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Thevenin equivalent circuit

  1. Jun 15, 2008 #1

    Defennder

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    1. The problem statement, all variables and given/known data
    I really don't see how the Thevenin equivalent circuit is obtained from the one on the left. The way learnt to find Thevenin equivalent circuit considers the circuit network in between 2 nodes. This one appears to have 3 nodes. And why is the thevenin resistance given as though the resistance were connected in parallel? I don't see why they are in parallel.


    2. Relevant equations/pictures
    Thevenin equivalence.
    [​IMG]
     
  2. jcsd
  3. Jun 18, 2008 #2

    CEL

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    To obtain the open circuit voltage you use a voltage divider with the 10V source and the two resistors, getting 3.33V.

    For the equivalent resistance you short circuit the source, so the two resistors are in parallel (the upper terminal of R1 is connected to ground).
     
  4. Jun 18, 2008 #3

    Defennder

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    Changed image

    Yeah, thanks, I think I can see it for the Thevenin resistance. I assume that the above gray box is also equivalent to this:

    [​IMG]

    But the open circuit voltage is between V1 and Vb or V2 and Vb?
     
    Last edited: Jun 18, 2008
  5. Jun 19, 2008 #4
    for this circuit there is an expression to calculate v(th)& R(th) which is:
    R(th)=R1//R2
    v(th)=R(th)[(v1/R1)+(v2/R2)]
    where v1=10v,v2=0v
    // means parallel
     
  6. Jun 19, 2008 #5

    Defennder

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    Where did you get that expression for v(th) from? And more importantly is the open-circuit voltage between V1 and Vb or V2 and Vb? That's all I need to be able to figure out the thevenin voltage.
     
  7. Jun 20, 2008 #6
    v(th) comes from the Voltage divider rule.

    v(th) is measured between vb and v2.
     
  8. Jun 20, 2008 #7

    Defennder

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    But why isn't it measured between V1 and Vb?
     
  9. Jun 20, 2008 #8

    CEL

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    Because V2 is ground and you want to polarize the transistor, so you are really interested in the voltage between base and emitter: Vbe = Vb - Ve.
     
  10. Jun 21, 2008 #9

    Defennder

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    Oh yeah you're right. It's the voltage across the Vbe that's relevant here. Thanks a lot!
     
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