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BJT impedance calculation

  1. Jan 22, 2012 #1
    A bit confused about direction of current flow in the BJT equivalent ckt.
    I'm trying to calculate the resistance seen, looking into the emitter terminal.
    In what direction does the current flow in the bjt output resistance ro?
     

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  2. jcsd
  3. Jan 22, 2012 #2
    Form + to -
    And direction of a current doesn't really matter.
    attachment.php?attachmentid=43007&stc=1&d=1327264695.png
    And also remember that

    Rin = Vin/Iin

    Iin = Iro+IRe + Ie

    Ib = Vin/(βre+RB)

    Ie = Ib + Ic = (β+1)*Ib
     

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    Last edited: Jan 22, 2012
  4. Jan 23, 2012 #3
    Why doesn't the direction of current really matter?
     
  5. Jan 23, 2012 #4
    Because we are dealing here with a BJT model not a real device.
    And this small-signal model is nothing more then mathematical abstraction.
    Our small-signal model describe the behavior of a BJT in "special condition".
    We need properly DC bias the BJT into the linear region and apply low-frequency very small AC signal. And in this condition we create a small-signal model.
    As for Rin = Vin/Iin our AC signal source can only see a resistance nothing more.
    Because our Vin source apply voltage and supplies Iin current.
    So our signal source see the BJT as a black box. So the only thing that our source is able to see is the load resistance .
     
  6. Jan 23, 2012 #5

    vk6kro

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    Science Advisor

    The input signal is AC, but for analysis purposes, we assume a direction for the input at one moment and then look at the directions of the currents in the circuit at that moment.

    It is usual to take the input voltage and current as positive and then the other voltages and currents in the circuit depend on this.
     
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