Sedra Smith Example 5.12 (Base current question)

Discussion in 'Homework Help' started by yitch, Aug 30, 2010.

  1. yitch

    Thread Starter New Member

    Jun 27, 2010
    8
    0
    Hi guys,

    I am also having a little difficulty in understanding the example. However, my concern lies more with why is there an extra 101 * 1 in the denominator for the calculation of the base current Ib?

    Shouldn't the Ib = (5- 0.7)/10k

    ?

    Thanks!
     
  2. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    Even though I have the book it would be nice if you posted a scan or transcribed the problem...

    Anyway, it's because of the load resistance in the emitter.
    You need to keep in mind these are always loop equations, meaning they always go around a closed loop.
    In this case, you have:

    -5 + Ib*10k + 0.7 + Ie*1k = 0

    Ie = 101*Ib

    That 101*1k is the 1k load resistor, as reflected to the base.

    -4.3 + Ib(10k + 101*1k) = 0
    Ib = 4.3 / (111k) = 0.039 mA
     
  3. yitch

    Thread Starter New Member

    Jun 27, 2010
    8
    0
    Ah yes.. forgot about that...

    Thanks!
     
Loading...