BJT question

Discussion in 'General Electronics Chat' started by Joe24, Dec 27, 2007.

  1. Joe24

    Thread Starter Active Member

    May 18, 2007
    Hello all,

    I have a question regarding the BJT configuration that I have posted. The values I had to find were the current Ie and the voltage V3, which is the same as Vc.

    The answers are: Ie = Ve - (-10v) / 4.7K = 2mA

    Vc = 10V - IcRc = 10V - 2MA*(3.3K) = 3.4V

    these answers are correct. But what I am confused about is the following:

    When I first look at this simple circuit, I see that the base is tied to ground. So to me, that mean that there is no current going into the base, therefore the transistor is in cut-off. If it is in cut-off, then the transistor behaves as an open switch and there-for no current through the collector or emitter.

    i know I am wrong here, but this is my first assumption when I first look at it. Can someone shine some light for me here. Thanks
  2. kubeek


    Sep 20, 2005
    Ground doesn´t mean there is no current into the base.
    You can re-label the circuit so now the bootom connection is ground, the base is connected to +10V and the upper connection is +20V.
  3. Joe24

    Thread Starter Active Member

    May 18, 2007
    That's an interesting approach Kubeek, but it doesn't work out to the same Collector voltage.

    I get the same Ie ( Emitter current) as before, but for Vc I get 13.4V using your approach.
  4. Pootworm


    May 18, 2007
    That's 13.4V after renaming -10V as ground. If you re-adjusted to the ground in your pic, you'd subtract 10V and get to your 3.4V answer (relative your original ground).

    Incidentally, you can just treat the voltage at base as zero, run through the exact same arithmetic process and wind up at the same answers. The 4.7kOhm resistor has -0.7V on one end and -10V on the other, so it still drops 9.3V, etc etc etc.