Small-signal BJT analysis

Discussion in 'Homework Help' started by corsair, Apr 19, 2010.

  1. corsair

    Thread Starter Member

    Mar 6, 2010
    51
    1
    Hi guys,

    I attached the answer for the problem. I am having trouble understanding one part of the solution in finding Rinb.

    My original solution was:
    Rin = (10k + 10k) || (.0145 + 2k)

    When they solve for Rinb, they multiply the resistors in the emitter by beta + 1 (where beta = 100). Why is this?
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    The small signal emitter current ie is (1+β)ib.

    "Looking into" the Base one sees an effective small signal resistance of

    rB=vb/ib=ie(re+RE)/ib=(1+β)ib(re+RE)/ib=(1+β)(re+RE)
     
    Last edited: Apr 19, 2010
    corsair likes this.
  3. corsair

    Thread Starter Member

    Mar 6, 2010
    51
    1
    thank you very much!!
     
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