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Kirchhoff's Law and current Sources?

  1. Sep 21, 2009 #1
    When you are using Kirchhoff's voltage law and you come across a current source, what are you supposed to include that as? Pretending that it is an ideal current source, there will be no resistance and thus no voltage drop. So, do you just ignore it in the voltage equation? (Ex: V1+V2+V3=0)

    Because of the fact that you can re-arrange a resistor and voltage source in parallel to a current source and resistor in series, I'm a little confused.

    Any help much appreciated.
  2. jcsd
  3. Sep 21, 2009 #2
    A current requires there be an electric field, which in turn, requires there be a potential difference.
  4. Sep 21, 2009 #3


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    Staff: Mentor

    That's one of the reasons I use KCL to solve circuits. I never liked KVL...
  5. Sep 21, 2009 #4
    Gear, would you treat the current source as a separate voltage source altogether or combine it with the surrounding resistor and use it in the equation V=IR?
  6. Sep 22, 2009 #5


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    A current source does not project a voltage difference, just a current. You cannot replace it with a voltage source unless you re-imagine the circuit as a Thevenin equivalent circuit. Use the KCL rule for the branches that have current sources as the current source forces the current in that branch.
  7. Sep 22, 2009 #6


    Staff: Mentor

    This will get you the wrong answer almost every time. Current sources typically have very high impedance. The current source supplies whatever voltage is required in order to maintain the set current.

    Include it as what it is, an unknown voltage that you have to solve for. In such a problem you have one more unknown (the voltage across the current source) and one more equation (the sum of the loop currents through the current source equals the set current). You still have enough equations to solve for all of your unknowns.

    I agree!
  8. Sep 22, 2009 #7
    Wouldn't there be a voltage difference to induce the current source?
  9. Sep 22, 2009 #8


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    Not exactly, as DaleSpam stated, the sourced current would create a voltage offset through the source's load. It does not apply a voltage difference, it applies a current source, the resulting voltage difference is dependent upon the load on the current source.

    Take a look at a MOSFET current mirror. The voltage across the sourcing transistor in a current mirror is not created (in the ideal case) by the transistor but entirely by the load connected to the mirror. In fact, with a current mirror, your load needs to create a sufficient voltage drop to force the sourcing transistor into saturation for the mirror to work properly. Actually, the current mirror itself is a Norton source, with a large but finite Norton resistance due to what is called channel length modulation. Ideally, the Norton resistance would be infinite, in which case the current mirror would not have any voltage drop associated across it (since the resistance branch would become an open circuit).

    A simpler example is the Thevenin and Norton circuits. You can see that in the Norton circuit, the parallel load resistance creates the voltage drop.
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