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Bedini RPX Sideband Generator
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  #61  
Old 10-10-2017, 05:39 PM
dragon dragon is offline
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I realize some think that there is some detriment to charge and discharge simultaneously but I've never seen any problems caused by this. It would, however, become a problem if you were using HV spikes to charge a battery while connected to sensitive low voltage loads. It's not the case here. Also the series batteries are temporarily disconnected when the charge battery is in operation.

In reality we do it all the time - solar to battery to load - running accessories in the car while charging and maintaining the battery etc.

Quite often, with some systems, you wouldn't see a gain if the two were combined because the output of the charging portion is considerably less than the required input - in order to demonstrate the battery charging it has to be into a separate battery.
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  #62  
Old 10-10-2017, 07:49 PM
RoliK RoliK is offline
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Interesting thread. I want to make a replication. Because of the switch.
Is it possible to use a single change-over relay? Would be much easier.

Lg rolik
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  #63  
Old 10-11-2017, 12:01 AM
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Hi dragon, thanks for sharing the information, i will definitely be trying the 2 battery version.
Like i was saying, Bedinis 1984 free energy generator, discharged and then charged the same battery, very similar to this, with a mechanical switch and seemed fine, though I recall they said something changed within the battery also.
Though his version had capacitors charged by a rotating generator and yet it still used capacitors dumped into the battery, just like we're doing here.

Hi Rolik, yes, wistiti posted a video here, where he used a relay driven by a 555 timer.
peace love light
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  #64  
Old 10-11-2017, 02:15 AM
Wistiti Wistiti is online now
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:)

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Originally Posted by RoliK View Post
Interesting thread. I want to make a replication. Because of the switch.
Is it possible to use a single change-over relay? Would be much easier.

Lg rolik
Hi Rolik. As Skywatcher say it is what I use for switching and it work really well!

All, here is a video of the last circuit Sky an Dragon have share:
https://youtu.be/5iybsIiUecE
My load is 300w and the starting voltage of the battery is:
B1: 12,59v
B2: 12,53v

Until now, the load is really slow to warm this little glass of water... for sure I will not have a tea from it!! maybe the voltage going to it is to small. I am thinking to adjust it with a dcdc converter as Sky have already suggest... (see the image)

Any comments are welcome!
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  #65  
Old 10-11-2017, 02:24 AM
Wistiti Wistiti is online now
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Battery resting

Hey Sky!
When you leave your battery resting, do you disconnect everything or you keep them connect series....??
Thank you!
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  #66  
Old 10-11-2017, 03:39 AM
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Hi wistiti, thanks for sharing the video.
I disconnect all connections from the batteries when resting.
I'm using tractor batteries, which can give high amps briefly, more efficiently than gel batteries, i would think.
Also, Can the relay efficiently pulse high amperage, I'm not so sure.

After resting more than 24 hours:
Battery A = 12.54 volts
Battery B = 12.52 volts
Battery C = 12.63 volts charge battery

Also, water has the highest heat holding capacity of all materials i'm aware of, which might explain why it's heating so slowly.
peace love light
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  #67  
Old 10-11-2017, 03:15 PM
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Wistiti, have you measured the amperage moving through the heating element? I noticed in one of you other videos you have a clamp meter handy - set it on AC and clamp one leg of the element. That element, normally, would require somewhere in the range of 25 amps. Something in the range of 15 to 20 amps should be cycling through it with the pump circuit. If the amperage is low we need to find out why or what might be restricting it. I don't see a problem with the relay, I use a 30amp SPDT automotive relay in the demonstration video here and as long as it's running below its rated amperage it shouldn't be a problem.

SkyWatcher is right about the gel cells, they aren't as effective as the tractor battery - starting batteries have the ability to "give up" their energy very quickly... they can develop 100-300+ amps very quickly. Deep cycle batteries store a lot more energy but are designed to give it up at a slower rate. In any case, since you have the gel cells, they should be fine producing 5 to 10 amp bursts needed to drive the pump - I use them quite often in a lot of my tests even though they're a bit short winded ( 7ah ratings ), also, quite often because of their limited energy you can see if things are going good or bad much quicker than would be seen in a larger battery.

In any case, your looking to relieve any possible restrictions in the pump circuit to develop huge amp flows from the capacitive discharge.
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  #68  
Old 10-11-2017, 03:49 PM
Wistiti Wistiti is online now
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Hi Dragon
yes I have already mesure the amp going to the load and it is in the range of 5 amp... That why it is not eating fast... The fact Sky and you point about the CCA of the "Starting battery" vs "Deepcycle" might be the case why I have so low amp reading...

About the relay switching, I also think mine is rate for 30amp. so I do not see why it should eventually fail.? Anyway it is still working fine after the few test I have done. We will see with the time.

Ciao!
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  #69  
Old 10-11-2017, 03:49 PM
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Something a little more advanced running on 18650 li-ion cells using an IR2153 half bridge driver.
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File Type: jpg IMG_1517.JPG (93.5 KB, 31 views)
File Type: png igbt driver 2153.1.png (18.8 KB, 37 views)
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  #70  
Old 10-12-2017, 01:20 PM
dragon dragon is offline
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Wistiti, I don't believe your low amp problem is caused by the battery type. It has to be a resistance problem or voltage drop. If you connect the heating element directly to a charged cap it should instantly go to full amperage then taper off until the cap has been drained. Likewise, connecting to a battery - any battery should drive that element. I haven't tried the other arrangements charging a battery, I suspect there may be something there that is slowing the movement of current.

Additional Food for thought.... Let's say your using the double pass circuit ( fig 2 in the charge pump compare drawings ). The first cycle draws 25 amps from the battery through the load to charge the capacitor, the second cycle discharges the capacitor through the load. So you have a 25 amp load on the battery at a 50% duty cycle. If you run the system for 1 hour you are drawing 12.5 amp hours from the battery - 1/2 of the energy of a single pass circuit yet your still driving the load on every cycle. The charge pump circuit (fig 3) becomes more complicated because we are creating a voltage divider with the ability to double the current... a capacitive transformer of sorts. We then complicate it even more by charging the capacitors which increases the energy surges. Because of the voltage drop it becomes more important to use loads that are much lower in resistance ( hence my demonstrations of multiple loads in parallel ) or increase the working voltage.

Essentially, the single pass circuit drives a 300 watt load at a cost of 300 watts, the double pass circuit will drive a 300 watt load at a cost of 150 watts - the charge pump circuit does the same at a cost of 75 watts.
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Last edited by dragon; 10-12-2017 at 04:16 PM.
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  #71  
Old 10-13-2017, 04:55 AM
Wistiti Wistiti is online now
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Thank you Dragon for the reply!
I will do more test in the next few day.
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  #72  
Old 10-13-2017, 05:29 AM
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Hi all, have not been making any tests, not feeling too well, though my head is starting to clear up.

Thanks for sharing that food for thought dragon.
So what about the variation i've been testing, the drawing i posted.

If it works out the same, 300 watts output, while only using 150 watts, then what does charging the battery on the second pass contribute.
Maybe it works out the same, 75 watts input, if charging the battery is only 50% efficient.
Hmm, have to think more about this.
peace love light
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  #73  
Old 10-13-2017, 02:15 PM
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Once again we jump to the conclusion that the output is higher than the input... I don't believe this is the case. When we charge a cap it takes a certain amount of time, during that time you have a peak energy exchange at the beginning and as the cap fills the voltage rises, reducing the potential difference and the current flow decreases - this gives us an average energy from start to finish.

Then we have the resistive load which is said to be "using" the energy. I tend to question the amount the resistive load actually uses or converts vs what is generally wasted during the neutralizing process of an energy source such as a battery. The ohms law and BTU standards don't always coincide with each other so in order to explain the phenomenon we introduce a COP value that allows us to acceptably bend the rules without really breaking them.

If you take a higher potential and dump it into a lower potential ( thus storing the energy not neutralizing it ) the resistance between them acts as a regulator of time - so then what is really lost?

If you take a fully charged capacitor of a given value and dump it into another capacitor of equal value that is not charged you loose 1/2 of the energy. But... if you charge both caps one higher than the other the losses are reduced drastically and the resistance in line plays a very small role in those losses while balancing the pair.

As an example I'll use the 1F caps - one charged to 24 volts and the other to 12 volts. The 24volt cap contains 288 joules the 12 volt cap 72 joules. If we connect them in parallel they will balance at around 18 volts each. Each now contain 162 joules. We lost 36 joules of energy during the balancing - with or without a resistive load this proves to be accurate. The exchange shows a 10% loss between them, considerably lower than the 50% loss of the first example - one charged one not charged.

Now consider as an (extreme) example the same caps charged to 112 volts and 100 volts using the same potential difference as the first example. We have 6272 joules in the higher potential and 5000 joules in the second a total of 11,272 joules stored. They would balance at around 106 volts each containing 5618 joules or a total of 11,236 joules, again a loss of 36 joules because of the 12 volt difference but the actual loss of energy is less than 1% - with or without the resistive load. 6272 joules - 5000 joules = an exchange of 1272 joules... is it possible to drive a 1200 watt resistive load with only a 1% loss?? This is what I'm trying to find out...

The circuit itself isn't anything special - not overunity nor even unity - simply a circuit that allows us a small view of the possibilities.... if your willing to look beyond what's on the surface and question how and why the accepted rules apply.
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  #74  
Old 10-13-2017, 03:53 PM
bistander bistander is online now
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Cap energy

Quote:
Originally Posted by dragon View Post
...
As an example I'll use the 1F caps - one charged to 24 volts and the other to 12 volts. The 24volt cap contains 288 joules the 12 volt cap 72 joules. If we connect them in parallel they will balance at around 18 volts each. Each now contain 162 joules. We lost 36 joules of energy during the balancing - with or without a resistive load this proves to be accurate. The exchange shows a 10% loss between them,
...
Hi dragon,

When I do the calculation of connecting two 1F caps (one charged to 12V and one at 24V), I get a resulting voltage of 18.98V. Assuming ideal capacitors and circuitry, there are no losses or missing energy. Assuming an average voltage when connecting the two caps is incorrect.

Regards,

bi
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  #75  
Old 10-13-2017, 04:01 PM
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Quote:
Originally Posted by dragon View Post
Then we have the resistive load which is said to be "using" the energy. I tend to question the amount the resistive load actually uses or converts vs what is generally wasted during the neutralizing process of an energy source such as a battery. The ohms law and BTU standards don't always coincide with each other so in order to explain the phenomenon we introduce a COP value that allows us to acceptably bend the rules without really breaking them.

If you take a higher potential and dump it into a lower potential ( thus storing the energy not neutralizing it ) the resistance between them acts as a regulator of time - so then what is really lost?

If you take a fully charged capacitor of a given value and dump it into another capacitor of equal value that is not charged you loose 1/2 of the energy. But... if you charge both caps one higher than the other the losses are reduced drastically and the resistance in line plays a very small role in those losses while balancing the pair.

As an example I'll use the 1F caps - one charged to 24 volts and the other to 12 volts. The 24volt cap contains 288 joules the 12 volt cap 72 joules. If we connect them in parallel they will balance at around 18 volts each. Each now contain 162 joules. We lost 36 joules of energy during the balancing - with or without a resistive load this proves to be accurate. The exchange shows a 10% loss between them, considerably lower than the 50% loss of the first example - one charged one not charged.

Now consider as an (extreme) example the same caps charged to 112 volts and 100 volts using the same potential difference as the first example. We have 6272 joules in the higher potential and 5000 joules in the second a total of 11,272 joules stored. They would balance at around 106 volts each containing 5618 joules or a total of 11,236 joules, again a loss of 36 joules because of the 12 volt difference but the actual loss of energy is less than 1% - with or without the resistive load. 6272 joules - 5000 joules = an exchange of 1272 joules... is it possible to drive a 1200 watt resistive load with only a 1% loss?? This is what I'm trying to find out...

The circuit itself isn't anything special - not overunity nor even unity - simply a circuit that allows us a small view of the possibilities.... if your willing to look beyond what's on the surface and question how and why the accepted rules apply.
I agree with dragon, this is not OU, you cannot get more energy than what you put in!! BUT!!! you can use the energy WISELY to do MORE "WORK" (which can also be to generate another potential to supply the source).

I guess this is what the circuit can teach us (the REAL TRUTH about energy).
its an experiment, a TEST.. to study "Energy" and its behavior more.
the "Usage" "Dissipation" "Consuming" "Transformation" "Conversion" of energy is it real or myth, not just believe what we are taught or what we learned..
Been doing some test lately...
..to see how much I can use the same "Energy",
..to find out If your really just "Transforming" energy into other forms,
.. to see If the "Resistance" really was dissipating "Energy"
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  #76  
Old 10-13-2017, 05:01 PM
dragon dragon is offline
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Quote:
Originally Posted by bistander View Post
Hi dragon,

When I do the calculation of connecting two 1F caps (one charged to 12V and one at 24V), I get a resulting voltage of 18.98V. Assuming ideal capacitors and circuitry, there are no losses or missing energy. Assuming an average voltage when connecting the two caps is incorrect.

Regards,

bi
bistander, that's interesting... can you show and explain the math? I tend to use averages for quick reference and extrapolate data from experiments to get an overall view of what I'm seeing. Processing the end results of my circuits is sometimes perplexing based on all the variables. ( for me anyway )....
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  #77  
Old 10-13-2017, 05:39 PM
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Hi all, before even reading the recent replies, of course i didn't mean higher output than input.
I meant, we might be getting more power out from an external source, not associated to what we input.
So yes, i meant coefficient of performance, i just assumed you all knew that is what i meant.
peace love light

Well, take for example, the collapsing field of a coil, does that give us more than we input, not usually, though it gives us back some to use from the environment, external to what we input.
I think your logic with this is sound dragon and your original circuit is probably the best design to use, based on the low losses and multiple recycling passes.
Though dragon, you still didn't respond to what you thought the benefits of the charging battery might be, or how it compares to your 4 pass circuit.
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  #78  
Old 10-13-2017, 05:47 PM
dragon dragon is offline
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Originally Posted by SkyWatcher View Post
Hi all, before even reading the recent replies, of course i didn't mean higher output than input.
I meant, we might be getting more power out from an external source, not associated to what we input.
So yes, i meant coefficient of performance, i just assumed you all knew that is what i meant.
peace love light
I knew what you meant, I just wanted to be clear to those that might interpret what I was saying as some type of overunity effect.
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Old 10-13-2017, 06:09 PM
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Quote:
Originally Posted by SkyWatcher View Post
Hi all, before even reading the recent replies, of course i didn't mean higher output than input.
I meant, we might be getting more power out from an external source, not associated to what we input.
So yes, i meant coefficient of performance, i just assumed you all knew that is what i meant.
peace love light

Well, take for example, the collapsing field of a coil, does that give us more than we input, not usually, though it gives us back some to use from the environment, external to what we input.
I think your logic with this is sound dragon and your original circuit is probably the best design to use, based on the low losses and multiple recycling passes.
Though dragon, you still didn't respond to what you thought the benefits of the charging battery might be, or how it compares to your 4 pass circuit.
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Though dragon, you still didn't respond to what you thought the benefits of the charging battery might be, or how it compares to your 4 pass circuit
To be honest Skywatcher I haven't tried that set up. Anything I might say about it would simply be speculation. There is no doubt in my mind it will work reasonably well. The extent of its performance would need to be measured and compared.

When time allows I'll be setting up a few different innovations you've brought to the table. Until I get caught up on some other projects I can only watch and drool over what others are doing...although I'm sluffing off a bit today - to many 16 hour days, my motivation level is a bit lacking today...
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Old 10-13-2017, 06:10 PM
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Math

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bistander, that's interesting... can you show and explain the math? ... ( for me anyway )....
Sure. Two 1F capacitors + to + and - to - are parallel connected between + and -. So equivalent capacitance is 2F. Total capacitance of capacitors in parallel is the sum of the individual capacitances.

You were correct in the energy calculations of 288J and 72J so the total energy of the two capacitors is the sum or 360J no matter how they are wired.

We're assuming zero resistance in the capacitors and zero resistance in connecting wires. So just solve the energy equation of .5 * 2F * V^2 = 360J for V. V = square root 360 = 18.98V.

When you have non ideal components like real capacitors and copper wires and sparks upon connections, the math becomes more complex using integrals because the V, I functions are not linear with respect to time. Also the reasons that simple averages are inaccurate.

bi
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Old 10-13-2017, 06:37 PM
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Thanks bistander, while testing I was showing around 18.3 volts balanced so were not to far off from the calculated vs real world numbers.
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Old 10-14-2017, 12:54 AM
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Hi all, ok thanks for the reply dragon, i wish you could have more time to experiment as you wish to, then I await any results you might share down the road.
Just for reference, i checked the battery resting voltages again today, no testing, still not feeling that great.
So I think it has been almost 3 days now resting:
Battery A = 12.54 volts
Battery B = 12.52 volts
Battery C = 12.60 volts charge battery

Keep in mind, these batteries started out resting, before any testing at all at:
Battery A = 12.52 volts
Battery B = 12.53 volts
Battery C = 12.55 volts

peace love light
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  #83  
Old 10-14-2017, 02:46 AM
ricards ricards is online now
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2 12V 1F in series is 24v 0.5F which results to 18.4v much closer to the actual,

I was using the same formula bistander has provided on connecting 2 capacitor with different values and voltage.. compared "work performed" vs "initial Energy" (Charge). and still manage to come-up with COP greater than 1 Only on assumed scenario:

87% Energy Transfer (connecting capacitor to capacitor)
90% Energy to Work (load)
10 Cycles Passed

Charge_Calc.zip
This Calculation is actually adjusted to match the actual voltage measure on each cycle, meaning if we could attain close to 100% efficiency we would get higher COP...

my intuition tells me that there really isn't any "Energy Conversion" happening at all, It's more like Its just cause and effect!

I'm speaking my mind out again..
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Old 10-14-2017, 12:45 PM
Wistiti Wistiti is online now
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Interesting results Sky!
Hope you feel better soon!
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Old 10-14-2017, 01:05 PM
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Very nicely done ricards !!! I lean toward similar conclusions from the work I've done finding you can get very close to unity with different senario's.

SkyWatcher, get a healthy dose of sunshine and shake the ill's - sending lots of good thoughts your way!
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Old 10-14-2017, 02:52 PM
ricards ricards is online now
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Very nicely done ricards !!! I lean toward similar conclusions from the work I've done finding you can get very close to unity with different senario's.

SkyWatcher, get a healthy dose of sunshine and shake the ill's - sending lots of good thoughts your way!
hi dragon,

thanks.., how did you manage to switch 160v @ 3800uf? I remember you posted something like? if i remember correctly.

the toy relay rated at 10 amps didn't last 2 switches @ 100v 940uf..
now its dead.
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Old 10-14-2017, 03:31 PM
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hi dragon,

thanks.., how did you manage to switch 160v @ 3800uf? I remember you posted something like? if i remember correctly.

the toy relay rated at 10 amps didn't last 2 switches @ 100v 940uf..
now its dead.
Yes, The relays would really need to have the capability of high amp surges to survive switching higher voltages, I think this is the one your referring to https://www.youtube.com/watch?v=l3oI5nAIT7w

The caps are 6800uf 350volt. The IGBT's are QM300HA-2H, 1000 volt 300 amp. I'm planning to replace them with a half bridge module of similar rating to reduce the parts count. This way the clock, driver and IGBT module can be consolidated into a small package. The IR2153D is like a 555 timer on one side with the half bridge driver on the other side - seems to fit the bill quite nicely. Still playing with it to make sure there are no surprises. I have a small one set up with the mosfets that I can play around with late in the eveining until I have more time to play with the larger stuff. ( schematic shown a few posts back ).
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Old Today, 04:47 AM
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Hi all, thanks for get better wishes.
I ran the 3 battery setup yesterday, for an hour total.
So that's about 2 hours run time so far with this setup.

I think this needs more run time to be more conclusive, though I have no explanation at the moment, for why the voltage seems to be increasing on the batteries, overall.
I am aware of the 4 battery tesla switch, that a company was testing, using mechanical rotating switches and apparently, the batteries never discharged, but gained energy.
I don't know yet, if that may be happening here, we shall see.
I did place the 2 input batteries in parallel with the charge battery for a few hours today, to transfer some charge.
So the batteries are now resting at:
Battery A = 12.56 volts
Battery B = 12.61 volts now in charge battery position
Battery C = 12.56 volts

peace love light
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Old Today, 05:59 AM
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Hi all, thanks for get better wishes.
I ran the 3 battery setup yesterday, for an hour total.
So that's about 2 hours run time so far with this setup.

I think this needs more run time to be more conclusive, though I have no explanation at the moment, for why the voltage seems to be increasing on the batteries, overall.
I am aware of the 4 battery tesla switch, that a company was testing, using mechanical rotating switches and apparently, the batteries never discharged, but gained energy.
I don't know yet, if that may be happening here, we shall see.
I did place the 2 input batteries in parallel with the charge battery for a few hours today, to transfer some charge.
So the batteries are now resting at:
Battery A = 12.56 volts
Battery B = 12.61 volts now in charge battery position
Battery C = 12.56 volts

peace love light
Hello Sky,

one more test that is worth looking is if you charge the battery through a load (ala 3 BGS) as I think that is what you are heading.. which is actually good..

One thing that makes this circuit more attractive to me is the capacitive discharge method which at larger scale can deliver much more power than a battery.. aside from that its lightweight and small in size..
BUT of course only if we could make a successful loop.
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Old Today, 04:13 PM
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The 3 battery system is a remarkable piece of work - the work Dave and Matt are doing is quite amazing. While playing with that arrangement my only complaint was the limitations based on the restrictions of the batteries ability to absorb high current. This meant I needed a large bank of batteries to provide the power and assure the batteries weren't being damaged in any way. Capacitors remove the limits of voltage and current. They still provide many challenges along the way but the energy I need/want is there. It's a matter of how to manipulate the charges to provide the work I want from the system.

I've seen mentions several times about the possibility of "looping" the charge back to the initial source battery. Maybe I can add a twist to the thought of replacing the energy to instead - reducing the input requirement. Once maximum efficiency is reached replacing the energy becomes a much simpler task.

I'm sure many of you have built circuits that charge capacitors and found that once the cap is charged the circuit goes into an ultra low power low level oscillation. If you pulse the charged cap you can use the HV to produce a high current surge without draining the cap to a large degree thus leaving the oscillator in a low power state while only having to provide a maintenance charge.

Since I don't believe there is energy in excess of the original input source I'm forced to think in terms of finding alternative sources (free energy sources) or manipulate what is available by finding more efficient ways to use it which includes recycling in many forms. The easiest way to start is to manipulate energy over time. An extreme example, lets say your heating a meal in a 1500 watt microwave, an inverter connected to a 12v battery bank as the source. If we draw 125 amps for 1 minute how much energy did we remove from the batteries? The answer is 25 watts (w/h) or 2 amp/hours removed from the source. With a 25 watt solar panel I could run a 1500 watt load for 1 minute every hour.

I like to think in terms of how can I run things I normally use - making coffee, toast, running the frig, computer, TV or even simple lighting. So ask yourself... If it could be done... How would I do it? I don't claim to have all the answers, little by little finding creative solutions that reduce the need to purchase the energy for things we do every day isn't that far out of reach.
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Last edited by dragon; Today at 04:50 PM.
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