NPN with a floating emitter

Discussion in 'Homework Help' started by psycoadam, Nov 17, 2011.

  1. psycoadam

    Thread Starter New Member

    Oct 18, 2011
    19
    0
    The question given is to solve for the collector current. The emitter is floating and the collector is biased at 5V with respect to the base.

    The givens are Is = 10fA, BF = 100, and BR = 1

    I understand the problem, I am just having trouble figuring out how to set it up properly. My assumptions are that iE = 0, so therefore Vce = 0

    using the equation
    ic = Is*e^((VBE/VT)(1+ VCE/VA))

    Then VBE would be 5V, VT = 0.025, and VCE = 0?

    Since iE is equal to zero then I know that iE = (BF +1)iB = 0, so then iB = 0 which does not make sense because ic = BF*iB and ic is what we are solving for.

    Any help would be appreciated. Thanks in advance.
     
  2. jimkeith

    Well-Known Member

    Oct 26, 2011
    541
    100
    Note that in this case, Ib = Ic = Is
    Simple series circuit problem
     
  3. psycoadam

    Thread Starter New Member

    Oct 18, 2011
    19
    0
    is big IB = iB? and Ic = ic?
     
  4. jimkeith

    Well-Known Member

    Oct 26, 2011
    541
    100
    Yes, assuming Big I is quiescent current and Little i is incremental current.
    In this setup, the transistor base-collector junction is simply a reversed biased diode.
    Hope I am not getting you into trouble with your prof.

    A more difficult question is: What is the emitter voltage? a little trivia that I cannot answer--would have to measure it myself
     
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