BJT

Discussion in 'Homework Help' started by andrewharbert, May 1, 2013.

  1. andrewharbert

    Thread Starter New Member

    May 1, 2013
    2
    0
    Hello, I think I have this correct but confirmation and or corrections would help. Here is the circuit. βdc = 100. I need to find Ic and Vce. [​IMG]
    My calculations: Vrb = 10V-.7V = 9.3 V Ic = 100(9.3V/680kΩ) = .0136mA Vce = 10V-(.0136mA)(2.7kΩ) = 9.96328V Is this correct? My next question is would I use the same calculations for the next two circuits? [​IMG]

    [​IMG]
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,758
    4,800
    None of your images is showing up. Try attaching them to the post. Use the Go Advanced option and the Manage Attachments button from there.
     
  3. andrewharbert

    Thread Starter New Member

    May 1, 2013
    2
    0
    [​IMG]

    [​IMG]
    [​IMG]
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,758
    4,800
    You approach is correct, but your math isn't.

    BTW - Thank you very much for tracking your units throughout your work. Doing so is as valuable as it is rare.

    In addition to religiously tracking your units, which will catch the majority of conceptual and algebraic errors you make, you need to get in the habit of also asking if the answer makes sense, which will catch most of the math blunders you make.

    In this case, consider your calulation for the collector current.

    Ic = 100(9.3V/680kΩ) = .0136mA

    Does this result make sense?

    Well, 100/680 is about 1/7. 9.3/7 is a bit more than 1, call it 1.2. V/kΩ is mA. So the result should be in the range of 1.2mA. The answer clearly doesn't make sense. But, without touching a calculator, I can pretty safely predict that the problem is that you didn't multiply by the 100 for the beta.

    Fundamentally, the second circuit is the same as the first (different component values, of course) except the polairity is swapped.

    The thrid circuit is different. But if you understand WHY you approached the first circuit the way you did, you will have no problems figuring out a valid approach for it.
     
    screen1988 likes this.
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