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  • Turion
    replied
    The BatterySaver model takes the remaining low-voltage output from dead batteries and raises the voltage to a usable level by discharging the remaining energy at a high rate. It does this by operating at a lower voltage than standard voltage regulators, allowing it to extend the life of almost any battery type, capacity and chemistry.

    This unique approach avoids expending energy when boosting is not required, as this technology uses low power switching to autonomously engage the boosting circuitry only when necessary. No other boosting method can recover this remaining energy in a dead battery without sacrificing response time or power output, both of which make other methods unusable in most electronic applications.

    How does this work? It is essentially rewiring a circuit board that can fit on top of an existing battery or inside the battery casing; It’s small, low-tech, and cheap enough to be retrofitted or built into a device from the get-go, and smart enough to tap into energy stores that regularly go to waste.

    After completing four years of prototyping, the team has bootstrapped $100,000, and is exploring licensing offers.

    At Udell’s location in the Santa Rosa hills, he shows me a working prototype, about the size of a penny, a strip of wires to connect to a battery terminal. BatterySaver prototypes are small, inexpensive, flexible, and can be attached to terminals or housing, installed on the battery itself, or retrofitted after manufacture.

    Gabriella Hasbun

    Udell removes a dead battery from a radio, connects the prototype, reinserts, and the machine comes to life.

    “We have an electronic solution to a chemical problem,” Bill Seidel, 75, a longtime Udell collaborator with a gray ponytail, tells popular mechanics, Seidel is QVC turned out to be a hit, Developed a US military product, and now serves as Vice President of Marketing for BatterySaver.

    Udell says the cost to manufacture is just 12 cents per unit, with retail plans to package two BatterySaver units for 99 cents per package.




    https://biz.crast.net/this-91-year-o...to-30-percent/








    Last edited by Turion; 07-31-2022, 06:28 PM.

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  • Turion
    replied
    Not sure why you aren’t just using the two kilowatt meters you have shown and measuring watts in and watts out and comparing.

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  • BroMikey
    replied
    I have made calculations that show 94% eff or better. .175 x2=.325+booster light and circuit = .400

    1.4+ .725 = 2.15 next max draw is 2.25

    eff = 2.15 / 2.25 = 95% at these low power levels.

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  • BroMikey
    replied

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  • BroMikey
    replied
    Okay test coming. I have this one? https://www.fridayparts.com/linear-d...30v-0-3a-5v-3a

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  • Turion
    replied
    Save yourself hours of time. Use a power supply with your meter between it and the boost module input. Then take the output of the boost module straight through another meter to a load. You can measure watts in and watts out and calculate efficiency and have a CONSTANT source of supply rather than a declining one as batteries discharge.

    Alternately, take a Variac with a bridge rectifier attached set to low AC voltage. Run the kilowatt meter off the DC side of the bridge with the boost module attached. Again, take the output of the boost through a meter to a load. Watts in vs watts out. Efficiency.

    Test Circuit. jpg.jpg

    Once you know the efficiency of the boost module you can set up the attached schematic. I forgot to show the kilowatt meter hooked directly to the battery and the two wires I show "connected to the battery" should be connected to the kilowatt meter output. This will test whether a boost module, in and of itself, is a free energy device. It allows you to compare watts into the boost module to watts into the load. If your boost module is 87% efficient and MORE than 87% of the watts you input to the boost module arrive at the load, where did they come from? Especially when you consider LOSSES in the TWO meters, the boost module and the two diodes. Silicon diodes have a forward voltage loss of approximately 0.7 volts, and there are two of them in the circuit. so that's 1.5 volts. What % of your input is that, and how should that ALONE affect the output. Does it?

    This is the simplest process and simplest test setups I could come up with to prove the point. Lots of folks have most of this stuff on hand.
    Last edited by Turion; 07-31-2022, 06:45 AM.

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  • BroMikey
    replied
    3 hrs into my first baseline run battery pack 3 blew out a brand new battery running 2 watts. Can you imagine? Can't trust any of these packs now, without running them down. I used over 5wh and bang.

    Test was restarted at 12:30am battery 3 voltage 11.60v @170ma right at 2.0 watts. Burning the midnight oil.

    Load voltage 18.3v @80ma right on 1.4 watts


    I am looking to have good batteries THEN the fun begins thanks for all of your support.
    Last edited by BroMikey; 07-31-2022, 06:11 AM.

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  • BroMikey
    replied
    Here are the rest of the details. Meter bouncing between 1.8 and 1.9watt on first watt meter say 1.85 and same at the end watt meter bouncing between 1.2-1.4 watt so 1.3watt is average. See video of idling draw.

    11.34 X 25ma meter bouncing as you watch 20ma -30ma multiple 11.34 X .025 = .2835 for both the watt meter and booster idle. No load condition only 1 watt meter.

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  • BroMikey
    replied
    Originally posted by OrionLightShip View Post


    66% efficient....not good....better results available with an active clamped boost convertor. I know this to be true but I've never had a need to shop for one so I have no idea where to find something like that

    Could be some inaccuracy with your meters as well......swap places and see if you get a better or worse result.

    Orion
    It is in the 95% range afte you see what I see. 1.85watt and then 1.4watts at the end. Each meter .15 x 2 and the boost takes .15watt also just idling. it ain't to bad

    Here are the circuit Dave let me post 4 years ago and he must have said don't forget to add caps if you want.
    I hear him say add diodes too but I am unsure.

    http://flyer.thenetteam.net/3battery...ryDiagram1.jpg

    http://flyer.thenetteam.net/3battery...ryDiagram2.jpg

    http://flyer.thenetteam.net/3battery...ryDiagram3.jpg

    http://flyer.thenetteam.net/3battery...agramTest4.jpg






    ...............
    Last edited by BroMikey; 07-31-2022, 06:19 AM.

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  • OrionLightShip
    replied
    Originally posted by BroMikey View Post









    ....................

    66% efficient....not good....better results available with an active clamped boost convertor. I know this to be true but I've never had a need to shop for one so I have no idea where to find something like that

    Could be some inaccuracy with your meters as well......swap places and see if you get a better or worse result.

    Orion

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  • OrionLightShip
    replied
    Originally posted by Turion View Post
    A former friend used to say, find an effect and figure out how to take advantage of it. Let SOMEONE ELSE figure out how it works and WHY it works.
    HaHaHa Genius, I really don't care what happens to energy I have no way to control.

    However, I favor Tesla's method of building it in my head until I understand it completely, then bringing it to the physical. I'm just a whole lot slower on the go and life happens so it's taken me years and years to figure out what I now know. Now I just need to get motivated and fire up my 3d printer and start building it.

    Orion

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  • BroMikey
    replied
    Originally posted by OrionLightShip View Post

    YUP, it's all about potential, and like you said with batteries. Also, those ions move a lot slower than electrons so a problem exists there unless you place a capacitor in parallel with the battery to hold that higher voltage charge so the battery can have time to absorb it.

    Orion
    Originally posted by Turion View Post
    POST 3448

    I have theorized this many times in writing and I will put it forth again. What if when we charge a battery we are forcing the charged ions all to one side, so when we put a meter on it, we are measuring the difference between one side and the other.

    When you let a battery just sit, those charges make their way back to the other side and the two sides “equalize”, so our meter registers 0 volts eventually.

    when you put a load on the battery you give those charges a path of least resistance to follow but with the SAME result. You measure less in the battery because the two sides have equalized. The “load” didn’t “use up” the energy, it ran because the energy moved through it. Yes, there are losses in the load so you end with less than you started with.

    To charge a battery we use a tremendous amount of energy to force those charges back to one side. Consider this possibility when you look at what you saw happen.






    ....................
    Last edited by BroMikey; 07-31-2022, 01:03 AM.

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  • Turion
    replied
    A former friend used to say, find an effect and figure out how to take advantage of it. Let SOMEONE ELSE figure out how it works and WHY it works.

    Leave a comment:


  • OrionLightShip
    replied
    Originally posted by Turion View Post
    Right, otherwise you hit it with a pulse and MOST of it is wasted. Even when you understand the concepts it is the little things that make the difference between success and failure.
    Then, the question appears; what happens when all the electrons are not absorbed by the battery.

    I only have one theory. There is no such thing as an electron. It is an energy perturbation in the aether that we call an electron. Perhaps if it has no potential behind it and nowhere to go; it simply is absorbed back into the aether.

    Orion

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  • Turion
    replied
    Right, otherwise you hit it with a pulse and MOST of it is wasted. Even when you understand the concepts it is the little things that make the difference between success and failure.
    Last edited by Turion; 07-30-2022, 09:48 PM.

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