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Super Capacitor Charging/Discharging

  1. Nov 28, 2016 #1
    Hello, cap.jpg

    I am designing a phone charger with super capacitor with capacitance C=100F. The max voltage of cap is 2.7 V. When I charge the capacitance and then charge my phone, it is charging really slowly with low current (around 700mA). However, as I read, capacitance should output high current. The other thing is, when I charge my capacitor while phone is connected to it, it is charging itself then charging my phone slowly. Please let me know how I can make my cap output more current to charge my phone. The schematics are below. I used both . Please share your experience and give advice.
     
  2. jcsd
  3. Nov 28, 2016 #2

    phinds

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    No, not at all. A capacitor in an open circuit output no current at all. A cap in a high impedance circuit outputs very little current. A cap in a very low impedance circuit outputs high current (briefly).
     
  4. Nov 28, 2016 #3
    Yes you are right. Supercap has a low ESR .
     
  5. Nov 28, 2016 #4

    phinds

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    It has nothing to do with whether it is a "super" cap, it just has to do with the fundamental characteristics of caps.
     
  6. Nov 28, 2016 #5

    berkeman

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    I'm not sure all supercaps have low ESR. Check the datasheet to be sure...

    https://en.wikipedia.org/wiki/Supercapacitor
     
  7. Nov 28, 2016 #6

    phinds

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    Sure, but what I'm trying to get him to realize is that the ESR might well be a red herring in this case, because he doesn't seem to understand Ohm's Law and expects a high Farad cap to produce lots of current regardless of what it's hooked to.
     
  8. Nov 28, 2016 #7

    Averagesupernova

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    I believe most super caps have a relatively high ESR. The trend has been that the ESR is dropping, but from one cap to the next super caps do not typically have a very low ESR.
     
  9. Nov 28, 2016 #8
    When I said low ESR, i compared them to batteries so caps outputs higher current. Please give a feedback the circuit I am using to charge my supercap.
     
  10. Nov 28, 2016 #9

    Averagesupernova

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    Ok, the circuit you are using to charge the cap will destroy it. There is nothing to prevent the voltage from exceeding 2.7 volts when being charged from a 5 volt source.
     
  11. Nov 28, 2016 #10
    Oh sorry, I forgot to mention that I am using two of caps connected in parallel. SO they form 5.4V together.
     
  12. Nov 28, 2016 #11

    Averagesupernova

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    No they don't.
     
  13. Nov 28, 2016 #12
    I meant in series connection.
     
  14. Nov 28, 2016 #13

    Averagesupernova

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    Probably best to draw it that way then. And do you have anything to guarantee that they will charge equally? What is the spec on the caps for tolerance? If their capacitances differ then one will charge to a higher voltage than the other.
     
  15. Nov 28, 2016 #14

    berkeman

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    That's probably not a good idea. As ASN points out, there are symmetry issues with it, and the way you usually handle that is with resistors in parallel around the caps, which robs charge and power in the overall circuit. Do they not make supercaps with a rating > 5V?
     
  16. Nov 28, 2016 #15
    There are caps with 5 v but with lower capacity.
     
  17. Nov 28, 2016 #16

    berkeman

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    And what is the straightforward way to increase that capacity and keep the 5V (plus margin) rating?... :smile:
     
  18. Nov 28, 2016 #17
    I gues adding the caps in paralel.
     
  19. Nov 28, 2016 #18

    NascentOxygen

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    A separate issue to consider: when you power anything from a capacitor that was initially charged to 5V, after the capacitor has given up just 10% of its charge then its voltage will have dropped by 10%. So, I'm wondering is 4.5V still going to be any good for charging a phone?

    Or maybe you have this solved .... ?
     
  20. Nov 28, 2016 #19
    4.5V is giving out about 800ma to charge but it is varying (going down). But I think when the current is over 600ma ,the phone can be charged
     
  21. Nov 28, 2016 #20

    Averagesupernova

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    An interesting thing to note is that you were planning on putting higher value caps in series. And you avoided the higher voltage caps because they had smaller capacitances. Do you know what happens to total capacitance when capacitors are placed in series?
     
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