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  #31  
Old 06-10-2016, 08:16 AM
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For those of you dangerous enough to build your own
555 timer based BOOST CONVERTER look at these
engineers.


http://electronics.stackexchange.com/questions/152432/555-timer-boost-converter-doesnt-meet-spec



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  #32  
Old 06-10-2016, 08:30 AM
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Boost converters or boost regulators?









op amp - Is it possible to construct a voltage doubler that both powers an opamp and is driven by it? - Electrical Engineering Stack Exchange





Constant current regulation with boost converter, PWM and 555 timer. | Electronics Forums

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  #33  
Old 06-11-2016, 03:25 AM
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The windings can still be clamped into a vice to gently flatten
them as the resins harden.plenty of room to clear the windings.
To flatten windings put a block of soft wood such as pine on each
side and apply pressure. Or just strike the boards with a heavy
hammer against the table and check progress with each tap.

I don't show it here in the picture but I did have plastic spacers
thru the rotor housing keeping the winding away from the bearings.














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  #34  
Old 06-11-2016, 05:10 AM
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Potential Difference charging.

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  #35  
Old 06-11-2016, 11:12 PM
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Here is a converter that operates from 3.5v up on the input
but I have no idea if it shares a common ground. This might
be cheap enough for someome to use only as a small test.
Or maybe these people already have a bigger one. I use one
to step up voltages now but the one I have it only 1 sq inch.



http://www.ebay.com/itm/6A-100W-DC-Converters-Power-
Regulated-3-5V-30V-5V-12V-Voltage-Step-Up-led-Volt-R-/190898609868?hash=item2c72719ecc:g:jHgAAMXQ0pNRvDL c




Features :
Dimensions: 67x42x18 mm (L*W*H)
Input voltage: DC 3.5V-30V
Output voltage: DC 3.5V-30V (boost,input voltage < or = output voltage)
Continuous current: 6A (long-term work)
Maximum input current: 10A (peak)
Output power: Max 100W, (U-in * I-in * Efficiency = U-out * I-out)
Voltmete display color: red
Voltmeter accuracy: (0.5% +1 digit)
Static power consumption: typically around 15ma
LED indicator:Have
Anti-reverse protection: Have
IN+: Positive input
IN-: Negative input
OUT+: Positive output
OUT-: Negative output
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  #36  
Old 06-11-2016, 11:23 PM
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Also beware that China Man speak bad English sometimes
he calls this a BUCK CONVERTER but I don't think so.

It looks like a simple circuit? But i don't know.

Chip number XL4016


http://www.ebay.com/itm/8A-DC4-40V-
To-DC2-36V-Buck-Power-Converter-200W-XL4016-
PWM-Volt-Step-Down-Moudle-/262159873990?hash=item3d09f237c6:g:FgkAAOS
wHQ9WVYQj





Description
Main parameters:

Product name: DC voltage converter
Conversion method: PWM
Input voltage: DC 4-40V
Output voltage: DC 1.25-36V
Max current: 8A
Max power: 200W
Conversion efficiency: 94%
Switch frequency: 180KHz
Appearance size: 61*41*27mm

Features:

This product uses XL4016 switch voltage conversion chip, and
realizes double rectification by collaborating with MBR10200.
Max input can reach 40V, continuously adjustable
for 1.25~36V output.

Max output current can reach 8A.
Need a fan.
Recommend no more than 5A for long-time work.
Support panel installation:
1. Drill a hole for 7mm on the panel.
2. Extend potentiometer handle.
3. Tighten the screw and cover the knob.

Package List:

1 x 200W XH-M401 DC-DC 4-40V To 1.25-36V Buck Module 8A XL4016E1 Voltage Converter
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  #37  
Old 06-12-2016, 12:03 AM
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LTC3780 Constant voltage and current



http://www.ebay.com/itm/DC-to-DC-
converter-Boost-buck-power-supply-Constant-voltage-
current-Regulators/351755769339?hash=item51e647
15fb:g:RRsAAOSwqv9V6Mrz





LTC3780 Constant voltage and current DC to DC converter 4-32V to 0.8-32V adjustable Automatic Step-up/step-down Boost/buck power supply Module Solar Wind Energy charging Vehicle Voltage Regulator led driver #1300201

Module parameters:

Module name: 8A Automatic Step-up/step-down Constant voltage and current module

Module Properties: non-isolated synchronous rectifier Automatic Step-up/step-down module

Input voltage: DC 4-32V

Input Current: 8A (MAX),Peak 10A(Long natural Radiating Working within 6A)

Quiescent Current: 5mA (working voltage not simultaneously be errors)

Output voltage: DC 0.8-32V continuously adjustable

Output Current: 8A (MAX) Peak 10A (<=6A long natural cooling inside)

Constant current output range:0.2-8A

Operating temperature: -40 to +85 C

Operating Frequency: 200KHz

Conversion efficiency: Up to 98% (Efficiency is related to Input
and output voltage, current, voltage difference)

Over current protection: yes

Short circuit protection: (input 10A fuse) Double circuit
protection, use more secure

Over-temperature protection: Yes (over-temperature
automatically reduces the output voltage)

input Reverse Polarity Protection: None, (if required in the input string into the diode)



Installation: 4pcs 3mm screws

Output preventanti-irrigation: None, when used for charging
coupled blocking diode, otherwise it will damage the module.

Wiring: none welding wire terminal output,
(IN as input, OUT is the output)

Module dimensions: length 60mm X width 51mm X height 20mm

PACKING INCLUDING:

1x power supply Module
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  #38  
Old 06-12-2016, 11:44 PM
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For Your Info (FYI) This is what I found out about the scooter motor
commutator design for clamping the winding into it. I am afraid to
press it back down but I only raised each tab 2mm or less. Still
I might solder the wire on instead.


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  #39  
Old 06-16-2016, 04:07 AM
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Boost Converter 10AMP 400Watts

It's gonna Bee Sweet. Sposta bee here in 2 weeks









Specification:

Input voltage : 6V-40V

Input Current: 15A(max)

Output voltage : 8V-80V(adjustable)

Output power:400W(max)

Output current: maximum output current of 10A ( adjustable )

Output power: the effective power P = input voltage V * 10A

Conversion efficiency: up to 95% ( input voltage, current;
output voltage and current impact of conversion efficiency )

Short circuit protection : Fuse
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  #40  
Old 06-16-2016, 09:58 PM
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You wil have to let us know if it will work wired as Matt has shown us. Some of the better boost converters I have purchased will NOT run that way. I have purchased 8 different boost converters and two of them would not run the way Matt showed us to wire them. Unfortunately, they were the more efficient ones.

Dave
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  #41  
Old 06-17-2016, 01:47 AM
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Quote:
Originally Posted by Turion View Post
You wil have to let us know if it will work wired as Matt has shown us. Some of the better boost converters I have purchased will NOT run that way. I have purchased 8 different boost converters and two of them would not run the way Matt showed us to wire them. Unfortunately, they were the more efficient ones.

Dave
I sure will and I also remembered that when I bought it, I
would like a boost that is a tiny bigger than 100 watts so
I decided to take the chance. I can always buy more.

I used 20awg wire and uses 30 extra turns hoping to run
a tiny bit more power to charge my 2000 ah 12 battery set.

I would love to charge those big batteries off this small
Pulse Motor Tesla switch rewound Kool charger.
Tell Matt I appreciate all the hard work handed over on a
silver platter, I understand now that the motor switch
is an off shoot and people don't realize how hard it is to
duplicate the motor switch action with a circuit.

When I get it running I'll post my success story for all to see.

I don't know why I never picked this experiment up? I see it
so clear now but 2 years ago? I don't know.

Keep splitting the pos.... everytime you want to run a load
Humm... I am thinking about it. I am charging batteries.

I will do what I said first, run a light bulb on pure lectric current
and watch the battery burn down. Record time and joules.

A car light bulb as a load is a great way to start because a bulb
will stay at the same amp draw all the way down to 12V + 12V
=24v to stop at. Bulbs are nice to work with, you can measure
the current once or twice say at the end and the draw will be the
same.

I will use all three batteries in parallel to run my light till the three
12volt batteries reach 12v or maybe 11.5v something like this.

I don't need my magic switching motor to do that nor a converter
yet this must be done to give me an idea how many joules my
batteries under test can hold and how many joules is consume in
conventional (Burn them amps to the ground) operation.

With an idea how many joules I have in those three batteries
I will then connect all three batteries up to the light bulb as
suggested with a split pos....... arrangement.

That I can tell you will far surpass the previous connection.

I see all of the information, the diagrams, connection paths.

The part that is making my brain go cross-eyed is how a load
with a voltage drop doesn't cost anything in joules to run.

And I know the answer too. The answer is that instead of power
going to ground and burning up it is going to another battery to
charge it. My conventional horns are sticking out all
over.


$18

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  #42  
Old 06-17-2016, 01:30 PM
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Quote:
Originally Posted by Turion View Post
Post 969 In the Basic Free Energy Device thread has a link to the pdf of building the motor as well as a link to the schematic of the boost converter.
... if you know what dropbox is and have an account:
Basic Free Energy Device
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  #43  
Old 06-17-2016, 07:24 PM
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For beginners who have many questions look here
this video tells you about boost converters.

Longer wires or longer coils produce higher voltages.


Working in 105 degree temps is kicking my other end.
Extreme high temp advisory in effect now.

http://www.energeticforum.com/289423-post1072.html






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  #44  
Old 06-18-2016, 08:58 AM
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Okay I did my very first ever SPLIT THE POSITIVE runs
And I have to say "IT WORKS!!! Or I am loosing my
marbles?

Here is what happened. I couldn't wait any longer so I
pulled out some of my 2ah Lithium ION (Choice Batteries)
Batteries and this diagram shows you how I did it.

I ran the system 4 times and each time the current draw on
the modified household LED light bulb was always 400 mil-amp
at 3.15 volts.

These batteries are A-number one charged. All batteries started
at 3.85 well one was 3.90 volts. Now after running my load 4 times
for 10 minutes and the last time 15 minutes before switching I
am reading all batteries higher than the start so there must be
some kind of a mistake. I had to say that because it really
does seem to work. I'll keep doing it to see where I messed up.

I have this crazy feeling as I watch these batteries run this
light with no drop in battery capacity. Something is not right.

I am pretty tickled about it. Of course in the back of my mind
it can't work so I must of made a mistake right?This is to
easy. I don't want to start dancing to soon but it really look like
we have our foot in the door here.

I'll shake these meters to see if they are acting up. Nope no
rattle there.

I am liking this recirculated power. It is a regular LED bulb from
China with the aluminum heat sink and it lights the whole room
up. I didn't want to much pull on these little batteries so I only
enabled 4 LED's to match the power draw I was after.


PS I am back after letting batteries sit for 30 minutes
and a final 5th run trying to get these batteries to drop.
Battery Bank A = 7.87 v and Battery Bank B = 3.93 v

PS the is the second return, I see the loaded voltages now
changing first run 7.35 v and 4.05v now after 6 runs the
first bank loaded voltage is 7.25v so now miliamp draw
went down to 300 ma from the original 400 ma.draw.

So little by little the batteries are dropping slightly. I will hook
it direct now.


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  #45  
Old 06-18-2016, 08:35 PM
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Last night I kind of jumped the gun so to speak getting
ahead of myself so here it is, the conventional curve (No Curve Yet)
that is needed to establish a norm with these batteries.

To maintain the 400ma draw I have installed a 1.5 ohm
resistor as shown. Let us see what these batteries can
do over a period of say an hour or two, then we will
split up the same three batteries to run off positive terminals
only.

Just like the regenx, everybody throws a coil on a magnet
wheel and speculates but what about the measurement during
transition? I mean some sort of establish norm must set the
precedent.



It was fun to watch the light run and battery charge the other way.

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  #46  
Old 06-18-2016, 10:04 PM
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Calculations in Joules

Without a precedent of conventional operation this deduction
as to whether or not power is being recycled leaves us guessing.

If anyone has done this comparison and has it posted online
please let me see your results. The calculation is done with grade
school math using times tables to multiply volts X Amps per
second or Volts X Amps X seconds = JOULES

So many Joules per run. Each minute has 60 second so the value
of the number 60 is multiplied by your watts. This is watt/minutes

So if you are making a run like me at 400ma and 3.15volt for ten
minutes your calculation will look like this.

.400ma X 3.15volts X (TIME IN SECONDS)

Time in seconds for 10 minutes = 60 sec X 10 Min = 600 sec

So this means for my experiment

(.400ma X 3.15volt = 1.26watts) X 600 sec = 756 Joules

Now I can clearly see how many nuggets I ate up over
a 10 minute run time. Next looking at battery curves.

As a battery runs down it will hold steady at one place for
extended periods not moving many points on the scale then
all at once it will decline slightly more rapidly and next it will
decline very quickly over a short space and fail.

I will be back to show you my curve for these batteries.


(NOTATION)
To be fair conventional runs must be done the same as is required
with split positive runs. Split positive runs for my batteries need
changing around after 10-15 min. This constitutes a rest interval
however small say 2 minutes between runs, therefore each
conventional run should follow the same pattern.



For those of you who have never seriously considered all of the
efforts this man has made to show battery curves I suggest
you take advantage of it.

God Bless you JOHN B.



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  #47  
Old 06-19-2016, 01:26 AM
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Here are a few numbers crunched in for my Lithium ION
2ah batteries connected in parallel and conventional mode.

I have extended the chart. I will be updating this chart
every so many minutes as I collect data.


PS notice I have included the amp hour capacity of the
batteries at the C20 rate with calculation. Look at the diagram.

Later we will need to recalculate what a split positive C20
rate would come to to be fair. In the split positive arrangement
only 2000mah are considered. A C20 rate for a 2000mah battery
comes to a load of 100ma so I would have to run a split positive
system 3 times as long to collect data.


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  #48  
Old 06-19-2016, 03:54 AM
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See the previous chart

This chart is bigger so the curve may be seen with
greater ease. 270 minutes into the C5 discharge for
Lithium ION batteries.

The graph has been expanded with a 1 volt area of 3 to 4 volts.
Also notice the ever dropping current from over 400ma on
start up until now at less than 300ma the run in progress.

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  #49  
Old 06-19-2016, 05:47 AM
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Joule Count for Chart above

As an average over all mili-amp draw from 405ma starting
down to 300ma ending a figure of 350ma is found. Yet each
segment might be calculated separately each having a specific
number of joules and then add them all together.

Each time frame for a given ma draw varies.

Segment Number

(1) 3.90v-3.76v= average =3.83volts X .405 X (sec 900)
Joules = 1396 j

(2) 3.74v X .400 X 1800sec = 2693 Joules

(3) 3.70v X .380 X 2700sec = 3796 Joules

(4) 3.67v X .375 X 1800sec = 2477 Joules

(5) 3.64v X .360 X 1800sec = 2359 Joules

(6) 3.60v X .340 X 1800sec = 2203 Joules

(7) 3.56v X .325 X 1800sec = 2083 Joules

(8) 3.52v X .305 X 1800sec = 1932 Joules

(9) 3.49v X .300 X 1800sec = 1885 Joules

10) 3.44v X .280 X 1800sec = 1734 Joules

11) 3.33v X .300 X 1800sec = 1798 Joules


Total Number of Joules? Nuggets I call them = 22,960 J
There is no grey area/kind-a-sorta happy horsezhit this
is the number. I will be letting you know. All testing is
carried out to the second.
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  #50  
Old 06-19-2016, 06:05 AM
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These Lithium ION batteries have been charging for an
hour now after reconfiguring them into a series string. I use
the C5 charge rate, just as I use the C5 discharge rate
please make a note of it.

Battery voltage must reach 12.6v in the 3 battery series
line up with each of the three batteries needing to top
out at 4.2v.

I bought some impulse relays and am brainstorming a possible
switching madness to keep me from changing batteries every
30 minute for quite probably days to finish the counter part
test to finalize this comparison run.

If you have an easy circuit to switch batteries please post
here ASAP.

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  #51  
Old 06-19-2016, 11:28 AM
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Split Positive First Run Under Way.

This is where I will stop till the next day as I have no switch gear.
For those of you who have not switched batteries
every 30 minutes may be over charging the charging
battery, and burning up energy as witnessed by no increase
in charge voltage when the max charge is reached.

This is unacceptable therefore batteries MUST be rotated
before energy is thrown away.

In the last 2 hours of this first split positive run 2 batteries set
at 3.85v and the charge battery at 3.96v

The amp draw is closely watched and averaged, the start
voltages shown are measure after 30 seconds. Each start up
run for 30 minutes is preceded by a 2 minute rest during that
time batteries are changed around.

I am anxious to continue


PS it is day two and I have updated this chart (See Joule Count)

We have almost hit the halfway mark for joule counting after 3.5hrs

The chart is being updated every so many minutes please
stay tuned. It's looking pretty good. I wasted joules I won,t
count, not to many. Gotta be sure watt's up with this experiment.
I see everyone's post, thank you.



SECOND ENTRY DAY TWO

It looks like it's going to be on into the night before I reach
the 22,000 joules but I can tell you this, the batteries all get a rest
and some charging are along the way. I like the way that the
battery has time to recover when being charged. To give you
an idea of what I mean battery C or the charge battery
always stays at 3.95v at 200ma being fed into it.

At 300 minutes into this run

Battery A = 3.65v
Battery B = 3.45v
Battery C = 3.95v

This is the voltage while the circuit is running and circulating
200ma thru my LED light bulb. To keep a C20 rate I should be
down to 125ma, instead am going with a C10 rate. Using a
C10 rate will cost me joules as compared to the C20 rate
yet these Lithium ion batteries are made to handle that.

These experiments are more than meets the eye and very
little information in video form or otherwise is available on
the subject.

No one who has the time has taken a systematic approach
showing video footage of how energy can be recirculated.


Third entry: New graph below, I am running out of space.
See 2nd plot below.













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  #52  
Old 06-19-2016, 11:33 AM
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Hey Bro M,

Some switching info for you from James MacDonald (posting problems?) in the following:

I wanted to post some information that Bro Mikey is looking for.

It appears that the company called Blue Sea Systems makes Lead Acid battery switches which can be wired up to a few batteries to switch them In and Out of the charging circuit. They have Isolators that allow 1 battery to run the engine while another battery runs the boat electronics. They also have automatic switching for charging a battery or manual switching for adding Battery 1 and 2 to the charging circuit. The products are not cheap but they will get the job done he wants to do. I just wanted to be able to post. Maybe you can send him the information. A combination of the two below or just one of them would help in making battery switching in and out easier.


https://www.amazon.com/Blue-Sea-Syst...lue+sea+system


https://www.amazon.com/Blue-Sea-Syst...lue+sea+system

Happy switching,
Yaro
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Old 06-20-2016, 01:54 AM
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Quote:
Originally Posted by yaro1776 View Post
Hey Bro M,

Some switching info for you from James MacDonald (posting problems?) in the following:

I wanted to post some information that Bro Mikey is looking for.

Maybe you can send him the information. A combination of the two below or just one of them would help in making battery switching in and out easier.



Happy switching,
Yaro

Thanks Yaro

Good to know what can be purchased across the counter.

Also look at my updated graphs above.
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  #54  
Old 06-20-2016, 04:54 AM
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This split positive battery system I built has past the equal
mark hrs ago. After 9hrs plus the light is burning along and my batteries voltages are far from 3.25v which is considered a dead battery.
The conventional system killed the battery in 5.5hrs extracting
22,000 joules to keep the light going. Now after 9 hrs of run time
I have recovered or recirculated some energy and the joule count
is over 24,000 J with no end in sight.

Battery A = 3.48v
Battery B = 3.62v
Battery C = 3.82v

It;s-a-lookin pretty dern good from over here.

See updated graphs above.


...
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  #55  
Old 06-20-2016, 05:38 AM
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BroMikey,
I learned a long time ago that the only one you need to prove this to is yourself. Make sure you do that beyond ANY doubt, because you will spend the next few years (if you are active on the forums) defending yourself and your "research methods" from those who believe they know better.

Remember that both the inverter OR the 12 volt brushed DC motor add something to the circuit that aids in the charging of the third battery, and that is a PULSE. If you want performance with larger batteries and larger loads, one of those two is essential.

Matt's motor adds something that other motors don't do quite as well, and that is add the generated voltage of the motor (acting as a generator) to the circuit which increases the voltage you are moving around rather than decreasing it.

When you take THIS CIRCUIT, that allows you to run loads without using up the energy in the batteries and add Matt's motor which generates additional voltage and run the RIGHT generator, I believe you will begin to see where we are coming from.

You CAN get your batteries to run loads without losing any energy. It requires TIME and PATIENCE and PERSISTENCE. (And pay attention to what Bob French is posting on the forum because he is spending hours every single day on this and is applying this to working situations) If you learn what it takes to do it with batteries, you can do it with capacitors. That's the next step, but those who think they are going to skip the step of learning to do it with batteries and leap to the next step of doing it with capacitors will most likely find out that their shortcut cost them in both TIME and MONEY. Once you can do it with capacitors you are ready to learn how to do it without either.

Have fun
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  #56  
Old 06-20-2016, 05:53 AM
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BroMikey BroMikey is online now
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Hey Dave

I will listen to the best of my ability to understand what Bob
is saying. To split the positive without a capacitor or battery
might be done by creating a potential difference using say motor
coils that would require some form of rectification.

After rectification there is a positive and a negative.

That is what I needed, I been trying to go beyond this present
circuit to apply it to other things, this answers my question.

Keep at those exotic motors with magnets.
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Old 06-20-2016, 06:16 PM
bistander bistander is offline
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Test question

Quote:
Originally Posted by BroMikey View Post
See the previous chart

This chart is bigger so the curve may be seen with
greater ease. 270 minutes into the C20 discharge for
Lithium ION batteries.

The graph has been expanded with a 1 volt area of 3 to 4 volts.
Also notice the ever dropping current from over 400ma on
start up until now at less than 300ma the run in progress.

Hi there,

Good to see you posting data. I assume these results are from this set up:

[/SIZE]


And where you calculated 22,960J. And ran for 5.5 hours. However you were attempting a C20 discharge. You were running a bit higher current, so maybe it was more like a C15. But that means you should have run for 15 hours, not 5.5. Something is very wrong with that test. Perhaps the cells were not fully charged.

I think if you want to use that as a baseline, you need to explain the discrepancy or repeat the test.

Thanks,

bi
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Old 06-20-2016, 08:02 PM
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BroMikey BroMikey is online now
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hello Bitstander

Yes I agree that the C20 and C15 or the C10 rates do
not seem to apply with these Lithium ion batteries.

This is my first test ever with this kind of battery.

Also Lithium ion batteries are not generally charged by
such low rates like 100ma so I really am at a loss to
apply any such standards as we do with lead acid wet cells.

However I have concluded that each time I charge these
batteries I get a certain voltage for a full charge at about
4 volts after 20 minutes.

During my recirculating split positive runs I speculated
that since as little as 60ma was being returned at the end
that these batteries eventually ran down.

Also I have concluded that if I let the batteries set on the
table for a day that the same amount of power is still in them
as when I had left them.

I have also concluded that I don't know much about the
charge, discharge rates or capacity for true baseline
standards for lithium ion batteries. I maybe that the 2 ah
rating on these cells is not the actual value. This would not
surprise me as lead/acid batteries are regularly only good
for 60 percent of their joule count.

Whatever the joule count is splitting the positive is not suppose
to care. The claim is that energy is recirculated back to
battery C while running loads off of the elevated potential
difference from series batteries B & C.

I have concluded that the system does recirculate energy.

I have concluded so far with test #1 that these batteries charge
up to 4.2volts and discharge to 3.2volts as specified by the
manufacturer of these batteries.

I will have to re-figure the real amp hour ratings and their "C"
rate that is probably designed for a C5, you are right., but then
I am not sure yet about Lithium ion batteries.

This leads us to consider the possibility that internal resistance
from loading batteries at a great rate might waste some power.

Based on my observations in this simple split positive first
run I can get more energy out of this set up with the same
batteries. And it looks like I just got 50 percent more before
the system went dead.

Whatever the joule count is suppose to be the claim is that
all power can be recirculated and if correctly done will result
in a circuit that runs loads for free.

My test of course is inconclusive, so a self runner is out, THUS FAR!!

This testing has finished and I will be back with a joule count.

This document says C2 and C1 rates apply


http://powerelectronics.com/site-fil...g/504PET23.pdf


So I don't know what I am doing with these batteries. Yeah these
batteries were a poor choice.
............

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Old 06-21-2016, 04:53 AM
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BroMikey BroMikey is online now
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I have been reading all comments and the one about
limiting charge current to the charge battery is a good
one.

Let me give the joule count for this latest experiment.

See setups in charts above.

Conventional mode = 22,100 Joules

Split Positive mode = 29,900 Joules

So we can see a huge difference in available joules to
light a bulb even with my crumby quick setup not very
well thought out. I couldn't wait, had to get my feet wet
you understand.


This experiment has shown me that the principle splitting
the positive offers extra power right away. However in my
build the batteries are unknown to me or I am in unfamiliar
territory by using them. Probably the miliamp usage is to high
and the charge battery is constantly over charging and burning
up energy as the max has already been reached.

I remember now watching the charge battery over charge for
several 30 minute periods where no increase in voltage occurred.

I have some bigger batteries. And that Matt
motor/switch in the works.


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Old 06-21-2016, 07:38 AM
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3 battery splitting the positive basic

Quote:
Originally Posted by BroMikey View Post
Battery voltage must reach 12.6v in the 3 battery series line up with each of the three batteries needing to top out at 4.2v.
BroMikey,

You're doing some great work here and I hope you keep posting your results.

What was the highest voltage batt 3 (single reversed batt) went to when it didn't climb any higher?

Lithium lifepo4 and lithium ions (usually cobalt based) are constant voltage batteries and not constant current batteries (like lead acids) so although they can show a higher voltage for longer, it can be deceiving to see a voltage last so long.

However, with that said, you are right about joules vs joules so whether it is a constant voltage or constant current battery, work is work from a certain voltage down to a certain voltage and that is what you want.

The Peukert Effect doesn't apply that much to lithium batteries like they do with lead acids. A lead acid 20 hour discharge will give you a certain amps discharge for 20 hours. At a 10 hour discharge, it will be 90% of that and at a 5 hour discharge, which is the most any lead acid deep cycle should be discharged at will be about 80% of the 20 hour discharge. So with lithium batts, this effect is supposed to be non existent so whether you discharge a lithium at a 20 hour rate or 5 hour rate, you should still have the full capacity.

If this is the case, then a 1 or 2 hour discharge rate should be an honest test with your little lithiums and won't take too long before you have to rotate them.

With your conventional discharge chart - 6000ma you're showing 300ma for 20 hours. If there is no Peukert effect, you should be able to hammer them with a 1 or 2 hour discharge rate and get the full real capacity.

6000ma for 2 hours is 3 amps * 3.x volts = about 10 watts. It would be a good test with all 3 batts charged and discharged at a 1 or 2 hour rate to see if you get the same as the 20 hour rate. If so, you validate the claim that lithiums have no Peukert effect and therefore, you can accelerate all your tests with multiple rotations in one day.

Do you have a factory charger for the batts that charges them up and a green light comes on when they're done? If so, what voltage are they at when they're supposed to get done?

Also, if you charge it conventionally and then discharge it at x hour rating, what voltage does the battery drop to when you get 100% of the rated amp hour capacity? If you know that number, then charge all 3 batts with a conventional charger until they're full. Then discharge one battery until you get 100% of the ah rating and it is considered dead. Put that battery in position #3 and have 1 & 2 in series and run the 3 battery test to calculate how many joules you windup getting until they're all dead. Starting with 2 full charged batts and one fully discharged, you should not be able to get more than 4ah's of capacity at the 3.x volt range until all 3 are dead.

This is the old original concept and has nothing to do with pulsing, recovery, anything other than just the simple basic 3 battery schematic John had posted for ages.

I may have missed something here, but just wanted to get a few clarifications.
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