Coin Cell Eliminator Does More Than Save Batteries

Coin cells are useful things that allow us to run small electronic devices off a tiny power source. However, they don’t have a lot of capacity, and they can run out pretty quickly if you’re hitting them hard when developing a project. Thankfully, [bobricius] has just the tool to help.

The device is simple – it’s a PCB sized just so to fit into a slot for a CR2016 or CR2032 coin cell. The standard board fits a CR2016 slot thanks to the thickness of the PCB, and a shim PCB can be used to allow the device to be used in a CR2032-sized slot instead.

It’s powered via a Micro USB connector, and has a small regulator on board to step down the 5 V supply to the requisite 3 V expected from a typical coin cell. [bobricius] also gave the device a neat additional feature – a pair of pads for easy attachment of multimeter current probes. Simply open the jumper on the board, hook up a pair of leads, and it’s easy to measure the current being drawn from the ersatz coin cell.

If you’re regularly developing low-power devices that use coin cells, this tool is one that could save a lot of mucking about in the lab. [bobricius] has them available on Tindie for those eager to get their hands on one. We’ve seen similar designs before too, albeit pursued in a different way!

32 thoughts on “Coin Cell Eliminator Does More Than Save Batteries

    1. Coin cells have their place due to a very slow self-discharge. Some did power SRAMs in game cartridges for 30 years. A gold cap or similar can’t keep up with that. Not powerful lead batteries, even. They die of deep dis-charge years before.

      1. I forgot to mention.. It’s necessary to wear gloves or clean the coin batteries with isopropyl alcohol or similar before use. A tin film of fat will already ruin the longevity of coin cells. Because, the fat of the fingerprints will accelerate self-discharge.

        1. Self-discharge not quite … if anything it will increase the parasitic series impedance making power extraction less efficient but it cant increase self-discharge as a) it’s not in parallel with the battery terminals and b) it’s not conductive enough to allow current to flow directly between the terminals even if a were true

    2. You’d need 864F to match a CR2032.

      The Energizer CD2032 datasheet specs energy storage of 240mAh from 3.0V to 2.0V:

      http://www.cr2032.co/cms/prodimages/energizer_cr2032_datasheet.pdf

      Amperes have units of Coulombs per second, so multiplying current by time gives you units of Coulombs. 1mAh == 3600mAs == 3.6Amp-seconds == 3.6C. 240mAh == 864C.

      Volts have units of Joules per Coulomb, and multiplying Volts by Coulombs gives you units of Joules. 864C times 2.5V (the average between 3.0V and 2.0V) is 2160J.

      You can expect to get 2160 Joules of usable energy from a CR2032 as it discharges from 3.0V to 2.0V.

      The equation for energy in a capacitor is V^2F/2, where V is the voltage and F is the capacitance. The Farad has units of Coulombs per Volt, so F*V*V immedately cancels to units of C*V. Then the Volt expands to Joules per Coulomb, so C*V becomes C*J/C, which cancels to Joules.

      For a capacitor discarging from 3.0V to 2.0V, the average voltage will be 2.5V. The 2160J we calculated for the CR2032 is 2.5 * 864, so dividing 2160 by 2.5 gives us a capacitor value of 864C.

      Digikey has four 850F supercaps in its product database:

      https://www.digikey.com/en/products/filter/electric-double-layer-capacitors-edlc-supercapacitors/61?s=N4IgjCBcoEwAwBYCcVQGMoDMCGAbAzgKYA0IA9lANogAcArHAAQBiIAuqQA4AuUIAqgDsAltwDymALKFs%2BAK4AnQiAC%2BK0jCoh5nQgrTZO2NKLIL2KoA

      whose volumes range between 112cc and 173cc. A CR2032’s volume (from the datasheet) is 1cc.

      Supercaps are neat, but still have much lower energy density than chemical batteries

      1. There is a mistake in your calculation. The voltage in the capacitor is not 2.5V in this case. It is 1V. You discharge vrom 3 to 2. The rest of the energy from 2 to 0 stays in the cap and can’t be used in this case. Your cap would have to be 2*5 times larger than what you calculated.

        1. No. Despite the tortured (and incorrect) new-math thought process (Seriously? Is that how they teach arithmetic in school now? That’s awful.), [mike stone] arrived at (close to) the correct answer.

          Energy in a cap is 0.5*C*V^2. Energy difference between two voltages is 0.5*C*(V1^2-V2^2) (not the 2.5 V average). Only a 4% error here, but still an error. Blame the education system.

          1. No, new math has nothing to do with it, and while as ridged and inflexible in its own way as “leaning your ‘memorizing you times divides tables’ was on back in your day, there is a lot of wisdom in trying to teach kids ‘maths’ by a method that required that they actually understand what they are doing rather than just memorizing and endless and meaningless set of rules.

            What that is however is a typical answer your find on a basic physics test. It receives partial credit because he clearly understands the idea that the units for voltage and capacitance and current are composed of other quantities and tries to use dimensional analysis to get to the answer. This is a valid technique to find a solution.

            Confusing dimensional analysis and now math suggests that you have studied neither.

            If your not taking an exam there is a simpler solution however. Google energy in a capacitor, or look it up in a book. That gets you to (1/2)CV^2

            It reminds me of a classical dynamics exam. I looked up the trig identity that made things easier and then used Maple to solve the system of equations. The professor shoes the class mine as the example of how to do it and every one was all “your allowed to do it like that?”

      2. I just remember the energy in a capacitor is (1/2)CV^2. Its really easy to remember. Energy is generally 1/2 times some inertal thing times some forcey thing. For example. (1/2)MV^2 for kinetic energy or (1/2)kx^2 for a spring.

    1. The problem is that the diode drop is highly dependent on current, temperature and part to part variations. i.e. very crude. At very low current in uA range (typical for memory backup), your diode drop is very low – much less than the 0.7V you speculate. Try read up a datasheet once in a while.

      A low quiescent regulator on the other hand can regulate the voltage, so not going to overvoltage the device. You can get them down to 1uA very easily (before the chip shortages).

  1. Reminds me some tinker my teenager self done in the 90s with some washers, plastic disk and metal sheet to connect to a AA pack back when my TI80 eat coin cell in a month of full programming on it…

    1. Why did you insert the word normal? Making assumptions? Maybe you should also tell us to switch on the device under test before measuring. Should also warn about battery doors not fitting, should also warn about using a fresh battery for testing. If you feel compelled to warn us then do it right.

    2. Hunh? What do you consider ‘normal’? Even my run-of-the-mill multimeters measure to a precision of 0.1 microamps: If shelf life didn’t get it first, a CR2032 would last about 250 years at that rate. With 1% precision, I can measure 20 microamps, which is well over a year of life.

  2. Years ago, we used to run a “computer club” for Autistic adults (the actual club was actually for social skills)… and of course we had a fleet of PCs.

    The machines were typical of the mid-late 90s… Pentium 200MHz, 32MB? RAM, Windows 95 (the OEM code I was able to recite from muscle memory for years). The RTC used a CR-2032 cell for battery back-up, and we got fed up of replacing them, so one of the volunteers (a marine engineer) took a few of the duds home, drilled a hole into them, cleaned out all the dud electrolyte, then soldered the flying leads from a 2×AA cell battery pack to the terminals.

    We could then just plug this “cell adaptor” into the motherboard CR-2032 socket and position the battery pack where ever it was convenient. Saved lots of trips into the case.

    1. I think that’s more expensive than just getting a new CR2032… they should last ages in a PC, plenty of alkaline AAs will self-discharge to death faster than a CR2032 will run out when *doing its job*.

      Maybe this is worth it if you’re paying $8.50 a battery at a drug store, but you can easily buy them for much less. IKEA even has an 8-pack them right now for $4 CAD (just over $3 in USD).

      1. I’m reading this and thinking this was 20-25-ish years ago, at that time perhaps cheap, dodgy cells were more common, perhaps the CMOS were more power-hungry, perhaps working in a computer was even more of a pain, with proprietary cases using weird caddies, with zip-tied running cables everywhere from the PSU as hidden cable management wasn’t really a concern, with IDE large flat cables floating around, with VGA cables that need to be unscrewed… Ugh. I can see doing that for a fleet of tower computers to be a pain.

  3. Inside something like this to an old RCA VTVM that used a D-cell for a bias voltage. I tossed in a 2volt regulator on the filament supply and tossed on a diode to get around 1.5v. for giggles I did it to another VTVM but used a 2032 and a voltage divider. It’s been two years and the battery still works.

  4. Neat idea,but an improvement might be to add some castellations around the edge so that it will work with coin cell holders that only contact the edge of the cell for the positive terminal.

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