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

 

After 10 years or even 15 if possible,

That can be difficult to predict accurately without actually testing it for 10 or 15 years.

I think the 30% degradation for the 1000h test @60°C would be a reasonable value after 10 years @30°C, but that's just a WAG.
Don't know where you would find a more definitive value.

 

If the device is nothing critical, unplug the power and see how long the backup lasts. Or if you want peace of mind - replace the cap.

 

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.

 

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

 

The first question would have been "how critical is the backup protection?" I have come across capacitors that hold a charge for quite a while.

 

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.

 

It doesn't specify if there's any special quirk of it.
But there's graphs for the 0.22 F version:

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.

 

So now you only have to discover, to be accurate, how low the voltage can go while maintaining the data. It should be in the RTC spec.

 

Why did you choose a super capacitor over a coin cell?

 

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.

 

The information you have that you can rely on is this:

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.

 

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?


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.

 

I think 1000 minutes would be great. Thank you very much for your answer. Sorry for the late response.
Displaying it on the manual should be enough, it's not necessary to guarantee that.
Do you think that there may be supercaps that are more compatible with such a function without being too expensive?

 

For the 0.2 F capacitor, the ideal calculation is

t = C x (Vinitial − Vminimum) / I

So, for example, from 3.3 V down to 2.0 V at 5 micro A, the nominal 0.2 F gives about 14.4 hours. Using the −30% capacitance limit gives about 10.1 hours.

However, the 60 degree C/1000 h endurance test does not mean that −30% will occur after 1000 h at 30 degree C. It is simply the manufacturer's specified endurance test condition. Extrapolating that result to 10–15 years requires an actual degradation model, including temperature and voltage acceleration data.

There is another important issue here: capacitor leakage current. With only a 5 microA load, even 1–2 microA of leakage materially changes the backup time. You therefore cannot safely calculate this from capacitance degradation alone.

 

There is another important issue here: capacitor leakage current. With only a 5 microA load, even 1–2 microA of leakage materially changes the backup time. You therefore cannot safely calculate this from capacitance degradation alone.

That's why we use the datasheet's graphs to discover how much time it will take to discharge, given a load and the self discharge. These graphs include it, and it's the only way of getting to know what you need, since there's no other mention to something related to self discharge that we can use to our benefit.

 

Do you think that there may be supercaps that are more compatible with such a function without being too expensive?

Supercaps are leaky. The better ones are expensive, for what I know. If I were you, I'd buy other supercaps (near the same value of capacitance) and test them under the same load condition. It would give the answer to if it can provide longer supply to the application. You don't need to implement them on the PCB if it's not worth the price, but buying 3 is somewhat cheap from a development perspective. Also, test at least 3 of them, to account for deviations from the typical value.

Another way of testing, but this shouldn't make you don't test the first one, is to charge the supercap to a middle value (with a stable/regulated power supply) of operation and putting a amperemeter (µA range) and let it settle. After 1 hour or so it should be stabilized. The value you read on the multimeter is the leakage, so, the lower, the better. This way you can evaluate which supercap is better and by how much.

 

An additional suggestion is to form the new capacitor by connecting it to the rated voltage supplied by a separate supply, so that it can do its initial forming on cheaper power, instead of on expensive batteries. That may not be possible, though.

 

An additional suggestion is to form the new capacitor by connecting it to the rated voltage supplied by a separate supply, so that it can do its initial forming on cheaper power, instead of on expensive batteries. That may not be possible, though.

Very unlikely that it's possible. Never seen such scenario. Nowadays 90%+ of products that have batteries are rechargeable, the common li-ion. The energy might be critical, but it's certainly cheap, because you have to charge it someday.

 

Very unlikely that it's possible. Never seen such scenario. Nowadays 90%+ of products that have batteries are rechargeable, the common li-ion. The energy might be critical, but it's certainly cheap, because you have to charge it someday.

OBVIOUSLY you shop in places different from where I am. NOTHING CHEAP about anything rechargable.