy (does this energy go to heat or something?) and travel out through the wire. But with this view of the barrier potential, the battery will not give them enough energy to get back up into the conduction band of the N type, then back up into the conduction band of the P type to start over again. As far as I can see, the only way you can get a continuous current is if the battery gives the electrons an amount of energy equal to the energy difference between the conduction band and valence band in the P region. So is this the barrier potential :confused:

    Those are my two main questions for now. Sorry for being a bit long-winded, but hopefully I explained my confusion clearly.

    Thanks for any help!
    Jon.

    Oh and Merry Christmas!!!! :)
     
    Last edited: Dec 24, 2011
  2. crutschow

    Expert

    Mar 14, 2008
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    It involves some quantum theory that I don't pretend to understand but the carriers can only move in the conduction band, not the valence band, so they have to jump the barrier between the two bands.

    The energy required to move between conduction bands is indeed provided by the battery potential and this energy is dissipated as heat across the junction. That is why a diode temperature increases when conducting. The power dissipated is equal to the current times the forward voltage drop across the diode junction.
     
  3. epsilonjon

    Thread Starter Member

    Feb 15, 2011
    65
    1
    Okay thanks. I guess I will have to wait until I have read more quantum theory before I can understand it properly.

    One other question I have: is it correct to visualize the holes moving across the depletion region in the opposite direction to the electrons? The reason i'm not sure is that I thought the electrons were coming from the conduction band in the N region, so are not really leaving holes (as these exist only in the valence band)? So I struggle to see in what sense the holes move through the N region and round the circuit? This point also bothers me when I try to think of the PNP BJT as functioning the same as the NPN BJT, just with the current carriers being holes instead of electrons.

    Thanks again! :)
     
  4. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    When looking at circuits at the silicon level, you must think of electron flow rather than"hole flow". Current moving from negative to positive.

    Once outside the silicon (IC, Transistor, etc) and in a standard circuit schematic, using "conventional flow" or "hole flow", where current goes from positive to negative.

    In the macro world of schematics, either method works if you use the same method consistently, such as this forum consistently uses "conventional flow" so the diodes and transistors look like they are pointing the correct direction.

    In the nano world, where quantum physics rules behavior, the two cannot be simply interchanged to get the same result, thus electrons/current flow from a high density negative area to a less negative area (more positive area). This also defines the majority carrier and minority carrier materials p or n.
     
  5. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    Depending on the type of semiconductor, you probably have to address both since they both contribute to how it works.
     
  6. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Ok, Majority Carrier and Minority Carrier (not always holes or electrons), but I didn't want to make the thread too confusing.
     
  7. epsilonjon

    Thread Starter Member

    Feb 15, 2011
    65
    1
    Thanks for the replies guys.

    So with the NPN BJT electrons flow from emitter to base, but since the base is very thin and lightly doped, not many of these electrons fall into the holes and most of them get swept through to the collector instead. What is the correct way to think about this at the silicon level for the PNP BJT? All the information I can find just explains it for the NPN, then says the PNP is basically the same but with holes, which is not very satisfying.
     
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