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Supercapacitor charging voltage?

  1. Apr 7, 2012 #1
    Is it safe to charge a supercapacitor from a source with more than it's rated voltage, provided that the voltage across the cap never exceeds the rating?

    For example, the cap is rated at 2.7V, and i charge it from two AA batteries giving just over 3.2V in open circuit.
    Once the battery, which gives vastly less current than the cap can draw, is connected to the empty cap, the voltage across it stays near zero, and is slowly climbing.

    What i can't figure out - is the "never-to-exceed" limit of 2.7V apply to the voltage across the cap, or the voltage of the source that charges it?

    In the example above, the first is true, but the second is not.
    How does it actually work?
     
  2. jcsd
  3. Apr 7, 2012 #2

    phinds

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    If your cap can draw enough current for long enough, it will burn out the battery since it will act like a short circuit across the battery.

    If the battery can supply enough current and not burn out, then it will charge the cap to 3.2 volts and if the cap can't stand that voltage, it will arc over and burn out.

    Best you learn more about this stuff before hooking anything up.
     
  4. Apr 8, 2012 #3
    So the important part is the voltage across the capacitor?

    If i can provide a reliable way of switching off the charging part before it goes over the voltage limit, would it be safe to use a source with higher than rated voltage capability?
     
  5. Apr 8, 2012 #4

    vk6kro

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    Yes, that would be OK.

    As long as the dielectric of the capacitor does not have to experience more than its rated voltage, it doesn't matter where the current comes from.

    In this case, most of the voltage is being dropped across the internal resistance of the battery.
     
  6. Apr 9, 2012 #5
    And it well may require hundreds of batteries in parallel where now is one to charge the cap.
     
    Last edited: Apr 9, 2012
  7. Apr 9, 2012 #6
    You can charge that super cap up to 3 V but the lifetime will rapidly decrease as you overvoltage the electrolyte (whatever it does to it). Excessive overvoltage won't actually make the cap arc but will boil the electrolyte instead, causing the cap to swell and eventually explode. Don't do that :). DC ESR of ultracaps varies with model and size but it will be in the range of mOhms -> an Ucap can sink or source thousands of amps (for a second or so).

    Source: work for ucap manufacturer
     
  8. Apr 10, 2012 #7
    That overvolting them is Bad Thing i know, just had a practical question of common-sense-applicability in this case.
    Got a variable power supply to charge them, so it no longer apply directly.

    So, in case you want to charge one with no special equipment and is careful, connecting a power supply with higher voltage and disconnecting it before the voltage across the cap would cross the limit should be safe?

    Nice, can you provide some concrete info or documents on the failure modes of big supercapacitors?
    I.e. what would happen if one got wedged between metal plates thick enough to conduct over 9000 amps the spec lists as the "maximum possible current"?
    What would happen if it is mechanically damaged?
    What would happen it is repeatedly subjected to 500-1000 amps while rated for 200 A continuous and 2000 in peaks?

    Etc.
    Realistic worst-case scenarios to be on guard against?

    Not exactly - one cheapest salt AA battery gets it up to 1 volt, then 2 AA NiMh batteries (2.4V) get it up to 2.2V (mind the heating).
    Haven't actually tried two regular AA batteries.
     
  9. Apr 10, 2012 #8
    Essentially all power supplies are current limited. Thus you can use them to charge ucaps. Just be careful when it comes to large (10 kW) devices. Those tend to have huge output capacitance and direct connection of two large capacitors results in a something that could be mathematically described as Dirac pulse current. It's loud and dangerous.

    Regarding the 9,000 A question - you could possibly *weld* the bar to the terminals. Nothing bad would happen but of the cell is really large (3000 F+), there could be enough energy dissipated that the cell would overheat and get damaged.

    Max currents are typically specced for 1 s.
     
  10. Apr 11, 2012 #9

    sophiecentaur

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    It's as well to calculate the total energy stored in all these cases. Also it's available 'all in one go' so it's much more of a potential problem. No prizes for burning your eyebrows off.
     
  11. Apr 13, 2012 #10
    That's what it says in the specs, but what about the currents between the max. continuous and max. pulse ratings?

    I.e. for BCAP3000 it says that at 130A continuous it will heat up 15C, and at 210A up 40C, both labelled as maximum continuous current.
    Then, 2200A is named as maximum peak current for 1 sec, and short circuit current is up to 9300A.

    As i understand it, the problem is the heating.
    Then, as the current go up it would:
    0-200A: dissipate enough heat to stay within operational temperature indefinitely
    200A-2000A: won't be able to dissipate enough heat and eventually overheat with "bad consequences"
    2000A+: heats up internally fast enough for immediate "very bad consequences" (rapid electrolyte boiling? There is something that looks suspiciously like a pressure release valve on it's side)

    If so, the question is what exactly would these "bad consequences" be, and how long should it cool off to avoid them?
    At 3000F one such cell can supply 2kA for about 3 seconds, comfortably above the rated 1 sec.
     
  12. Apr 13, 2012 #11
    Good questions but way too specific for a particular cell. At this point you will need to conduct experiments or perform a thermal analysis on a cell to determine appropriate current duty ratio so that the temperature rise is reasonable.

    Rated means guaranteed. At the end of cells life, the ESR can go up by 200 %. Hence 3x more heat.
     
  13. Apr 13, 2012 #12
    Ok then.
    Thank you all for the info.
     
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