Coin Cell Jump Starter

Starting a car with a CR2477 coin cell

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The goal is simple: replace my car battery with a coin cell (plus clever circuitry), and get the car to start at least once using only the energy from the cell.

OK, it's not all worked out, yet, but here's the back of the envelope calculations.  I found that Mehdi on electroboom has already started a car with super capacitors. He connected 6x 400F capacitors in series to get 66.6 F.  This was enough (charged to 14V) to start his car.  How much energy is this?

Sounds like a lot.  How much energy is in a CR2477 coin cell?  They're rated at 1000mAh at 3V:

So, if I can create a boost converter to charge the capacitors to 14V from the cell that's at least 6527/10800 = 60% efficient, I should be able to start a car from a coin cell.

The first rough plan is to make a simple inductor-based boost converter to charge a similar (possibly identical) capacitor bank over perhaps 24 hours, then start the car with the capacitors in place of the usual battery.

Of course, the devil is in the details, but at first look it seems like it could work.

  • More Coin Cell Discharge Data

    Ted Yapo19 hours ago 0 comments

    I tested a CR2032 cell at continuous 10mA discharge.  The results aren't very good.

    If we take 1.5V as the cutoff voltage, around 275J can be drained from the cell in a little over 3 hours.  Draining down to the mV level gets you to 325J in just under 5 hours.  This is roughly 12 and 14.5% of the cell's capacity, respectively.

    I tried the same experiment with a CR2477 cell:

    This cell manages about 850J to the 1.5V level in 10 hours.  I stopped the test about 15 hours in, when the energy reached 1000J; by this time the cell voltage is below 500mV.  This isn't very good either, although it is roughly 3x the energy of the CR2032.  The datasheets would have you believe the CR2477 can supply 4x or so at very low discharge rates (1000 vs 225 mAh), so I guess the relative performance of the cells I am seeing roughly makes sense.

    I also tried a longer test with a CR2477 at 5mA continuous drain.

    In this case, the cell takes 43 hours to reach the 1.5V level.  At this point, it has yielded around 1900J.  If you can make used of the cell down to mV levels, you can get around 2100J at this current, but it takes 55 hours.  Again, not very good at all.

    Looking at the four data points I have for CR2477 cells, we can see what effect the drain current has on the yield (I also add the single data point for CR2032):

    Yield is a very strong function of drain.  I wondered how it would look on a log-log scale, so I plotted it, and it is roughly linear, hinting at an exponential relationship between capacity and current drain, similar to Peukert's Law for lead-acid batteries.  Here are the data points along with a best fit line in the log-log space:

    The exponential model is of the form:

    where E is the energy yield of the cell in Joules, I is the discharge current in mA, and k and m constants.  For the CR2477, a least-squares fit gives k = 35594, and m = -1.704.

    Using this model, I would estimate that at 80mA, you should get 35594*80^-1.704 = 20J, while at 2.5 mA, you should get 35594*2.5^-1.704 = 7469 J.

    Just like Peukert's Law, this model predicts you can get infinite energy by draining at lower and lower currents, so at some point it is no longer applicable.  Just for fun, I re-arranged the equation to solve for the current you'd need to use to get 9000J (approximately full capacity) from a CR2477.  The answer: 2.24 mA.  This sounds like an interesting test, and if it wouldn't tie up my meters for five days, I'd start it now.  Maybe after the contest :-)

  • DC-DC Converter Efficiency

    Ted Yapoa day ago 0 comments

    I must have figured this out while sleeping, because when I woke this morning, I knew I could use the cell discharge tester to measure the DC-DC converter efficiency.  The discharge tester is an adjustable current load.  By connecting this load to the output of the DC-DC converter and manually sweeping the current, I could record the converter efficiency as a function of output voltage.

    The plot shows efficiency and output current vs output voltage for the homebrew converter tuned to draw 10.7mA from a 3V source.  A 1.7Ah LiSOCl2 cell can supply this current without any decrease in capacity.

    The efficiency peaks at 65% between about 2.5 and 3.5V output.  This is a good match to charging a series battery of two AA NiCd cells (nominal 2.4V).  To ensure good NiCd efficiency, I'll want to limit the cells to 70% charged, so I will probably use two sets of NiCd's in parallel, making a 2S2P pack of 2.4V / 2Ah capacity.

    I ordered some 1000mAh NiCd cells, since I haven't had any around here in over a decade.

    The efficiency of the converter is not very good compared to commercial offerings, but the input current is easily adjusted to a fixed value, an odd requirement for a power supply.  I am not sure how easy this is to do with commercial switcher offerings.  Does anybody know about this?

    I will have to re-test this converter with a higher input current, because if the efficiency curve is the same, this converter is unsuitable for charging capacitors to 14V.  At only 9V output, the efficiency has dropped to below 35%.  The efficiency peak around 2.5V might make it OK for charging 2.7V capacitors in parallel (they would then be connected in series once charged).  Even then, the steep drop-off of efficiency at low voltages isn't very good.

    I still have to investigate commercial DC-DC converters for charging the supercapacitors.

    The problem with the 10.7mA current is that it would take a week to get all the energy out of an LiSOCl2 cell at this rate.  The cell is rated for 10mA continuous/50mA pulse, but I'm not sure how far I'm willing to push it.  Going to 30mA would drain the cell in 57 hours, which seems more reasonable.

  • Capacitors? Why not NiCd's?

    Ted Yapo2 days ago 5 comments

    So @EricH gave me an interesting idea.  He asked if using an intermediate step with another set of capacitors could help with the energy transfer problem.  If you could find large-valued capacitors with low self-discharge, you could take a long time to charge those with a coin cell, then charge the supercapacitor quickly from the intermediate caps. It sounds like it could work, but I don't think the right capacitor exists for this intermediate step.

    What about using a rechargeable battery as the intermediate energy storage?  This gets interesting.  Everyone seems willing to allow an electrochemical capacitor as intermediate storage, so why not a rechargeable battery? (I'll refer to the coin cell as a "cell" and the rechargeable battery as a "battery" in the discussion below).

    Let's forget about any technical problems for a minute and consider the contest judges and spectators.  You have to convince them somehow that you're not running anything from energy pre-stored in the battery.  Since state-of-charge is very difficult to measure accurately, I'm not even sure I wouldn't be cheating with most battery chemistries.  The exception is NiCd, which can and should be stored with the terminals shorted and at a zero state of charge.  It's how NASA stores their NiCd cells, as detailed in this technical report on NiCds.  So, if I take a couple of AA NiCd's that have had their terminals shorted for a few days, then verify there is 0V across them, I think I can make a convincing argument that there's no energy hidden up my sleeve.

    OK, so there's a way to verify that all the energy is coming from the coin cell.  What are the properties of a NiCd battery? 

    • Nominal voltage 1.2V with a flat discharge curve.
    • Can sustain very high rates of discharge (think of a cheap cordless drill).
    • Self discharge rates quoted as 10% per month (wikipedia) or 1% per day (NASA TR).
    • Tolerant of varied charging methods (C-rate and end-of-charge detection)

    Overall, they sound like a good intermediate reservoir for energy storage.  They have a much lower self-discharge rate than supercapacitors, so can be charged slowly from a coin cell without terrible losses (a DC-DC converter is still required).  Then, once charged, they can be drained very quickly to charge the supercapacitor before supercap self-discharge becomes an issue.

    What are the drawbacks?  First, the energy will be going through two DC-DC converters, so losses get compounded there.  Also, there's the charging efficiency of the NiCd's. Wikipedia mentions that at a C/10 charge rate, you have to apply around 1.5C of charge to fully charge a NiCd (equivalent to a 33% loss of energy).  The NASA TR, however, shows that this ratio is a strong function of temperature (P. 13).  With battery temperature near 0C, the ratio approaches 1, so much less energy is lost in charging.

    So, can I take a 1.7Ah LiSOCl2 cell, charge some 1000mAh AA NiCd's, then use the NiCd's to charge a 67F capacitor to 14V?  Here's how everything stacks up:

    StorageCapacity (mAh)Energy (J)
    2x NiCd AA10008600
    3x NiCd AA100012960
    4x NiCd AA100017280
    67F capacitor @14V -6566

    From the TL-5935/P datasheet, it looks like I can get the full 1.7Ah from the cell at 10mA, which is the maximum recommended continuous drain.  I also estimate that the cell voltage will remain stable at around 3V for the entire discharge.  Discharging at this rate will take 7.08 days.  Assuming the 1% per day self-discharge rate for NiCd's, I might lose 600J during this week. Assuming a 33% loss due to NiCd charging inefficiency (which might be improved by cooling), and a 70% DC-DC converter efficiency, I end up with 8961 J in the NiCds, which almost fits in 2 AA's. I'll call it 8600...

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  • Capacitor Self-discharge

    Ted Yapo2 days ago 0 comments

    The numbers are in on the self-discharge test.  I charged a 400F capacitor to 2.33V with my bench supply, and let it soak at that voltage for about 2 hours.  Then, I recorded the capacitor voltage over the next two days while the capacitor self-discharged.

    To estimate the self discharge current, I fit a series of lines to the local voltage-vs-time curve using least-squares regression with a window of 2001 points wide (about 45 minutes of elapsed time). Some of the noise in the curve is due to the quantization of the voltage steps.

    The datasheet specifies a 1mA maximum self-discharge after 72 hours.  The capacitor meets the specification, with the leakage current dropping below 1mA after about 12 hours.  After about 16 hours, the self-discharge current levels off at around 0.5 mA. The bad news is that the self-discharge current starts at around 5 mA, so 10% of the capacitor energy is lost in the first 5 hours, extending to 20% lost in 24 hours.

    I don't know how this curve would look if the capacitor had been "soaked" for a longer time at 2.33V.  It may be that if held at a specific voltage for an extended period of time, the initial self-discharge would decrease.

    So far, I'm also not sure exactly how to apply this data to the charging problem.  For instance, what does the leakage current look like during charging?  Does it increase or decrease as the capacitor charges?  In any case, the initial self-discharge current doesn't look good.

  • This is my hack...

    Ted Yapo2 days ago 0 comments

    ...and it's likely to spark some debate.  Let me introduce the idea by way of an analogy.  Let's say you're tired of the city and want to visit the country.  Your friend agrees to lend you her car for the weekend so you can get away and clear your head.  When you pick up the car from her, it has 3/4 of a tank of gas.  The first thing you do is stop at the gas station to fill the tank and buy some snacks for the trip.  The trip is great, and you return on Sunday evening completely refreshed.  When you return the car, it has a little more than 3/4 of a tank of gas - say 7/16.  Your friend notices you haven't used any of her gasoline, and says you can borrow the car again whenever you want.

    After a few trips like this you decide it makes sense for you to buy your own car.  Unfortunately, you are taken in by the slick salesman and end up with a car which has trouble starting.  It turns out that the real issue is not the battery but the alternator - for some unknown reason, the alternator will keep the car running once it has started, but never re-charges the battery, not even a little.  Luckily, your friend has a 67F capacitor she can lend you to start your car.  When you pick up the capacitor it is charged to 12V, so contains 4824J of energy.  Just like you did with the gasoline, you first "fill up" the capacitor to 14V (you happen to have a little device which does this) - now the capacitor contains 6566J.  You connect the capacitor in place of your car battery, and start the car.  Starting the car takes 1500J, so afterwards, the capacitor contains 5066J of energy, and is charged to 12.297V.  Just as you did with her car, you return the capacitor to her with a little more energy than when you borrowed it.  Again, she notices that you haven't used any of the energy in her capacitor, and offers to lend it to you whenever you need.

    Unfortunately, the little device you have won't charge the capacitor from 0 to 14V, only from 12 to 14.  So, if your friend lends you a fully discharged capacitor, you can't start your car.  If, on the other hand, she lends you the same capacitor you began with, you always leave it with a little more energy than the last time you used it, and you can continue starting your car indefinitely using just the energy from your little device.

    The Proposal

    I don't know if I can extract 6566J from a coin cell fast enough to charge a supercapacitor, but I may be able to extract 1500J.  In this case, I propose to begin with a capacitor charged to 12V, charge the capacitor to 14V, then start the car with it.  By monitoring the voltage and current during starting with an oscilloscope, I can verify that the starting took less energy than was deposited in the capacitor by the coin cell.  The numbers used here are just examples: the actual values will probably differ somewhat.

    Ideally, I'd modify the car so that the alternator would not charge the capacitor once the engine is started (imagine a beefy ideal diode made with MOSFETs).  The problem is that cars depend on the battery to filter the alternator voltage - with older cars, you could disconnect the battery once the car was running, but with modern computerized engines, this is asking for a whole lot of trouble.  I am willing to believe that a 67F capacitor can stand in for the battery as a power supply filter, but don't think it is wise to have the capacitor isolated from the alternator once the engine is running.  So, the next best thing is to analyze the voltage and current waveforms during starting to verify that none of the original charge of the capacitor was used in starting the engine.

    Will the judges accept this argument?  Will they disqualify my entry?  What do you think?

    Then again, maybe I can still find a way to get 6566J from a coin cell :-)

  • Coin Cell Discharge Tests

    Ted Yapo4 days ago 12 comments

    Early on I figured I'd need to test some CR2477 cells to see how much energy they could practically supply in a short time.  I finally got around to building the tester and putting it to work (not to spoil the surprise, but things don't look great so far).  Here's the tester:

    The circuit is basically what I described in an earlier log, except with a few R's and a C added for stability.

    The first low-offset op-amp I found in the parts bin was an MCP6V01 chopper.  It's overkill for this application but contributes a ridiculously small error to the result.  The cell is discharged at a constant current set by the potentiometer, while serial-port-enabled DMMs log the current and cell voltage (not shown).  I didn't have any 1% 10-ohm resistors with a decent power rating, so I paralleled 10x 100-ohms.  The MOSFET and op-amp are mounted on #Ugly SMD Adapters.

    My initial plan was to discharge the CR2477 at a constant 40mA.  This represents a C/25 discharge rate, which is completely reasonable for any battery other than a coin cell.  For the CR2477, it turns out, this is an enormous drain that wastes most of the energy in the cell.  Here is a plot of the voltage during the 40mA discharge as well as the total energy (in this case dissipated as heat in the 10-ohm resistor):

    After 10 minutes of discharge, the cell voltage has decreased below 1.5V, which I think is as low as I can reasonably expect to go using a PIC12LF1571 in the DC-DC converter (the PIC is specified down to 1.8, but I will push it).  At the 10-minute mark, the energy extracted from the cell is around 55J.  Terrible.  Integrating the current to this point yields 6.6 mAh, which is 0.66% of the cells capacity. Awful.

    I let the drain continue for about 9 hours, even though after 25 minutes, the cell voltage dropped below the 400mV required to maintain a 40mA discharge in the load resistor.  At the end, 157J had been extracted.  Ghastly.

    Lesson: don't discharge CR2477 cells at 40mA.

    What about 20 mA?

    Undaunted, I discharged a CR2477 cell at 20mA (note the different time scale).

    In this case, around 260J and 35 mAh have been extracted from the cell before the voltage drops below 1.5V (lousy). This 5x as much as the 40 mA discharge, but still represents a small fraction (3.5%) of the capacity of the cell.  Atrocious. When I finally stopped this test after about 8 hours, 392 J had been extracted from the cell. Dreadful.

    Lesson: don't discharge CR2477 cells at 20mA.

    Are you seeing the pattern yet?  Lower currents will be required to extract more energy from the cell.  Of course, these discharge curves don't come anywhere near the 25-hour estimate I originally used for charging (the 20mA discharge is essentially done in 3 hours). Still, the total amount of energy I have been able to obtain from the cells is very low.  Abysmal.

    What I Got Wrong

    So, in my initial estimates of current drain, I assumed that a 1000mAh CR2477 was like a 225mAh CR2032 scaled up by 4x, so I could draw 4x the current (I found data fro CR2032s on the web).  The differences between the cells are more subtle than that.  The maximum current you can draw is proportional to the electrode area, and assuming the electrode size is proportional to the cell diameter, a CR2477 only has 24^2/20^2 = 1.44x the electrode area.  The total amount of reactive chemicals inside may be 4x, but the maximum current you can reasonably draw is likely to be around 1.4x.

    I also based my initial estimates on numbers in this article, which says you can use 88% of the available energy in a CR2032 when draining it at 10 mA.  Unless CR2032s are much better at higher currents than CR2477s, I just don't see it.  I should test a CR2032 for comparison.

    What's Next?

    There is a trade-off between the length of time taken to drain the cell and...

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  • How much current does it really take?

    Ted Yapo6 days ago 2 comments

    I took some measurements on my car today.  I hooked the 650A current clamp to an oscilloscope and measured the battery current during starting.

    The engine is a 4.7l V8, which was relatively cold, only having been started a few times briefly before I got the scope triggering right. (You can also see how I fixed my cracked fender with a drill and zip ties at the bottom right).

    The battery current looks like this:

    The current peaks at just above 500A for about 20ms, then drops below 200A within 50ms.  It stays around 150A until about 300ms into the start, then drops to about 100A for the remaining 500ms.  The total start time is around 850ms.  This doesn't seem too bad.

    To simulate what this discharge current would do to the capacitor voltage, I made a quick LTspice simulation using a piecewise-linear current source to model the starter current.  I assume the capacitor starts at 14V.

    The simulation shows how the capacitor voltage would drop if the starter drew the same current:

    In this simulation, the capacitor voltage ends at 12.4V, so (12.4^2/14^2) = 78% of the energy remains in the capacitor.  This is very interesting.

    Of course, the current profile will probably look different with the capacitor in place of the battery, but this data is encouraging.  It doesn't really take that much energy to start a car.  In this simulation, it only took around 1500 J.

  • Help: ideas wanted

    Ted Yapo6 days ago 16 comments

    I need to come up with a good way to connect the capacitors in series with a low-resistance connection.  I've gone round and round in my head, and haven't come up with the right way yet, so I figured I'd ask you, dear readers, for your input before I just slap something together.

    Here is a picture of two of the capacitors as they will be wired.  There are three positive terminals and one negative on each capacitor.  The "outside" positive terminals (smaller pins) seem to be there just for mounting reinforcement, so I'm going to ignore those and only connect the rectangular tabs.  The tabs are 5mm wide by 6mm long by 1mm thick.  Coincidentally, 0.2" Faston connectors seem to fit, but they aren't suitable for the large currents involved.

    So, I need to come up with a way to connect those two terminals.  I have thought of machining a copper bar with slots to fit the tabs then soldering, or cutting short pieces of solid copper ground wire to fit in-between them, but no ideas so far feel right.

    Ideally I'd like a secure mechanical connection first, then add solder to that.


  • Supercapacitor Testing

    Ted Yapo12/10/2017 at 19:13 0 comments

    My supercapacitors arrived yesterday, and I've had a chance to do some tests with them.  I set up a test rig where I can log data while charging the caps.  I don't have the charge balance boards back from OSH Park yet, so I've been testing a single capacitor.

    The test setup includes three serial-port-enabled DMMs and a steel ammo can to contain the coin cell just in case it bursts during high-rate discharge. 

    Here's an interesting bit of trivia: a CR2477 coin cell is rated to deliver 10,800 J of energy.  The 7.62mm NATO rounds originally stored in the ammo can have a muzzle energy of 3,304 J, so the coin cell can deliver as much energy as three .30-caliber rifle bullets.  Just much less quickly.

    Here are six of the capacitors, 400F each:

    I bought eight of them, figuring that a few extra to play around with could be fun :-)  I'll probably wire them together in this in-line arrangement, which will place the output terminals close to where they would be on a car battery; this will make it easier to connect the car's battery cables to the capacitor pack.  I might print a case for them just for fun.

    I arranged a simple test overnight to measure the capacitance and self-discharge rate of a single capacitor.  I charged the capacitor for a while, then logged data while it "self" discharged.  This discharge was due to a combination of true self-discharge, leakage through the 1N5817 diode, and the over-voltage protection circuit, which is a 10k trimmer and a TL431 shunt regulator. As a fist approach, I've chosen to model all the leakage as a parallel R, treating the whole thing as a single RC circuit.

    If I knew the actual value of the capacitance, I could determine Req from the discharge curve.  So, in order to find both the Req and capacitance value, C, I added an extra 10-ohm resistor across the capacitor and continued to log the voltage.  The result is a pair of exponential curves:

    For each curve, I estimated the RC time constant by linear regression; To do this, you can re-arrange the equation:

    to get:

    which is easily recognized as the form (y = a x + b), so can be fit with linear regression.

    For the first part of the curve, this yields an estimate of 330,914 for the RC time constant.  For the second part (with the 10-ohm resistor added in parallel), the RC time constant is now much lower, 4298.  Combining these two measurements allows us to determine Req and C, since we have two equations for the two unknowns:

    Solving these equations simultaneously (I used maxima) yields C = 435F, and R = 760 ohms.

    The capacitance seems reasonable for a 400F -10/+30% part, but the equivalent resistance seems low - this would mean a leakage current of 3.5 mA at a fully-charged 2.7V.  The capacitor itself has a maximum 1mA leakage, the Schottky diode 1mA (at 20V reverse voltage, so it should be lower here), and the potentiometer and regulator should be only a few tens of uA maximum.  Looking at it another way, with RC=4298, the capacitor will lose 10% of its energy in 3.7 hours, or half of its energy in 25 hours.

    I will measure the leakage current of the TL431 and Schottky diode separately, and see what those look like.

    This isn't necessarily terrible news, though, since I don't intend to leave the capacitors connected across the charger when it isn't running.  Now, I need to re-do this experiment without the charger connected to get a better estimate of the self-discharge of the capacitor alone.

    EDIT: some of the "extra leakage" I detected is probably due to dielectric absorption issues.  I see the datasheet specifies 1mA maximum leakage after 72 hours - I interpret this...

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  • Bad capacitors lead to bad measurements: now 62% efficient

    Ted Yapo12/09/2017 at 02:40 0 comments

    My initial estimate of 33% efficiency was due to a bad supercapacitor from the junk box.  I still haven't fully diagnosed the problem with that capacitor (high ESR might be part of the issue), but measurements with a new, verified, one look much better.

    The 400F capacitors won't arrive until Monday, so I did some new measurements with a 10F/2.5V aerogel capacitor that I bought some years ago (it is now obsolete).  Despite its age, it was still in the sealed DigiKey bag in my parts bin. To check its capacitance, I charged it to 2.0V, then timed how long it took to discharge to 1.0V when loaded with a 10 ohm resistor: 73 seconds.  The voltage follows an exponential curve:

    Using this equation, you can solve for C.  It's a fun little exercise; I did it in 8 lines on a scrap of paper, but this maxima code will do the same thing faster:

    solve(V[t] = V[0] * exp(-t/(R*C)), C);

    Either way, you get the following:

    Using this formula, we calculate a capacitance of 10.5F. Not bad for a -20/+80 tolerance part.

    New Efficiency Estimate

    So, armed with a validated capacitor, I timed how long it took to charge this new capacitor to 1V from a CR2477 cell with the converter.  The converter was drawing around 34mA for 82 seconds.  I will assume the battery voltage was 3V, because I didn't measure it (more about this below).  This equals 8.4J out of the battery.  Charged to 1V, the capacitor holds 5.25J, so the transfer was 62.5% efficient.  This is much closer to what I would have expected.

    So, there's a subtlety involved here, because under load the battery voltage is less than 3.0V.  I've chosen to use the nominal 3.0V in the efficiency calculation because this will approximately account for losses due to internal resistance in the cell: the energy actually delivered by the cell is somewhat less, but this should give a rough approximation of the capacity of the battery that has been consumed.  The 62.5% number is closer to an end-to-end efficiency measurement rather than just the converter itself.

    So far, I haven't connected enough DMMs to monitor the cell voltage and current at the same time.  I have three serial-port-enabled DMMs that I'll connect up once I clear some space on the workbench so I can log data during the entire charging process (cell voltage, cell current, and capacitor voltage).

    Next Steps

    So, the efficiency at low output voltages looks reasonable now.  I'd still like to improve it, and will be looking more closely at where the losses are.  But, losses are likely to increase when stepping up to higher voltages (like the 14V required on the big caps).  So, until the final caps arrive, I'll find the largest capacitance I can that's rated for 16V or more, and time some higher-voltage charges.

    Although the caps are still in transit, my 650A current probe did arrive (thanks, Amazon), so I can check out the voltage/current curves when starting my car (with the normal battery - for now).

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Enjoy this project?



rafununu wrote 5 days ago point

It looks that's faisable, theorically ! The energy given by the battery isn't 10800J because when the manufacturer measures the overall current  during discharge (1000ma), it says the voltage drop to 2V at the end. But anyway, there's far more than the 1700J your simulator gave you. Good luck.

  Are you sure? yes | no

Philip wrote 12/11/2017 at 11:58 point

As an alternative to charging the capacitors in series, which requires a DC/DC converter that can't be 100% efficient, may I suggest charging the capacitors one at a time, since the per capacitor voltage is near your coin cell voltage. The capacitors are still wired in series, with suitably robust wire since the series path must handle the engine cranking current. The coin cell is connected to each capacitor in turn, using relays (or a more complex circuit could be done with MOSFETs (they don't have to big, since the charging current is in milli amps)).

If the self discharge rate of the capacitors is low, then just one pass of charging each capacitor for 1/6 of the total charge time would work. If self discharge is high such that by the time you have charged capacitor 6, capacitor 1 has discharged too much, an alternative would be charge each capacitor for less time, and go around multiple times. For example 1/10 charge to cap 1, then 2 .... cap 6, then back to cap 1. Repeat cycle 10 times for total equivalent charging time.  With only relay (or saturated MOSFETs) , the losses in the charging process might be far less than the DC/DC converter.

Or,  now hear me out....  I could be full of it.

  Are you sure? yes | no

Ted Yapo wrote 12/11/2017 at 16:59 point

Charging in parallel/discharging in series is a reasonable idea, and switching the charging circuit seems easier and cheaper than switching the discharge circuit (200A battery switches are $7 each).  I considered a charge-in-parallel scheme at first, but finally settled on the charge-in-series.

The sneaky problem with charge-in-parallel (all at once or individually) is limiting the charge current to something the cell can handle efficiently.  Just hooking the cell across a 400F cap (which acts like a dead short) will charge it somewhat, but will introduce huge losses because of the internal resistance of the cell.  Also, the cell voltage will droop significantly under this load, so the final capacitor voltage will be low.

You can think of this as an impedance matching problem.  To get maximum power transfer, you'd want to match the impedance of the 400F capacitor (near zero) to the impedance of the coin cell (tens of ohms).  But you actually don't even want maximum power transfer, you want maximum energy transfer (minimizing the losses in the cell's internal resistance), so you want to present an even higher impedance to the cell, and drain it more slowly.

So, in order to charge even a single cap efficiently, you still need a DC-DC converter.  You might think you could get away with a buck converter, since the capacitors are 2.7V rated, and the cell is 3.0V, but because of the voltage droop, you actually need a boost converter for most of the energy. @jaromir.sukuba had to build such a converter to charge a single cap for his spot welder.  So, you have to build a DC-DC supply anyway, and it seems simpler to avoid switching the capacitors.  One snag may be maintaining the converter efficiency once the output voltage gets large; it's easier to design a converter that's efficient over a narrow range of output voltages.  No data on how bad this will be yet.

The multiplexed "parallel" charging is an interesting idea. I keep coming back to a "total discharge time" argument thinking that you don't gain anything (self-discharge-wise) by multiplexing the charge, except that the capacitors end up more balanced at the end.  And, of course, you don't have to switch the discharge path. I'll have to think about it some more.

Thanks for the switched parallel charging idea; I'll keep it in the back of my mind.  Who knows what might be needed in the eleventh hour?

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esot.eric wrote 5 days ago point

Somehow, intuitively, to me anyhow, this'd also result in more stored charge. I understand that e.g. two series caps will charge with half the overall capacitance, but... if charged in muxed-parallel like this, it seems like the overall stored charge would be higher... I'll have to contemplate this further.

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Ted Yapo wrote 5 days ago point

@esot.eric there are a bunch of thoughts/facts/theories about capacitors that always make for an interesting few hours with pencil and paper (or SPICE simulators or computer algebra software).  Like charging a capacitor through a resistor always wastes half the power.  Even if the "resistor" is just the series resistance (ESR) of the capacitor itself.  Then, there's the capacitor paradox - two ideal capacitors: one charged with charge Q, one empty.  Connect them together, and each now has charge Q/2.  Sounds good until you realize that each one now has one quarter of the energy of the original one charged to Q.  So half the energy is gone, now, too.

I think capacitors should be the new over-unity free-energy fake video subjects; everyone is tired of magnets and red enameled wire.

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esot.eric wrote 5 days ago point

@Ted Yapo, interesting math/paradoxes I don't recall from my studies over a decade ago.

Simulation of ideal components shows exactly the same discharge curve, when series-discharged, for both series and parallel-charged. ... as would be expected from training.

Here's a thought-experiment: say two capacitors are in series, the two capacitor plates which are connected together, when in series, (along with the interconnecting wire) have been somehow previously stripped of all their "free" (valence?) electrons... what, then, is attracted by the most-positive plate, when the series-capacitors are connected to a battery? And the most-negative? Am thinkink SPICE doesn't handle that... otoh, maybe such a situation would destroy the capacitors. Otooh, surely there's an inter,mediate state which wouldn't, and would affect the devices' capacity...?

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Ted Yapo wrote 4 days ago point

@esot.eric That's a good one.  It strips through several layers of abstraction.  "Capacitors" and "inductors" are idealized abstractions (the so-called lumped elements) of Maxwell's equations.  Maxwell's equations are idealized abstractions of something else, I guess quantum electrodynamics.  Who knows what that's a simplification of? Each layer of simplification probably introduces some apparent paradoxes (paradoxen?).

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oberstdackel wrote 12/09/2017 at 07:34 point

I doubt a coin battery can generate enough current 60-200 ampere requried for 3+ seconds, approx 1kW. ...

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Ted Yapo wrote 6 days ago point

No, the coin cell can't do that (alone).  If you charge a bank of supercapacitors from the cell over a longer period of time, you can use all the energy from the cell (excluding losses during charging) much faster than using the cell directly.

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BUT wrote 12/08/2017 at 18:25 point

I would think it would be easier to open a fuel injector and fire a spark plug.

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Vije Miller wrote 12/05/2017 at 04:23 point

Perhaps the 3V cell turns a crank-gearing-device that generates current - ie - the cell may not need to directly provide the power - uh - directly - 

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Ted Yapo wrote 12/05/2017 at 13:55 point

The problem is one of power vs energy.  Calculations show that there's enough energy in the cell to turn the engine over, but you can't extract it quickly enough (i.e. the cell can't provide enough power, energy/time).  So, the idea is to extract the energy from the cell slowly then release it all quickly to start the engine.

I can imagine doing this with a mechanical device - maybe the cell runs a small DC motor that's geared way down to wind a very strong spring over a turn or two.  You let the cell run the motor for a day, transferring most of the energy into the large spring.  Then, you connect the wound spring to the crank-starter of your Ford Model T, and let the spring go to start the engine.  The spring releases the energy much more quickly than the cell can, so produces much more power for a short time, which is what you need to start the engine.

But, alas, I lack the tools - and quite frankly, the skill - to build such a mechanical device (and more importantly, I am fresh out of Ford Model T's).  I'm OK with a soldering iron, though, so I'll substitute supercapacitors for the huge spring and do the whole thing electronically.

At least, that's the plan.

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Vije Miller wrote 12/08/2017 at 18:10 point

[whispers] cheaty idea [whispers] it says a SINGLE coin but only at a time - using a magazine loading mechanism - you drop - drain - drop - drain - drop - drain - drop - drain - drop - drain - drop - dra--well you shd get the point. One cell technically at a time - draining it in a cap series - so all you need is a Mentos like stack and a Pez dispenser. [unwhispers] BUT THAT WOULD BE CHEATING [whispers] don't listen to me - I think that could work [unwhispers]

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Ethan Schwartz wrote 12/08/2017 at 16:57 point

Assuming that getting enough electrical energy from a 3V cell to directly operate the electric starter in a car engine is just not going to happen, I'm wondering if it would be considered cheating to use a gravity assist to generate power and the power anticipated from a running engine to reset the gravity assist?

What if you had a weighted flywheel attached to a generating device where the 3V cell's job is to nudge that weight from a balanced at-the-top position to the point where the flywheel is being spun by the force of gravity acting on the weight? 

If you use energy later generated by the engine to move the flywheel back into position then you have a device (admittedly finicky and not very portable) that can repeatedly start an engine using a supply of 3V coin cells.

Is that cheating?

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rafununu wrote 12/04/2017 at 15:04 point

3W isn't enough energy to run a starter whatever you can do, it's only 3W !

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davedarko wrote 12/04/2017 at 15:52 point

if there's one person who can, it's Ted ;)

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Ted Yapo wrote 12/04/2017 at 20:20 point

Thanks for the vote of confidence.. I'll try not to let you down :-)

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Clayton G. Hobbs wrote 12/04/2017 at 15:55 point

Watts aren't energy, joules are.  The calculations in the project's details show how much energy is in a CR2477, which should be enough energy to start a car assuming the linked video isn't fake.

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Ted Yapo wrote 12/04/2017 at 19:54 point

The video might be fake; but the idea is plausible.  I found ranges of 130-225A to crank 4 to 8-cylinder engines on automotive sites.  At 14V, that's 1820-3100W.  Assuming you need that power for 3 seconds (a conservative guess), that's 5460-9300J.  I also found links that indicate the true cranking time is substantially less than 3s.

I have a 650A oscilloscope current probe on order so I can collect my own data.

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rafununu wrote 12/07/2017 at 15:03 point

You're playing with words but know what I mean. I've got such a starter box for cold mornings, it contains supercapacitance (several Farads) and a loading circuit to charge these, it is only able to give 4 to 5 "starts". It takes more than an hour to fully load the box ! A normal starter needs 40A/12V during of few seconds. That's obvious 3W cannot fit else if why manufacturers still use batteries, they're permanently reducing their costs.

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Clayton G. Hobbs wrote 12/07/2017 at 17:00 point

"Using words correctly" is "playing with words" now? 🙃

Honestly, I didn't really know what you meant because of the confusion between power and energy.  I still don't feel confident that I know what your point is.  It's true that 3 W can't run a car's starter motor, but I don't think anybody here said it could.

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Ted Yapo wrote 12/04/2017 at 19:43 point

Yes, your statement is true (if you meant "power" instead of "energy"), 3W is nowhere near enough power to run the starter.

I intend to use 2000-3000W instead (depending on which car I end up starting (4, 6, or 8 cylinders).  That is plenty.

@Clayton G. Hobbs  has the explanation above.

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rafununu wrote 6 days ago point

Yes I wrote power instead of energy, and english isn't my mother's language, which isn't an excuse but doesn't indeed help. Anyway, the goal was clearly defined "The goal is simple: replace my car battery with a coin cell (plus clever circuitry), and get the car to start at least once using only the energy from the cell.". I only gave my opinion writing this is actually impossible with only one coin cell and a standard starter. Who knows what will be the future, maybe coin cells will deliver hundreds of amps, maybe Ted will discover something totally new and unpredictable as Dupont's chemists did when they found Nylon ?.

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Ted Yapo wrote 6 days ago point

@rafununu I don't think there's anything new here - I intend to use the coin cell to deliver tens of milliamps over many hours to charge a supercapacitor, which can deliver hundreds of amps for a few seconds.  The supercapacitor and charger are the "clever circuitry" in the original statement.

You're right, there's no way a coin cell can deliver hundreds of amps.  But, if drained slowly enough, you can theoretically extract enough energy to start a car.  The trick is storing all that energy where it can be released quickly, and supercapacitors are perfect for that.

Will it work?  Maybe. You have to charge the capacitor very efficiently to end up with enough energy to start a car.  This may not be possible.

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Riyadh Elalami wrote 5 days ago point

3W is power, what he is intending to do is store the energy he has in that coin cell in capacitors which may very will take hours to charge. and discharge them all in a few fractions of a second, producing a huge amount of power momentarily.

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