Supercap internal resistance & discharge speed

Discussion in 'General Electronics Chat' started by RogueRose, Apr 21, 2016.

  1. RogueRose

    Thread Starter Member

    Oct 10, 2014
    189
    4
    These super and ultracaps seemed to make a lot more projects possible if they work the way they seem to be explained.

    These are all in the $4 range which I am having a tough time believing their ratings. One is made by Maxwell the other 2 by Nesscap.
    5v 1.5F 143mOhm
    5v 2.5F 69mOhm
    2.7v 15F 30mOhm

    2.7v 3000F .29mOhm $60 Maxwell 2.4" diam x 5.5" high
    2.7v 600F .64mOhm $$ ?? Nesscap 2.4" x 1.1"

    I'm very skeptical of these numbers, not just because the top 3 are under 1" by 1" in size but because of the pricing. I'm looking at making a capacitive discharge welder and the ideal plans call for 4F @ 14-16v so a number of welds can be made without a "long wait" for recharge. I've looked at 16v electrolytic caps and it would cost quite a bit for 1F total let a lone 4F so I'm trying to see where the catch is here.

    I'm also unsure of the ESR and how that effects the cap. Does a lower number mean a faster discharge rate? Again these numbers seem pretty incredible as well.

    Can anyone verify the use of these and how they compare with traditional Aluminum electrolytic caps?
     
  2. wayneh

    Expert

    Sep 9, 2010
    12,154
    3,061
    Yes, but you need to pay attention to the specs for peak current. I think for your application the ESR is not terribly important. When you have large currents at high frequency, the internal resistance matters much more. If I understand your application, the current may be large but it's intermittent. You're not charging and discharging at 100kHz.
     
    Roderick Young likes this.
  3. RogueRose

    Thread Starter Member

    Oct 10, 2014
    189
    4
    Thank you, that is correct. It also seems that the ESR is dependent upon frequency. I was reading that the measurements in the mOhms are usually at 100kHz so that makes a big difference and seems to be an irrelevant number (from my current knowledge level, lol...). I was looking at those caps as being able to discharge their entire capacitance in milliseconds which seems entirely unlikely for the size and price - I'm assuming that isn't possible.

    So is the ripple current the important factor in how quickly caps discharge if looking for very quick pulses of high power? Is that the same as ripple current? I guess what I need to know is what value determines how quickly a cap can charge and discharge.

    If I have a 15v .25F cap (~28 joules) and I want to discharge in .1 seconds what would spec would I need to look for?
    1 watt hour = 3600 joules :-: 1 watt second = 1 joule :-: 28 watt (secs) / .1 = 280 watts/second 280/15v = 18.6Amps.. Is this correct?

    So I would have to find a cap(s) that had capable of this but what is the rating I am looking for?

    I see so many caps listed on Mouser & Digikey and they often don't have amperage listed even for the large "can" caps.
     
    Last edited: Apr 21, 2016
  4. wayneh

    Expert

    Sep 9, 2010
    12,154
    3,061
    The load is or should be the rate limiting part of the circuit. The caps will be a factor but hopefully a small one.

    What do we know about the load?
     
  5. tsan

    Member

    Sep 6, 2014
    41
    4
    Internal resistance makes rc circuit with the capacitance. I took the second capacitor (5v, 2.5F, 69 mOhm) as an example. For 15 V three capacitors are connected in series. Capacitance is 2.5/3=0.83 F. Resistance is 3x0.069 = 0.27 ohm. Rc time constant is 0.83x0.27 = 172 ms. This is without external resistance. Could be that four capacitors is required in series connection.
     
    RogueRose likes this.
  6. RogueRose

    Thread Starter Member

    Oct 10, 2014
    189
    4
    Well that it is usually different for every application be it the item being welded, the welder design or be it a different project. That is why I was I wrote out the energy equation/comparison to see if I was looking at it properly.
     
  7. wayneh

    Expert

    Sep 9, 2010
    12,154
    3,061
    Just double checking.
    Stored energy = 1/2CV^2 = 1/2•0.25•(15^2) = 28J
    Stored charge Q = CV = 0.25•15 = 3.75 coulombs
    Discharging 95% of that in 0.1s requires an average amperage 3.75•0.95/0.1 = 35.6A
    However the peak current will be much higher and towards the end will be much lower.

    Your analysis above failed because you cannot use a constant 15V as the average discharge voltage. 15V is the starting peak. The voltage (and current) follow the RC decay curve.
     
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