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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.

  • How much current does it really take?

    Ted Yapo2 hours ago 0 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 Yapo11 hours ago 11 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.

    Ideas?

  • Supercapacitor Testing

    Ted Yapo2 days ago 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...

    Read more »

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

    Ted Yapo4 days ago 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).

  • First Prototype DC-DC Converter

    Ted Yapo6 days ago 9 comments

    I wired up a first prototype converter last night and coded up a quick test this morning.  So far, it "works," but not great.

    The converter is based around a PIC12LF1571 that drives a simple MOSFET/inductor/diode circuit.  Here's the schematic:

    The PIC code (attached at the end of this log) generates a 17us pulse every 94us to magnetize the inductor by switching on the MOSFET.  When the magnetic field decays in the coil, the voltage rises to maintain the same current flow, charging C1 through the Schottky diode.

    I chose this topology because I had better NMOS transistors in stock than PMOS. I first saw the IRLML6244 when @Elliot Williams used it in his version of #TritiLED, and it has quickly become one of my go-to parts.  I might end up substituting something else later, but for now, this seems fine.  Two output pins from the PIC are paralleled to increase the drive current and promote faster switching in the FET.

    I originally modeled the circuit with a BAT54 diode, because everyone has a million of them, but the 1N5817 is more efficient in this circuit due to the large current pulses.  The downside to the 1N5817 is the larger reverse leakage (up to 500uA, even 1mA at 20V on some datasheets).  Quick estimates make it more efficient, but I'll have to examine this in more detail.

    I wanted a low-ESR capacitor, and the best I could come up with was 10x 10uF 1206 ceramic caps in parallel.  I'll probably add another few hundred uF of low-ESR polymer caps when I find the missing bag of them around here.  This large capacitance averages out the current drawn from the cell so that it doesn't see the hundreds of mA drawn by the inductor during the pulses. There are some other miscellaneous bypass caps on the board, but they probably don't do anything interesting :-)

    I didn't have any beefy inductors of around 100uH in the junk box, so I hand-wound one on a core salvaged from an old power supply.  Nearest I can figure, the core is a T131-26 (yellow/white paint).  I wound 35 turns of 20AWG wire on there, which ends up measuring out to 115uH, as predicted by the formulas (113.9).

    I've only had a chance to test the converter briefly, and so far it "works," charging a 10F capacitor to 1V in a little less than 2 minutes, while drawing 43mA from a CR2032.  A quick calculation shows that this is 15.5J out of the cell, with 5J stored in the capacitor for an efficiency of around 33%.  I need to at least double that to hit my goal of 67F charged to 14V.  In the circuit's defense, I didn't precisely time the charging, so the measurement might be off.  Another possibility is that my 10F capacitor is more than 10F.  It came from the junk box, and isn't marked with a tolerance, so who knows.

    Next, I need to instrument the circuit to see what's going on - I'll add a 0.1-ohm resistor in series with the inductor so I can see the current waveforms, which will probably be the most interesting.

    I also need to add an over-voltage crowbar to the output.  With the 67F capacitor, there probably won't be any danger of over-charging, but the smaller capacitors I have to play with for now could easily be damaged by a few minutes of inattention.  My plan is to add an adjustable shunt regulator, a TL431, directly across the output.  The power output of the converter is low enough so that the TL431 can just dissipate it all once the target voltage has been reached.

    Here's the trivial code running on the PIC:

    ;;;
    ;;; coin_cell_cap_charger.asm: DC-DC converter for charging supercapacitors
    ;;;                            from coin cells
    ;;;
    ;;; Copyright (C) 2017 Theodore C. Yapo
    ;;; Licensed under MIT license (see file)
    ;;; 
    
        LIST        P=12LF1571
     #include    <p12lf1571.inc>
    
      ERRORLEVEL -302
      ERRORLEVEL -305  
    
    ;;;
    ;;; I/O pin configuration
    ;;;
    GATE_DRIVE_A  equ   4
    GATE_DRIVE_B  equ   5
    
     __CONFIG _CONFIG1, _FOSC_INTOSC & _WDTE_OFF & _PWRTE_OFF & _MCLRE_OFF & _CP_OFF & _BOREN_OFF...
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  • Extracting energy from coin cells (quickly)

    Ted Yapo12/05/2017 at 02:35 0 comments

    I need to test some CRxxxx coin cells.  So, I'm going to build a tester.  Here's the rough idea.

    The actual parts will depend on what I find first in the parts bins.  The op amp maintains a constant discharge current from the cell through R1, while the cell voltage is monitored by a serial-port DMM.  With this setup, I can measure the discharge characteristics of the cell at various current drains.

    Why do I feel the need to do this?  The amount of energy you can extract from a cell depends on the rate of discharge: faster discharge means less energy available.  This is especially true for coin cells.

    Jack Ganssle wrote an interesting article about his experiments with CR2032 cells on embedded.com a while back.  I read it before when playing with TritiLEDs but quickly decided that his findings didn't apply to the really low current drain of those devices.  I do agree with him that it's pointless to estimate run-times in excess of cell shelf life, though.  (Shelf life for CRxxxx cells is generally taken to be 10 years.)

    In any case, the article describes the problem with high-drain use of coin cells - the internal resistance increases quickly as capacity is used, and this resistance consumes more and more of the power as the cell ages. One data point from the article is that at a 30mA drain, you can only extract 39% of the energy from a CR2032 cell.  This equates to a C/7.5 drain rate (225/30), which is very high.

    I'm planning to use CR2477 cells, which have about 4.4x the capacity of CR2032s (1000 mAh vs 225 mAh).  My initial thought was to discharge the cell (and charge the capacitors) at C/25 (25 hour rate), which for the CR2477 is 40mA.  If we assume a CR2477 is just a 4.4x larger CR2032, then this would be equivalent to around a 9 mA drain on a CR2032.  The article has curves from a 10 mA drain on a CR2032, which show that 88% of the energy can be extracted from a CR2032 at this rate before the cell voltage hits 2.0V.  So, I roughly estimate that I can extract 88% of the energy from a CR2477 discharging it at a constant 40 mA.

    What does it all mean?  I might be OK with a 25-hour discharge/charge cycle.  But, I think I should build a version of the circuit above and test some cells.  I'll probably test some CR2032s initially, since they're so much cheaper than CR2477s (and I have a *lot* of CR2032s around for another project :-)

    You may be wondering why I don't just charge the capacitors over a week (or more) to extract most of the energy from the cell.  Other than having to wait for results (before the contest ends), the other problem is the self-discharge of the supercapacitors.  The capacitors specify a maximum leakage of 1mA, which I'd like to keep to a small fraction of the charging current to maintain efficiency.  There's probably a cross-over point where the extra capacity extracted from the cell is lost to capacitor leakage.

  • Proper care and feeding of supercapacitors

    Ted Yapo12/03/2017 at 23:15 0 comments

    Strings of capacitors have similar issues to strings of lithium ion cells: differences between cells can cause one or more cells in a string to become overcharged.  In the case of capacitors, leakage current and capacity differences are the culprits.

    To keep all the capacitors happy, you need a charge balancing circuit, so I designed a PCB around the ALD810022/ALD910022 Supercapacitor Autobalancing MOSFET IC's.  These devices are amazingly simple in operation, using V/I curve of specially trimmed MOSFETs to shunt excess charging current around fully-charged capacitors.  You simply need to connect the board pads to each node in the capacitor string. The Eagle design files for the PCB are in the GitHub repo.

    ALD markets similar boards, which I will diplomatically call "high-margin items".  You can get an un-populated PCB from DigiKey for around $20.  The PCB above will cost you $5.35 for three copies at OSH Park.  Even splurging for Super Swift service, I'm only paying $10.70 for three copies.  Not bad.

    The circuit is pulled right from the datasheet; the 8100 device contains 4 MOSFETs and the 9100 two of them, so both packages neatly handle six capacitors.

    To see how these devices work, I ran a quick LTspice simulation using MOSFETs in the library - of course, these devices don't match the capacitor voltage I'll be using, but the qualitative behavior is similar.

    C1, C2, and C3 represent capacitors with a spread of capacitance and leakage current (modeled with parallel resistors).  The three MOSFETs start to shunt current around each capacitor as it reaches about
    1.3V.  For voltages above 4.0 for the string, the capacitor volatges become balanced, which you can see in this DC sweep:

    With the ALD8100xx devices, ALD has accurately trimmed the V/I curve on each MOSFET instead of the random choice of devices I made in this rough simulation.  In practice, you choose the version of the ALD ICs based on the maximum leakage current and maximum voltage for your capacitors.

    Capacitor Selection

    I chose to use a string of 6 SCCY68B407SSBLE capacitors from AVX.  These caps are rated 400F at 2.7V, so the string of 6 will have a maximum 16.2V rating with 66.7F of capacitance.  To select the proper ALD charge balancing IC's, I note from the datasheet that the capacitors have a maximum leakage current of 1000uA.  From the ALD81-9100xx family datasheet, I see that the ALD{8,9}10022 parts have  a forward voltage of 2.64V at 1000uA of current, which will keep each capacitor safely under the 2.7V limit.  Perfect - DigiKey had them in stock, too.  They're on the way.

    The capacitors have a maximum ESR of 2.2mOhm, so the string of 6 will have 13.2mOhm resistance, plus whatever wiring I add.  They are also specified for a 245A peak discharge current.  This is probably enough to start most 4-cylinder engines, and maybe some 6-cylinder,  but probably not my daily ride, which has a 4.7l V8.  I may have to find another vehicle to experiment with.  Then again, the ESR alone will allow currents of around 1000A, so if I use beefy wiring, maybe I can push the caps a little.

    Either way, I'll have to make sure the engines are properly warmed up for the attempt(s).  Cold oil is tough to get moving.

  • Buck, boost, or bust?

    Ted Yapo12/03/2017 at 05:00 0 comments

    Astute readers will have noticed I incorrectly said I needed a boost converter to transfer energy from the 3V coin cell to the 14V capacitor bank.

    This is true once the capacitors are charged above the cell voltage, but from 0 to 3V or so, I need a buck converter.  Unfortunately, I can't just connect the coin cell directly to the capacitors to get them initially to 3V.  If I did that, the internal resistance of the cell would waste too much power - to get the maximum energy from the cell, it has to be transferred slowly, probably over the course of many hours.  So, I need a buck/boost converter.


    Since I don't really care that the cell and the car share a common ground, I can use the simplest buck/boost topology, which incidentally inverts the output voltage.  The car won't mind as long as I get the polarities right at the battery connection.  I made a first pass at a LTspice simulation for the converter, as shown above. I chose a P-channel MOSFET from the library basically at random; none of the components are set in stone yet.  R1 simulates the internal resistance of the CR2477 cell - this changes over the life of the cell, so I'll have to account for that at some point.  C1 serves to smooth the current drain from the cell, providing a reservoir for the current pulses into L1.  M1 switches current into the inductor when turned on, then switches off to allow L1 to drain into D1 and C2, the B.F.C. (note the "67" with no units  - that's 67 Farads).

    Here is what the cell current and output capacitor voltage look like in the first 100ms of simulation:

    The circuit draws about 40mA from the cell, which would deplete the cell in roughly 25 hours - I may have to wait a day for the capacitors to charge, but this should extract a decent amount of the available energy from the cell.  Once I have a stock of cells to play with, I can look into decreasing the charge time (or possibly extending it, if necessary).

    During these 100ms, the capacitor voltage has risen to around 150 uV.  Not very impressive, is it?  But, if that rate were continued, it would charge the capacitor to 150e-6 * 10 * 60 * 60 * 25 = 135V in 25 hours.  Of course, this would greatly exceed the capacity of the cell, so can't happen.  In reality, the output of the converter will be roughly constant power, causing the dV/dt on the capacitor to decrease over time.  So, the last 100ms of charging will actually see much less than 150 uV of voltage change.


    Oh, the magic pulse source, V2, driving the MOSFET is probably a PIC15LF71.  In the simplest case, it can just generate a constant driving waveform, but if there's something to be gained, I can monitor the cell and/or capacitor voltages with the PIC ADC and make adjustments.  I haven't really thought that part through, yet.

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Philip wrote 2 days ago 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.

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Ted Yapo wrote a day ago 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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oberstdackel wrote 4 days ago 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 11 hours 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 4 days ago 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 4 days ago 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 4 days ago 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 6 days ago 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 5 days ago 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 13 hours 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 11 hours 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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