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Determining maximum runtime: 17.6 to 20.2 years (CR2032)

A project log for TritiLED

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Ted YapoTed Yapo 01/06/2018 at 21:3424 Comments

OK, so for the coin cell contest, we want to know the maximum runtime.  I have considered estimating this three different ways.  As an example, I'll take the minimum pulse rate of the V3.0, 0.5 Hz (the light blinks once every 2 seconds).  I measured the current drain to be around 1.15uA at this rate. 

How long will this run on a CR2032?

First, you need to decide on the cell's capacity.  I'll use 225mAh, which is on the Panasonic CR2032 Datasheet, since I'm using Panasonic cells.  Other manufacturers quote other numbers.

Capacity / current

In this method, which you see most commonly used, you divide the capacity of the cell (in Ah) by the current drain of the device (in A), to yield a lifetime in hours. To get years, we divide by 24 hours/day, and 365 days/year (if you want to get technical, you can use 365.25).  In this case, we get:

0.225 / 1.15e-6 / 24 / 365 = 22.3 years

Nice, but this exceeds the "shelf life" quoted as 10 years. 

WTF is shelf life, anyway?  It certainly doesn't mean the battery is dead at that point if left on the shelf.  Instead, it's a measure of self-discharge.  Elsewhere in their Lithium Handbook, Panasonic quotes a 1% self discharge rate per year.  So, at the end of the 10 year shelf life, the cell should retain more than 90% of its initial capacity (depending on 1% of what: see below).  Considering this, the 22.3 years calculated above doesn't seem correct anymore - at that point, the battery would have lost some of its initial capacity sitting on a shelf.  It would be nice to take this into account.

Considering Self-Discharge

To include the effects of self-discharge, I first assume the 1% per year means 1% of the remaining capacity.  This means the capacity obeys the differential equation:

where s is the self-discharge rate, in this case 0.01/year.  The solution of this equation is an exponential decay describing the capacity of a battery on the proverbial shelf, which is interesting, but doesn't answer the question.  To figure the runtime, we also need to include the circuit's current drain.  For the moment, assume the drain is a constant current of d amps.  The equation now looks like:

In 2018, there's no need to be intimidated by (simple) differential equations like this: just throw them into Maxima.

I got maxima to solve the ODE, selected a specific solution by giving the initial conditions (fresh battery at t=0), and finally solved the result for the time when C=0.  The result: 20.2 years.  Two years shorter than the earlier estimate. (Note: earlier, I posted 19.4 years here due to a typo in the maxima code).

Linear Self-Discharge

It's also possible that 1% discharge per year means 1% of the initial capacity per year.  This would mean a linear decay of capacity obeying:

The solution to this one is trivial: we just add (0.225 Ah * 0.01 / 365 / 24) = 257nA to the device current.  For the lifetime, we get 0.225 / (1.2e-6 + 257e-9) / 24 / 365 = 17.6 years.  Two and a half years shorter still.

So, there you have it: the TritiLED V3.0 can run for somewhere between 17.6 and 20.2 years on a CR2032.

There's a subtle problem with each of these estimates.  When the cell voltage falls over time, the circuit current also falls.  This would extend the run-times, since the current is continually dropping.  Ignoring this effect would tend to under-estimate the run-time, so the above estimates may be slightly conservative.

Discussions

electrobob wrote 01/16/2018 at 10:57 point

There is a difference between the self discharge on a shelf and in a circuit. 

The passivation layer mentioned by energizer is the one needed to keep the battery at low self discharge. If the battery is used, you can expect the self discharge to be much higher than 1% per year. 

A long time ago at a TI seminar about MSP430 they were mentioning that a coincell out in the weather can be expected to last ~5 years if your circuit only consumes 1/3 of that capacity during this time. The rest of 2/3 is lost in self discharge and increased resistance in the cold. 

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Ted Yapo wrote 01/16/2018 at 12:28 point

This is an interesting point.  I wonder how much current drain is required to prevent formation of the passivation layer?  Does a few uA do it?

Also, the document says the passivation layer forms after a few years of storage, but doesn't say that the self-discharge is significantly higher during those early years - the self discharge graph is linear at 1% per year right from the start.

I agree on the cold weather issue.  Everything I've read says that this is a problem.  Better for LiMnO2 than some other chemistries, but still a problem.  For serious lifetime and temperature durability, you're probably better off going with LiSOCl2 cells.  Tadiran claims 40-year lifetime in some applications.

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electrobob wrote 01/16/2018 at 13:34 point

Unfortunately, I do not know any numbers. Obviously a lot of experimentation is required. 

The ballpark rule that i base my estimation on is that there is 50% of the battery capacity gone within the specified shelf life, as self discharge. Don't forget that humidity and dirt/grease may significantly increase the external leakage current. 

The thing is that we never really had circuits so optimized that they could work on such long periods of time on a coin cell. Might be why manufacturers were never concerned with such battery usage. If there is sufficient juice left after 5 years on a shelf to still power your device for a while, the manufacturer is happy. 

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crjeder wrote 01/12/2018 at 09:59 point

Does self discharge apply to batteries that are used? I. E. if there is an constat discharge of a few nano amps (for the mcu) do some additional nA disapear into space?

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Ted Yapo wrote 01/12/2018 at 12:00 point

I don't see any reason it wouldn't. It seems like the model is just an additional tiny drain on top of any application current use.  For high-power projects, you don't think about it because it has a negligible effect on lifetime, but for really long-lived circuits, you have to consider self-discharge.

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crjeder wrote 01/12/2018 at 12:22 point

 All of the references I found state that self discharge is a "phenomenon in batteries .. without any connection between the electrodes" (Wikipedia)

Chemistry migth be different when current is flowing. But I don't know. 

To be on the safe side your approach is good, of course.

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Ted Yapo wrote 01/12/2018 at 13:02 point

I guess the chemical species inside the cell are seeking lower energy states by reacting together, just as they do during external discharge.  The energy gets released as heat, which literally disappears into space.

But, lets say for the moment that you can stop self-discharge with an external load.  Then, we should all have resistors across our batteries during storage, right?   It seems counter-intuitive.

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crjeder wrote 01/12/2018 at 13:21 point

Re-reading the sources I found.. I was wrong, the chemical process in self discharge is the same as discharge with external load. So my assumption would be that electrons are set free and they either flow as a current throug external load or find an internal way to lower energy state producing heat (just as you said). That woul imply that you can use the "self" discharge to power external loads. Or put differently: use it or loose it. (???)

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Ted Yapo wrote 01/12/2018 at 14:11 point

Maybe.  You could probably test this with some cells that have higher self-discharge rates, so you don't have to wait forever for the answer.  Set up an experiment with some high-capacity (not LSD) NiMH cells, which can lose 30% per month.  Have some discharged through accurate Coulomb counters, and others just sit on the shelf.  At the end of the test, discharge both sets to see what capacity was left.

But, I'm sure there are differences between the different battery chemistries, so this might not tell you anything about LiMnO2 cells.

In the end, battery lifetime is affected by so many things: temperature, humidity, salinity, current drain, manufacturing QC, phase of the moon, etc.  You can plan and analyze all you want, but in the end you get what you get for lifetime.

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Vishnu M Aiea wrote 01/08/2018 at 20:03 point

As per Energizer, 1% figure is of fresh capacity. http://data.energizer.com/pdfs/lithiumcoin_appman.pdf So for CR2032 (225 mAh) internal self-discharge current is around 256 nAh.

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Ted Yapo wrote 01/08/2018 at 20:08 point

Interesting, I see their linear graph.  That's the 257nA figure I got above, so 17.6 years it is.

The energizer document also says 5-8 years shelf life, which doesn't fit with losing only 5-8% of capacity.  There's something else going on that degrades the cells earlier.

If that's really the case, maybe trying to use nA-level currents is silly.  Shoot for 5-8 years on a CR2032, and use all that extra power to do something cool!

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Vishnu M Aiea wrote 01/08/2018 at 20:18 point

Yeah. I think the shelf life they recommend is a guarantee, just to be on the safe side. Because the operating environment can influence the life too but they didn't want to consider countless varying parameters. They also mention the change in internal resistance (IR) that occur over long storage of the cell. They say the IR will increase at long storage periods. If that's correct, then the self-discharge current must decrease too. What do you think ?

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Ted Yapo wrote 01/08/2018 at 21:14 point

@Vishnu M Aiea I think it depends on exactly what the self-discharge mechanism is.  It's convenient to model the self-discharge as a current sink connected to the battery terminals, but the reality is probably much more complex.  I can imagine a process where the chemical changes in the cell contributing to the self-discharge don't depend on the internal resistance to regulate them.

Assuming Energizer knows their battery, it might be best to assume the linear self-discharge curve they show continues to be linear at longer time periods.

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Nikos wrote 01/08/2018 at 21:43 point

I'm on board with the "design for shelf-life" approach @Ted Yapo, there must be some chemical degradation. However I'm wondering if the sealed cases will extend the shelf life or not. I have some no-name CR2032s which in a permeable package are almost useless after a couple of years on my shelf.

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Ted Yapo wrote 01/08/2018 at 23:25 point

@Nikos I think it depends on your shelf: temperature extremes (or cycling), humidity, salt, and probably a few other things. The cases may protect from moisture, so that could help.

The no-name brand brings up another point: cell quality control. I also have purchased no-name cells which turned out to be lousy.  The Panasonic cells are actually relatively cheap, and seem pretty good.

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Nikos wrote 01/08/2018 at 23:46 point

@Ted Yapo that's a very good point, those no-names were for cheap LED key-chain lights, I wouldn't put a no-name in my data loggers :) . I've found Vartas, Energizers, and Panasonics at a good bulk price through a European supplier, so I'll order a bunch of those once ready for deployment. The logger can also log battery voltage so I'll set a couple of those to do that, and see how long they last. It'll be a while before reporting back though... I hope!

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Vishnu M Aiea wrote 01/07/2018 at 11:42 point

1.15 uA is the average current right ? I mean I know there's no way an LED will consume only that much current. What's the peak currents ? and what's the current consumption of your MCU ?

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Ted Yapo wrote 01/07/2018 at 13:54 point

Yes, 1.15 is the average current.  It's averaged out by the capacitance on the PCB.  I don't have a good idea of the peak cell current - that's more difficult to measure.  At slow pulse frequencies, it's also helpful to add some external (low-leakage) capacitance to smooth out the current even more so it can be measured with a DMM.

The peak current to the LED is around 23 mA (not measured directly) for this particular LED.  The inductor ramps up to this peak over a time of 7.6us, so that's around 11.5 mA average current drain over those 7.6us.  At 0.5Hz, that's around 44 nA average that finds its way to the LED.

The MCU consumption is around 1.1 uA (the best measurement I have, regression fit from many tests is 1.08uA).

So, at this "brightness setting", almost all of the energy is going into the MCU, not the LED: maybe it's 4% efficient.  At brighter settings, it's much better, like 85% at a 2-year brightness.

That's the price you pay for running a long time.

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Vishnu M Aiea wrote 01/07/2018 at 14:00 point

Nice. How do you measure currents in 0.01uA range ? What equipment do you have ?

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Ted Yapo wrote 01/07/2018 at 14:08 point

I have a few DMMs that read to 0.01uA on a 20uA scale, but measurements at this level are tricky.  Even a 10M-ohm voltmeter will draw 0.3uA at 3V, so I end up doing a lot of juggling to try to get accurate measurements.

To get a decent idea of the individual pulse current, for example, I take measurements of the current over a range of pulse frequencies (like 10-1000Hz), then regression-fit a line.  The slope gives you the current per pulse, and the intercept is the baseline (MCU) current.

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Vishnu M Aiea wrote 01/07/2018 at 06:52 point

This is interesting. I've thought about this while building my Birthday Alarm project but didn't want to delve into that much detail. But now I feel I should do it. Thanks :)

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Simon Merrett wrote 01/06/2018 at 22:46 point

Should we be reconsidering whether the battery is the critical point of failure with these runtimes? No electrolyte to dry up in a TritiLED but what about the controller memory? 

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jaromir.sukuba wrote 01/06/2018 at 23:36 point

40 years to temperatures up to 85 degrees C. I assume centuries at room temperature.

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Ted Yapo wrote 01/07/2018 at 01:43 point

Even the battery lifetime might be shorter due to leaky seals, corrosion from moisture ingress, temperature cycling, being stolen by an angry dog, etc...

Likewise, RAM state could get flipped by a cosmic ray, decays from the 210Pb in my non-isotopically-pure leaded solder, or a direct strike with an errant comet (microseconds before vaporization).  The PIC resets every 256s and re-runs all initialization code to mitigate this, but it's probably possible that the right bit flip turns off the INTOSC entirely and kills the thing dead.

Planning for these kinds of contingencies is very interesting, but unfortunately, not something I know too much about.

Maybe the answer is to make them as cheap as possible and use massive stupid redundancy, like Rodney Brooks' 100x 1-kg robot space probes vs a single 100kg one.

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