Coin cell supercapacitor degradation

Thread Starter

bonzer

Joined Oct 16, 2026
3
Hello guys,

Considering a backup CSK621 3.3V 0.2F supercapacitor for a load that would consume 5uA @30°C during a blackout. I was wondering after years or 10 years of normal operation at 30°C on average, what could be the minimum backup time if the main supply is suddenly gone? I tried to look into the datasheet but they always show a short time reference case of ±30% capacity degradation with 1000h of test @60°C. I tried to do some calculations but they are too generic and I don't know how far should I go with the worst case scenario. Could you please kind of guide me through that? I was also wondering, does the result depend a lot on the single sample?

I just want to use that component model as an example to figure out for the future how to deal in these conditions. In case you want to bring up another example, feel free if you think this would be easier for the explanation.

After 10 years or even 15 if possible, would be really nice to know what happens after such a long time within e certain decent estimation considering how long this time is.

https://www.elna.co.jp/wp-content/uploads/2022/05/0-tech-note_EDLC_e.pdf
https://www.elna.co.jp/wp-content/uploads/2022/05/0-tech-data_EDLC_e.pdf
 

Thread Starter

bonzer

Joined Oct 16, 2026
3
Thanks for the answers. It is not critical but some notifications would rely on the clock settings related to the heat pump, even though the "reset" clock window would appear on the HMI in case even the backup power of the supercap goes off, having at least a clear idea of when would it go to zero like for replacing would help. I believe replacing the supercap after 10 years could be enough. I think that according to how they are designed they happen to be easily replaceable.
 

MrChips

Joined Oct 2, 2009
35,247
The backup cell has to supply 5 μA @ 3.3 V.
That puts the load at 3.3 V / 5 μA = 0.66 MΩ
RC time-constant = 0.66 MΩ x 0.2 F = 132,000 s
Using 0.1 RC, run time = 13,200 seconds ≅ 3.6 hours
 

kiroma

Joined Apr 30, 2014
272
The backup cell has to supply 5 μA @ 3.3 V.
That puts the load at 3.3 V / 5 μA = 0.66 MΩ
RC time-constant = 0.66 MΩ x 0.2 F = 132,000 s
Using 0.1 RC, run time = 13,200 seconds ≅ 3.6 hours
In this case the self-discharge would be far greater than the consuming.
 

kiroma

Joined Apr 30, 2014
272
1792413057976.png
It doesn't specify if there's any special quirk of it.
But there's graphs for the 0.22 F version:
1792414065862.png
5 µA would be a displacement of the 10 µA to the right the same spacing as 200 µA to 100 µA (half of it). So it should sit somewhere near 2500 minutes to reach 0 V for the right curve. Better calculation would be necessary if you need accuracy, don`t trust my eye and mind pair.

Edit: change image because I had put 2 models, but TS is using only 1.
 

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Thread Starter

bonzer

Joined Oct 16, 2026
3
Thanks for your answers. I forgot to say that the RTC would survive up to 2V from 3.3V. In the past I've already come across the graphs brought up by Kiroma, thank you for posting them, but I didn't know how to apply them to my case because I believe the capacity would also be a lot smaller after 10 years thus that time I guess would become even shorter. I think it would be even less than 1 hour in the end which I could afford but 0 would not be good.

to BobTPH, it wasn't me chosing it, I guess it had to do with the small size.
 

kiroma

Joined Apr 30, 2014
272
The information you have that you can rely on is this:
1792498241534.png
So in the worst-case scenario, it's about 12% reduction in capacitance. As energy is proportional to capacitance, you'll have 12% less energy. And as you're consuming constant 5 µA, it means 12% less time.
From the graph from my post (#8), I'd put some room to have certainty of other various varying parameters and have certainty that 1000 minutes will be handled.
Is that enough for you for the worst-case scenario? Do you have space for another supercap if that's not enough? Maybe it can be of a smaller capacitance.
 

kiroma

Joined Apr 30, 2014
272
Forgot to mention this graph posted here is the endurance test of 85 °C. If you look closely, they measured 3 points, and it seems it settles down, it converges, as a RC/RL circuit does (it's probably going to converge on -20% or so). And I think this is extreme, 85 °C at 5.5 V is the limit, I'd expect to vary much less in good environment.

Do you have to guarantee the time, or just display it on the manual?

1792503580475.png
This equation tells that for every 10 °C drop, it will survive 2 times more. So, from 85 °C to 35° there's 50 °C, so 2^5, which is 32 times more.
32 times more than 1000 hours is 32000 h, which is about 3.65 years to reach that point of 85 °C and -12% capacitance.
Also, 32 times the lifetime expectancy is unknown to me, since I haven't found the L0 parameter in any of these datasheets.
 
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