## CR2032 longevity napkin calc over time

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CR2032 @ 225ma hr,  -1% per year, but lets ignore that.

Hypothetical 225ma pulse for 2ms, 288 times a day (every 5 minutes).

So current usage is 576ms @ 225ma per day.

1hr = 3,600,000 ms.

3,600,000 / 576 = 6250 hrs.

6250/24 hrs= 260.4 days of power usage.

Is the math correct?

~William

Last Edited: Wed. Nov 11, 2020 - 08:22 PM

Well, other than the fact that practically, pulses of 225ma, even if only 5ms, would really challenge it ...

You also need to be careful because most real circuits that draw 225ma when the battery is fully charged probably will not  draw 225ma when it gets discharged.

BUT, if it really is a constant current pulse that stays constant over the full range of battery voltage, the arithmetic looks OK. It is really more informative to deal with energy rather than current in a real application, though it is a lot harder, calculation wise.

Jim

Until Black Lives Matter, we do not have "All Lives Matter"!

Iv yet to short a cr2032 through a meter, but a search online showed one cell @ 70ma draw. Ill go try it.

I have an IR sensor circuit pulsing 20-25 ma through the battery. With a narrow beam IR diode and a photo transistor I can get  hand length signal sensing at a 63us pulse length.

The higher the current the better the sensitivity and shorter on time required, so win win.

I am looking at the ATtiny 214, if im reading the data sheet correctly the RTC @ 1,024khz in Standby @3v draws 0.71uA.

So a perfect(fantasy) cr2032 without losses would run the Attiny214 RTC for 36+ years.

Tadiran does make 40 year batteries.

Attiny214 cpu at 32khz Idle draws 4uA at 3v. Could theoretically sip away a cr2032 for 6+ years.

~William

No the math is not quite right. I come up with 6250 days.

Can you draw 225ma from a button cell without the cell voltage dropping?  These things normally power devices that draw .2ma or less!

Don't know,

Jim

FF = PI > S.E.T

I hardly think you'd get more than 100ma, or even 50ma , without a very severe impact (in terms of  effective mAH loss, since the voltage would dip so low, you'd  quickly reach effective end of life (the Vdrop, makes the battery appear empty to your circuit).

You need a capacitor to smooth out the draw as seen by the battery; the biggest drawback might be the long-term leakage effect.  If using a cap, you also wan to ensure it is not too large, otherwise ,when "off" all that stored energy just gets lost in your circuit & you end up having a high peak recharging the cap...so it is a balancing act.  I've seen too many  "shut down the supply" designs, where they think they will save loads of power, but end up just  recharging the capacitor bucket (and at a high peak the I2R losses may mean you waste even more power than the savings).

I have an IR sensor circuit pulsing 20-25 ma through the battery. With a narrow beam IR diode and a photo transistor I can get  hand length signal sensing at a 63us pulse length.

two things:

Maybe use a flyback coil or charge pump arrangement to pulse your led (somewhat like a photoflash)..then you can spread the energy required from the source.  I used to have a box of custom coils made just for this purpose...but they were several AMPS.

A pin diode may be better, since it is more suited for detecting fast pulses.  I don't recall what our pulses were, maybe a few us (4 or 5?)...of course that isn't super speedy either.

You can detect weak signals if you know the freq you will be sending at.  If you have electrical access to the pulses, you can do it synchronously & really go to town (like detecting nanowatts in presence of room light.  You can also get close by using a PLL to regenerate the pulse clock if you don't have electrical access.

Why not use a better battery?

found something: https://www.dmcinfo.com/Portals/...

When in the dark remember-the future looks brighter than ever.   I look forward to being able to predict the future!

Last Edited: Wed. Nov 11, 2020 - 10:57 PM

Some CR2032 are worse than others.

"Dare to be naïve." - Buckminster Fuller

vertamps wrote:
Tadiran does make 40 year batteries.
Though only from distributors; sourcing from local supply is usually fast if can fit the cells into the enclosure within the mass requirement (lithium iron disulfide)

l92.pdf (Energizer)

The Art of Electronics 3rd Edition | by Horowitz and Hill

9.12 Energy storage: batteries and capacitors

[page 116, bottom of right column]

"Dare to be naïve." - Buckminster Fuller

A 2032 cell should not be used with more than 20mA. draw 200mA for even a very very short time and the battery will die on you almost immediately.

#1 Hardware Problem? https://www.avrfreaks.net/forum/...

#3 All grounds are not created equal

#4 Have you proved your chip is running at xxMHz?

#5 "If you think you need floating point to solve the problem then you don't understand the problem. If you really do need floating point then you have a problem you do not understand."

Might the way to solve this be to use the coin-cell to charge a capacitor via a resistor, and use the capacitor to supply the current pulse?

Without some form of help the internal resistance of the cell will stop your use dead in it's tracks. It looks like a typical figure for a CR2032 is 9 Ohms...

9R x 225mA = 2.025V.

3V - 2.025V = 0.975V.

Will your circuit work on less than 1V?

#1 Hardware Problem? https://www.avrfreaks.net/forum/...

#3 All grounds are not created equal

#4 Have you proved your chip is running at xxMHz?

#5 "If you think you need floating point to solve the problem then you don't understand the problem. If you really do need floating point then you have a problem you do not understand."

Last Edited: Thu. Nov 12, 2020 - 07:51 AM

charge a capacitor via a resistor

Maybe, but you know what a resistor does.

When in the dark remember-the future looks brighter than ever.   I look forward to being able to predict the future!

avrcandies wrote:

charge a capacitor via a resistor

Maybe, but you know what a resistor does.

I do. They cost around 1p.

Not sure what your point is.

First size the capacitor....

C = At/V....A = .225A, t = 2ms, V = 1V (assuming we can stand a 1V drop at the end of the 2ms period)...so a 450uF capacitor will power the circuit as required. Use a 1000uF.

Now we need to charge the capacitor back up...

We have 5 minutes to do it = 300 seconds. To play safe set the RC timeconstant to 10RC. C is 1000uF, t is 300/10 = 30s so R = t/C = 30/10,000uF = 30k.

And finally test it...

Current in R1...

#1 Hardware Problem? https://www.avrfreaks.net/forum/...

#3 All grounds are not created equal

#4 Have you proved your chip is running at xxMHz?

#5 "If you think you need floating point to solve the problem then you don't understand the problem. If you really do need floating point then you have a problem you do not understand."

Maybe, but you know what a resistor does.

I do. They cost around 1p.

Not sure what your point is.

I'm sure you know they turn battery power into heat & loss..ban the resistor.   I'm marching through town with my sign ...not only is resistance futile, it is wasteful :)

I suspect soon there will be a global law/ban on resistors in all circuits...we will have to switch to the new Ohm's law (law: all resistance must be <0.001 ohm).  I think all these daggone resistors are responsible for global warming.

Well, time to snack on some AVR chips & a bubbling bowl of solder soup.

When in the dark remember-the future looks brighter than ever.   I look forward to being able to predict the future!

Last Edited: Thu. Nov 12, 2020 - 11:00 AM

Brian Fairchild wrote:
Use a 1000uF.
Polarized capacitors are leaky; so are MLCC though much less.

SimSurfing (Murata) (SPICE model)

IIRC, KEMET has similar

"Dare to be naïve." - Buckminster Fuller

Brian Fairchild wrote:

Paper from TI...

https://www.ti.com/lit/wp/swra34...

This TI paper is great, they have longer pulses than i am attempting but that Cap in parallel did wonders for some of the batteries tested.

I had no idea pulse currents were so damaging to longevity.

~William