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Fastest way to charge a super cap? — Parallax Forums

Fastest way to charge a super cap?

mctriviamctrivia Posts: 3,772
edited 2010-04-28 13:07 in General Discussion
I would like to charge a cap to 5V as quickly as possible from a 24V battery.

My thought was to use a 3A switching regulator to get down to 5V
Use a 3ohm 1W resister in series with the 0.33F cap.

This resister would be over spec for a bit:
0.0s: 800%
0.2s 560%
0.4s 370%
0.6s 250%
0.8s 170%
1.0s 110%

Could the resister handel this abuse for such a short time?


Now if I placed a 0.82ohm resister with a poly switch set to trip at 2.5A it would start to conduct after 0.9s from the start causing the cap to reach 4.5V at about the 1.5s mark


This scheme would cost about $2 not including voltage regulator. Any problems with it? suggestions for better solution? Keeping in mind space and money is in short supply.

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Comments

  • James NewmanJames Newman Posts: 133
    edited 2010-04-28 01:48
    Your probably safe with the resistor, just make sure your supply components can handle the load.
  • Martin HodgeMartin Hodge Posts: 1,246
    edited 2010-04-28 03:50
    Would a MOSFET or IGBT with a zener controlling the gate, powered directly from the 24v work? I've seen it done this way in photographic flash units. When the capacitor voltage reaches the zener threshold the [noparse][[/noparse]insert silicon here] switches off.
  • mctriviamctrivia Posts: 3,772
    edited 2010-04-28 04:29
    You would burn a lot of power off in the form of heat. I think my hair brain scheme probably will not work. the poly switches have a habit of staying on for a bit before initial turn off. this delay time is probably to long and would cause the power supply to cycle on and off slowing the charge time. 28.7R 1W is a safe method but takes 30s

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  • Phil Pilgrim (PhiPi)Phil Pilgrim (PhiPi) Posts: 23,514
    edited 2010-04-28 07:01
    You'll get a faster charge using a constant-current source than you will with a constant-voltage source and a current-limiting resistor. With the former, the cap voltage will rise linearly to full capacity; with the latter, the charge rate tapers off the closer it gets to a full charge. A comparator on the cap can be used to turn the current source on and off.

    -Phil
  • mctriviamctrivia Posts: 3,772
    edited 2010-04-28 12:15
    you know i must have been drinking last night(don't have any so not really possible). for some reason i was thinking since my power supply put out 5V i can't afford the voltage drop across a transistor. However I can just increase the powersuply voltage to 5.7V and problem solved. Yes a constant current source would be the fastest way and would probably be cheaper also.

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  • TubularTubular Posts: 4,718
    edited 2010-04-28 12:18
    Presumably for 5v you have 2 cells in series. You may need some cell balancing if you're going to repeat the charge cycle lots of times.

    Whats the internal resistance of the Caps?

    I think Phils approach is a good option, though I have to admit I've used a 5w resistor to limit the charge current. I'm not proud of that.

    You need to check the failure modes for the particular cap you have in mind, but in general supercap failures are benign. "venting" or going open circuit, unlike big electros. Have a read here

    What is the failure mode of a supercapacitor?
    If a supercapacitor is exposed to excessive voltage or temperature for extended periods it gradually degrades to essentially an open circuit. The time taken for this to occur depends on how much over voltage or over temperature is applied. Other than physically puncturing the supercapacitor there are no short circuit failure modes. There are no catastrophic failure modes.
    from www.cap-xx.com/resources/faq.htm
  • mctriviamctrivia Posts: 3,772
    edited 2010-04-28 12:27
    The cap is rated for 5.5V. If I remember correctly there is 2 caps in series inside the package. I realised for my application a 28.7R 1W resister with a shotkey diode in reverse will be perfectly fine. It would take 16s from initial startup to charge to minimum voltage needed to protect the circuit from brown outs but would be very cheap and not take up to much space on my board.

    raising the power supply voltage to 5.7V and using a schematic like this:
    Op-amp_current_source_with_pass_transistor.png

    would definetly be fastest though.

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  • Dr_AculaDr_Acula Posts: 5,484
    edited 2010-04-28 12:58
    The *fastest* way is to put as many volts on it as possible. I looked at a few specs and they have internal resistances from 30 ohms to 120 ohms, so the model for that is an ideal capacitor in series with a 30 ohm resistor. Heat will be dissipated the same as a resistor. Say you use 30V and the internal resistance is 30R then 1 amp flows and if the cap is 1F, then roughly 1V per second so 5 seconds to charge. W=I^2*R so that is 30W and if you look at the size of a 10W resistor maybe you might get away with that for a few seconds but I wouldn't count on it. The cap will be getting more than warm. It may well blow up.

    With either constant current charge, or charge via a higher than 5V (which due to the internal resistance tends to approximate constant current charge) there had better be a good cutout mechanism when it is charged at 5V.

    I wouldn't be game to do this at all. I'd be using a constant voltage 5V regulator and a dropping resistor.

    There may not be a need to use a 3A switcher. If you have a standard 7805 rated at 1A, and an internal resistance of 30R, the most current that will ever flow is 1/6th of an amp (5/30). So there is no point having a regulator that can supply more than this. So this saves some money too. Bog standard 7805 and if you like, maybe a protection resistor 10R or something in series.

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  • mctriviamctrivia Posts: 3,772
    edited 2010-04-28 13:07
    I never quiet understood the super cap internal resistance rating. For example I have seen caps listed with 10R internal resistance but with a short current of 4000A. Ohms law on 3V says this is impossible if the internal resistance rating is an ideal with a series resistance.

    I have a few super caps lying around that I can abuse so I will try placing one on a power supply with a tiny fuse to see if the current draw from empty will exceed the predictable current draw from the internal resistance.

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