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Unbranded 10F Supercapacitors 2.7V - leakage

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Undweeber:
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.

amspire:
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

thm_w:
What is the actual leakage?
As amspire is suggesting at:
- Measure the capacitor voltage
- Disconnect everything from the capacitor
- Wait a defined amount of time (say 15 minutes)
- Connect voltmeter again and measure voltage
- Calculate the leakage current

According to Murata, you might see 1-10uA of leakage: https://www.murata.com/~/media/webrenewal/products/capacitor/edlc/techguide/electrical/c2m1cxs-053.ashx
But 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.

Undweeber:
@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 slightly

I 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!

T3sl4co1l:
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

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