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  • wantomake
    replied
    Sorry BroMikey I've no idea what that means. But how many spikes per second will safely charge a battery? Can batteries of any type be charged and used as a source of energy at same time? How do car batteries last for two years or five while being used and charged at same time?
    ​​​​​
    I'm showing my ignorance here but don't care. Are lithium or sulfur batteries any different? I don't know. I read articles today where some said yes, some said no to that question. Do batteries need to rest after being used, or can they be charged immediately?

    I don't have enough time with these experiments to write books on the subject. It seems capacitors are the perfect fit for what we want to accomplish here. If you can afford the high Farrad caps. Pricey from Amazon.

    In the last 12 years I've heard more different opinions on batteries than free energy ideas.

    Leave a comment:


  • BroMikey
    replied
    Originally posted by Turion View Post



    How do we design the circuits to be MORE efficient?
    What parts of the "basic circuit" could be eliminated or replaced with more efficient parts?
    Is sending the energy directly back to the source the BEST use of what comes out of the secondary of the boost module? [/COLOR]
    Problem with taking out all the snubers is you kill all your chips with back spikes you are wanting to collect.


    https://images.app.goo.gl/6r92KhkqkexGFwit5
    Last edited by BroMikey; 08-17-2022, 11:00 PM.

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  • wantomake
    replied
    Well this is running better now. The 25 watt 12v dc bulb is pulling 2.2 amps on my fluke meter. The second boost converter (not dc to dc) is not running hot. The LAB is holding at 12.3 volts. Can't make long run until get back from Asian markets and such today. That marine deep cycle battery will run that dc bulb for long time I'm guessing. So no proof of "looping back". Unless the source battery charging up is that proof.

    Interesting I'll say for sure. Sorry I can't offer any technical jargons because I'm new to All this. I'll do what I can.

    I may try a lithium 12 volt pack later today. Will need to put it together.

    Leave a comment:


  • wantomake
    replied
    Originally posted by Turion View Post

    Yeah,
    You should listen to someone who knows what they are doing and that's obviously not ME. I guess I wasn't thinking when I drew that schematic. Here is what I am ACTUALLY running. Instead of a second boost I am running a DC to DC converter, which acts much the same, but is more efficient. And I have a cap across the input. It probably isn't necessary, but that DC to DC is like $130 rather than $30, so I take no chances. I am changing the picture on the previous post to the correct one. The main ISSUE is which side of the diode the LOAD is on. It changes the relationship between potential differences. No wonder bistander asked if I had ever run that circuit before.




    Ok. I'll give this a shot.

    Leave a comment:


  • Turion
    replied
    Look at what the SINGLE BATTERY SETUP does:

    Can we send energy from a higher potential to a lower potential through a load "running" the load
    Can we take energy out of the source and take (at least some of it) in a loop back to the source?
    Can we recover some of the "A" input energy (Some % of A) put into or through the primary side of the boost module?
    Is energy GENERATED in (B) the secondary side of the boost module? (Look at how an ISOLATED DC to DC converter works) "an isolated power converter isolates the input from the output by electrically and physically separating the circuit into two sections preventing direct current flow between input and output" NO ELECTRICAL CONNECTION EXISTS BETWEEN THE TWO SIDES. WHAT COMES OUT OF THE SECONDARY SIDE IS NOT PART OF WHAT WENT INTO THE PRIMARY.
    Is % of A recovered + B greater greater than input A?
    Is % of A recovered + B greater greater than input A + other losses in the system and enough to overcome impedance in the battery?
    Is a boost module the most efficient means of creating energy?
    How do we design the circuits to be MORE efficient?
    What parts of the "basic circuit" could be eliminated or replaced with more efficient parts?
    Is sending the energy directly back to the source the BEST use of what comes out of the secondary of the boost module?

    Last edited by Turion; 08-17-2022, 10:45 AM.

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  • BroMikey
    replied
    . Cheap boost modules are enough to demonstrate EVERY SINGLE ONE of the principles of success. Then its build it bigger, better, more efficient, higher voltage

    When I get it to loop, I will let you know. Until then, it's just extended run times. That's all it will ever be with what we are playing with here. If I'm really, REALLY lucky I might get THIS SETUP to loop a 10 watt load.[



    That is what we need, to begin. I like slow

    Leave a comment:


  • Turion
    replied
    ISOLATED DC to DC converter. I was mistaken. Found it for $149.00, not $130.

    If you don't know what to do with it, it does you absolutely no good. I'm not advising people to spend that kind of money on parts when they don't yet understand how this works, why it works, or what you have to do to make it work more efficiently. Cheap boost modules are enough to demonstrate EVERY SINGLE ONE of the principles of success. Then its build it bigger, better, more efficient, higher voltage and knowing what to do after you learn how to do all that. But all THAT comes first. And these "toys" we are playing with are never going to put out the kind of power you really want to see. But if you learn all the principles, you can look around and figure out what other things to apply them to.

    When I get it to loop, I will let you know. Until then, it's just extended run times. That's all it will ever be with what we are playing with here. If I'm really, REALLY lucky I might get THIS SETUP to loop a 10 watt load.
    Last edited by Turion; 08-17-2022, 07:44 AM.

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  • BroMikey
    replied
    Here is a gutsy module 98%
    https://www.ebay.com/itm/29493212882...Bk9SR_KXk6nVYA

    Leave a comment:


  • BroMikey
    replied
    If anyone is interested (don't buy this one) Aaron pointed out years ago that we might use mppt converters

    https://www.amazon.com/DPX800S-Boost...s%2C248&sr=8-1

    Leave a comment:


  • BroMikey
    replied
    Originally posted by Turion View Post

    Yeah,
    You should listen to someone who knows what they are doing and that's obviously not ME. I guess I wasn't thinking when I drew that schematic. No wonder bistander asked if I had ever run that circuit before.
    Hello Mr Bowling

    I found this on the internet. plz be more specific on the box costing you $130 oh and thank you

    Introduction:
    Both buck boost converters are DC to DC converters with different voltage and current at the output compare to input. There are two types of topologies viz. inverting and non inverting. In Inverting type Output voltage polarity is different than input where as in non-inverting type output voltage polarity is same as input voltage. They are used as switching regulators which use switching element (typically one or two MOSFETs) and an energy storage device (such as inductor) to efficiently regulate input voltage to lower or higher output voltage.



    Leave a comment:


  • Turion
    replied
    Originally posted by wantomake View Post

    Hey Turion,
    I connected everything exactly as your schematic. I'm using lead acid deep cell not lithium battery. I put a simple on/off switch between the negative IN on the second boost module that connects to the returns to positive of the battery. Of course there's a diode in line there also.

    When I flip the switch to connect that negative IN to that return , it makes the 25 watt bulbs go almost out and the module heats up without a fan on it. But if I cut that switch off the bulb lights up to normal.

    I've checked all connections, got both modules set to same 26 volts.

    Any ideas? I tried different valued diodes (new ones). I'm trying but cannot get any run times.
    Yeah,
    You should listen to someone who knows what they are doing and that's obviously not ME. I guess I wasn't thinking when I drew that schematic. Here is what I am ACTUALLY running. Instead of a second boost I am running a DC to DC converter, which acts much the same, but is more efficient. And I have a cap across the input. It probably isn't necessary, but that DC to DC is like $130 rather than $30, so I take no chances. I am changing the picture on the previous post to the correct one. The main ISSUE is which side of the diode the LOAD is on. It changes the relationship between potential differences. No wonder bistander asked if I had ever run that circuit before.





    Attached Files
    Last edited by Turion; 08-17-2022, 04:15 AM.

    Leave a comment:


  • wantomake
    replied
    Here is the second boost module I'm using.
    https://a.co/d/bw8NwTb

    My first boost module is the one like yours with digital readout and adjustment buttons.
    Last edited by wantomake; 08-17-2022, 01:39 AM.

    Leave a comment:


  • wantomake
    replied
    Originally posted by Turion View Post
    I would suggest you take all those schematics and throw them in the trash. They all have changes or additions that I did not put on ANY drawing I created. Or if I did, they are incorrect. I know I NEVER used the term "UPS" or "MPPT" nor do I have that little picture of a "light bulb" that appears on so many of these drawings. These drawings were someone else's version of something I posted. They are way too complicated for a beginner. A beginner needs a basic setup and needs to learn the principles involved. Some of these schematics show what NOT to do, which is why I know they were altered because they do things I would NEVER do. Like running a load directly on a "UPS" instead of running that energy THROUGH the load and collecting it on the other side. That's a BASIC mistake.

    There are basic principles I have been trying to show.
    bistander claimed on his thread that I SAID a boost module is a free energy device. I actually said "I figured out YEARS ago that the boost module was a free energy device." What I SHOULD have said is, a boost module USED CORRECTLY is a free energy device. There's a HUGE difference. If you don't get that, you do not understand the most basic concept of using potentials. And not EVERY boost module will work on the system the way it needs to. And not EVERY boost module is efficient enough to give you good results.

    If you run a boost module connected directly to a battery you might put in 100 watts and expect to get around 85 watts out the secondary on an "average" boost module. There is nothing "free energy" about that now is there? FAR less out than in. But if you run the boost module on the 3 battery system, or between potential differences, you input 100 watts and still get 85 watts out. But THIS time you recover some of what went THROUGH the primary of the boost module to the charge battery. If you recover LESS than 15 watts, you have nothing special. But if you recover MORE than 15 watts, you have a gain, and that GAIN is free energy. You are GENERATING energy here. You put some in and you get MORE OUT than what you put in when you combine the "generated" energy (85% of input energy) with the "recovered" energy (more than 15% of input energy). If NO energy was being recovered, battery 3 WOULD NOT CHARGE.

    So how do you make sure that you get that gain? Well, the first thing is testing the crap out of a bunch of different boost modules to see which is the most efficient. There is one out there, very expensive I might add, that "CLAIMS" 95% efficiency, but when run under higher input voltages actually shows a return of 99.7%. So how much of the "input" do I have to recover to achieve that gain? If you couldn't do the math, the answer is .3%.

    So now we look at means of recovery. A capacitor is best because it basically takes everything you want to give it up to its limit. Unfortunately you have to have a BIG cap. A cap in parallel with a battery will also work, and is better at collecting the energy that the battery cannot immediately accept . But you are still limited by the impedance of the battery. This is where the lithium batteries come in or the lithium/graphine hybrid. Their ability to accept what they are given without FIGHTING BACK can be the difference between success and failure. Battery impedance is a KILLER. More batteries in parallel to receive the "charge" energy the better.

    I look at some of those circuits above and I see loads running on outputs. NEVER. NEVER. NEVER run a load unless it is between potentials. NEVER!!!! Now it can't be avoided if you only have one battery, but if you have THREE, eliminate that boost module running off a single battery to give you a higher potential source. You eliminate one BIG strike against success right there.

    Think of a river flowing from the mountain top to the ocean. You can put hydroelectric dams along the river, but each dam produces less power than the one upstream (at least in this analogy). The Colorado in Arizona is an example. There must be four or five dams on the Colorado, and by the time it gets to Mexico, it is a stream not a river. Your system is going to flow downhill from the source, and you can tap it along the way to produce energy Each time you tap it you get out less until you get out very little. The HIGHER the source ( the farther up the mountain you go) the more times you can tap the source to produce and recover energy. Start with a 12 volt battery and you have to BOOST it to at least 24 volts to get your downhill flow. Start with 48 volts, and you have the difference between 48 and 36. The difference between 36 and 24. The difference between 24 and 12. From my count, that is 3 times you get to tap the flow of energy. YES, there will be losses. But if you use the RIGHT boost modules (three in this example) and the right batteries (ten in this example) and you run the outputs from the boost modules THROUGH loads back to the individual batteries at the top of the mountain, interesting things happen.

    I would NEVER recommend starting with something this complex. Most people don't even understand how this really works and are incapable of putting together a functioning circuit. And then they will blame me because it did not work. Once you DO understand the principles and can build a simple circuit on your own, you certainly don't need ME to show you the benefits of a potential difference circuit. You will SEE the reasons on your bench. Or lighting up your shop.

    HERE ARE THE TWO SIMPLEST CIRCUITS YOU CAN RUN
    If you have only one battery, the simplest circuit is to run a boost module directly on it, positive to positive and negative to negative. Take the positive output of the boost module set to about 26 volts and run it to the positive IN of a SECOND boost module. Take the negative IN of that boost module and connect it back to the positive of your source battery. Now take the positive output of that second boost module(set to 26 volts also) and run it through a load back to the positive of that source battery. DON'T forget the diode!!! It gives you a voltage loss, but you may need it to protect your boost module. Then again, you may not. Want to fry a $30.00 boost module to find out? It may be worth it in the long run, but certainly NOT the short term. I wanted to add that with a single battery system, the capacitor across the battery is INCREDIBLY IMPORTANT, and it should be as huge as you can find. Because it will collect ANY energy it receives and outputs whatever it has in it, it is the lake being fed by a river at the top of the dam, and can provide as much over the top of the battery as is required while your battery slowly fills it up. If your system is outputting MORE than what is being consumed by the circuit and the load, you can ELIMINATE any "battery" issues of impedance with that capacitor. Set BOTH boost modules to around 26 volts.

    If you have 3 batteries, put two in series and connect the boost module positive in to the positive output of the two batteries in series. Connect the negative in of the boost module to the positive of the third battery. Connect the negative of the third battery to the negative of the two batteries in series. Take the output of the boost module set to 26 volts and run it through a load to the positive of the third battery. You will have to rotate your batteries to keep up the potential difference and keep the batteries charged. Again, do not forget the diode!. Set boost module for 26 volts.

    I have included a schematic for both circuits.

    I am currently running something similar to the single battery circuit shown here. Just not using these same components, and because my output voltages are higher, I have to buck the voltage down before the load. But the principles are the same. And I am running 30 watts of load, and have run it with 70 watts of load for short periods to see what would happen to my battery voltage. It's called research.

    Hey Turion,
    I connected everything exactly as your schematic. I'm using lead acid deep cell not lithium battery. I put a simple on/off switch between the negative IN on the second boost module that connects to the returns to positive of the battery. Of course there's a diode in line there also.

    When I flip the switch to connect that negative IN to that return , it makes the 25 watt bulbs go almost out and the module heats up without a fan on it. But if I cut that switch off the bulb lights up to normal.

    I've checked all connections, got both modules set to same 26 volts.

    Any ideas? I tried different valued diodes (new ones). I'm trying but cannot get any run times.

    Leave a comment:


  • Turion
    replied
    So, when I did my run last night with the single battery setup, I was only able to run the system for 2 hours with the battery I am using. That's with a 30 watt load at full bright. I noticed that even though the unloaded voltage was 13.38, the loaded voltage was 12. 87. That means the most I could be returning to the battery was 12.87, or it would show the higher input voltage as the voltage across the battery. Or at least that's my assumption. Someone who knows please correct me if I am wrong.

    My belief is that 30 watts is just TOO MUCH for this setup to maintain and keep the single battery charged. So I dropped the load down to 10 watts for this next test run, and if it is successful, I will go back up to 20. But if 10 doesn't work, no use trying 20. Anyway, the voltage under load only dropped to 13.23, so that is a really good sign as far as I am concerned. Time will tell. At 3:20 my two hour run will be up, and I am going to let it just run until the BMS kicks it off from low battery voltage. I realize 10 watts is not much, but if the system self sustains, it is proof of concept, and can be scaled up.

    I came up with several ideas to deal with the issue of inefficient boost modules that I am going to try. Every day is an adventure.
    Last edited by Turion; 08-17-2022, 12:20 AM.

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  • wantomake
    replied
    Took apart the 4 battery switch setup and using one battery and one DPDT switch with the two boost converters set to 26 volts each. I have 3 12v 50 watt bulbs and 1 25 watt 12v bulb. Only have two light sockets. Home Depot road trip time. Don’t know if 175 watt load is good or bad. Will find out with the new setup and some research. Have many values of diodes. Will research that too.
    Much to do.
    Last edited by wantomake; 08-16-2022, 09:36 PM.

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