Unbranded 10F Supercapacitors 2.7V - leakage - Page 1
Forum section: electronics / beginners

author: Undweeber
author_rank: Regular Contributor
timestamp: September 20, 2019, 11:19:52 pm
content: I charged it at 1A in 17 seconds, but then it started to drop voltage. These supercaps do not have any brand on them apart from specs, came from you know where.. I got 6 of them and already soldered one of them to a 16mm x 35mm calculator solar cell (Assume its amorphous silicone, its from casio).It looks really cute, the 10F standing and holding a small solar cell, when the sun comes out tomorrow ill see how long it takes to charge, but what worries me is that the cell will not be able to charge it faster than it LEAKS, is there a way to stop the leakage, or is this inevitable with this OneHow Low brand, is there the same issue with the Maxwell made 10F supercaps? I have Maxwell 350F and its so big you don't really notice it, definitely negligible at that size, but takes 15 minutes to charge from 0V to 3.8V, meanwhile this takes 17-20 seconds (I used Apple's 1A brick)I thought it would be fun to have a photoresistor controlled LED that would glow at night, sort of like a garden light, just not relying on batteries that will die.

author: amspire
author_rank: Super Contributor
timestamp: September 20, 2019, 11:48:25 pm
content: Are you actually measuring the current charging the capacitors?I am not sure what you mean by an Apple 1A brick. If you are using an Apple 5V 1A supply, it is not really designed for charging supercaps. Initially, the supercap appears as a short, and I have no idea how the Apple designers decided to handle a short. A 1A power supply may limit at significantly over 1A. Some power supplies have foldback current limiting which means that a 1A supply with a short on the output may be delivering only 100mA.If you have a genuine 1A constant current supply, then a 2.7V 10uF capacitor would charge in 2.7 x 10/1.0 seconds = 27 seconds. 17 seconds at 1A constant current would mean you have a capacitor that is actually about 6F. If the capacitor has a high leakage, then the actual capacitance would be much less then 6F.The 350F 3.8V capacitor at 1A would charge in 3.8 x 350/1.0 seconds = 1330 seconds (just over 22 minutes).If the voltage is dropping quickly on a supercap, that is a real problem. A quick voltage drop with no load would mean that there is significant heat developing in the core of the capacitor, and that means it will not live for long. Do you have anything connected to the capacitor as the voltage is dropping other then a multimeter?Richard

author: thm_w
author_rank: Super Contributor
timestamp: September 21, 2019, 12:57:27 am
content: What is the actual leakage?As amspire is suggesting at:- Measure the capacitor voltage- Disconnecteverythingfrom the capacitor- Wait a defined amount of time (say 15 minutes)- Connect voltmeter again and measure voltage- Calculate the leakage currentAccording to Murata, you might see 1-10uA of leakage:https://www.murata.com/~/media/webrenewal/products/capacitor/edlc/techguide/electrical/c2m1cxs-053.ashxBut maybe the large caps are worse.Their method is charging the cap and measuring the current, but you'd need a quite reasonably sensitive meter for this.

author: Undweeber
author_rank: Regular Contributor
timestamp: September 21, 2019, 01:47:28 am
content: @amspire,Yes Apple 1A 5V, it does auto-regulate current, I noticed that at the start when it detects it as a short, it seems to be giving bursts of voltage, and my analog multimeter on the lowest voltage was showing that very clearly, it was struggling this way until it got to 0.4V, after 0.4V the charging sped up slightlyI am not sure if those calculations are the correct way to measure the capacitance, the Maxwell is genuine with serial numbers, its rated 2.7V but I charge it to 2.8V.10F one charges in 20 seconds, unfortunately, I have no idea how to test the capacitance in a better way, charge times are not really a good indicator I would say.The voltage on the 10F drops too fast, like if I had to guess it would be 1V/30-40 minutes, I would have to test and see, and yes nothing was hooked up to them apart from the charging USB cable which was not plugged into the outlet of course.@thm_w,will do, thanks for the PDF!

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 21, 2019, 02:15:56 am
content: Leakage cannot be measured on short time scales. While all capacitors exhibit dielectric absorption, supercapacitors practically harness it as a dominant mechanism.The physics behind it is, the electrodes are porous (activated charcoal), and the dielectric is an ionic electrolyte soaked into those pores.It simply takes a long time for any charge to reach the deepest pores. Charges move by ionic diffusion, so the impedance also has a diffusion characteristic, Z ~ sqrt(F).It takes about a week for charges to equalize enough to be able to measure leakage current.Generally, leakage increases exponentially in the 2.2-2.7V range, with gas evolution (electrolysis) occurring at higher voltages (the 2.7V limit is due to breakdown of the electrolyte itself; do not exceed!).The important questions to answer will be:- Is this unknown capacitor actually the value it states? (Again, capacitance needs to be measured at very low frequencies, otherwise you aren't getting the capacitance of the full active electrode surface area.)- Is the diffusion acceptable? (Does the leakage stabilize at a shorter time scale, say a few days, or even hours, or does it continue to fluctuate over weeks?)- Is the leakage acceptable at typical voltages and room temperature? (Say 2.5V for 1 week, 25°C.)- Is the leakage acceptable at rated voltage and temperature? (Say 2.7V and 65°C or whatever.)Probably the leakage at 2.7V sucks regardless; brand name types aren't rated for anything remarkable (though they also take the measurement at 3 days, where diffusion may still be relevant). It's how good the capacitor is, at lower voltages, and even more importantly, in typical use, that determines its suitability.Tim

author: Undweeber
author_rank: Regular Contributor
timestamp: September 21, 2019, 03:02:58 am
content: @T3sl4co1l,wow thanks for explaining everything, learned a lot of new things, I noticed that charges voltages are not real, they seem to "dampen", like I shorted out the 350F cap and got it down to 0.4V, and hooked it up to charge again, but the charge IMMEDIATELY jumped up to 0.8V, so i shorted it again to maybe 0.35V, plugged the charger back in -- and it shot up againit sounds like the capacitors i have might have the true capacity written on them, the problem is i am not slowing down the charge to allow the electrolyte to saturate (?) am I getting this correctly? Like if I want full capacity I should charge it slower or at different specific amperages, (to be honest for my experimentations its not very critical)as per voltage dropping, I will conduct a test tomorrow, it will be an 18 hour long test, I don't really want to leave 350F charged, it can discharge like 50A and turn any metal white hot, the 10A just makes a spark and most of the charge is gone

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 21, 2019, 04:56:59 am
content: Doesn't much matter how you charge it, as long as you deliver the charge you want without overvolting it.Say you use a CC/CV supply set for 1A and 2.5V. If the cap is fully discharged, then when charge begins, voltage will step up by ESR * 1A (not too much, short-term ESR is usually reasonably small on these; assuming they're the power kind, not the memory-backup kind), and on top of that, ramp at about 1A/10F = 0.1 V/s. So, it'll reach nominal voltage in around half a minute. Maybe shorter than this, because again, effective capacitance is lower on shorter time scales.Once nominal voltage is reached, it sits there, and current decays gradually as charge soaks in. It should drop to, say, 100mA after some minutes, 10mA after some hours, etc.Or if it's set to a lower charge current, more charge will be absorbed on reaching nominal voltage; the effective capacitance is higher at the slower charge rate. And obviously, by the time it reaches nominal voltage, the absorption current is at most the charging current. Basically, it's already further down the soaking curve. Which is to say it's absorbed more charge already, which is another way to say the effective capacitance is higher.Variable capacitance probably sounds weird. Capacitors are usually introduced as a fixed quantity. It's still some total capacitance period, but you can only measure that at very slow rates (microhertz). It's one of those thermodynamics versus kinetics problems. Thermodynamics is what happens at infinite time ("thermostatics" might make more sense, but it's "dyn" because it deals with energy, go figure.) Kinetics is what happens in motion (finite time).Think of it as an equivalent circuit of many R+Cs connected in parallel, of different time constants. The total is the rating, but each individual piece may take days to charge (large RC constant). Or a ladder network where there's some immediate C, and in parallel with that an R+C. And in parallel with that C, an R+C, and... So you can think of it as a really long low-pass filter, with no output, you're just looking at it from the input end.Tim

author: Undweeber
author_rank: Regular Contributor
timestamp: September 22, 2019, 04:42:17 pm
content: Thanks again for the useful info Tim,I have conducted the tests, one cap was connected to the solar panel from calculator, another was just charged, the one with solar panel kept charge better than one disconnected, the panel was facing the dark so it wasnt charging, after 8 hours i stopped the test and left the solar panel pointing towards my table lamp, in the morning the voltage remained unchanged, so the panel kept being topped off.In the morning I discharged it by shortening down to ~0V and burning my finger, and left it to charge on the sun, after 2 hours on direct sunlight it is charged to 1.8V exactly, 0.9V to go (the sun started low and only recently got the full noon sunlight), the cap is spec 10F.UPDATE:It took almost all day to charge, and it seems to have stopped charging at about 2.7V I cannot tell for sure because its an analog multimeter, maybe it does have resistance above the saturation level.Anyway, now I can proceed to building a night light with it using light sensitive resister and joule thief circuit.