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Noise Across Capacitor

  1. Oct 23, 2009 #1
    Hi,

    Suppose you have a simple capacitor, where the potential on one side is held constant and the other is allowed to vary randomly.

    What is the capacitive current?

    thanks,
     
  2. jcsd
  3. Oct 23, 2009 #2
    AC signals pass through capacitors as a function of their wavelength, I don't know the function off the top of my head however...

    EDIT: Maybe this equation?

    i = C de/dt

    'i' is the current
    'C' is the capacitance
    'de/dt' is the voltage differentiated with respect to time, I don't know any fancy latex
     
  4. Oct 23, 2009 #3
    James, thank you for your reply.

    With my limited understanding of electricity, I presumed that to calculate capacitive currents we had to take the derivative. ie.
    [tex]
    I_C=\frac{dQ}{dt}
    [/tex]
    But since the potential is random (noise) on one side then [tex]Q[/tex] is random as well, and this would lead us to calculate the derivative of a random number.

    Any idea?

    Thanks
     
  5. Oct 23, 2009 #4
    Yes exactly, but in this case e - (the voltage differential) is random.
    ... Any clue?

    thanks again,
     
  6. Oct 23, 2009 #5
    I dunno, take the root mean square of the voltage 'noise' and use that as your AC 'input'?

    Random in random out, you can only quantify the relationship here... which we have already done I think.
     
  7. Oct 23, 2009 #6
    True thermal "noise" power across a capacitor may be an oxymoron, because the voltage is in phase-quadrature with the current, and therefore is not real power. Sometimes Johnson-Nyquist thermal noise power (e.g., kTB) formulas for low-bandwidth systems end up with a C in the formula, but this may be just due to a cancellation of R in R/RC.
    Bob S
     
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