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Battery charging


jdg600

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I recently bought a little 12v 7Ah battery and a charger, just for running the cooling fan on my dob and the dew preventer on my finder (when I get round to making it). However, on reading the istructions that came with the charger I find it is not for use with batteries less than 20Ah so it looks like I need to get another charger. I've had a look around on the net and seen several chargers with an output current of less than 700mA which look like they might be suitable, most of these appear to be just a plug-in psu with croc clips attached. So this got me wondering... I have a 350mA 0 - 15v stabilized bench supply, could I use this to recharge the battery?

As far as I can see the main problem is that it won't reduce the output as the battery charge increases but I'm not sure if the plug in ones do. If I did use it for this purpose what charging voltage should I use and how long should I charge it for? If I accidentally left it charging too long is there a risk of the battery exploding?

I would be most grateful for any advice on this.

John

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I use a 21w car sidelight bulb in series with the battery; this limits the charge to about 0.5A.

It also acts as a charge indicator by glowing brightly at first and then going dimmer as full charge is reached.

It is not good to leave gel cell on charge all the time, unless you are using a special (float) charger.

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Thanks for your reply Keith but I'm a little confused. Would the bulb not need to be in parallel with the battery to reduce the current going to it? I had wondered about doing something like that myself but wasnt sure what sort of load I would need. My charger is 4A and I have a 12V 24W bulb and holder lying around somewhere, would this doe the trick?

John

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The use of a light bulb in series with the battery is quite an old trick that a few of the 'older engineers' ( myself included :grin: :grin: :grin: ) have used for some time.

If your PSU is set to around 15V, then the resistance of the bulb will restrict the current flowing into the battery. When the battery voltage is quite low the positive terminal of the battery will cause a voltage drop across the bulb causing it to light up. As the battery charges, the voltage drop becomes less and the bulb dims until there is no appreciable voltage across it.

Hope this helps.........

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Thanks for your reply Keith but I'm a little confused. Would the bulb not need to be in parallel with the battery to reduce the current going to it? I had wondered about doing something like that myself but wasnt sure what sort of load I would need. My charger is 4A and I have a 12V 24W bulb and holder lying around somewhere, would this doe the trick?

John

Yes the bulb you have will be fine, definitely in SERIES.

AstroTux is quite correct in what he says.

An old trick, but a good'n.

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Forgot to add that if you choose the correct bulb, it also acts as a fuse.

Lightbulbs blow really quickly ( faster than quite a few of the 'custom' fuses available ) and can be used to give an indication of current whilst protecting systems when used in series. I have a friend who works with DC/AC inverters and this is probably his favourite trick for protecting circuitry.

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Investing in a cheap multimeter is never a bad idea... You can pick up the really cheap and nasty ones for £5-£10 from maplin or the like. I assume the battery you have is a sealed lead acid (SLA) type... it will also likely list charging voltages on it's side, 13.8v for float charging and 14.4v for cyclical charging (NEVER let a sealed lead acid batter go above 14.4 volts or it will start gassing heavily and pop the safety caps)

Using the bulb as a current limiter is a good idea as generally SLA batteries don't like being charged with a current over C/10 (one tenth on their Amp hour rating) so it's best to charge a 7Ah battery at no more than 0.7A (700ma). If your charger isn't an electronically controlled one that shuts off automatically, keep checking the battery with the multimeter every hour until it's close to 14.4V (or whatever the manufacturer recommends as a cyclical charge voltage)...

Hope this helps!

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Thanks for all the advice. I'm definitely going to try this trick with the bulb but I'd be happier if I understood how it worked a little better, I wonder if someone could explain.

My understanding is that in order to charge the battery I need to supply a current of something less than 700mA at a little over 12v. With the setup suggested ie 12v bulb in series with 15v supply then I would have thought there would be a 12v drop across the bulb so that the battery would be getting about 3v and, with a 24w bulb, a current of about 2A. Is this correct?

Obviously this is a tried and trusted technique and I don't doubt it works its just my knowledge that is deficient. I'd be very grateful if someone could give me an idea how it works or point me in the direction of an explanation.

John

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Thanks for all the advice. I'm definitely going to try this trick with the bulb but I'd be happier if I understood how it worked a little better, I wonder if someone could explain.

.......

Suppose the battery in its discharged state is showing , say 11V. When the bulb is in series with the charger and the battery the voltage difference across the bulb is 4V (15V from the charger minus 11V from the battery)

The resistance of a 21W bulb is about 7ohms. If a voltage of 4 Volts is placed across a 7ohm resistance then a current of 4/7 Amps will flow (approximately 0.5A)

As the bettery charges the voltage difference between it and the charger decreases so the voltage across the bulb decreases and the current through the bulb also deceases as it will be equal to the voltage difference divided by the resistance (which is still 7 ohms).

Because the current through the bulb is decreasing, the bulb glows dimmer. When the battery voltage equals the charger voltage there will be no current through the bulb as there is no voltage difference.

Hope that helps

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Hey john, I hope that you have your maths hat on... it's all to do with voltage drops and resistance.

The key here is Ohms law V(oltage) / I(current in amps) * R(resistance in ohms)... For a given voltage only a certain current will flow across a set resistance.

When a battery is flat it'll be about 10.8 volts when it's fully charged it will be at 14.4 volts, Your 24w 12v lamp has a resistance of 6 ohms, Your charger assuming it's a transformer with a bridge rectifier will be putting out about 16.9V

Now when your charger is connected to a flat battery 10.8v there will be a voltage drop of (16.9 - 10.8) = 6.1 volts across the bulb... 6.1v/6ohms = 1.01A (this is a quite acceptable charging current for a 7Ah battery)

As the battery charges and it's voltage increases to 14.4 volts things change... the voltage drop is now(16.9 - 14.4)= 2.5 volts 2.5v/6ohms = 0.41A (notice the charging current has tapered off? I wont go in to detail but this is a good thing)

IN A NUTSHELL:When the battery is more deeply discharged the bulb will allow more current to flow, As the battery charges the resistance in the bulb with reduce the amount of power reaching the battery by a greater degree.

I've likely explained this badly...

Matt

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Thanks Matt. Yes I've got it now. Got myself into a bit of a muddle over it before but now that you and others have explained it seems straightforward. I'm going to give it a go tomorrow. I'm glad that the charger I bought won't now go to waste.

Thanks,

John

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Thanks to all for your helpful responses. I connected the bulb in series with battery this morning and the battery seems to be charging nicely :hello2:

Now, my next question is how often should I recharge it. At the moment I'm just going to be using it to run my cooling fan, so 20-30min say at 900mA at the beggining of each observing session. The time between usages will be variable but often on the order of weeks ( :clouds2: ). So should I recharge after every use? If it is not used for several weeks at a time should I recharge it anyway and if so after how long?

Again, any advice would be much appreciated.

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The best way to judge this is by calculating how much you are taking out of the battery, and not let it drop below about half-charge ( depending upon the type of battery ). I'm guessing that you are using a Valve Regulated Lead Acid (VRLA) battery similar to the 7Ah Yuasa NP7-12 that I use......

If you take 700mA out of this type of battery for 10 hours, then you must recharge it ( 700mA x 10 Hours = 7Ah ), however if you take 7A out of it for 30mins then you also need to recharge it ( 7A x 30 mins = 3.5 Ah ). From this you can see that, unfortunately the discharge rate has an effect on how long you can use these batteries before they need to be recharged, and it is not a simple Ah calculation. The voltage that they can be discharged to also varies with the current you are drawing from the battery ( OMG !! ).

Consulting the datasheet for my battery it would indicate that for a current draw of 900mA it can run for about 7-7½ Hours and will discharge to about 10.5V before it needs to be recharged. Be aware that this is the absolute maximum, so I would advise that you recharge it before it gets this low.

Hope this helps .........

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The Yuasa NP7-12 is rated for either float or cyclic discharge use. What does this mean?

  1. Float is where the battery is typically used in a standby situation and is used when main power fails. It is normally being 'trickle-charged' from the apparatus it is connected to.
  2. Cyclic discharge means using the battery to power the equipment until it is nearly exhausted and then recharging it.

From this it can be seen that you can really charge the battery whenever you wish. I tend to recharge mine after about 8-10 Hours of operation, but my system only draws about 350mA, so I am in a lower league than you :grin: :grin: !!

I you want more information on these batteries, the datasheet is here http://www.yuasabatt..._literature.php . This will give you the self discharge rates and an idea of how often to recharge the battery if you have not used it for a while.

A very useful publication explaining a lot about these batteries ( called The Little Red Book of Batteries ) is here http://www.norbain.c...O40FFD244D6899/ .

Hope this helps.......

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If your 15V supply is regulated and stable you could put diodes in series with the battery each diode would drop 0.6V, battery could then be charged at a constant Voltage. Put an led + resistor in parallel with 2 diodes as an indicator.

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