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Capacitor discharge

Discussion in 'General Electronics Discussion' started by tokamak, May 2, 2011.

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  1. tokamak

    tokamak

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    0
    May 2, 2011
    Hi, I have a voltage profile that is sampled every 64 seconds.
    Its initial voltage is V_0, final voltage is V_f, and V_0 > V_f.
    I am assuming there is an instantaneous step down from V_0 to V_f, but the voltage response I see is an exponential decay according to:

    V(t) = V_f + ( V_0 - V_f ) exp( -t / RC )

    assuming it can be modeled as an RC circuit.
    V_0 and V_f are known, plus some points in between along the curve spaced out 64 seconds from each other. The time of the step down in voltage is unknown (but it is no more than a known quantity of time, 'x', prior to the first data point along the response curve); RC is unknown.

    Is it possible to calculate the time constant (RC) and the time of the step-down in voltage?

    here's a plot (keep in mind it is a line strung together with data points every 64 secs):
    [​IMG]
     
  2. tokamak

    tokamak

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    May 2, 2011
    See also

    This probably gives a better representation of what I'm dealing with:

    [​IMG]

    Thank you.
     
  3. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    Could you explain how a timescale of 50 hours in the first graph comes to be represented by a timescale of 500 seconds in the second graph?
     
  4. tokamak

    tokamak

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    May 2, 2011
    The 2nd graph is just a small portion of the 1st. Only including a few points along the curve of interest.
     
  5. poor mystic

    poor mystic

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    Apr 8, 2011
    I don't understand what I'm seeing.
    What are the circumstances of this measurement?
    What forces make the voltage charge and discharge? Why do you sample the voltage at such a long interval?
     
  6. tokamak

    tokamak

    6
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    May 2, 2011
    Well my question is more mathematical than hardware related, but this is from a spacecraft's telemetry system. Its data rate is very limited, hence the 64 second sampling frequency.

    It's for all intents and purposes an unknown circuit. I was hoping there was a way to determine the time constant, RC, even if the exact time of the trailing edge of the voltage step was not known.
     
  7. poor mystic

    poor mystic

    1,061
    30
    Apr 8, 2011
    If you know the rate of change of the discharge and the final voltage that will be reached there is a simple geometric solution to the problem of finding the time constant. I have illustrated the method in the attachment.
    The red line starts at origin at slope given by dy/dx at t=0. If memory serves me correctly, the time at which the red line crosses the asymptote is equal to the time constant. :_D
     

    Attached Files:

  8. tokamak

    tokamak

    6
    0
    May 2, 2011
    Thanks for the alternative method of determining the time constant. It doesn't seem feasible with the sparse data and lack of an accurate time=0, but I'll keep it in mind as I try to figure it out.
     
  9. poor mystic

    poor mystic

    1,061
    30
    Apr 8, 2011
    Hi tokomak
    I'd be quite interested to look at the complete data series including whatever you know about loads and charging details. I don't promise I could come up with anything but I'd like to take a look!
     
  10. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    Even though there is no accurate time=0 where Vo=max, any point on an exponential curve can be treated as time=0 where Vo=whatever V is at that point on the curve. No matter where you place time=0, it always takes one time constant to lose 63% of Vo. Of course when you are working farther down on the discharge curve, sampling noise can be as much of a problem as not knowing when the discharge initially begins.
     
  11. poor mystic

    poor mystic

    1,061
    30
    Apr 8, 2011
    Quite so, Laplace. And it follows that providing our curve is exponential, and providing that there is a Vfinal for our asymptote, we can always derive a time constant.
     
  12. tokamak

    tokamak

    6
    0
    May 2, 2011
    Thanks, I considered that and it didn't seem right at first, but after thinking about it more it makes complete sense that the voltage will always drop 63% from its current value after 1 time constant.
    Unfortunately, implementing that has now convinced me that either the time constant is not constant in my case, or Vf is slowly dropping also, or both, so this may not be solvable or not worth the effort.

    Thanks everyone for your input.
     
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