Finding Voh for BJT inverter

Discussion in 'Homework Help' started by hitmen, Oct 28, 2008.

  1. hitmen

    Thread Starter Active Member

    Sep 21, 2008
    I know that to find Voh, the solution is to get
    (Rb/(Rb+Rc))(Vcc-0.75) + 0.75.
    However I do not know why this is done. I have always thought that electrons flow from the emitter to either the base or collector. So how it is possible for an electron to flow through both Vrc and Vrb. I don't understand:confused:

    Similarily I dont understand why I(IH) for BJT inverter is (Voh-0.75)/Rb. I thought that the collector current and the base current are separate entities. So where does Voh come into the picture?
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    There is no schematic to work from, but electrons are always the charge carriers. Find the point with the most negative potential, and electrons will always move from that point to every other part of the circuit with a more positive potential. Both the collector and base circuits satisfy that requirement, so the emitter sources electrons to both.

    Collector current and base current are only separate entities until you work back to the emitter. All the transistor's current goes through the emitter.
  3. Wendy


    Mar 24, 2008
    What is Voh? I suspect the .75V is the Base Emitter drop, I use .7 myself if so. The current through the Base Emitter controls the current through the Collector Emitter, which is the basis of a BJT.
  4. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    disregard this line in the posting. it was done to accomandate the need for a message longer than ten characters.
  5. hitmen

    Thread Starter Active Member

    Sep 21, 2008
    Voh is V out high.
  6. hitmen

    Thread Starter Active Member

    Sep 21, 2008

    I understand that electrons flow to the collector AS WELL AS the emitter. But in calculating Vout high, why are we assuming that the SAME set of electrons go through the emitter AND the collector. It just doesnt make sense to me.