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Conceptualling understanding transistors

  1. Jan 16, 2010 #1
    Okay, so my basic understanding of an NP junction (diode) is as follows. When no external voltage is applied, some of the electrons in the N type material jump over to the "holes" in the P type material. The holes then end up moving back into the P type material further, and you end up with negatively charged ions in the P type and positively charged ions in the N type. Since there aren't many free electrons around in the depletion zone, it contains a rather high resistance.

    When you apply a forward biased voltage to the NP junction, the electrons get pushed into the positively charged ions and cancel with them, and get pulled out of the negatively charged ions, therefore shrinking the depletion zone and lowering the resistance so current flows.

    When you apply a negative biased voltage to the NP junction, the electrons fill even more of the holes in the P type material and get pulled out of the N type material, so the depletion zone or charged ions grows, as does the resistance, allowing a negligible amount of current to flow.

    Now I'm confused over NPN bipolar junction transistors. In particular, if a forward biased voltage is applied between the base and emitter, and a larger reverse biased voltage is applied between the base and the collector. I understand that the emitter-base depletion zone breaks down and current flows, but I don't understand why the base-collector depletion zone doesn't grow as a reverse biased diode would. Could someone try and explain this?
     
  2. jcsd
  3. Jan 18, 2010 #2

    sophiecentaur

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    If the b e junction is not forward biased, then the c b junction will not pass current and acts just like you say. Forward biasing the b e junction produces carriers in the b layer which prevent the cb depletion layer from forming, I think.

    The actual geometry is important if you want a transistor to work best - thickness of the base layer is an important factor. They started off with Betas of less than 1, I believe!
    A transistor will actually work (badly, though) connected the other way round.
     
  4. Jan 18, 2010 #3
    The reason collector current flows is as follows. When the b-e junction is forward biased, holes are injected from the base to emitter, and electrons from the emitter to base. This is for npn type. The holes from the base recombine in the emitter region. The electrons from the emitter transit through the base, a small fraction recombine in the base, and almost all keep moving into the collector. They are drawn to the collector due to the electric field in the reverse biased c-b junction.

    Thus, the emitter current due to forward biasing the b-e junction, consists of electrons from emitter to base, which transit through the base, arrive at the collector, and become collector current. Emitter current charge carriers cross the junction and become collector charge carriers, i.e. collector current. Current is just charge motion.

    The math equation for this transistor action is Ic = alpha*Ie. A good device has alpha close to 1, i.e. 0.98-0.998. Did I help?

    Claude
     
    Last edited: Jan 18, 2010
  5. Jan 18, 2010 #4

    sophiecentaur

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    Well yes. But the original question was about the possibility formation of a depletion region between n Collector and P Base, though. I think this is what happens without base current.
     
  6. Jan 18, 2010 #5
    Yes this does happen, with or without base current. The c-b junction is reverse biased, and a depletion region exists. When the b-e junction is forward biased, the c-b depletion region remains, as the c-b junction is still reverse biased (except if the bjt saturates).

    Does this help?

    Claude
     
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