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Equivalent circuit for lumped capacitance with dielectric losses

  1. Apr 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Draw the equivalent circuit for a lumped capacitance with dielectric losses that would be appropriate at RF frequencies.

    2. Relevant equations

    Possibly

    Xc=-j/ωC

    3. The attempt at a solution

    I do not really know what the question is asking. I know that a lumped capacitance is a capacitor with no parasitic elements, ie, no electrical resistance or inductance. It is just a pure capacitance.
    As for dielectric losses, I guess this means that not all of the energy "used" by the capacitance is stored in the electric field created by it, some of this energy is lost due to ohmic heating caused by the electrons "leaking" through the dielectric (which has a resistance). I'm not sure if this is correct though.

    My best guess for the answer to the question will be to draw a capacitor in parallel with a resistor.

    The problem is that I feel that this contradicts the "lumped capacitance" part of the question. I also don't know why it was mentioned that this capacitor will be operated at RF frequencies. This question is from a revision sheet, so the actual answer doesn't matter as long as I know exactly what the question means, and I know how would answer similar types of questions in the exam, which is why I think the statement about it being at RF frequencies is important.

    Please help!

    Thank you
     
  2. jcsd
  3. Apr 27, 2013 #2

    rude man

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    "Lumped" capacitance is in contradistinction to "distributed" capacitance. A coaxial cable has distributed capacitance. You can't take a length of cable say 2m long and say, OK, capacitance is 100 pF per m so I can model the cable as a lumped capacitor of 200 pF. So your model can add resistors and whatnot and still be lumped.

    You are on the mark in suspecting that a resistor is needed to model dielectric loss. But a parallel resistor will not demonstrate one of the characteristics of dielectric loss, viz. you short a charged cap, then release the short & lo & behold some voltage comes back on it. Hint: you are very close though! In addition to adding a resistor, which is frequency-independent, what kind of component has little effect at low frequencies but is substantial at RF frequencies?

    PS - there is really no one right answer to this question. Models can be as complex as desired or as simple as practical.
     
  4. Apr 27, 2013 #3
    Thank you for answering. Your explanation of "lumped" capacitance has helped a lot.

    I suspect that a very high resistance resistor will be in parallel with the capacitor, as this will represent a small amount of charge passing through the capacitor (the dielectric losses). I'm then guessing that then you add an inductor in series after the cap and resistor. (R||C)+L

    I'm guessing this because an inductor is a component that has little effect at low frequencies but some at RF frequencies. However I do not fully understand what this other characteristic of dielectric loss that you mentioned is. Are you saying that because there will be a changing current inside the dielectric it will cause a changing magnetic field and hence an opposing voltage, so it has an inductance?
     
    Last edited: Apr 27, 2013
  5. Apr 27, 2013 #4
    I've just had a thought. Is it the "displacement current" which causes the magnetic field? and hence why you model the capacitor with a series inductor?

    The displacement current is not the current that flows through the capacitor, but the current that appears to flow through the capacitor because of the electric fields that go through it. It appears in amperes law (maxwells form of the magnetic circuit law)
     
  6. Apr 28, 2013 #5

    rude man

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    You idea of a high-value resistor in parallel with an ideal capacitor is OK since that can represent leakage. And the series inductor is also good since it obviously represents lead inductance plus any internal inductance.

    But you need to add one more component to generate the 'bounceback' effect of returning voltage after the capacitor is temporarily shorted.

    And no, displacement curent is the fundamental current you get with an ideal capacitor: i = C dV/dt.
     
  7. Apr 28, 2013 #6
    A resistor in series after the inductor? This will represent the impedance mismatch and hence the 'bounceback' effect.

    I'm not really sure what this 'bounceback' effect is though, or how it arises. can you please explain?

    I also don't really know what you mean when you say:

    Are you saying there is more than one property of dielectric loss? so a leakage current is one, but there is also another property that I need to represent with another component?
     
  8. Apr 28, 2013 #7

    rude man

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    There's dielectric hysteresis which I probably equated to dielectric loss.
    You have the right idea in putting a resistor in series with the inductor. Now, when you short the capacitor temporarily, not all the charge will be removed from it. Instead it will diminish as exp(-t/RC) where R is your series resistor. (I am ignoring the inductor which is very small).

    I can't explain the micro reason for this behavior. Just realize that without the series resistor there would not be a 'bounceback' effect. The parallel resistor would not cause this at all.

    As I said, there are many better models than this out there. Here's a more complex one by a leading mafr. of semiconductors attached.
     

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