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Poppy
Average Member


USA
418 Posts

Posted - December 25 2011 :  10:16:04  Show Profile Send Poppy a Private Message  Reply with Quote
Hi Everybody...

Here are the voltage doubler and the voltage quad diagrams...





~~~Life's greatest FAILURE is not learning for our MISTAKES!~~~

Good Experimenting...
Poppy

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USA
Mountain View


IrishDave
Senior Member



Ireland
850 Posts

Posted - December 26 2011 :  09:09:00  Show Profile Send IrishDave a Private Message  Reply with Quote
Hi Poppy.

I have seen a few videos of peeps using them, but I know very little about their applications.
I do have some questions -
Are they caps in the schematics?
Are they powered by AC or DC?
What types of diodes are needed?
How do you determine the Pos and Neg of the AC?
What are they best used for?
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49er
Administrator



USA
4442 Posts

Posted - December 26 2011 :  10:28:52  Show Profile Send 49er a Private Message  Reply with Quote
Hi Poppy

To go with Dave's questions also how many amps can be put out with this system and can it be controlled with frequency

Doug
The sky is not the limit...There are footprints on the MOON.
Your only as DUMB as where your standing.
No matter where you go there you are.
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Poppy
Average Member



USA
418 Posts

Posted - December 26 2011 :  14:30:14  Show Profile Send Poppy a Private Message  Reply with Quote
Hi IrishDave...

1. The capacitors are electrolytic.
2. The circuit is powered by both AC or Pulsating DC.
3. I use 1N4007.
4. The polarity of AC changes with each half cycle so it makes no difference which side of the input is connected to the circuit.
I connect the voltage pump to the ST and SR of the SSG circuit.
5. They can be used to supplement the run battery or used to charge a second battery.

To 49er...

Don't Know.

~~~Life's greatest FAILURE is not learning from our MISTAKES!~~~

Good Experimenting...
Poppy
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49er
Administrator



USA
4442 Posts

Posted - December 26 2011 :  14:58:13  Show Profile Send 49er a Private Message  Reply with Quote
Hi Poppy

I think I will try the first one. Thanks



Doug
The sky is not the limit...There are footprints on the MOON.
Your only as DUMB as where your standing.
No matter where you go there you are.
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IrishDave
Senior Member



Ireland
850 Posts

Posted - December 27 2011 :  06:25:56  Show Profile Send IrishDave a Private Message  Reply with Quote
Cheers Poppy
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TinMan
Advanced Member



4082 Posts

Posted - December 28 2011 :  04:35:13  Show Profile Send TinMan a Private Message  Reply with Quote
I would say 49er that the amps avalible would depend on the amps of your power supply and the amps the caps could put out.In saying that if it was used in a puls system say with a 20% on time then the amps avalible would be 5 times the amps avalible from the ac input divided by 2 in the case of the voltage doubler and divided by 4 in case of the quad.
EG-24 volt ac supply at 5 amps through the doubler with 20% on time pulse motor would give around 48 volts at 12.5 amps with a 20% on time-oh minus the voltage drop through the diodes.Those big orange caps i have that you would have seen in a lot of my vidio's will put out an 87 amp pulse-these could be well suited in this situation.

Remember-when you make a mistake you have just learned how not to do it.This only brings you one step closser to success.The only way to fail is to never try>>>>TinMan<<<<
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ron_o
Moderator



United Kingdom
1052 Posts

Posted - December 31 2011 :  08:00:28  Show Profile Send ron_o a Private Message  Reply with Quote
Voltage doubler or tripler circuits etc do so at a cost. You cannot gain from these circuits, if you double the voltage then you will half the current ( in actual practice the available current will be less than a half because of losses )
Also the output will be lower if the circuit is driven by a squarewave waveform compared to a sinewave waveform of equal voltage potential.

ron
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TinMan
Advanced Member



4082 Posts

Posted - December 31 2011 :  09:59:29  Show Profile Send TinMan a Private Message  Reply with Quote
Hi Ron-i thought it would be the other way around as a square wave has a longer peak voltage time than an ac sinewave.Also an ac sinewave has a longer 0 volt time than a square wave although its not much.This is just my thoughts, but i would like to know how it realy is-im all up for learning lol.The picture below is why im thinking it is the other way around.Wouldnt the sections that i have scribbled in white be the current lose of a ac sinewave to a squarewave?


Remember-when you make a mistake you have just learned how not to do it.This only brings you one step closser to success.The only way to fail is to never try>>>>TinMan<<<<

Edited by - TinMan on December 31 2011 11:21:11
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ron_o
Moderator



United Kingdom
1052 Posts

Posted - January 02 2012 :  15:10:17  Show Profile Send ron_o a Private Message  Reply with Quote
Hi TinMan

Ok , i can see where your comming from , the squarewave waveform will indeed provide greater power per unit time.

This perhaps will explain where i was comming from.

If you measure ( using you standard multimeter ) a pure sinewave its RMS value ( as displayed ) will be 0.707 times its true actual peak voltage . A squarewave waveform voltage will indicate its true peak voltage on the same meter.

So when rectified the actual voltage from the sinewave will be displayed voltage / 0.707 ( minus rectifier losses )
The squarewave will be the original voltage minus rectifier losses

So i guess i'm guilty of not fully defining my reply , sorry

ron
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TinMan
Advanced Member



4082 Posts

Posted - January 02 2012 :  17:13:08  Show Profile Send TinMan a Private Message  Reply with Quote
Ah ok-all good now,Thanks Ron

Remember-when you make a mistake you have just learned how not to do it.This only brings you one step closser to success.The only way to fail is to never try>>>>TinMan<<<<
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TinMan
Advanced Member



4082 Posts

Posted - January 04 2012 :  08:51:54  Show Profile Send TinMan a Private Message  Reply with Quote
Ok here's one for you all-(vidio posted in general discution-WTF-HELP)I have a large 24 volt transformer-26.2 volts ac befor rectification.I built one of poppy's voltage doublers useing my big caps(10000uf-63 volts,87amp) and from that 26.2 volts ac i get 71 to 72 volts accross the caps?????
By rights we should only end up with double 26.2 which = 52.4 - diode loss.So what go's on here then???

Remember-when you make a mistake you have just learned how not to do it.This only brings you one step closser to success.The only way to fail is to never try>>>>TinMan<<<<
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TinMan
Advanced Member



4082 Posts

Posted - January 07 2012 :  07:06:23  Show Profile Send TinMan a Private Message  Reply with Quote
Hi All-SD was looking for a simple invertor,so here it is SD.Being in america you will ofcourse have to get a 110 volt with a center tap of 12-0-12. I would probably double up on the transistors aswell to handle the extra current due to your lower voltage, as we use 240 volts here in oz-so we only need half the amp's you guys need.


Remember-when you make a mistake you have just learned how not to do it.This only brings you one step closser to success.The only way to fail is to never try>>>>TinMan<<<<>>Were's the vidio??<<<
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ron_o
Moderator



United Kingdom
1052 Posts

Posted - January 10 2012 :  03:13:44  Show Profile Send ron_o a Private Message  Reply with Quote
Hi TinMan

The answer is due to the discussion we had above regarding peak and rms readings

so :- 26 x 1.414 = 36.764 for first stage then 36.764 x 2 = 73.528 volts for the second stage.

ron
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TinMan
Advanced Member



4082 Posts

Posted - January 10 2012 :  17:06:50  Show Profile Send TinMan a Private Message  Reply with Quote
Hi - i knew you would have the answer lol.
Cheers

Remember-when you make a mistake you have just learned how not to do it.This only brings you one step closser to success.The only way to fail is to never try>>>>TinMan<<<<>>Were's the vidio??<<<
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Magneticitist
Senior Member



USA
681 Posts

Posted - January 10 2012 :  18:58:07  Show Profile  Visit Magneticitist's Homepage Send Magneticitist a Private Message  Reply with Quote
"26 x 1.414 = 36.764 for first stage then 36.764 x 2 = 73.528 volts for the second stage."

hey Ron could you explain that for me? totally lost me there. need some education in this department so i dont know if thats a *Not So Smart* question. where does the 1.414 come from thats like the square root of 2? also how does the method of voltage measurement in a meter vary in readings when measuring a cap? i mean wouldn't that only matter when measuring the pulsed current? thanks

really? $tupid is censored? =)
[on the gravy train with biscuit wheels]

Edited by - Magneticitist on January 10 2012 18:58:55
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kcarring
Moderator



Canada
1057 Posts

Posted - January 10 2012 :  22:29:21  Show Profile Send kcarring a Private Message  Reply with Quote
@TinMan

If voltage increase comes at the cost of current, how do you figure it will take more current to produce 110V, than 220V? Or were referring to the opposite direction.

Also, have you built this? How's the heat on those 2n3055's? I have built the Flip Flop, and it works pretty good, 2XMPSA06 feeding 2XFETS in a standard flip flop circuit. I do notice though, sooner or later, one side gets hot (on the FETS), I figure this may be due the fact that the circuit, impedence cannot be truly 2 equal mirrored halfs? Not sure.
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TinMan
Advanced Member



4082 Posts

Posted - January 11 2012 :  03:08:10  Show Profile Send TinMan a Private Message  Reply with Quote
Hi kcarring-I think you misunderstood what i was saying lol What i was trying to say is at a lower voltage you need more amps to obtain a set amount of watt's out
ok lets say you have a load that requires 110 watts to run it-so this would be 110 volts at 1 amp or 220 volts at half an amp.So the higher the voltage the lower the amps have to be to get the same watt's of power.-Volts x amps= watts.The second thing to remember is the wire size can be smaller the higher the voltage to drive a set load because the amps required to be carried by that wire is less.
Hope this explains what i was trying to say.

Remember-when you make a mistake you have just learned how not to do it.This only brings you one step closser to success.The only way to fail is to never try>>>>TinMan<<<<>>Were's the vidio??<<<
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ron_o
Moderator



United Kingdom
1052 Posts

Posted - January 11 2012 :  03:32:15  Show Profile Send ron_o a Private Message  Reply with Quote
Hi Ryan

Yes it is the square root of 2 , when you measure the ac voltage with a multimeter it displays the rms value ( in the case of a sine wave , when this is rectified the voltage now obtained is the peak voltage ) so thats where the increase comes from BUT as with all things the scales have to balance so the available current is reduced by a corresponding amount.

A square waveform voltage will not give an increase in voltage because it displays as the peak voltage in the first instance.

Hope that helps

ron
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TinMan
Advanced Member



4082 Posts

Posted - January 11 2012 :  03:44:14  Show Profile Send TinMan a Private Message  Reply with Quote
Hi Mag-im a bit new to this aswell ,but i,ll give it a shot.With ac voltage there is it's rms (root mean square)voltage and it's peak to peak voltage.Most meters will read your rms voltage.So at 1 volt ac(rms)the peak voltage must be the 1.414 volts he is refering to.So you have 26 volts rms that our meter reads but in actual fact it is 1.414 x that voltage.So though your meter cant read the peak to peak ac voltage-once that has been converted into a dc voltage in the caps your meter can then read the dc voltage that came from that peak to peak ac voltage your meter couldnt read befor,minus the diode losses ofcourse.This is just my guess as im still learning at the moment lol.So why not just make the meters to read that peak to peak voltage?-well this is just my guess and only a guess,but i believe that because this p-p voltage is at the top and bottom peaks of the ac wave it would have very little current-so it would be lost with the smallest of load's anyway-i would think even the inductance of a transformer would kill this p-p voltage-but just my guess.

Remember-when you make a mistake you have just learned how not to do it.This only brings you one step closser to success.The only way to fail is to never try>>>>TinMan<<<<>>Were's the vidio??<<<
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ron_o
Moderator



United Kingdom
1052 Posts

Posted - January 11 2012 :  05:05:10  Show Profile Send ron_o a Private Message  Reply with Quote
Hi TinMan

I think you should check out the difference between capacitive , resistive and inductive loads on rectified ac regarding the way meters read voltages.

ron
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ron_o
Moderator



United Kingdom
1052 Posts

Posted - January 11 2012 :  05:14:57  Show Profile Send ron_o a Private Message  Reply with Quote
Hi kcarring

A "standard" mains transformer being used in reverse can cause these problems. Basically a tranformer works on the turns ratio , not wire length so if the first winding was say 12 volts and then the seecond 12 volt winding was wound over that then the length of wire will be greater and hence of slightly high resistance which leads to one transistor / FET carrying more current than the other.

True inverter transformers have the "drive " side winding either wound bifilar (side by side) or a thicker wire is use on the outer winding to compensate for the resistance.

You could test this by swopping the two transformer wires over and see if the other device now gets warm.
If not suspect the drive level to the output device(s)

ron

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TinMan
Advanced Member



4082 Posts

Posted - January 11 2012 :  05:51:22  Show Profile Send TinMan a Private Message  Reply with Quote
Hi Ron-yep i still have a lot to learn in reguards to voltage under diferent loads.As i have started at scratch with no teacher i am having to learn myself how things work reguarding electronics.But in saying that-8 years ago i knew that if you turned on a radio-noise came out,but today i can modify and make my own circuits.AC current is something that i havnt had to deal with much other than in generators,but as the time come's i will teach mysef all about it.It's people like yourself that has help'd me get to where i am today-and for that i thank you all.

Remember-when you make a mistake you have just learned how not to do it.This only brings you one step closser to success.The only way to fail is to never try>>>>TinMan<<<<>>Were's the vidio??<<<
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kcarring
Moderator



Canada
1057 Posts

Posted - January 30 2012 :  20:53:02  Show Profile Send kcarring a Private Message  Reply with Quote
I'm still rather lost on all of this voltage doubling. I just fail to see the point... you have X amount of volts to work with, if you want to double, you pay in current. So? Where do you gain?

I am wondering if it has to do with doing a better job of desulphating... if so, that'd be interesting?

~~~~~~~~~~~~~
You wouldn't laugh at my igloo if you knew how cold my beer is!
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ron_o
Moderator



United Kingdom
1052 Posts

Posted - January 31 2012 :  03:13:29  Show Profile Send ron_o a Private Message  Reply with Quote
Hi kcarring

NOTHING !..... You are correct , what you gain in voltage you loose in current. You must remember its how you apply this advantage in voltage thats important , for example a 12 volt supply will not charge a 12 volt battery BUT if you "double" the voltage you can now charge the battery ( but at a lower supply current ).

Voltage doubler / tripler etc were developed a long time ago for use in such things as oscilloscope power supplies ( crt eht voltage ). It also allowed lower voltage working components to be "stacked" to be used at higher voltages.

High voltage pulses are know to dislodge sulfation buildups,

Hope that helps

ron
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49er
Administrator



USA
4442 Posts

Posted - January 31 2012 :  10:00:10  Show Profile Send 49er a Private Message  Reply with Quote
Hi Ron

With that said can this be used like it is for a cap pulser for batteries or do we need to switch it and a second cap?

Doug
The sky is not the limit...There are footprints on the MOON.
Your only as DUMB as where your standing.
No matter where you go there you are.
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ron_o
Moderator



United Kingdom
1052 Posts

Posted - January 31 2012 :  11:20:23  Show Profile Send ron_o a Private Message  Reply with Quote
Hi Doug

Basically you need a switch that isolates the cap(s) from the charging source then it connects them to the battery and then disconnects the cap(s) from the battery and reconnects it to the charging source. Example , Bedini patent circuit to drive a load via a pulse ( solar charger )
or a simple multiple contact relay.

ron
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49er
Administrator



USA
4442 Posts

Posted - January 31 2012 :  12:22:30  Show Profile Send 49er a Private Message  Reply with Quote
Hi Ron

Thanks that is what I thought but I am not up on caps wasn't sure Thinks again

Doug
The sky is not the limit...There are footprints on the MOON.
Your only as DUMB as where your standing.
No matter where you go there you are.
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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - January 31 2012 :  13:08:50  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
quote:
Originally posted by ron_o

Hi kcarring

NOTHING !..... You are correct , what you gain in voltage you loose in current. You must remember its how you apply this advantage in voltage thats important , for example a 12 volt supply will not charge a 12 volt battery BUT if you "double" the voltage you can now charge the battery ( but at a lower supply current ).

Voltage doubler / tripler etc were developed a long time ago for use in such things as oscilloscope power supplies ( crt eht voltage ). It also allowed lower voltage working components to be "stacked" to be used at higher voltages.

High voltage pulses are know to dislodge sulfation buildups,

Hope that helps

ron



I try really hard 'not' to jump in on these things and just read and learn because I'm such a novice... but... You just hit a spot that confuses what I thought I learned.

"for example a 12 volt supply will not charge a 12 volt battery"

I use a 'constant 12.10v supply', thus giving me constant wattage.
My belief has been, if your adding a constant amount of wattage you can then go beyond the voltage (maybe not wattage??) of the source.

So, with that ideal in mind, I ran this test.

This is a run from my 205ah (high wattage) battery to my 230 cca (low wattage) battery.



Voltage gain higher than source.

I attributed this to wattage not voltage.



Old in age, not in mind, so
'Teach me something new'!

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kcarring
Moderator



Canada
1057 Posts

Posted - February 01 2012 :  01:07:22  Show Profile Send kcarring a Private Message  Reply with Quote
@olddawgsrule

You say that you use a constant supply, yet you show using a battery. Do you mean you take the big battery and hook it up to a contant voltage / constant current circuit, and then use that to power your Bedini?

I think ron was generalizing when he stated you cannot charge from the same voltage - we all know that given the inductive collapse spikes - it can be done - I think people just don't do it much - and opt - instead - for a voltage doubler or constant voltage, to do it? MrTwalley stated that he uses a linear voltage regulator, which I do not understand despite listenting to him explain it in the video, #1. inefficient lossy devices that impose an initial voltage drop. #2 You'd need to have two batteries, and thus 24 volts, it'd drop to say 22, and you could maintain 18 the whole time or even twelve, but the further you drop it, the more you lose in heat. I tend to stay away from them. I think simple boost converter with some smoothing caps on the end would be far more efficient? -- Or a voltage doubler - as shown, because at the cost of current, at least you always have enough voltage.

I suppose it all comes from people charging one battery from another - myself, I just don't see the point of charging one battery from another?! If i have a battery charged, I use it. LOL It's as simple as that, to me? To me its like taking the peanut butter of one piece of toast and putting it on another, and dropping 30% on the floor. LOL Consequently, I am using either the wall, or a solar panel, I never use one battery to charge another?




~~~~~~~~~~~~~
You wouldn't laugh at my igloo if you knew how cold my beer is!

Edited by - kcarring on February 01 2012 01:21:08
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ron_o
Moderator



United Kingdom
1052 Posts

Posted - February 01 2012 :  03:12:03  Show Profile Send ron_o a Private Message  Reply with Quote
Hi Olddawgsrule

Kcarring is correct in his assumption."I think ron was generalizing "



Watts are a product of volts X amps therefore to provide constant wattage if voltage increases then current must reduce to compensate.

The statement should read ( or was ment to read as :- )"you cannot charge a 12 volt battery from a 12 volt source".... in order to charge a battery you need to provide a charging source that has a higher voltage than the battry to be charged ( reason .. the higher voltage "overcomes the resistance ( not ohms )" and allows the current to flow )

Sorry if i have caused confusion by not clearly stating the information.

ron
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ron_o
Moderator



United Kingdom
1052 Posts

Posted - February 01 2012 :  03:15:57  Show Profile Send ron_o a Private Message  Reply with Quote
Hi Doug

Take a look at "Solar Charging/Pulsing Circuit ( with desulfator )" REvision 1.4 by Farmhand on EnergeticForums
Focus on the circuitry around the SCR BT151 on the bottom righthand side , this would do the job.

ron
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49er
Administrator



USA
4442 Posts

Posted - February 01 2012 :  08:45:49  Show Profile Send 49er a Private Message  Reply with Quote
Hi Ron

I have never had any luck with the SCR except for the 6" fan that I got from Rick and that is because I don't use IT. LOL TO much noise and it eats the front battery alive it does charge but I think that latch slam open and stays and that is why it eats the front battery.

Doug
The sky is not the limit...There are footprints on the MOON.
Your only as DUMB as where your standing.
No matter where you go there you are.
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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - February 01 2012 :  08:47:16  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
quote:
Originally posted by kcarring

@olddawgsrule

You say that you use a constant supply, yet you show using a battery. Do you mean you take the big battery and hook it up to a contant voltage / constant current circuit, and then use that to power your Bedini?

I think ron was generalizing when he stated you cannot charge from the same voltage - we all know that given the inductive collapse spikes - it can be done - I think people just don't do it much - and opt - instead - for a voltage doubler or constant voltage, to do it? MrTwalley stated that he uses a linear voltage regulator, which I do not understand despite listenting to him explain it in the video, #1. inefficient lossy devices that impose an initial voltage drop. #2 You'd need to have two batteries, and thus 24 volts, it'd drop to say 22, and you could maintain 18 the whole time or even twelve, but the further you drop it, the more you lose in heat. I tend to stay away from them. I think simple boost converter with some smoothing caps on the end would be far more efficient? -- Or a voltage doubler - as shown, because at the cost of current, at least you always have enough voltage.

I suppose it all comes from people charging one battery from another - myself, I just don't see the point of charging one battery from another?! If i have a battery charged, I use it. LOL It's as simple as that, to me? To me its like taking the peanut butter of one piece of toast and putting it on another, and dropping 30% on the floor. LOL Consequently, I am using either the wall, or a solar panel, I never use one battery to charge another?




~~~~~~~~~~~~~
You wouldn't laugh at my igloo if you knew how cold my beer is!



Thank you and this does clear things up for me.

Typically, I use my 'constant supply', but in this case I was draining the 205 so I could run again in my attempts to re-condition the 205.
I just figured, if draining, why not use it as supply to charge another battery.
I'm into 1/2 dozen cycles now and slowly gaining on the 205.

Now, if again I'm understanding, then my 12.10vdc 'constant' supply needs to be upped in voltage or I may never get the 205ah battery beyond the 11.94v I reached on last night's run (or introduce one of these voltage pumps to my circuit).



Old in age, not in mind, so
'Teach me something new'!

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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - February 01 2012 :  09:13:36  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
quote:
Originally posted by ron_o

Hi Olddawgsrule

Kcarring is correct in his assumption."I think ron was generalizing "



Watts are a product of volts X amps therefore to provide constant wattage if voltage increases then current must reduce to compensate.

The statement should read ( or was ment to read as :- )"you cannot charge a 12 volt battery from a 12 volt source".... in order to charge a battery you need to provide a charging source that has a higher voltage than the battry to be charged ( reason .. the higher voltage "overcomes the resistance ( not ohms )" and allows the current to flow )

Sorry if i have caused confusion by not clearly stating the information.

ron



No problem!
As I said, I'm still a novice and learning.

I attributed that run (purpose was to drain the 205ah and continue the re-conditioning of both batteries) to the fact that the 205ah is 25x that of the 230cca (figuring the 230cca to be equal to a 6-8ah battery).

As I said to Kc, it now looks like my 12.10v 'constant supply' needs an update to increase the voltage or I may never get the 205ah beyond the 11.94v I reached on last night's run.

Thank you for clarifying.



Old in age, not in mind, so
'Teach me something new'!

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49er
Administrator



USA
4442 Posts

Posted - February 01 2012 :  10:30:29  Show Profile Send 49er a Private Message  Reply with Quote
Hi ODR

1 quick question how many amps on discharge OK 2 questions how low are you going with the voltage??

Doug
The sky is not the limit...There are footprints on the MOON.
Your only as DUMB as where your standing.
No matter where you go there you are.
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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - February 01 2012 :  11:14:10  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
quote:
Originally posted by 49er

Hi ODR

1 quick question how many amps on discharge OK 2 questions how low are you going with the voltage??

Doug
The sky is not the limit...There are footprints on the MOON.
Your only as DUMB as where your standing.
No matter where you go there you are.
You must be logged in to see this link.



Now that I've started this new record keeping this is easier to get back to.

I dis-charge 2 ways;

First using my circuit that takes 980ma to run
Second is thru the inverter using 88ma (LED's), 210ma (cfl) plus the unknown (what the 400watt inverter is using)

Either way, I figure I'm under an amp.

I have not run the battery below 11.3v ever.
The 400w inverter screams at me at that level, so I stop.






Old in age, not in mind, so
'Teach me something new'!

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kcarring
Moderator



Canada
1057 Posts

Posted - February 01 2012 :  17:27:49  Show Profile Send kcarring a Private Message  Reply with Quote
@olddawg,

Similarily I am repairing a battery that (conventionally) could not be charged past 12.4V and would drop in 30 days to 12.0V. No matter what i did, conventionally, it was a poor performer. Thus the reason I found it in the dump. The battery is a 2006 and by the wear and tear, it looks like it got used, didn't just "sit" for years. After many different experiments I found this battery to be the closest of many I had found to "an ideal candidate". I figure if a battery will go to 12V, conventionally - but not much further - then at least you likely do not have a dead short, or, any shorts are dendril/sulphate based. When I found it at the dump it was 10.5V. Now, my new rule of thumb is, until I get a bigger charger/desulphator built, I wouldn't consider taking home anymore 2V or 4V reading "dead batteries". Often they turn out to be a waste of time. With the exception of the battery that was a very expensive batter, I might try that, "just in case" - for example if I found a Rolls single cell, a transmission tower battery, or something worth in excess of 1500 bucks.

Anyway a lot of people would say, you can't do that with a simple i coil wimpy SSG starter kit... well you can actually when you have a good candidate (a reasonable candidate).

It's been on the charger for about 4 days straight now, inputting about 80 mA. When I'm desulphating with this scenario - tiny circuit - big bat.. I don't go for efficiency, I tend to run the tune at just under max. So, if MAX output was 90 mA at the cost of 180 mA input, I'll run it at 140 mA to get 85 mA.

The battery rose fairly quickly to it's "realized ceiling" of 12.4V, but then it REALLY slowed down. It only went up .1v in the first 24 hour period. But as of this morning, the previous 24 hours - before that - it went up .2V. So all in all it might take a week, it's sitting at 12.95V right now, was 12.77 last night.

My input is a wall adapter that says 12V, but in reality its a bit wonky (maybe why it was in the dump). It actually reads 17V open circuit, and if u measure the rails of the Bedini PCB, it says about 13.85, loaded, running.

My feeling is similar to how The MAgneticitist stated it: "You need the spikes to hit a certain voltage, but not hugely over that. You need about 2-300 volt peaks, but you do not need 500V". So, if you up your voltage on the input, you get two things, POTENTIALLY higher peaks (but this can only really be determined at the scope -> don't be fooled into thinking it is a linearly relationship. Any amount of playing with building inverters will teach you this... many things are in play for your final output voltage!) - but one thing is for sure, the voltage and current are mathematically related. So by upping your input voltage, you will inherently increase your current, some. As well all know if a circuit, (and a battery is actually a circuit within itself...) needs an amount of current to function, it will attempt to draw that current and if it cannot get the current it needs, it may not perform. OR, it may perform incredibly slow. Look at Rick Friedrichs demo, his latest - it took MANY hours to get the huge bank up a couple of tenths of a volt - but it did climb - and his source was only 12V... that tells me that the peaks do see through to the other end.. they make it through even a great deal of batteries put in series... If the battery bank, in this demo a 48 V MASSIVE bank, was not seeing "200-300 volt peaks" and thus averaging out to over 48V RMS, it simply wouldn't be charging. So when you are taking the voltage of the bank, 48.x volts, you are not seeing the output voltage, not even close. That's how I see it. We all know you can pulse 12v linear at a 12v battery all day and it won't charge, it must "see" over 12V. BUT, whether anything gets done in a reasonable time period, or efficiently... (charging, or desulphating) that is ultimately also dependant also on current flow. I do not believe for one second in any of these Bedini circuits "current is your enemy" as a few out there do... In my opinion you want to optimize for all things relevent. Do not use so much current as to heat everything up ridiculously... including the battery/pots/resistors/transistors - do not use so little as to not get the job done in a reasonable amount of time. That's what I say.

What I am finding, and learning, and trying to make clear to myself is something that Rick or others' do not often discuss - and that is the charging curve. And the desulphation curve - which will vary from on battery to another, hugely. All too often we see a video showing the top end of battery's voltage reaching its' peak, quickly. This states absolutely nothing, just check out the data sheet of any lead acid battery. Mag made a good video on this. In Ricks video he is charging on the top end, he is taking a 48 volt bank, from 48+ volts, it's peak may be only 51V, or less even if the bank is deteriorated or not yet "Bedini conditioned - I don't know - I don't follow that close) but the relevent thing is, it can be rather deceiving and I'm finding the only way to learn it is to see, play with it, and know your own gear, as to what it can do and can't. When you apply that to desulphating it all becomes even more vague; because we rarely know the history of used, sulphated batteries - unless we bought them, and took them to that state. You can believe whatever you want, some fantastical tale of new science and free energy, or whatever - as I see it - it's not free energy it's battery maintenance using alternative technique to what the industry wants: and that is for us to throw it away and start over. Even Tom Bearden stated "it warrants further investigation by the highest order of chemists". You cannot intelligently make bold statements as to some new scientific process occuring, when A. You do not even know the higher order of chemistry you are attempting to claim you are defying, and B> you are not even willing to back your claims with known scientific measurement such to support the claim.
In the end though, it does work, and that's good enough for me. Saving batteries is saving money. They are expensive.

~~~~~~~~~~~~~
You wouldn't laugh at my igloo if you knew how cold my beer is!

Edited by - kcarring on February 01 2012 17:54:44
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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - February 01 2012 :  17:49:31  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
Kc, thank you, this helped!
I am going to have to re-read and break it down a bit to understand 'all', got quite a bit of it.

The best part was "In my opinion you want to optimize for all things relevant".

Being a 'carpenter' excuse the pun when I say "you hit the nail on the head"!

I listen, read, watch what you all are doing and learn.
I use (should say try to use) "what is relevant" to what I'm doing.

You might think about somehow adding that into your signature, rather profound!

Old in age, not in mind, so
'Teach me something new'!

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ron_o
Moderator



United Kingdom
1052 Posts

Posted - February 03 2012 :  03:08:17  Show Profile Send ron_o a Private Message  Reply with Quote
Hi Doug
Was a little surprised by your latch up problems with the SCR circuit . Did you actually a BT151 type as these are known to be very latch up resistant ?

ron
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49er
Administrator



USA
4442 Posts

Posted - February 03 2012 :  09:36:16  Show Profile Send 49er a Private Message  Reply with Quote
Hi Ron

I used what Rick sells, haven't look what type they are.

Doug
The sky is not the limit...There are footprints on the MOON.
Your only as DUMB as where your standing.
No matter where you go there you are.
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Poppy
Average Member



USA
418 Posts

Posted - March 08 2012 :  08:14:42  Show Profile Send Poppy a Private Message  Reply with Quote
Hi Everybody...

Here is the Voltage Double Quad.



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kcarring
Moderator



Canada
1057 Posts

Posted - March 13 2012 :  16:57:24  Show Profile Send kcarring a Private Message  Reply with Quote
How & Why The Voltage Doubler Works



~~~~~~~~~~~~~
You wouldn't laugh at my igloo if you knew how cold my beer is!
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shakamuni01
Average Member



USA
213 Posts

Posted - March 17 2012 :  08:12:39  Show Profile Send shakamuni01 a Private Message  Reply with Quote
Thanks KC. I think you report on the recharging process will come in quite handy. I have been trying to get a feel on the what to expect.
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49er
Administrator



USA
4442 Posts

Posted - March 17 2012 :  10:50:06  Show Profile Send 49er a Private Message  Reply with Quote
Hi kc

That help me also Thanks

Doug
The sky is not the limit...There are footprints on the MOON.
Your only as DUMB as where your standing.
No matter where you go there you are.
You must be logged in to see this link.
SKYPE bxx49er
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kcarring
Moderator



Canada
1057 Posts

Posted - March 17 2012 :  13:10:56  Show Profile Send kcarring a Private Message  Reply with Quote
I've learned a lot from AllAmericanFiveRadio on youtube, he's got a great way of teaching stuff, really puts a lot into his presentations. Nice fellow too.

~~~~~~~~~~~~~
You wouldn't laugh at my igloo if you knew how cold my beer is!
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