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Operation of a P-N Junction

  1. Jul 11, 2014 #1

    I was just having a hard time understanding how reverse bias works for a p-n junction. Put simply, why don't electrons just diffuse back into the n-type semiconductor during reverse bias, thus generating current? I was thinking maybe it's got something to do with energy levels, but I'm not entirely sure.

    If it helps, here's what I know about the p-n junction. Maybe there's a flaw in my reasoning somewhere that complicates things:

    P-type semiconductor = Neutral, but has holes due to doping by a group III element. These holes are neutral as they are simply "empty space," but relative to the surrounding electrons, they can be classified as "positive charge."

    N-type semiconductor = Neutral, but has extra electrons due to doping by a group V element.

    Setting Up the P-N Junction = When a p-type semiconductor and n-type semiconductor are joined together, the extra electrons within the n-type will diffuse into the holes into the p-type. When this happens, the n-type semiconductor will contain positively charged ions since electrons have been lost, while the p-type will contain negatively charged ions since electrons have been gained. This results in the formation of a depletion zone, since the free electrons have entered holes, and are now immobilised (although I'm not too sure why). These positive and negative ions set up an electric field, which stops further electrons from diffusing from the n-type to the p-type.

    Forward Bias = Connect positive terminal to p-type and negative terminal to n-type. The battery draws electrons out of the p-type semiconductor (but I thought they were immobilised?). Thus, the negatively charged ions are now neutral. Similarly, electrons enter the n-type semiconductor, and therefore the positively charged ions are now neutralised. This has effectively narrowed the depletion zone, thereby reducing the strength of the electric field that was set up. Electrons can now once again, diffuse from the n-type into the p-type, and the entire process repeats. Thus current flows.

    Reverse Bias = Connect negative terminal to p-type and positive terminal to n-type. The battery now draws electrons out of the n-type semiconductor. This increases the number of positively charged ions in the n-type semiconductor. Simultaneously, electrons are entering the p-type semiconductor, hence increasing the number of negatively charged ions. Both these events result in the depletion zone becoming wider, hence increasing the internal electric field strength. This somehow prevents current from flowing. But why don't the electrons that have filled the holes in the p-type diffuse back into the n-type, where they originally came from, thereby breaking down the depletion region and allowing for a current to flow?

    If anyone could help me, it would be greatly appreciated.

  2. jcsd
  3. Jul 12, 2014 #2
    quote: " But why don't the electrons that have filled the holes in the p-type diffuse back into the n-type, where they originally came from, thereby breaking down the depletion region and allowing for a current to flow? "

    If you put a large enough voltage, they will.

    It seems you understand have a qualitative understanding. The question you're asking requires applying quantitative reasoning to what you already qualitatively understand.
  4. Jul 13, 2014 #3
    That's the thing I don't understand though. How come the voltage from the battery is not enough to break down the depletion zone during reverse bias? Both the internal electric field from the depletion zone and external electric field from the battery are in the same direction during reverse bias, so shouldn't it be easier for a current to flow in reverse bias than forward bias?
  5. Jul 14, 2014 #4
    Draw a picture of how you understand things and post it here. That will make it easier to debug.

    Make sure to include the depletion zone, the ions, the electron density (just treat it as a giant blob or liquid).
    Use the semiclassical approximation to label all the "forces" acting on an electron.
  6. Jul 21, 2014 #5


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    Here's a hint. When you deplete part of the pn junction of free carriers is anything left behind? Is what is left behind charged?
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