Motor Generators

[BroMikey]

And the fighting continues. Want no part of it. I get that from my extended family, so I stay away from them.

Taken care BroMikey. When you succeed, you may have to do it alone.

I been on my own since day one with these guys haggling over who is the king pin. Calling me and each other everything under the sun when we should be like little children, forgiving one another in brotherly friendship. Maybe we can finally move past the friendly fire and get back to work. Matt is right about David s weaknesses and Turion is right about Matt but who cares if we never patch it up and go ahead and do what we do best.

Gotta keep trying gentlemen. Do not harbor ill will, it is a burden heavy to be borne. Bitter grief is a cancer so let it go.. yeah yeah he is this and he is that is a good thing not a bad thing. It is the last time little children. Including me, I remind myself. Get it all out and be free. Winners . Oh shut up bm? You know I am right. Do it and live, don't be a fool, you can not bear the strain forever.

What we are doing is standing against the scientific hogwash ingrained inside the masses without knowing what we were going to face. It is not easy nor our choice. Shall the clay say to the potters hand, "Why have you made me like this"?????

Repeat

[wantomake]

And the fighting continues. Want no part of it. I get that from my extended family, so I stay away from them.

Taken care BroMikey. When you succeed, you may have to do it alone.

[wantomake]

Ok thanks BroMikey.

It's seems things have moved over to the standby thread.

[BroMikey]

Sorry BroMikey I've no idea what that means. Amazon.

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

No this is not opinion, Dave pointed out converters in general, buck or boost are running pulses thru glass parts so a snubber circuit is used to burn all of the extra energy to ground. We need to collect that energy. Here is a simple snubber


There is an arrow forming a circle. Do you see that? It is a cap sucking up the back spike and the resistor burns off the cap energy thru heat. That is what resistor do. Each time a Mosfet, scr, transistor issues a pulse the cap and resistor burns the back spike. If we could collect that energy and keep the damaging back lash away from the glass semiconductors we would be miles over unity.

[wantomake]

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.

[BroMikey]

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

[wantomake]

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.

[wantomake]

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.

[Turion]

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?

[BroMikey]

. 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

[Turion]

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.

[BroMikey]

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

[BroMikey]

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

[BroMikey]

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.

[Turion]

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.