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Simple DC Circuit Analysis with Transistor

  1. Apr 17, 2017 #1
    1. The problem statement, all variables and given/known data
    317e7fk.png

    2. Relevant equations

    xdv0c7.png 2z7g9p4.png
    2zf4jk5.png
    3. The attempt at a solution
    All loops are clockwise;

    KVL Loop around 1V, 120kOhm and VBE (I1 is assigned to the loop):


    -1V + 120kOhm * I1 + VBE = 0
    I1 = IB = 0.3V / 120kOhm = 2.5 * 10-6 A;

    Loop around 10kOhm, vCE (I2 is assigned to the loop);

    KVL Loop around 10kOhm, vo, and 20V (I3 is assigned to the loop):


    10kOhm * I3 + 20V - vo = 0
    vo = 10kOhm * (I2 - I3)
    10kOhm * I3 +20V + (I3 - I2) * 10kOhm = 0 (1);

    At the node connecting the collector part of the transistor, Io and 10kOhm:

    I3 + (I2 - I3) + IC = 0
    since Io = I2 - I3

    IC = B * IB = 80 * 2.5 * 10-6 = 2 * 10-4 A

    I2 = -2 * 10-4 A (2);

    Subbing (2) into (1):

    10kOhm * I3 +20V + (I3 + 2 * 10-4 A) * 10kOhm = 0

    20kOhm * I3 = - 20V - 2 * 10-4 A * 10kOhm

    I3 = -22 / 20k A = -1.1 * 10-3

    I2 - I3 = 9 * 10 -4 A ! = 6 * 10 -4 WRONG

    And I've been stuck redoing the same equations for a day now... Would be nice if someone could help find the problem in my "solution" so I can finally move on haha.
     
    Last edited: Apr 17, 2017
  2. jcsd
  3. Apr 17, 2017 #2

    NascentOxygen

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    Staff: Mentor

    Hi Dan97. :welcome:

    clockwise Σ of voltages: vo + 10k × I3 + –20 = 0

    EDITED
     
    Last edited: Apr 17, 2017
  4. Apr 17, 2017 #3
    Hi @NascentOxygen ,

    If loop 3 is clockwise, and it goes through the passive configuration, then it should be +20V - vo +I3 * 10kOhm?

    I'm not sure what you're trying to say...
     
  5. Apr 17, 2017 #4

    cnh1995

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    Homework Helper

    The given values of Vo and Io do not follow Ohm's law.
    I am also getting Io=9*10-4 A.
     
  6. Apr 17, 2017 #5
    Hi @cnh1995 ,

    Could you show your working (if it's not too much of a hassle)? :)
     
  7. Apr 17, 2017 #6

    NascentOxygen

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    Staff: Mentor

    Sorry, typo. Corrected.

    Following your path, it will be +20V - vo – I3 * 10k = 0

    Current through a resistor (in fact, through any element that's not a source) flows from its more-positive end to its less-positive end. If you draw an arrow showing this PD across the resistor, the arrow's tail is at the less-positive end and the arrow head is at the more-positive end.
     
  8. Apr 17, 2017 #7
    @NascentOxygen ,

    Why would it be - I3 * 10kOhm? Isn't that breaking Ohm's Law?
     
  9. Apr 17, 2017 #8

    cnh1995

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    Homework Helper

    The given answers are 12V and 600uA, which are incorrect. They don't follow Ohm's law.
    Collector current Ic=2x10-4A.

    Using KVL you can write,
    104*Io=20-104*(Io+2*10-4).

    Solving this for Io, we get Io=9×10-4 A.
     
  10. Apr 17, 2017 #9
    @cnh1995 ,

    Oh actually I've made a big fuss for nothing! I've just look at the author's errata and the answer is indeed 9 * 10 -4!

    1zlddt0.png
     
  11. Apr 21, 2017 #10
    what book is this anyway?
     
  12. Apr 23, 2017 #11
    Alexander's Fundamentals of Electric Circuits
     
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