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Thevenin equivalent question

  1. Feb 3, 2015 #1
    1. The problem statement, all variables and given/known data
    Determine the Thevenin equivalent with respect to the terminals a,b

    2. Relevant equations
    Rth=Voc/Isc

    3. The attempt at a solution
    I'm doing this practice problem and I calculated the the short circuit current below:

    500uA*100ohm/(100ohm+1310ohm)= 35.460 uA

    since 80*ib=current in v2, 35.460 A * 80=2836.87 uA

    So Isc=2836.87

    Now for the open circuit voltage..
    80ib* 50kohm= 40.0*10^5 ib
    Now to solve for ib is where I'm lost.
    In the solution they have 4*10^-5v2=-160ib, but I have no idea where it came from?
     

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  3. Feb 3, 2015 #2

    NascentOxygen

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    Staff: Mentor

    A fraction of v2 is fed back to influence the input. That fraction is given as 4x10-5v2, so substitute in this your value for v2 (which you have in terms of ib).
     
  4. Feb 3, 2015 #3
    Thank you. I have a question about calcuating Isc vs. Voc, why isn't Voc simply Isc/50kohms?

    I know the circuit is suppose to be "open" but isn't it still the same thing?
     
  5. Feb 3, 2015 #4

    NascentOxygen

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    Staff: Mentor

    It would be if the Thèvenin resistance were 50kΩ, but it isn't. That's what you are trying to find! The feedback changes the output impedance. And it changes the input impedance, and it changes the gain.
     
    Last edited: Feb 3, 2015
  6. Feb 3, 2015 #5
    So is the Vth or Voc I'm calculating simply the voltage across the 80ib dependent current source?

    In the open circuit voltage I thought the 50kohm resistor would still be part of the circuit, and the voltage across that would equal the voltage across a-b?
     
  7. Feb 3, 2015 #6

    NascentOxygen

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    Staff: Mentor

    Yes, but the 50k remains in place, it's part of the model. (Without some resistive load, the voltage across a current source would be infinite.)

    Yes, and yes.
     
    Last edited: Feb 3, 2015
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