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12v Lead Acid Battery Charger aka I need to stop killing my batteries!

Hi

I have killed a 12v lead acid battery, and would like some help understanding why, and how to stop it:

I currently have a 'plug in the wall' style 12v lead acid battery charger that I use to recharge a sealed 7ah 12v gel cell.

The cell is generally used one afternoon a week to start model aircraft engines (i.e. high discharge rates for short periods) and is left in a cold garage the rest of the time, usually on charge. After less than a years use the capacity of the cell is less than half that of new.

The charger I have just researched a little more, no load it outputs

12v, loaded (ie charging the battery) the voltage is 15v. The current flow is a constant 400ma (all measured using a electric flight watt meter) and does not change between a discharged battery and a fully charged battery. I opened up the wall charger and it is simply a transformer and a rectifier - nothing else.

I don't understand the theory of above, how come a current is flowing when attaching the battery 'appears' to raise the voltage across the charger and the rest voltage of the cell is higher than the rest voltage of the charger?

So, the reason I am killing the battery is that I routinely over charge it for long period of time. Once I understand the theory above I hope to add something to the output of the existing charger to stop me killing the battery - be that turning it into a float charger by limiting the voltage (so all I would need to do here is to add a 13.x v regulator to the output?) or by something more sophisticated. Options welcomed.

Thanks in advance for the help :)

Cheers Kev

Reply to
Kevin Walton
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Hi Kevin. Lead acid batteries are to be charged with constant voltage. If you are using that constant current charger, they are being killed. A 7812 regulator with a couple of 1N4004 diodes in series to its ground reference pin will raise the regulation to about 13.3 V, good enough. Miguel

Reply to
Externet

Greetings Kevin, Though I'm no great shakes in electronics I think I can answer your questions. First, since there is no filter on the charger it is delivering pulsed DC. When measured with your meter it will show the wrong voltage. Connecting to the battery puts a filter in the circuit and your voltage reading is now closer to correct. Second, use a resistor to limit the current. 15 volts should be OK for the battery. It's the high current that kills the battery. Does the battery maker have a website or a phone number? If so, maybe you can ask them about the voltage and current requirements to keep the battery charged without overcharging. Cheers, Eric

Reply to
Eric R Snow

Hi, Thanks for the reply

I think you missunderstood my posting a little, it is not a 'constant current charger', it is a charger designed for 12v lead acid batteries, just that measuring the current flow from the charger during the charge cycle it is a consistant 400ma.

Sounds like a nice simple option to improve the wall charger that will also physically fit inside the wall charger.

Thanks Kev

Reply to
Kevin Walton

Actually, after just doing a discharge test it barely has 10% of its original capacity, so alhtough this thread is relevant as I need to not kill the next one, I dont think I will be geting much use out of this one again unless I can rejuvinate it

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Cheers Kev

Reply to
Kevin Walton

....

The current is pulsatile, but its average value is approximately constant under the conditions that exist while it is destroying your battery(s). See the post by Mr. Fields to get a better idea why that current is approximately constant.

It is a decent solution, but it will dissipate more power than the diodes and tranformer presently do, so you need to be sure that the circuit does not overheat in that location.

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Reply to
Larry Brasfield 

--
If your meter is average-reading and is reading 12VDC out of the
charger when the charger isn\'t connected to the battery, then the peak
voltage out of the charger will be about 19V even though your meter
only shows 12V.  However, as soon as you hook it up to the battery the
battery will soak up the peaks and your meter will then be reading the
steady DC voltage across the battery terminals which, when you
measured it, was 15VDC.  If you were to let the battery discharge to
the point where it had about 11V across it, under load, and then you
removed the load and started to charge the battery, the voltage would
jump up a little, maybe to 12V or so at first, but then it would take
much longer to get to its fully charged voltage if it was a good
battery.  If you have a discharged battery and and it reads 15V as
soon as you put the charger across it, then its internal resistance
has risen to the point where much of its capacity will be lost.
Reply to
John Fields

--- Yes but, unfortunately, if it's putting 400 mA into the battery when it's discharged and also putting 400 mA into the battery when the battery is fully charged then it is, in fact, a constant current supply!

Lead-acid batteries generally want to be charged using a constant-voltage supply because as the battery voltage rises when it's charging, the current being put into the battery tapers off with time because the difference between the battery voltage and the charger voltage gets to be smaller and smaller, so the current being forced into the battery gets lower and lower as the battery voltage goes higher and higher.

However, I suspect that what's happening in your case is that as the battery voltage increases when it charges, the diminished load on the transformer causes its output voltage to rise as well, forcing the voltage difference between the battery and the transformer to remain more or less constant, forcing the current to remain more or less constant. Also, I believe that as the battery charges its internal resistance decreases somewhat, which would aggravate the effect. With a regulated (or voltage-limited) supply, though, the change in the battery's internal resistance would be more than compensated for by the rise in battery voltage, so eventually the system would come to rest with only the trickle charge current flowing into the battery.

-- John Fields

Reply to
John Fields

Ahhh, ofcourse, RMS and all that.

If 12v = average, then peak = 12/0.637 = 18.8v, rms = 1.11 x 12 =

13.32v

Not quite the 15v seen but then without putting a scope on it, it does explain it.

Yup, I have now realised that this battery is pretty much dead and have been experimenting with a battery that does have 50% of it's capacity left. Charging the second cell (12ah) started with a lower voltage and a current of around 600mah, the voltage slowly rose through the charge to 15v and then sat at 400mah current.

What part does the Rs play in that circuit? I appretiate it sets the output impedance, which has an effect on the current flowing. If I know the internal resistance in a battery limits the initial charging rate, and the voltage is only set at the appropriate float charging voltage, when the battery is fully charged the voltages will equalise and the current flow will become zero?

Cheers Kev

Reply to
Kevin Walton 

--
What\'s your point?  The 15 volts you saw, you said, was when you read
the DC battery voltage with the charger hooked up to the battery.
That means that was the voltage the battery was limiting the 19V peak
input DC pulses to.
Reply to
John Fields

Kevin Walton formulated on Monday :

I keep an automatic motoecycle battery charger in my flite box. never killed a battery that way

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Reply to
Lawrence Oravetz

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