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[HELP]Thevenin Norton with dependent source in Transistor

  1. Jun 5, 2015 #1
    1. The problem statement, all variables and given/known data
    My teacher gave me a method to find Impedance of Thevenin and Norton. Here is her method: she removed Indepentdence source (voltage is closed, current is open) and did not put V external ( so the dependent source did not working because of no source). And she find resistance

    2. Relevant equations
    Is that method is true?

    3. The attempt at a solution
    I search google but nobody give that method
     
  2. jcsd
  3. Jun 5, 2015 #2

    gneill

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    Staff: Mentor

    It may depend upon the particular circuit and how any dependent sources influence the output.

    Can you post a diagram of the circuit in question?
     
  4. Jun 6, 2015 #3
    My circuit is AC equivalent circuit of BJT. She said that when she calculate Output impedance = Vo/io, she put Vin = 0. I dont know what she means, maybe she removed Vin give it away or it is like Thevenin-Norton circuit it replace voltage source or current source with resistance source. In this circuit, when she analysis she did not remove Rs. And I analysis this circuit by Thevenin-Norton I remove source put source resistance, Vext and calculate output impedance. She said I'm wrong :(
     

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  5. Jun 7, 2015 #4

    gneill

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    Staff: Mentor

    Your circuit diagram image is very small and hard to read the labels, but even so it looks a bit odd for a BJT AC equivalent. I've never seen one with the current source feeding directly into the base! Usually it flows into the emitter. Something like this:

    Fig1.gif
    In the above circuit the emitter happens to be shown connected to ground. Usually this means that any external emitter resistor is "bypassed" by a relatively large capacitor which shunts AC signals around the emitter resistor.

    Setting Vin = 0 simply means placing zero volts at Vin. This can be accomplished by hanging a 0 V voltage source there, or more simply, shorting Vin to ground. For the above circuit this would make the base current zero, since for AC the base and emitter would be at the same potential. That would turn the dependent current source into a fixed source of 0 A...

    Can you show the details of your instructor's analysis and your own analysis?
     
  6. Jun 7, 2015 #5
    upload_2015-6-7_14-59-54.png
    Maybe My analysis is as same as with her but I know that is still exist a independent source Vin and it change to Vth (thevenin-Norton equivalent circuit), but when she wrote her analysis on blackboard, and told she give Vin = 0, I dont know she give it away or it is still there but transform to Vth like my analysis
     

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    Last edited: Jun 7, 2015
  7. Jun 7, 2015 #6

    gneill

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    Staff: Mentor

    Setting Vin = 0 implies that you are determining the output impedance at the quiescent operating point. If you think about it, for small signal analysis it's really the only practical way to define the output impedance.

    Alternatively, consider applying the superposition theorem. If there is some input signal Vin and and external current source on the output (you can stick a current source or a voltage source externally, but a current source makes the analysis simpler). Then the voltage at the output will depend upon contributions by both. Showing just the output circuit:

    Fig2.gif
    The output impedance will be the ##ΔV_o/ΔI_{ext}## contribution, and will be independent of ##i_b##.
     
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