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Calculating transistor bias currents

  1. Sep 8, 2012 #1
    1. The problem statement, all variables and given/known data
    http://img846.imageshack.us/img846/955/29278644.jpg [Broken]


    3. The attempt at a solution

    Basically I want to know whether my working method was correct as I have doubts with some of it.

    I first did voltage division at the two base resistances, and I get 3V at the base. Since we know there's a base-emitter drop of 0.7V, the emitter voltage of Q1 is 2.3V which is also the base voltage of Q2, therefore the emitter voltage of Q2 is 1.6V.

    This means we can calculate the emitter current of Q2 easily. After obtaining the emitter current of Q2, using the Ie = Ib + Ic relation and substituting for Ib = Ic/B, we can easily getting the collector current ic2.

    http://img542.imageshack.us/img542/8632/80235300.jpg [Broken]

    Now here is where I'm a bit confused as to how to proceed and get ic1

    I thought I can get the current from Vcc to Vb, let's call it ix, going back to the Vcc node, can I use that as a KCL point? So basically if I have ix leaving that node, and let's call the other current il which is going through Rc, those two must add up to give 0.

    Once I have il, then il = ic1 + ic2 and therefore I can solve for ic1 and hence also find the voltage at Vc which is the DC voltage they asked for.

    http://img6.imageshack.us/img6/664/12595537.jpg [Broken]

    Thanks.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 8, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    For a given transistor, IB and IC, and IB and IE are related through the β; you should know these expressions.

    Since you have calculated a value for the emitter current of Q2, you can work back to its base current. But that base current is also the emitter current of Q1... carry on.
     
  4. Sep 8, 2012 #3
    Ah...I see now. Yes, I know the expressions, I didn't read the question correctly: "neglect base currents where appropriate" - I completely neglected it.

    Thanks for that.
     
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