Splitting The Positive

[BroMikey]

I doubt it, I just drained out a LITHIUM battery rated at 1300mah
and it only gave me about 700-800 mah Max. You have to run them
and see just like you are doing.

This sets a standard baseline as I am sure everyone understands.

Aaron told me to run battery C all of the way down and when i did
that I found out the actual amount the batteries can offer.

Charging the death "C" position battery is easy now. To give everyone
a perspective I discharged a single battery into a bulb at 750ma
for 1 hour burning up all of the power to ground.

NEXT:

I put battery "C" in it's rightful position to charge in the split
positive diagram and only 500ma down to 400ma running thru
the light bulb load for 1 hr.

Now it is charged again all the way to 4.19v

Now i will see how much power is in this same battery C
by removing it from the split pos..........system and burning
up all of the power once again to ground.

THIS is what I am doing

Gone see how much I get out this way.

I am back to double check all these results. C-U-ALL shortly
when I have another result. I can't wait to see that water
generator breaking all the records. That one you setup with
high energy ignition? Yeah that one, can't wait.

Yeah I am back and actually finishing that first charge that is at
around 4.18 but I think I'll go all the way to 4.2ovolt. When i left
it was below 4.20v. It started up at 500ma and after 1 or 2
minutes slowly dropped. It is running 400ma like before.

4.19volts now.

Yeah right that is what I am doing, I am charged up and gonna
runher down once again

Now it is 10 minutes later and the charge current is 380ma.

Yes this is the way to reuse power, I love it.

There it goes 4.20volts After 1 minute disconnected the battery is
still 4.05volts. I will wait till it stops. 4.03v now. so yeah it is
important to charge up the battery all the way.

It will sit at like 4.01 or 4.00 volts.

One hour later the light ran once again at the 700ma average
staring volts 4.2v and ending was 3.15v. Here is what I have
taken from these 3 batteries so far.

1 hour run time of light directly off battery "C"
1 hour run time of light while charging battery "C"
1 hour run time of light once again straight off "C"

So far three hours of light so remember the the conventional
discharge curve above shows that all three batteries in parallel
gave a 5.5hr run time of light.

So the goal is to see how many times I can charge and discharge
battery "C" running the light and charging in this split pos..........
diagram.

The run voltages of battery A & B while I am on this 4th hour of light
while charging at the same time is A = 3.60v and B =3.66v under a
load. Now I don't know if you understand the significance here but
this is very impressive. I only ran it for a couple of minute to get the
readings and then stopped. We will see next time how many hours
we have left.

Judging from my past measurements I will have to say that we will
get a lot of extra light doing it this way.

[Aaron]

stairstep

I guess I don't know what you mean that Paul can stairstep
batteries?

If one is charged and one is discharged, he can charge the one on the output faster that the input one goes down. Swap them and repeat and keep going and eventually, they're both back up to the top.

[BroMikey]

If one is charged and one is discharged, he can charge the one on the output faster that the input one goes down. Swap them and repeat and keep going and eventually, they're both back up to the top.

Well first let me say that I am grateful for this information
that Paul and others you mentioned have spent time on these
concepts. This encourages many of us who can not find any materials
whatsoever anywhere around the web such as YOUTUBE
experimenters might do.

The idea the Lithium batteries are very low impedance and this
can aid in the recovery process is something we all need to
remember. I remember John B saying that his huge batteries
were conditioned til they were 1 ohm.

So recovery is inhibited by internal resistance of batteries, this
makes me flashback to John K. statement that he likes a cap
dump to counteract this internal resistance. The high energy surges
both clean plate area and efficiently deliver energy on target.

So resistance of batteries is very important

I understand what you mean by stair stepping.

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

[BroMikey]

Hi Sky

This is what I am doing.

I am using up the power in C which is the charge battery
as you know conventionally discharging it nothing out of the
ordinary then I am charging "C" (The single charge battery)
with the split positive connection diagram while running a load
and repeating the process of emptying batteries A & B these are
the series connected batteries.

Here is what I said for this test yesterday then I will add today.

1st hour run time of light directly off battery "C"
2nd hour run time of light while charging battery "C"
3rd hour run time of light once again straight off "C"



Okay to add today so far making it another 2 hours of runtime
and charging this time. This being the 4th and 5th hour added
to the 3 hours of run time of the light.

4th and 5th hours charging "C" and running the light at 315ma ending
charge voltage was 4.10volts.

Batteries A & B voltages are 3.47v and 3.51v respectively under load.



I am now discharging "C" which is disconnected from the circuit
once again and this will make it the 6th hour of light.


I am using the split positive connection but the only time I
use it is when I have a dead battery to charge.

[BroMikey]

Referring back to this diagram where only standard discharging
takes place uses a resistor to slow current draw down.

All charging and discharging runs are done without that resistor.

In these tests "C" is being discharged without that resistor and
this is why I show a 700ma drain during this time of conventional
discharging. 6 hours have past and if you compare figures you can
see that under these conditions of using the energy we get more
light (More Mili-Amps/ time) longer due to battery "C" always being
completely dead before using energy from A & B in the split positive
system.

Previous runs kept battery "C" in an over charging condition at the
beginning which wasted power.


The new result is in for the 6th hour.

6th hour conventional discharge "C" battery 650ma ending 3.15v

[BroMikey]

Now it is time to calculate joules.

V x A x 3600sec. or average voltage x Average ma.

Under load figures after 1 minute connection time

1st hr = conventional discharge figures
3.85v + 3.2v / 2 = 3.525v
at 700ma average = 3.525v x .700ma x 3600sec = 8883 J

2nd hr = Split positive 3.20v x .425 x 3600sec = 5240 J

3rd hr = conventional discharge figures
same as 1st hr = 8883 J

4th & 5th hr = 3.05V x .315ma x 7200sec = 6917 J

6th hr = conventional discharge figures
3.75V X 3.15v /2 = 3.45v
at 650 ma average = 3.45V x .650ma x 3600sec = 8073 J

7th & 8th hr = split positive 2.90V x .150ma x 7200sec = 3132 J

I still have whats left in this battery that charged while the light
ran ending charge voltage only reached 3.90v with batteries
A & B now fully discharged at 3.23v each.

9th hr = conventional discharge figures
3.70V + 3.22V /2 = 3.46V
at 575ma = 3.46V x .575ma x 1200sec = 2387 J

This gives a total of 43,515 J

Conventional discharging of the same 3 batteries in parallel
only offered 22,900 Joules approx.

See previous charts above for verification.

I guess Mikey's Fluke meter must be to blame.



...........

[SkyWatcher]

Hi mike, thanks for sharing your results.
So it looks like about twice the original for the basic setup and you aren't even splitting the positive again, when you discharge C all by itself.
If you split the positive from C into another battery or back to A or B, maybe a boost circuit of some kind (joule thief), you'd push the output numbers up even higher.
peace love light

[BroMikey]

Hi mike, thanks for sharing your results.
So it looks like about twice the original for the basic setup and you aren't even splitting the positive again, when you discharge C all by itself.
If you split the positive from C into another battery or back to A or B, maybe a boost circuit of some kind (joule thief), you'd push the output numbers up even higher.
peace love light

Yeah I think so. The 22,990 conventional is far to low when
you are hooked on recycled joules.

We need to understand that in both split positive tests why
one is collecting more than the other. I said once and now I'll
say it again, the charging battery "C" is the critical variable.

In test one of the split positive runs the charging battery "C"
has a higher impedance being fully charged than a dead battery.

In test two the configuration was changed slightly to make sure
any time a split positive run was made battery "C" was dead.
This lower impedance is a more receptive state to reorganize
energy so with lower resistance energy transfer is more efficient.

22,900 J plus or minus 10 percent error = 20,000 J

29,000 J split positive run plus or minus = 26,000 J

43,000 J split positive run plus or minus = 36,000 J

I'll have to rerun tests to be sure I didn't do something wrong.

On the boost idea I think somebody needs to say that the
SG OSC was born for just this reason and you don't hear
anyone talking about using it.

Doesn't anyone think John B ever thought of using these circuts
to create a potential difference with his SG OSC, and then
call it a boost converter? Or call it a joule thief?

Come on guys. I am ranting again.

I know how this all sounds like it is too simple stupid.

[dragon]

Not necessarily the bedini stuff, but this joule thief is quite efficient in recycling energy back and forth....

[BroMikey]

Not necessarily the bedini stuff, but this joule thief is quite efficient in recycling energy back and forth....

Thanks Dragon

I'll consider it because you are leading.

[BroMikey]

BroMikey,

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

The Peukert Effect doesn't apply that much to lithium batteries like they do with lead acids.

That eliminates the 'C" rate, just one less excuse.

https://en.wikipedia.org/wiki/Peukert's_law

[Aaron]

charging peak voltage then drop in voltage

That eliminates the 'C" rate, just one less excuse.

https://en.wikipedia.org/wiki/Peukert's_law

It certainly makes it easier if you're using lithium batts. Are you satisfied with the results you posted that you got about 1/2 of the rating? That's not good news for the lithium batts your using but were they not fully charged?

I know for the 3.2v lifepo4 batts, 4 is 12.8 volts and ideally, they get pushed to 14.8 just like gels and AMG's, which is about 1/2 a volt under what flooded cell lead acids need.

Since Bedini's charging technology is not compatible with the normal lithium batts, I don't know - maybe ideally they need to be pushed a little higher than you quoted.

I don't know if the normal lithium batts do this, but lead acid and nicads have the charging curve where it goes up, peaks and then drops and the charge should finish after it drops showing the impedance dropped and the battery is fully charged. Maybe you can see if the charging curve is the same or similar with the non-lifepo4 lithiums. If you have some pc based volt graphing program, you can see it go up and come to a peak and then drop - if it is like so many other batts, that is your signature to end the charge AFTER the voltage drops after it peaks up near where you said it needs to go. This is not a guess of what will happen, but just thinking out loud if it would be the same. If so, then you know when it's reached full capacity for it's own particular condition.

[BroMikey]

It certainly makes it easier if you're using lithium batts. Are you satisfied with the results you posted that you got about 1/2 of the rating? That's not good news for the lithium batts your using but were they not fully charged?

I know for the 3.2v lifepo4 batts, 4 is 12.8 volts and ideally, they get pushed to 14.8 ...........................................

Hi Aaron

I re-did the previous sloppy calculation and found an error of
1000 Joules it should have been posted as 23,900 Joules

Now after a few runs and some harder discharges the joule
count is rising. Keep in mind I am only dealling with the
conventional discharge right now to see what's up with
these batteries. Some say 1200mah for these batteries
when they are brand new and we all know batteries give
an actual output that is lower.

The new measurements are for a faster discharge and
very closely watched on the meters. I will be happy with
70 percent of 1200mah. I didn't go all the way to 3.20v
but judging from my curve I wasn't going to make it much
longer than a minute or two so i stopped at the half hour
mark.

This new chart shows a 2.5hr run offering 25,700 Joules

I think they look good when you think about it. Maybe it is
the way I am calculating. Under load the new battery would give
the same average voltage of 3.5v and the 1200 x .7 (705) = 840ma
for 1 hour split that in half over a 2 hr and so on.

I got an average of 805ma over a 2.5hr run that makes these
batteries over 1600mah batteries plus. You picken on my
batteries.

I'll have to confess now that I have been charging this with
genmode so my batteries are getting bigger? What do you think?

I thought if a battery was rated at 1200mah that meant that
over a 2.5hr time frame I should only get an average of 500ma
per hr? Is that right?

Anyway give it some thought here is my chart.
........................

[BroMikey]

The way I am figuring may be the problem

I see what I did wrong hang on a minute.

805 ma average over a 2.5hr run = 2000mah for 1 hr
then split that between 3 batteries in parallel = 700ma
per battery. pretty bad I guess if they were 1200mah?

I have another set thank goodness maybe I'll see what a set
of 3 of these batteries give off. These are new ones. I found
the new set.

[BroMikey]

Here is what the new set looks like and some call it
a 18,600 2 ah and some call it a 5000mah battery
so I don't know I really don't.

The ones I am running right now are green 1200mah
and these are blue 2200mah i guess. Who knows right?