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What is the physical origin of gain in a BJT?

  1. Aug 9, 2008 #1
    1. The problem statement, all variables and given/known data

    "What is the physical origin of gain in a BJT?"

    I can figure out gain using gain equations but am having a hard time describing the physical origin of gain in a BJT.. can anyone help?
  2. jcsd
  3. Aug 9, 2008 #2


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    Think of a valve. How do you leverage the characteristics of the BJT to input a small signal and get a much larger signal at the output?
  4. Aug 9, 2008 #3


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    Bear in mind that despite the notion of a "gain" relative to desginated inputs and outputs, conservation of current still holds here. Are you asking how exactly do the semiconductor physics of the BJT cause a "gain" in output?
  5. Aug 9, 2008 #4

    correct, that is what I'm asking.
  6. Aug 9, 2008 #5


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    I'll try to be rather brief in my explanation (it took me a while to understand BJT's and I'm still working on it.)

    The key is that the emitter is doped very heavily compared to the collector and base. So most of the conductors are "created" at the emitter node. (electrons for npn types, or holes for pnp)

    Also, the base node lies at the end of a very narrow thin path sandwiched between the emitter and collector. So most of the conductors from the emitter just "spill over" into the collector instead, as they try to make their way towards the base. Only a few conductors do make it to the base.

    The result is that the base current is much smaller than the collector and emitter currents, hence the high gain given by the ratio
    ic / ib
  7. Aug 10, 2008 #6


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    First of all, if that is what you're asking you should note that this isn't something which can be explained in a few short paragraphs. It would require equations, a lot of variables with subscripts along with accompanying concentration profile graphs and a fundamental (read: basic) understanding of p-n junctions and how they behave under forward and reverse bias.

    In short, it's not something that can be done here. Moreover you didn't specify the mode of operation. There are 4 possible modes, and only two modes "forward-active" and "reverse-active" would yield a current gain. The forward-active mode is the much more commonly used one. Read up on the mechanism elsewhere. Personally I used Semiconductor Physics 3rd Edn by Donald Neamen.

    Wikipedia has good info on this:
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