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Solar cell charging a super capacitor - need an equation

  1. Aug 14, 2012 #1
    I have a very small solar cell that outputs 4 volts (open circuit) and 50 μA (short circuit). I'm trying to charge a 0.33 F capacitor that is rated for 5 volts. I'm looking to understand the charging relationship.

    1. What equations govern the charging of this capacitor? I'm already aware of T = R x C; where T = time in seconds; R = resistance in ohms; and C = the capacitance (in Farads).

    2. Assuming I want to use the charged capacitor to power a small circuit, how can I convert the stored charge to something like mAh?

    3. If I do not use a resistor between the solar cell and the capacitor, using the equation T = R x C, wouldn't the time to a full (4 volt) charge equal zero seconds?

    Thanks! [Edited to correct the capacitor value.]
     
    Last edited: Aug 14, 2012
  2. jcsd
  3. Aug 14, 2012 #2

    davenn

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    greetings Alligator

    330uF isnt a super capacitor, you sure its not 0.33 Farads or 3.3 Farads etc ?

    330uF wont run any circuit for any time

    cheers
    Dave
     
  4. Aug 14, 2012 #3
    Sorry, I should have said .33 Farads. I changed it above.
     
    Last edited: Aug 14, 2012
  5. Aug 16, 2012 #4
    Nobody knows this, really?
     
  6. Aug 16, 2012 #5

    uart

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    Up to about 60 or 70 percent of open circuit voltage the solar cell will charge close to constant current. At a constant 50uA the dV/dt is 0.15 volts per ksec, so you'll get to about 2.5 volts in about 17 ksec, which is nearly 5 hours.

    This ignores capacitor leakage current, so it might take longer (do you have any specifications on leakage current?).

    Above about 2.5 volts the charging will slow down. The time to go from 2.5 volts to 3.5 volts will probably take another 2 to 3 hours (guesstimate). Charging to the open circuit voltage is asymptotic, so you'll have to wait a long time if you want to get very close to 4.0 volts.
     
    Last edited: Aug 16, 2012
  7. Aug 26, 2012 #6
    Hello uart

    How do calculate the 0.15 volts per ksec? I have this cpc1822 solar cell that outputs 4volts at 50uA. So I was just wondering how you calculated this, so I can figure this out with other capacitor sizes. I know this seems like a newbie question, but calculus wreaks havoc on my brain :)

    Thanks!
    NeoTesla
     
  8. Aug 26, 2012 #7

    uart

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    Hi NeoTesla. It's just a simple application of [itex]dv/dt = i/c[/itex].

    At 50 uA and 0.33F you get the rate of change of voltage is dv/dt = 0.00005/0.33 volts per second, which turns out to be about 0.15 volts every 1000 seconds.

    This relation will only work while the PV-cell provides approximately constant current, typically up to somewhere around 2/3 of it's open circuit voltage. After that you'd have to use numerical techniques (with a model of the cell's V/I characteristics) to accurately model the increases.
     
  9. Aug 26, 2012 #8
    Cool that's easy! Thanks for your help!
     
  10. Aug 27, 2012 #9

    NascentOxygen

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    The leakage current lost in your electrolytic may be comparable to your solar cell's output.
     
  11. Aug 27, 2012 #10
    Yikes! I'll never get off the ground with a leaky cap. The cap I'm using has a leakage current of 6uA, so will I subtract this from the 50uA and then calculate?

    Thanks
    NT
     
  12. Aug 27, 2012 #11

    NascentOxygen

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    Is 6μA nominal, or have you measured it? Is it almost irrespective of voltage? Regardless, yes, it's lost from your calculations.

    An experiment you could try: you may be able to lower the leakage of your supercapacitor if you connect 2 AA's across it for a few days, then discharge it (slowly), then use it for your charging measurements. I don't know if this will work, just a possibility it may.
     
  13. Aug 29, 2012 #12

    NascentOxygen

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    More informative: leave 2 AA's connected across your capacitor for a few hours, then connect your multimeter (set on current) in series with the capacitor and reconnect the AA's. When the current has settled down to a steady level, that is your leakage current at 3 volts.

    You could repeat this with 1 AA, then again with 3 AA's, and you would then know exactly what leakage current to allow for when trying to predict how the capacitor would charge from your PV cell.
     
  14. Aug 29, 2012 #13

    uart

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    The 6 uA is probably at max voltage and temperature. It could be significantly lower under the given conditions, but it wouldn't hurt to measure it.

    The better super capacitors only have leakage current of the order of around 1uA per F.
     
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