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elias
12-19-2009, 06:22 AM
Hi there for all!

I have been building SSGs and charging batteries for about two years, and the problem I am facing now is the fact that it seems that although the batteries deliver energy more than usual, the voltage level of the batteries charged with Bedini charger decreases under load significantly making devices such as mp3 players, not being able to function.

Don't get me wrong, the sitting voltage is fine, but when It is put under load the voltage level goes down so much that the mp3 player says low battery, but I know that it can deliver good amount of energy, but the mp3 player thinks the voltage level is what says the amount of energy in the battery is.

I don't know if the batteries need more conditioning to build up more internally or not, but this is what I am facing.

Has anyone faced the problem and has any workaround for it?

Elias

SkyWatcher
12-19-2009, 06:31 AM
Hi elias, is it possible the device your using needs higher voltage cells like 1.5v primary batteries, because using 1.2v rechargeable's will seem to not last long in those devices and yet still have much capacity left. Otherwise if thats not the problem, then maybe the rechargeable your using are damaged or sulphated still in some way as not to provide current well, or, maybe the charge you are giving the batteries is fluffy. I'm using the "Big Boy Variant" solid state oscillator circuit posted by Asweth and it gives all my cells, even primary batteries good charge, though primary cells seem to only take so many charges until they leak. Let me know what you think of those thoughts.
peace love light
Tyson

ren
12-19-2009, 09:02 AM
Good point Skywatcher.

Also Elias, you can try charging up to 14.5v (if 12v) and then set your machine to a "trickle" style charge. For this set it to keep the battery around 13-14v, the longer it can slowly creep up in this range the better. This is similar to a conventional charger.

Regards

Inquorate
12-19-2009, 09:42 AM
And add a ground wire to the negative of charge battery; the dielectric stress the spikes place on the battery may suck electrons out of the ground.

Radiant_Science
12-19-2009, 10:28 AM
@Elias
I too have noticed odd "phantom charge" phenomena in batteries. Instead of batteries, try a large capacitor bank. The capacitor bank will charge up to maximum voltage given time.
Try running the capacitors through a voltage regulator feeding you device.

Unlike normal charging, increasing capacitance does not slow charging in a linear manner. Depending on your setup, you will want to experiment with larger and larger capacities until you reach a balance. So, the capacitor stays at some acceptable/constant voltage lvl.
You can also achieve this by lowering your input voltage lvl.

If you can balance the system, then you should be able to safely run the charger as you use the load. If you can carefully, and patiently tune it...you will be able to do neat stuff!


Hope it helps :thumbsup:

lamare
12-19-2009, 12:33 PM
And add a ground wire to the negative of charge battery; the dielectric stress the spikes place on the battery may suck electrons out of the ground.

I doubt if that will work. The problem is that you have to move the charge from one plate to the other.

Let me explain a bit further about what I think is happening. I posted about this quite a bit recently, so you might want to look back my previous posts at various places.

Anyway, what I (and Inquorate) think is happening is that a non-permanent electret can be formed inside a battery and a capacitor as well. Electrets are normally formed by polarizing a dielectricum (a polarizable insulator) using a strong electric field, while the dielectricum is heated, probably in a molten state. Once it is cooled down, meanwhile maintaining the electric field, you get a permanent polarized piece of material, which permanently provides a voltage, a potential. This effect is very similar to what is known as dielectric relaxation, which is normally an unwanted effect in capacitors. You can observe this effect by charging a capacitor and then short it out for a very short time. After that, you can see that there will still be a charge on the capacitor, something in the order of 10% of the original voltage.

This is because in between the capacitor plates (like an elco) you have a polarizable dielectric material, which increases the capacitance of the plates. When you short the cap out, it takes a while for the dielectricum to de-polarize, so there is still an electric field present in between the capacitor plates for some time. This is capable of recharging the capacitor.

It appears that this dielectric relaxation effect can be super-boosted by "charging" the capacitor with high voltage pulses. It appears that this way you can create a non-permanent electret in between the capacitor plates, and it appears the same thing is happening inside a battery as well.

According to this theory, you basically end up with capacitor plates with a semi-permanent electric field in between the plates. Since we have Faradays law, electrons will start moving inside and, if possible, between the capacitor plates, such that the electric field between the plates becomes zero. In other words: the fields created by the capacitor plates and that created by the electret will become such that they cancel each other out.

In other words: the semi-permanent field in between the plates delivers a force, which is capable of charging the capacitor plates. It is clear that, in order to do this, charge (electrons) must move somehow from one plate to the other.

A very interesting demonstration about this, is the video mentioned here:
http://www.energeticforum.com/renewable-energy/1631-peter-whatever-happened-eric-p-dollard-2.html#post74102

What you can see here, is that the capacitor plates themselves are not able to create a spark. Only when the dielectricum is placed back between the plates, enough "energy" is present in order for the plates to discharge trough a spark gap. Even though the experimenter did not try to discharge the naked plates in the same position as they would have with the dielectricum in between and he touched them with his hands discharging them, it is clear that the dielectricum gives the power to the cap, so to speak.

This brings up an interesting question. Since apparantly we get additional charge and additional energy when the dielectricum is put in between the plates, where does this come from? And, where does the charge go when we charge a capacitor, if it is not stored in the dielectricum and also not on the capacitor plates???

I have posted about this before: http://www.energeticforum.com/renewable-energy/5009-discussion-re-physics-behind-negative-energy-systems-radiant-spikes.html#post76020

Then, I said this:

This suggests that once you have a polarized electret in between metal (capacitor) plates, that the electret will charge the plates. Apparantly, some electrons are drifting trough the electret from one plate to the other, such that the plates becomes charged, and we can measure a voltage on the outside.


this:


Where do all our electrons go, when we charge a capacitor the normal way?

The only thing I can come up with, is that they are wasted. Leaking from one plate to the other, without doing any good. If that is the case, then the potential applied to a capacitor is what charges it for 99.999%, because that is what polarizes the dielectricum, while all the current we ram in is being turned into heat or something else we don't need.

and this:


On the other hand, this is still puzzling, because once you remove the dielectricum, the electron charge is still left on the metal plates, which are no longer opposed by the polarized dielectricum. Since the capacitance of the plates is much less then, one would expect a much higher voltage on the plates. From the MIT clip, one cannot really say much about that, since the experimentor touched the metal "plates" with his hand, and the capacitor plates were in a different position when they were attempted to be discharged. So, IMHO any conclusion about whether or not any charge is actually stored on the capacitor plates based on the MIT clip is unreliable at best.

On the other hand, the MIT clip does show that the dielectricum inside a charged capacitor can hold its charge (or polarization) for at least several seconds. In other words: this definately confirms that it is possible to create a non-permanent electret in between capacitor plates using high voltages.

Now let's see if we can solve this puzzle. Over here http://www.energeticforum.com/renewable-energy/5096-no-electrolysis-water-split.html I posted some things about the difference between dielectric strength and the dielectric constant:


The dielectric strength says something about how good an insulator it is. For Super Corona Dope, this is 4100 V / mil:
Super Corona Dope 55ML/2oz (4226-55ML) (http://www.web-tronics.com/sucodo554.html)

The dielectric constant says something about the polarization of the dielectric, and that's what matters most.

The insulation properties are nice and somewhat important, because you don't want current leaking, but the energy source you want to utilize is the electric field created by a polarized dielectricum. And that's why that is the most important parameter.

However, you also want the polarization to be easy (fast) to establish and to last as long as possible, after you remove the polarizing field, because you want to induce the polarization with a spike and use the field caused by the polarization to extract energy once the spike is gone. At this moment, I can't say a thing about how this relates to the dielectric constant and other parameters.


Over here we can find a table telling us something about the general relation between the two, in the case Dielectric Ceramics and Substrates, materials commonly used inside capacitors:
Lead Magnesium Niobate (PMN) On GlobalSpec (http://www.globalspec.com/reference/3784/Lead-Magnesium-Niobate-PMN)

One of the items shown here, is the Loss Tangent:
Loss tangent - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Loss_tangent)
The loss tangent is a parameter of a dielectric material that quantifies its inherent dissipation of electromagnetic energy. The term refers to the angle in a complex plane between the resistive (lossy) component of an electromagnetic field and its reactive (lossless) component.

However, that's just one way to look at it...

Let's take some typical values for a "lossy" capacitor from this table. For the "worst" case, we have a dielectric strength of 29V/m and a dielectric constant of 2,400 (and up).

Now what is dielectric strength?
Dielectric strength - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Dielectric_strength)
Of an insulating material, the maximum electric field strength that it can withstand intrinsically without breaking down, i.e., without experiencing failure of its insulating properties.


Let's get this straight: in a typical "lossy" capacitor, such as an elco, we have a dielectric strength in the order of 30V/m (yes, that's meters) before the dielectricum "brakes down", when its "insulating properties" fail. And that's "worst case", "better" capacitors have an even lower break-down value, down to less then 8V/m. :thinking:

In other words: the "better" the capacitor, apparantly, the lower it's dielectric break-down voltage. Hmmm.


Now consider a typical capacitor. What would be a typical value of the spacing between the plates? 1 mm, "worst case"??
Great, that means the dielectricum shorts out when there's more then 0.03 Volts across the plates, not counting the canceling out of the two fields created by the plates and the dielectricum, of course. And remember, this 0.03 V really is about the maximum limit you would be talking about in a typical capacitor, since the typical spacing between the plates is much less, and the 30V is in the case of the most "lossy" capacitors.....

This means one thing is very, very clear: the "insulation" properties of the dielectric inside a typical capacitor can be taken with a grain of salt, to say the least.....
In other words: it is now clear that it is way more than "some electrons" that drift from one plate to the other.....


If course, this "drifting" trough the dielectricum works in both directions. What really "charges" the dielectricum, is the resulting electric field, and since the fields created by the plates and that of the dielectricum oppose one another, you end up with a very small resultant field to polarize your dielectricum with, at the cost of a significant loss current right trough the dielectricum.

Basically the difference during "charge" and "discharge" is the direction of the resultant field in between the plates. During "charge" the field created by the plates is stronger then the field created by the polarized dielectricum, so the dielectricum polarizes some more to oppose this field, while the electrons all but freely move between the plates.

During discharge, the field created by the plates is weaker than the field created by the polarized dielectricum, so the resulting field is pointing in the other direction, depolarizing your dielectricum, while once again the electrons all but freely move between the plates.

Of course, there's always some resistance the electrons encounter when moving between the plates, which is what limits the current you can draw.


Now back to the batteries. It appears that you also create such a polarized dielectricum inside a battery when charging it with HV pulses. One of the things observed by Bedini is that the batteries are harder to charge with a normal charger, which points in the direction of an insulating layer having formed on the battery plates.
If that is the case, then of course the chemicals inside the battery can no longer reach the plates themselves, so the only "source" you would have left is this polarized dielectricum. And that has a significant resistance, so it would not be so strange to find out that you can draw significantly less current from a battery charged this way.

I don know how to solve this problem yet, but I would guess that Tesla Switch actually does solve this problem.

SkyWatcher
12-19-2009, 01:37 PM
Hi folks, interesting thoughts lamare, though with good condition lead acid, nimh, nicad, alkaline I have had no problems getting good amperage out of them. Well when I first started out with these circuits, I was getting fluffy charge with little current capability, but not anymore. So I think it depends on what circuit folks are using and condition of cells and that includes your theory of the layer being formed causing higher impedance, which we may or may not wish to avoid, who knows It may lead somewhere.
peace love light
Tyson:sun:

lamare
12-19-2009, 02:08 PM
Hmm. This is very strange. The dielectric strengths reported elsewhere are much bigger, like here:

RF Cafe - Dielectric Constant, Strength, & Loss Tangent (http://www.rfcafe.com/references/electrical/dielectric-constants-strengths.htm)

Polystyrene - 500 (V/mil), which is approx. 20 MV/m ( Millivolt Per Meter [mV/m] To Volt Per Mil [V/mil] Converter (http://www.translatorscafe.com/cafe/units-converter/electric-field-strength/calculator/millivolt-per-meter-%5BmV/m%5D-to-volt-per-mil-%5BV/mil%5D/) ), that's Mega Volt/m.

So, either at globalspec they made a mistake and they actually meant MV/m instead of V/m, or elsewhere something else is meant.

Somehow, I get the feeling the guys at globalspec made a mistake, and so I really f**ed up this time :(

elias
12-20-2009, 05:51 AM
Thanks for the insight Skywatcher,

The batteries are 1.2v AAA cells for my wife's mp3player, they worked well before charging with the Bedini Charger. I have also got this problem with my philishave.The battery delivers more current than before I can verify it, but the voltage goes down, when driving loads, and the device "thinks" that the battery is out. I have seen this effect in Bedini's videos also, his batteries have good amount of energy in them, but the voltage drop is more when driving loads, I wonder.

Hi elias, is it possible the device your using needs higher voltage cells like 1.5v primary batteries, because using 1.2v rechargeable's will seem to not last long in those devices and yet still have much capacity left. Otherwise if thats not the problem, then maybe the rechargeable your using are damaged or sulphated still in some way as not to provide current well, or, maybe the charge you are giving the batteries is fluffy. I'm using the "Big Boy Variant" solid state oscillator circuit posted by Asweth and it gives all my cells, even primary batteries good charge, though primary cells seem to only take so many charges until they leak. Let me know what you think of those thoughts.
peace love l ight
Tyson

elias
12-20-2009, 05:53 AM
Good point Skywatcher.

Also Elias, you can try charging up to 14.5v (if 12v) and then set your machine to a "trickle" style charge. For this set it to keep the battery around 13-14v, the longer it can slowly creep up in this range the better. This is similar to a conventional charger.

Regards

Hi Ren,

What do you mean by "trickle" charge? I will try that; thanks.

elias
12-20-2009, 06:09 AM
Thanks all for the nice responses,

Actually The batteries can give the current I need, but the devices are designed to not work in lower voltages, The Bedini charger somehow, I think compresses the voltage of a battery while giving more current, My laptop batteries also had this problem, I had managed to charge my laptop battery with the Bedini charger it ran around 1.5 hours in the Battery Low condition. I assume that more conditioning is needed for the batteries ot buildup more Radiant or Negative Voltage, I suppose.

SkyWatcher
12-20-2009, 11:10 AM
Hi elias, are all these different batteries your having issues with, were they not holding a load well and then you used the oscillator to try and rejuvenate them. I think that may be the biggest issue because the camera and possibly your shaver device need a higher voltage than 1.2v and the computer battery is either sulphated or its a type of battery that doesnt take that charge well. I bought some generic brand AA nimh 2400mah at the store one time and they were all completely sulphated and would not take a normal charge. Another time I bought 4 duracell nimh 2650mah and they have been taking a radiant charge very well and holding the voltage on loads just the same. And I also have about 8 - 12v 7AH lead acid gel cells and all have held loads well when charged from the bedini radiant oscillator, not the cap dump method though, just straight off the flyback diode. I'm in the middle of tests myself to see how many recharges I can get out of these energizer max AA alkalines I just bought and not let them sit after discharge to sulphate at all.
peace love light
Tyson:sun:

Shanjaq
12-20-2009, 11:27 AM
Maybe you could use the radiant charging method accumulating into a large 12v battery bank, then use that to run conventional chargers on the batteries you use in everyday equipment?

DavidE
12-20-2009, 03:40 PM
Two reflective points.

Consider Bedini's past observation relative to level of battery conditioning (new crystal structure - time).

Secondly, this discussion reminds me of the astronomer that once declared --- "we have discovered stars that are older than the universe itself."

Uhhhh. Maybe Mp3 players made in 2009 are not the best test device for exploring the value of a new energy source...

http://www.coralcastlecode.com/sitebuildercontent/sitebuilderpictures/code.jpg

lamare
12-21-2009, 11:36 AM
I found some very interesting papers on electrolyte capacitors:

Electrolytic capacitors (http://www.ami.ac.uk/courses/topics/0136_ec/index.html)
"The function of the aluminium cathode foil is to reduce the series resistance of the capacitor by making contact with the paper over a wide area."

Electrochemistry Encyclopedia -- Electrolytic capacitors (http://electrochem.cwru.edu/encycl/art-c04-electr-cap.htm)
"The spacer is generally made of paper, which can be of many different types, densities, and thicknesses, depending on the voltage and effective series resistance requirements."

How can the series resistance (== "loss" resistance) be reduced by lowering the resistance between the plates with the dielectrics, if the dielectric is supposed to be non-conducting???? :thinking:

--

Al right, fair enough. The charge goes trough the electrolyte from the anode to very close to the cathode, so the the anode can be considered to be very close to the cathode.

The upper article states that "a typical thickness of the oxide film is in the order of 0.2 m."
So, it would be very interesting to see what kind of leakage currents could go trough such a thin layer...

Will be continued....

--

Another interesting article:
Electrolytic capacitor - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Electrolytic_capacitor)

"The principle of the electrolytic capacitor was discovered in 1886 by Charles Pollak, as part of his research into anodizing of aluminum and other metals. Pollack discovered that due to the thinness of the aluminum oxide layer produced, there was a very high capacitance between the aluminum and the electrolyte solution. A major problem was that most electrolytes tended to dissolve the oxide layer again when the power is removed, but he eventually found that sodium perborate (borax) would allow the layer to be formed and not attack it afterwards. He was granted a patent for the borax-solution aluminum electrolytic capacitor in 1897.

The first application of the technology was in making starting capacitors for single-phase alternating current (AC) motors. Although most electrolytic capacitors are polarized, that is, they can only be operated with direct current (DC), by separately anodizing aluminum plates and then interleaving them in a borax bath, it is possible to make a capacitor that can be used in AC systems.

Nineteenth and early twentieth century electrolytic capacitors bore little resemblance to modern types, their construction being more along the lines of a car battery. The borax electrolyte solution had to be periodically topped up with distilled water, again reminiscent of a lead acid battery."

sucahyo
12-22-2009, 03:06 AM
I have been building SSGs and charging batteries for about two years, and the problem I am facing now is the fact that it seems that although the batteries deliver energy more than usual, the voltage level of the batteries charged with Bedini charger decreases under load significantly making devices such as mp3 players, not being able to function.I think you have a broken battery. Maybe because of over discharge. Bomb it with straight DC or something like 1Amp radiant output and then charge it again with purer radiant charger.

I experience that capacitor like behaviour in my 12V battery that over discharge to 8V. Reach 15V standing voltage but have 8V running voltage. It become 9V after a couple of days. I bomb it with a combined output of two different radiant charger which significantly heat it up in two hour. It hold voltage more than 12V now. Never know that SLA can heat up. The opened seal keep popping the rubber cap a couple of time during charging. Do it with extra care.

I have experience a broken battery but have high standing voltage because my radiant are not pure enough. I got better battery running voltage (low voltage drop) when I increase the sharpness of my transistor trigger. But this after restoring/heating the battery with normal electricity.

I think the best battery for testing your charger is alkaline. The test is to see if your charger is capable to fill the allkaline with real charge. The alkaline should have a low voltage drop, high standing voltage but also high running voltage.

Zooty
05-10-2010, 02:35 AM
I know this is an old post but some things mentioned in it got me thinking. One of Bedini's golden rules for batteries charged this way is NEVER draw more current from it than it's C20 rate. I have tested this many times. If i take one of my 1.2v radiantly charged AA's rated at 2000mah and run it through a 200ma load, it should theoretically be able to last for 9 - 10 hours. Well it does not. The voltage drops quickly and i get usable power for 6 hours if i am lucky. If i run a load of 100ma then i usually get 18 hours but i have a couple that will do 20 hours. Maybe the reason these batteries last longer like this is because the energy is being stored differently and it needs to be drawn differently. What Inquorate said about the capacitor and energy being stored in the dielectric makes me think that this is where the spike is going and its seeping out from there into the plates. Maybe it's transition to the plates is slower than a normal discharge. So if we draw the current slow enough to give the dielectric a chance to feed the plates then we get a constant current but lower like an electric buffer of some sort.

elias
05-10-2010, 05:30 AM
Hi

Well, after this post I realised that the "charge" inside a battery charged with negative energy is really different, and a word of caution, I found out that it can destroy flash memory, because now the mp3 player has lost its memory. I charged its battery several times with SSG.
Also my shaver now works more than 2 hours, on a single charge, which worked only about 1 hour at first, but it has lost its circuitry and I had to bypass it.

Elias

SkyWatcher
05-10-2010, 07:09 AM
Hi folks, good timing considering i am conditioning a couple 8 cell packs of 1700mah sanyo nicads ive had and have and am using them still for my electric rc plane. Now these packs were not that bad off so there responding nicely and when i alternate and give them a normal charge they have close to the same capacity for a given load, in this case a tailight bulb. I took the plane out today and it flew well though capacity is a little less than usual, still needs more rejuvenating. Also, keep in mind that at c20 youll get close to stated capacity, but it drops off big time at c1 and below, since this tailight bulb is just below c1 for my 1700mah's. So at C1, 50-60% capacity is probably doing pretty good. i'm getting around 70% capacity with the tailight load which is just below C1. Though in the electric plane amp draw can be up to almost 30 amps and so capacity drops off even more, probably down to 50% or less and i would imagine this applies also to new good condition cells. Though i am still cycling these packs and they seem to get better and better. I may have to buy some new nimh packs and i just bet I could make those brand new cells be better than new. Hope that helps.
peace love light
Tyson:sun:

sucahyo
05-10-2010, 08:47 AM
I may have to buy some new nimh packs and i just bet I could make those brand new cells be better than new.I thought you need high power? I suggest 2000mAh nicad from Auldey. Unless you don't mind a drop off power. Longer life means lower power. NIMH is no match against nicad in amp draw.

Zooty
05-10-2010, 12:07 PM
A quick question. What should the normal fully charged voltage be on a 1.2v AA rechargeable battery?

Zooty
05-10-2010, 02:55 PM
For the first time i decided to look at the input current curve of my Bedini solid state oscillator. It's a standard SSG circuit with standard bifilar coil, no core and high base resistance. It also has the diode going from emitter to base. I just put a 1 ohm resistor in series from the battery positive terminal to the coil and put my scope probes across the resistor. When i saw the waveform, it all became clear. This oscillator may create spikes but it is not efficient at doing it. I noticed a linear increase in current from 0ma up to 70ma for the on pulse then the sudden off where the spike appears. Now it seems to me that if the magnetic field strength is solely responsible for the spike intensity then we could generate the same spike intensity loading the coil much faster to the same level over a much shorter time. Even the briefest moment of high current will generate the same magnetic field but the on pulse will be much shorter and average less current over time. This is where the 555 may be able to help and i think using an scr with low internal resistance instead of a transistor makes sense. In theory, if we can load the coil enough to the point of saturation and switch off before the current has time to circulate the we get the spike for nearly free and i think this is what inquorates COP>1 setup is partially doing. I also think that to avoid coil saturation during the on pulse you would have to limit the input voltage as this would be much more efficient than using resistance. Coil saturation is a big wast of energy as the spike will be the same level from saturation onwards and any current over this point is generating heat in the coil. So to recap, Shortest, highest current pulse to coil to the point of saturation and also make sure the current draw at this point is in the C20 range of the input battery. I have also come to realize how important the rise and fall time of the input pulse is. A fast rise time means less current waste to reach coil saturation and a fast fall time means a higher voltage transient. As well as the magnetic field strength during the on pulse being responsible for the strength of the spike, the speed at which the current stops going in also affects the spike strength. If the magnetic field collapses too slow then voltage of the spike wont be as high and more current will be present in the spike. I also think that coil design plays a big part in this. When current enters a normal single strand multilayer coil, the current slows down due to the equal but opposite current induced in the coil. Tesla's bifilar spiral coil seems to address this nicely and what you get is a much faster current rise time in the coil. You can make the same thing by winding a bifilar single layer coil on a big tube. I'm not sure but i think the net magnetic field in a bifilar coil of this type is zero because there are 2 fields equal but opposite, one in each strand.

ewizard
05-10-2010, 04:59 PM
A quick question. What should the normal fully charged voltage be on a 1.2v AA rechargeable battery?

They should be about 1.40 to 1.45 volts. Most NiMH batteries not up to par are that way because their capacity is killed due to overheating during charging or discharge. Also some of the devices mentioned more often use Lithium-Ion batteries and those are a whole different ball game in how they need to be charged. I don't actually see any of Bedini's commercial chargers like the 1AU even mention Lithium Ion - mostly lead acid and NiMh although the site does say 'also other types'. Due to the danger of explosion with Li-Ion batteries I have hesitated to try charging any with this type of charger. Just my bit of knowledge for what it's worth. I do know a real battery guru who says "most batteries don't die, they are murdered". ;)

Zooty
05-10-2010, 05:06 PM
Thanks ewizard :)

EMCSQ
05-11-2010, 09:18 PM
...And I also have about 8 - 12v 7AH lead acid gel cells and all have held loads well when charged from the bedini radiant oscillator, not the cap dump method though, just straight off the flyback diode.


Do all Renaissance Chargers work with the 'cap dump method' ?
Which do work as a 'bedini radiant oscillator' as a solid state application and which
work with moving parts (I suggest SSG / Monopole).
Are batteries charged a different way using the 'cap dump method' or 'bedini radiant oscillator' ?
Which are recommended on a given application purpose ?

Xenomorph
05-12-2010, 12:03 AM
@Zooty: In my experience a certain current is needed to charge batteries well with that technology. If the charge is too "radiant" then it wont do anything.
But dont be hesitant to try to improve the current signal shape during the on-time as you describe, it could maybe yield a better result.

Zooty
05-12-2010, 12:11 AM
It would be interesting to see what the input current looks like on other peoples setups. I haven't seen anyone on the forums mention it.

eastcoasttinker
08-04-2011, 09:27 PM
Hi, I was charging my laptop batt using the end terminals. The batt is a lith-ion. It seemed to work for a while, then something funny happened. The battery shows a charge on multi-meter, but amps. Laptop shows the batt being charge and shows a % but it will not increase like it did before. At one time the batt kept the laptop on, but now it dies if I remove the power cord. What went wrong?

Web000x
08-04-2011, 09:45 PM
Hi, I was charging my laptop batt using the end terminals. The batt is a lith-ion. It seemed to work for a while, then something funny happened. The battery shows a charge on multi-meter, but amps. Laptop shows the batt being charge and shows a % but it will not increase like it did before. At one time the batt kept the laptop on, but now it dies if I remove the power cord. What went wrong?

Lithium Ion batts have regulator circuits. You probably fried it..

Slider2732
08-05-2011, 02:22 AM
Such protection circuits are a daily thing for myself, being as R/C is my main hobby.
The circuits protect Lithium Ion or Lithium Polymer batteries from dipping below a threshold voltage. When they get to that voltage, typically 3.2V on a 3.7V cell, the circuit switches the battery off.
Lithium Polymer (Li-Po) batteries need to be charged in a specific way, involving constant current and constant voltage at different times. The little circuit onboard is called an LVC (Low Voltage Cut-off).
You can snip the protection board off from such a battery and it will charge with the correct charger fine...BUT...you'll now have no cut-off of course. I've salvaged many a battery whose LVC had blown, though would never recommend doing so unless a strong metal storage tin is in place and it can be certain that the battery will always remain in sight when charging. As soon as a cell puffs up it's dead. Leave it charging when that happens and they can and do catch fire.

Li-Ion's will charge radiantly :) They are much safer than Li-Po. I have a set of 3.7V cells that came from a laptop battery, that charge fine with a simple transistor switcher type pulse motor. I'd agree that the charge rate has to be quite low, 200mA or so, rather than high rates or you get that 'fluffy' charge condition. About 2hrs of charging, while the battery sits at 4.2V is how mine seem to charge best. Also, if the battery gets more than slightly warm, any battery, then things aren't running correctly.
Before I knew anything about these circuits, I used to charge a 3.7V Li-Ion cellphone battery with a R/C hobby Ni-CD/Ni-MH charger ! Cut the LVC circuit off it and it all worked fine.

WeThePeople
08-07-2011, 03:14 AM
Couldn't the LVC be left in to do It's job protecting it,
and be corrected by changing a resistor or a zener ?

Then a second wire added that bypasses the LVC
for the charging function bypassing the LVC ?

I guess what I am asking is,
if your not running the charge through the LVC to fry it out,
and there is no load on the LVC during the charging operation,
are the radiant spikes still going to toast them ?

Protecting the battery cells from the damage that does occur
if they have their voltage dragged to low seems worth it to try.