Boiling capacitor!

[Standuhman]

Hello! I was wondering if anyone here could give me some advice as to how many amps a capacitor can handle, since they don't have an amp or power rating.

To make a long story short I'm trying to bring back a lead acid battery I salvaged from the recycle drop off. The battery started off having 2.44V, and after adding RO water to the cells it dropped to zero. Charging it with a power supply from an old printer brought the voltage up to 4V, but then after I read that the process worked faster the warmer the battery was, and after heating up the bottom with a battery pad, the voltage dropped back down to zero. I think that the lead sulfate in the bottom probably got disturbed and shorted the cells out. Since then I haven't had much luck in getting any charge into the thing, the voltage reads zero with the 'charger' plugged in or not. I thought I might have better luck adding a capacitor to the circuit, but doing so only resulted in the capacitor overheating, bulging out the bottom and getting electrolyte on my table.

Since I was well within the voltage rating I'm wondering how you know how much power a capacitor can take???

The capacitor is rated for 63V, 2200uF and 85degreesC and my power supply puts out 20.5V(18V when in circuit) at 2.23 amps.

What am I missing here?

 

[Raven Luni]

Did you connect the capacitor the right way round? I'm not an expert but from my understanding of capacitors, they should only draw as much current as they need to charge to the voltage across them, conversely they can dump their charge very quickly in a single high current burst.

Current isnt your issue. Also, I suspect the battery is completely knackered otherwise it might not have been thrown out in the first place

 

[Standuhman]

The battery is in very rough shape, and I'll probably end up bringing it right back to where I found it! But I figured since I'm getting into this electronic stuff I might as well give it a try, and I'm also interested in the bedini charger which can supposedly bring back lead acid batteries which are heavily sulfated.. (I might also try boiling off some of the water by overcharging and adding an epsom salt solution although I still need to research this idea a bit more as I'm not sure of the chemical reaction)

To answer your question I'm pretty sure it's connected in the right polarity however the conventional current vs actual current has me a little confused.

I started off by connecting the capacitor lead with the arrow to the positive lead on the battery and the other lead to the negative.(in parallel) But since that didn't produce any extra voltage I decided that perhaps I needed to connect the capacitor in series before the battery.(in parallel would just act as another battery would?)

So I then connected the capacitor terminal which is marked with an arrow to the positive side of the power supply, and I connected the other side to the positive side of the battery, but thinking about it now maybe I SHOULD have connected it the other way! (keeping the positive sides pointing the same direction as you would if you were connecting batteries in series)

Thanks for pointing that out, I think that you might have something there!

(although it sure did up the voltage!)

 

[Raven Luni]

The side with the 'arrow' is always negative. The reason you got more voltage when connecting in series was probably because the capacitor had already been charged from your source and that charge is what you were reading.

You dont want to connect in series anyway because a capacitor basically blocks DC. Once it is charged it acts as an open circuit. Connecting in parallel is usually done as a stablisation measure against transient loads or to smooth out AC ripple from the mains. Theres no harm in doing this but for your purpose I'd say its not really necessary.

Finally, I would suggest searching for information on charging profiles. Depending on the type of battery, the process can be quite different. Some need constant current, some need constant voltage, some need complex and carefully timed combinations of both or levels following a specific curve etc.

 

[Standuhman]

Thanks for pointing that out about polarity, with all the reading I've been doing somehow I had it in mind that the side with the stripe was the positive!(duh, that's why it has the - sign on the stripe..)
So now I suppose that I had it hooked up the wrong way in both parallel and the second series attempt

However I'm going to have to disagree with you on the point you made about capacitors blocking dc, to my knowlege they store an electrostatic charge which opposes change to voltage, so in a dc circuit it acts as a 'voltage storage device' that gets used up during a surge in power demand. I'm not as knowlegeable about caps in an ac circuit however I'm under the assumption that they both smooth out the power ripple and alter the phase of voltage and current by 90 degrees, and that's what makes an ac motor work. Or you can use a copper band around one half of the poles acting as an inductor to do pretty much the same.(but please don't quote me on that and any criticism on my current idea of how this works is certainly welcome)

(the reason your meter tells you that a charged capacitor is an open circuit on dc is that there isn't any power draw)

The main reason I tried adding a capacitor is that without it there isn't any voltage, probably due to a safety in my power supply. One thing that I did learn in this experiment is that if your battery cells are shorted out no amount of juice is going to recharge your acid and restore your plates because that's not the path that the current is flowing through.

With that said, I'm assuming that the capacitor overheated by drawing current inside itself through the action of removing the aluminum oxide layer, kind of like the cleaning action that occurs when tig welding aluminum with ac.(electrode negative doesn't clean, electrode positive does)

 

[CocaCola]

However I'm going to have to disagree with you on the point you made about capacitors blocking dc, to my knowlege

Your knowledge has to be expanded, as it's incorrect... Pay attention to series and parallel implementations of a cap while you expand your knowledge...

 

[jackorocko]

However I'm going to have to disagree with you on the point you made about capacitors blocking dc, to my knowlege they store an electrostatic charge which opposes change to voltage, so in a dc circuit it acts as a 'voltage storage device'

And when that storage device charges up it is essentially seen as an open circuit. This is why a series cap between amplifiers stages is common practice. To block the DC component from affecting the next stage of amplification.

 

[Standuhman]

Your knowledge has to be expanded, as it's incorrect... Pay attention to series and parallel implementations of a cap while you expand your knowledge...

This is a difficult area of study as every explanation/teaching I've so far come across only tells me that to add caps in parallel adds the individual capacitances while the voltage rating stays the same and that to add them in series increases the voltage rating but decreases the overall capacitance.

While this is very valuable information I'm still a little new on the 'implementation' of this in a circuit. From my battery experiment I can say that the capacitor once charged had the imput voltage available at the other side.. but without it there was none. When you test a capacitor that's not in circuit your tester is charging the capacitor, at first the cap is taking lots of charge, and when almost charged very little current is flowing as the voltage difference is ever decreasing. I can only 'imagine' that the circuit reads open because it actually is, there isn't any connection between the two terminals, but can the dc current still flow if there is a demand? I think yes, but I realize that what I need to do is actually build some circuits and test them.

The amplifier is another circuit that I'm interested in but don't yet fully understand.

 

[CocaCola]

This is a difficult area of study

Hardly there are easily 100s er 1000s of books on the subject as well as 1000s upon 1000s of website hits in Google...

Heck you can start at the most obvious, Wikipedia page, it's a good read...

http://en.wikipedia.org/wiki/Capacitor

Capacitors are widely used in electronic circuits for blocking direct current while allowing alternating current to pass, in filter networks, for smoothing the output of power supplies, in the resonant circuits that tune radios to particular frequencies, in electric power transmission systems for stabilizing voltage and power flow, and for many other purposes.

Because capacitors pass AC but block DC signals (when charged up to the applied dc voltage), they are often used to separate the AC and DC components of a signal. This method is known as AC coupling or "capacitive coupling". Here, a large value of capacitance, whose value need not be accurately controlled, but whose reactance is small at the signal frequency, is employed.

 

[KrisBlueNZ]

Adding an electrolytic capacitor isn't going to help you charge a worn-out SLA battery at all.
If you connect the capacitor BETWEEN the charger and the battery, the capacitor will charge up to the charger voltage in a very short time and current will stop flowing (unless you connect the capacitor backwards, in which case it will vent if the voltage and current are high enough, as you discovered).
If you connect the capacitor ACROSS the battery, or the charger, it will have no useful effect.
Many battery chargers consist of a transformer and diodes, without a smoothing capacitor; connecting a capacitor across the output of this kind of charger will cause the measured unloaded output voltage to increase, because the capacitor acts as a reservoir, but will make no significant difference to the charger's ability to charge a battery. If it did, the manufacturer would put a capacitor inside the charger. In your case you're not using a dedicated charger; adding a capacitor will make NO difference here.
If you want to make the charger charge the battery more gently, you need to reduce the charge current. You can use a current or voltage regulator to do this, but the simplest way is to use a resistor. You'll need to use a high power resistor because it will dissipate a lot of power.
You put the resistor between the charger and the battery. Some, or most, of the charger's output voltage will be dropped across the resistor. The amount of current that flows can be calculated using Ohm's Law, I = V / R where I is current in amps, V is voltage ACROSS THE RESISTOR in volts, and R is resistance in ohms. So you choose your resistance to give roughly the desired charge current, using a rearrangement of Ohm's Law: R = V / I.
In this case, since we're talking about the resistor only, V is the voltage across the resistor, which is equal to the power supply voltage minus the battery voltage (under charge). For example if you try a 33 ohm resistor, and measure 20V across the power supply and 4V across the battery, the voltage across the resistor is 16V so the current is I = V / R which is 16 / 33 which is about half an amp.
I don't know much about SLA batteries, but some batteries can be partly resuscitated by charging them for a long time at a relatively low current, so this might be helpful to you.
You'll need to consider power dissipation in the resistor. P = V I which in this example is 16 x 0.5 (approximately) which is 8 watts. You'll need a resistor rated at 10W or more, and it will get hot. Some big resistors can be bolted to a heatsink, or you could put it in water (the wires will eventually corrode, though).
So forget the capacitor. I hope this helps.

 

[davenn]

hi Standuhman,

I thinks its time for you to have a look at this article, it tells you all about battery sulfation and one way of possibly rejuvinating lead-acid batteries
It even has a circuit for doing such

cheers
Dave

 

[KrisBlueNZ]

That's an interesting article.
Deterioration of lead-acid batteries is caused by the deposition of crystallised lead sulphate on the plates, increasing the internal resistance of the battery and preventing current flow.
The article says that it's possible to dissolve the lead sulphate back into the electrolyte by feeding a high AC voltage at 2~6 MHz into the battery, causing "rhythmic beating (resonance) of the plates", which makes the crystallised deposits break up and dissolve. This AC voltage must be "high frequency, high voltage, but low power".
They don't have any test reports or links to testimonials.
I couldn't say whether this idea is plausible or not.
The circuit they give uses a 555 driving a FET driving a couple of inductors. It seems that the circuit is supposed to be connected between a positive supply and the positive terminal of the battery; a trickle charge current flows through the circuit and the circuit generates the high-frequency voltage at the same time.
If anyone tries it, please let us all know how you get on!

 

[Raven Luni]

Its also worth noting that you're unlikely to get that kind of frequency out of a 555

 

[KrisBlueNZ]

Yes, I think the 2~6 MHz figure was a typo. The components specified for the 555 give a frequency of about 1.3 kHz.

That page links to another page, http://home.comcast.net/~ddenhardt201263/desulfator/desulf.htm which has links to other pages and some testimonials. There are links to two commercial desulphators.

The first commercial desulphator (solar-electric.com) costs USD 150 and the web page has a disclaimer: "There is no guarantee that this will help your battery. Our own tests have shown that it seems to help sulfated batteries in otherwise good condition, but it is often hard to tell what a battery's internal problems are." It's claimed that it "uses a sharp pulse of current forced into the battery suddenly to 'jar' the sulphite crystals and cause internal resonances, both mechanical and electrical, to grind down the sulphite crystals that form so they can be re-combined into the battery acid."

The second commercial desulphator web site no longer exists.

The page I linked to above shows the first design using the 555, and a much larger design using a toroid to couple the pulses into the battery. There's a link to a page with the schematic and construction notes. The circuit is powered directly from the mains (110VAC) so it's potentially dangerous, but it looks like it could produce useful current pulses as described. Whether it has any effect on a sulphated battery, I couldn't say.

Someone has already asked about desulphators here. There were no replies to his question. I found the thread while googling battery desulphators. https://www.electronicspoint.com/battery-desulfators-do-they-really-work-t128372.html

More googling produced a thread on randi.org but no conclusive answers there.
Most opinions I've found seem to say that desulphators can help but only in some cases.

 

[Standuhman]

A good article indeed, however the way I interpreted the page is that it tells us to re-condition the battery using pulses of dc current, not ac. Which is exactly the result I've seen on some of the youtube videos I've been watching on the bedini magnet motor/charging circuit. Although this is seen as an overunity device by some I don't believe that it is, but could be a usefull tool in battery re-conditioning as it provides high voltage spikes at low current.

 

[(*steve*)]

Pulses of DC can be considered AC with a DC offset, so you can be right both ways.

And you're right, it has nothing to do with "over unity" at all. The battery is discharged during the process (for those devices powered by the battery)

Here's a couple of other threads you'll find by searching for "desulfator"

https://www.electronicspoint.com/le...ircuit-help-t219764.html?highlight=desulfator

https://www.electronicspoint.com/lead-acid-battery-maintenance-t3302.html?highlight=desulfator

There are others, but they are quite short and probably don't have a lot of information.

 

[KrisBlueNZ]

Although this is seen as an overunity device by some I don't believe that it is ...

Well I'm glad you think that, since overunity devices don't exist! BTW any "device" that claims to be a magnet-only motor also doesn't exist. Its only practical application is to part gullible, scientifically uneducated investors from their money.

... but could be a usefull tool in battery re-conditioning as it provides high voltage spikes at low current.

High voltage spikes at low current is a contradiction in this scenario. Ignoring inductive effects, which will be insignificant at the relatively low frequencies used, a battery under charge is a load, and will draw current in a defined way according to the voltage provided. If the charging circuit cannot provide enough current to apply its designed voltage across the battery, the battery will load it down and the voltage will drop. A simpler way to look at it is that the battery terminal voltage will increase by an amount related to the current being delivered to it; if the current is small then the voltage increase will also be small. It's not possible to deliver high voltage pulses at low current into a heavy load such as a battery under charge. Just a bit of basic Ohm's law for ya

BTW I'm not claiming that battery desulphators don't work; it's just your description of "high voltage spikes at low current" that I'm disputing.

 

[Mongrel Shark]

Got an acid content measuring device? Like a hydrometer, I have one like an over-sized eve dropper with a float in it. Got it from auto store.
I believe there are digital versions too.

Getting the junk off the plates is only half the story. Won't do a lot if the acid is gone...

You can get acid crystals. but you need to add the right amount or risk all sorts of bad stuff like acid eating holes in stuff it shouldn't and explosion.

I can also highly recommend INOX brand battery conditioner. Works best in new batterys. Although it does add some life to older batterys.


I've seen a few pulse type DE-sulfateing chargers. Most do do something. Some are better than others.

I've never seen a Bedini in action, but I believe they work well (as long as you don't want over-unity, because that's not going to happen). Incredibly difficult to make though. Not a beginners project!

Sounds like your battery is completely stuffed though. I am assuming it's 12v. if the most charge you have seen is 4v it's probably only had 2 cells working. Sounds like you killed them now Just remember there should be (aprox)2v to a cell.

It's a very interesting topic. good luck studying up.