Transistor base current

Discussion in 'Homework Help' started by Steve1992, Nov 24, 2011.

  1. Steve1992

    Thread Starter Senior Member

    Apr 7, 2006
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    0
    Hi,

    Calculate Ib? Vb?

    The answer given for Ib is 25uA, so the base voltage is 0.8V (33k x Ib).

    But how is 25uA found?
    Hfe is given, but I also need to find Ic?
    Ib is 25uA when Ic is 1mA.
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Using

    33k*IB+VBE+(1+HFE)*IB*8.2k=10

    33k*IB+VBE+(41)*IB*8.2k=10

    369.2k*IB=10-VBE

    IB=(10-VBE)/369.2k=9.3/369.2k=25.2uA with VBE=0.7V
     
  3. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    hfe is defined as the ratio of Ic/Ib.
     
  4. Steve1992

    Thread Starter Senior Member

    Apr 7, 2006
    100
    0
    Thanks t_n_k.

    You multiplied base current by emitter resistor, IB*8k2.
    Is this because IE = IB + IC?
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    That's correct - the voltage drop across the emitter resistance is RE*IE. As to the base voltage VB keep in mind that in this case the base current is flowing from ground potential into the base via the 33kΩ resistor.

    So VB=-25.2uA*33kΩ=-0.83V
     
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