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Examination of Charge Recycling/Splitting the Positive with a purely resistive load

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  • Examination of Charge Recycling/Splitting the Positive with a purely resistive load

    This is an experiment I intend to start working on soon and thought I would first see if anyone has already done this and also see if anyone wants to make any suggestions concerning the methodology.


    Some time back there was a discussion concerning the "two capactior paradox" which I resolved to my satisfaction by learning that the discharge of energy from one capacitor into another follows conservation of charge, which by definition means it may not also follow conservation of energy. A second infrequently noted behavior of discharging one capactior into another is the absence of change in the intial and final states of energy as a function of resistance between the capacitiors. A 10 V 1mF cap discharged into a second empty 1 mF cap yields two caps at 5 volts. If one places a 10 ohm resistor between the caps and repeats the above one again ends with two caps at 5 volts. If one places a 1k resistor between the caps in a few seconds one finds two caps at 5 volts, 100k resistor and repeat, shut off the volt meters and come back in a few minutes and one finds two caps at 5 volts. The initial and ending states of energy are unperturbed by multiple orders of magnitude difference in resistance. This demonstrates that for the discharge of electrostatic energy from one cap to another there is no such thing as a "resistive loss" as a dissipative loss of energy would by definition require a change in the ending state of energy in the caps, there is also no loss of charge. There is only a decrement of power caused by increasing resistance.

    Two capacitors placed in series leads to a doubling of the dielectric space between the series capacitor plates. Therefore the capacitance of capacitors in series follows the forumla C(total) = 1/(C1/1+C2/1 ... Cn/1), in the simple case of two capacitors of equal capacitance placed in series this equals a halving of capacitance. Electricity stored electrochemically as opposed to electrostatically does not follow this dielectric spacing driven change seen with capacitors. Therefore, the capacity of batteries in series is unchanged, two 1 Volt, 1000 mA/hr batteries in series yields one 2 Volt 1000 mA/hr battery, not one 2 Volt 500 mA/hr battery as might be expected if the energy were stored electrostatically. Given the above one might envision two batteries placed in series the negative of the series batteries connected to a single battery yielding a voltage difference across the split positive which might power a resistive or other load. This approach is far from new and has been referred to as a common ground, split positive or Tesla switch approach, for the purpose of this write-up it will usually be referred to as charge recycling as I feel the term is more descriptive and catchier. Given the above it might be conjectured that as is the obvious case with two capacitors, discharging two batteries in series to a single battery across the split positive with varying resistive loads also leads to no resistive losses affecting the intital and final states of the batteries. Further as batteries in series do not lose capacity, such a set-up might effectively allow for a recycling of the charge. However, just because electricity is not dissipated by resistance does not at all mean that the electricity is performing the work we might like it to. I.e. if the battery being discharged to is full to start with the electrical energy, though not lost, is also not driving a reaction that alters the chemistry of the battery to store electrical energy it is instead driving a reaction that leads to off gassing and or other phenomena that might damage the battery. It is still reasonable to guess that a portion of the charge after drving the resistive load would lead to recharging of the battery and this charge would be effectively recycled.

    This was looked at roughly with three niMH batteries switched by hand. A three volt incandescent flashlight light bulb which remains lit down to about 1,2 volts was used as the resistive load. The measured resistance across this bulb indicated that for a single battery the discharge rate would be approximatley C rate 0.5. For the "control" arm the three batteries were charged fully, bled off down to about 1.4 volts and placed in parallel. The bulb was connected across the batteries for ten mintutes and voltage measured. Load was disconected for ten minutes, voltage measured, this was repeated for two hours. As the three bateries in parallel would have a C discharge rate of 0.17 as opposed to 0.5 for a single battery, secondary to Peukart's law, the null hypothesis is that discharging three batteries in a charge recycling set-up would do worse than simply putting the three in parallel. For the first experimental arm two batteries were again set to 1.4 volts while the third battery was at 1.3 volts (the same batteries are used each experiment). Again the load was placed on for ten minutes, this time across the split positive and voltage measurements taken, followed by a rest period of ten minutes, this repeated (with battery rotation) for two hours. For the batteries in parallel at the end of two hours they had lost 0.082 volts, for the charge recycling set up the three batteries had lost 0.055 volts a 1/3rd improvement in energy utilization. For the second experiment the charge recycling setup was again used however this time all batteries started at 1.4 volts. In this set-up after two hours the batteries had lost 0.08 volts as opposed to the 0.082 volt loss seen with all three in parallel.

    At this point I am 90% convinced there is something to this splitting the positive charge recycling. One might argue that in the case that showed a large improvement the average starting voltages were different and so one was on a different part of the battery discharge curve, still it is a pretty large change. It might have been a cleaner experiment if I had used a plain resistor as opposed to a light bulb as the temperature variation would likely be less and resistance of a resistor varies as a function of temperature, still it seems worthwhile to see if one is actually doing something useful such as lighting a bulb. The big one to me is perhaps the battery in the charge position is acting as further resistance as compared to just simply going to battery negative. In this case one has less voltage loss simply because less power flowed, this should be reflected in the brightness of the light. Eyeballing it, they appeared roughly equal, though the charge recycling set-up varied from very bright to less bright as the charge battery filled. Maybe there is 1/3 less voltage lost and 1/3 less light? It also seems to me if the charge battery is a source of resitance, in the case where all three batteries all started at 1.4 volts there would be the least power flow and this set-up would have shown the smallest decrease in voltage over two hours, which was not the case, so I am optimistic about this. To try and look at this more rigourously I am considering the following.


    Obtain a data logging lux meter, place the bulb in a box the ceiling of which houses the lux meter sensor. Obtain two datalogging volt meters, am looking at this Redfish meter Use four batteries in the Tesla switch set-up and use an arduino with four relays to swith between parallel and series, a fifth relay for if you want to disconnect the load and let batteries rest. Repeat the above experiment and log all the data. Once you know the lumens produced one can say more definitively whether there is an improved utilization of battery energy or if it is all smoke and mirrors. If there is something interesting, maybe try switching the batteries with no rest period, try one minue on one minute off, 6 seconds on, six seconds off, six seconds switch with no rest period, you get the idea. Once it is all automated and being logged you just set it up come back in a few hours and see what you've got. If there is something to it, there are likely sweet spots in terms of timing and discharge rate. Depending on results, eventually one might look at inductive loads or harvesting magnetic flux with diodes as per the Tesla switch, however this is a more simple question, is there charge recycling going on with such a set-up? If not the Tesla switch would be just a very complicated way of doing what you could do more simply with a cap dump. If there is charge recycling, then for the cost of a few relays maybe you could increase the size of a solar/battery set-up by 1/3rd.

    Will likely order parts this week-end or sooner, so if someone (_RS?) has already done this or someone wants to say that's a stupid set-up because ... please let me know in the next few days.

    Last edited by ZPDM; 05-02-2019, 10:40 AM.

  • #2
    If anyone is interested in this I am moving along and making progress. I have everything I need to look at this switching batteries manually, ie I have a "lightbox" I am able to track lux in real time and voltage in realtime. A few things. 1) the bulb draws about 0.8 watts, on the one hand this is good in that chargers are about 500 mA. On the other hand if I have 3000 mAh batteries and if I want to compare 4 of them in parallel (12,000 mAh) to a charge recycling set-up it would take 7.5 hours to run them down half way, 15 hours if you gave a 50% rest period after each bit of on time, and you would have to sit there waiting to switch everything on or off (for the rest period) every ten minutes for 8 or 16 hours on end. Now you could add two or three bulbs, which I may probably end up doing but charging at too fast a rate is inefficient and potentially damaging.

    Fortunately I have some very very crappy recycled batteries with 800 mAhrs capacity left. With these, even ten minutes on, ten minutes off it should only be two hours before depleted. So I will look at this first with these. I will proceed to arduino driven relay switching but a pain to optoisolate them, so compare them to manual switching and see if similar(i.e. however unlikely no leakage from laptop->arduino->batteries as I should but don't feel like trying to optoisolate the relays).

    Alright this just occured to me now, if you want to simplify this to as basic as I can think of 1) charge a battery through say 50 ohms of resistance from a power supply, measure watts 2) place two batteries in series, charge the battery though 50 ohms of resistance, measure watts and what you got (Faradaic efficiency) into charging the battery. 3) place two batteries in series, charge the battery though 50 ohms of resistance from an incandescent light or any other load.
    Last edited by ZPDM; 06-11-2019, 10:39 PM.


    • #3
      Yes, interested in your progress. Never did see that first post of yours.
      Aaron Murakami

      You never change things by fighting the existing reality. To change something, build a new model that makes the existing model obsolete.” ― Richard Buckminster Fuller


      • #4
        Hi Aaron and All,

        Doesn't work, Zilch, zero, nada. Aside from any benefits which may be found from extremely rapid pulsing which was not examined there is no beneift to this set-up, yes there is charge recycling but it is of no benefit. What confused me is if you have a say 4 volt 1000 mAhr battery and use it to light a lightbulb with 10 ohms resistance you measure a 400 mA draw. If you put two 1000 mAhr 4 volt batteries in series and place the light across the "split positive" to a third 4 volt battery that is connected by a common ground you have a 1000 mAhr 8 volt battery and again a 4 volt difference across 10 ohms resistance, the lightbulb lights the same, there is again a 400 mA draw, and one watches the third battery go up in voltage. Conservation of Charge, what can possibly go wrong?

        Not being an engineer it took a silly amount of time to gather some decent data. The 30 dollar battery tester had two of its four bays bad, and for weeks I thought I was being sent half used batteries from two different suppliers, while each capacity test takes ten hours. Three light bulbs burnt out mid test, I realized behaviour of Li ion batteries is quite dependent on state of charge and whether they have recently seen a load, can't solder to a flashlight bulb well etc. Eventually I was able to get good repeatable data, around the nominal voltage of the batteries and the series batteries went down twice as much as I expected them to. Why oh why was that? Then the "terrible" truth dawned on me. If you have a 4 volt battery 1000 mAhr battery and it loses 10% (100 mAhrs) of its capacity it will be at 3.6 volts (not correct with the discharge curve but you get the idea) if you put two 4 volt batteries in series you have an 8 volt 1000 mAhr battery. If this battery loses 10% (100 mAhrs) it is now at 7.2 volts. Yes nearly all the charge was recycled but it is a game of voltage and you lost 0.8 volts and recovered 0.4 as opposed to just losing 0.4.. Interestingly if you split the series battey back into two batteries it is as though they had lost 200 mAhrs, each 1000 mAhr battery went from 4 to 3.6. I should have seen this coming and feel stupid I went through all this effort to discern what should have been obvious from the start, a 100 volt 1000 mAhr battery loses 10 volts for a 100 mAhr loss, a 10 volt 1000 mAhr battery loses 1 volt for a 100 mAhr loss. I'll lick my wounds on this one for a bit and keep track of gongs on here, but as for the Tesla Switch all the "magic" is from 1) capturing induction and possibly 2) behaviour of batteries that are pulsed with extremely short duration rapid pulses.
        Last edited by ZPDM; 07-06-2019, 02:29 PM.


        • #5
          Read the Benitez patents - those run steady for long time intervals with no pulsing or short periodic switching.

          You seem to be equating a loss in capacity with the proportional loss in voltage, but it doesn't work that way. You mention the curve and that we'd get the idea, but they are really are two different things. And lead acids are different from lithiums.

          Lifepo4 for example is a constant voltage battery that runs and runs. You can lose 25% of the capacity and the voltage hardly changes - it hangs on until it falls off the cliff at the end.

          Lead acids are constant current, which is inverse - voltage will go down on the expected curve while being able to deliver fairly constant current.

          It's also difficult to see anything with such tiny batteries that have a lot of resistance, relatively speaking. 1 amp hour batteries are tiny. This kind of test needs to be done with 100 amp hour batteries and up where the internal impedance is at a bare minimum.

          Benitez's method does have distinct differences from the Tesla switch, which is from Ronald Brandt and it probably was inspired by Benitez's work since there is no evidence that Tesla used this method.

          If you haven't seen Peter's presentation on Benitez, I'd recommend it and would recommend studying Turion's 3 battery discussion in Energetic Forum.

          There is something to the fast pulse switching and Bedini experimented with that as well. And then there is Tibor's method of powering a load between the negatives - that was in my Cold Electricity presentation.

          The 3 battery method is all about potential differences as you probably know by now. What about a potential charge? If you charge a cap to 18 volts for example and dump it on a dead battery, you can do this so the cap may only go down to the level of the battery. You only have to make up the difference, but on each discharge, you are hitting the battery with the full potential of the voltage in that cap at whatever capacitance. If you do it really fast, the cap won't discharge down to the level of the battery and you hardly have to put anything back in it but are still able to benefit from the impulse of the full potential in the cap.
          Aaron Murakami

          You never change things by fighting the existing reality. To change something, build a new model that makes the existing model obsolete.” ― Richard Buckminster Fuller


          • #6
            Hi Aaron,

            Will see if I can take a look at those patents maybe sometime this summer, thx, with a 5 second glance I did see coils.
            Yes, Li ions have an S shaped discharge curve, dropping rapidly from 4.2 volts to about 3.9 befor plateauing ovr 95% of thier discharge between 3.9 and 3 volts then dropping off rapidly again, while lead acid have a more linear discharge. I am specifcally not equating loss of capacity with proportional loss of voltage which is why I stated "not correct with the discharge curve". Nonetheless the concept is obvious. To state it again, let's consider a three battery system powering a purely resistive load such as an incandescent light, here is what one would observe. If the batteries are near nominal voltage and on the linear part of the discharge curve the two series batteries go down 0.4 volts while the charge battery goes up 0.19 volts for a loss of 0,21 volts (these figures after the batteries have rested). If one simply powered the light from a single battery one would have seen a 0.2 volt drop. Again, this is obvious once realized, if a 4 volt and 8 volt battery are of equal capacity x loss of capacity leads to y loss of voltage in the 4 volt battery while x loss of capacity in the 8 volt battery leads to 2y loss of voltage in the 8 volt battery. Put even more simply a 1000 maHr 4 volt battery drained to dead gives a 3 volt battery. Two 1000 mAhr batteries in series yields an 8 volt 1000 mAhr battery this battery drained to dead give a 6 volt battery. 1000 mAhrs consumed in each case but a one volt versus two volt loss. Should have reasoned it out before I looked at it I thought some of the magic of the Tesla switch was the series to parallel discharge it is not, all the magic is the rectification of magnetic flux and capture of the displacement current, along possibly with the rapid pulses. Actually one is at a disadvantage as compared to going simply to a capacitor (what Bedini did put in his patent if I am not mistaken) as Faradaic losses from poor charge acceptance lead to losses going from battery to battery. In the case of lead acid this would be expected to be about 20% while 1 to 10% Coloumbic losses are seen with lithium ion.

            While I would have to dig out the link, the diagram I saw of the TS showed coils all over the place, six of them and six secondary coils as well, so I realized for some time that capturing induction was a large part of the TS I just guessed that the series to parallel battery discharge was an aid. It isn't, it makes it more difficult as opposed to simply placing the spike in a cap.

            While I am not ready to start on this at the moment it should be possible to determine the effect from pulsed DC and the possible effects on ion momentum that Bearden and Bedini postulated. 1) Charge a battery at constant current 2) charge a battery with pulsed current DC at same voltage and power 3) charge a battery with DC radiant spike at same power. I would guess you might only see something with the radiant spike as that voltage will be 10-20x higher.