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

    Active Member

    Oct 26, 2011
    539
    99
    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

    Active Member

    Oct 26, 2011
    539
    99
    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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