hi there I have to of this battery's
they are linked in series to give 24v out this are used for back up use only , what I wont to do is make a charging circuit, does any one have any suggestions on a good and simple way to do this?
thanks Joe
Lead acid battries should be charged with a current limited to one tenth of their amp hour capacity until the voltage reaches 28.5 volts.
You will find suggestions for charging regimes in the datasheets for the battery.
I find bench power supplies with independent voltage and current limit settings
very handy for charging the odd battery now and again (not very power efficient
though).
Lots of chargers are available for larger 12V battery pairs - golf carts and
wheelchairs and mobility scooters often use 2 x 12V deep cycle batteries,
but usually more than 7Ah... eBay for golf-cart charger perhaps?
thanks for your replies
I had come across this and think it will suit what I'm doing
the only issue I can see with this is R2 ( the currant limiting resister) when mains failure occurs the neutral path will be back through R2, so I was thinking of putting it just after the lm317 would this be ok to do this?
thanks Joe
Joes:
the only issue I can see with this is R2 ( the currant limiting resister) when mains failure occurs the neutral path will be back through R2
Sorry, that makes no sense to me at all.
When mains failure occurs, the LM317 output will drop to zero, and D2 will prevent the battery from discharging into the circuit.
yes but when we are in mains failure we will be discharging from the battery's , there is a Diode on the + line to stop feed back that is fine, but my neutral will have to go back through R2. do you see what I am saying?
It doesn't matter, the diode stops current. You must have a complete loop for current, just having the - lead attached to circuit ground will not discharge the battery.
And R2 -must- be exactly where it is to limit current. It CANNOT be moved.
ok so when I am pulling 3-6A from the battery's, R2 will be fine with that?
Ah, I see what you are talking about. That is a very good point. See that ground symbol? That is only there to illustrate that node as circuit common.
Do you need to connect that to ground? I don't think you do, as the rest of the circuit is isolated from everything else electrically. So ignore that ground symbol in the circuit. Use the side on the (-) terminal of the battery as ground for the circuit you are supplying.
To clarify: Leave everything else connected exactly as it is. You are only moving the ground connection to the right side of R2, so that battery discharge current does not flow through R2.
so the black bit's I have put on is where I would pick up from for my out going?
I presume by "my out going" you mean the circuit that is to be powered with battery backup. Why not just connect it across the battery?
btw I think using a 35V transformer is a bad idea. The voltage across C1 will be about 47.5V peak when fully loaded, and more than 50V when the battery is charged and the current has dropped to a low level. When the battery is charging, the LM317 will dissipate a lot of power and need a good heatsink. If you accidentally short the output or connect an over-discharged battery, you will probably blow the LM317. I suggest you use a 28V or 30V transformer instead. You don't need a 3A transformer, because the current limit is set at less than 1A. A 50VA transformer will suffice. Here's one possibility: http://uk.farnell.com/pro-power/ctfc50-15/transformer-50va-2x-15v/dp/1780858.
Also, make sure that C1 has a ripple current rating if at least 1A.
I under stand about the transformer I was going to is a 24v but I'm getting out 35v so I'm going to be tring a 18v to see what I get out.
But if I connect my load over to batters when I am pulling 3-6a I don't think the lm317 will take it lol
Joes:
I under stand about the transformer I was going to is a 24v but I'm getting out 35v so I'm going to be tring a 18v to see what I get out.
18V from the transformer will definitely be too little. Your 24V transformer may be OK, or it may be not quite enough. I presume you measured the 35V across the capacitor, with no significant load. While that may appear to be more voltage than you need, it will drop under load. You need 27.3V to float-charge the batteries, plus 0.7V for D2, plus nearly 3V for the LM317. That makes 31V. You will get around 5V peak-to-peak ripple on the capacitor at full load. If you want to use that 24V transformer, I suggest you increase the capacitor to e.g. 4700uF.
Remember, when you buy a "24V" transformer, that is VRMS. But the filter capacitor charges to the peak voltage, minus 1.4V (double the 0.7V diode drop when using a bridge rectifier), so 24VRMS x ?2 - 1.4V = 32.5V. Unloaded, that may indeed rise to 35V.
Ok cool thanks for that will Baer that in mind
Still still have not decided how we are going to connect this as I can not connect me load through lm717 as it's max load is like 1a and I'm going to be puting 3-6 amps through it?
Why do you want to charge the battery with such a high current? Do you really need to charge it that fast?
No that's that is what the load is going to be on the out put lines as shown on pick I put up, so this is not charging load
First off, unless somehow your load is in danger of supplying power back to the battery, get rid of both of the top diodes and connect directly to the battery. As it sits now, you'll get raw unregulated 32 to 35V to the load from the charger supply.
And picking power off where you are now, you'll blow that 2A fuse as soon as you draw more current.
Take power right off the battery. If you need to fuse the load, add a fuse between the battery + and the load of whatever current you want it to blow at. And remember - a fuse is not there to protect the charger, or the load, it is not fast enough. A fuse is there to prevent a fire in the event of a failure or short circuit.
The LM317's output current is limited, even if you didn't have the add-on limiter. So say it is set for 1A, and you draw 6A from the battery directly. So the LM317 supplies 1A, and the battery itself supplies 5A.
It is that simple.
You are putting a large(ish) heat sink on the LM317, correct? A small fan wouldn't hurt.
It's possible to increase the output of the LM317 using external transistors - see Fig. 39 of http://www.ti.com/lit/ds/symlink/lm117.pdf. However, this will further increase the minimum voltage you need on C1. I think it would be better to make your own regulator using a dual op amp such as MC34072 (which is OK up to a 44V supply), a voltage regulator chip, and a few transistors. You can use one of the op amps to regulate the voltage, and the other to regulate the current. I suggest you also implement foldback current limiting.
Alternatively, use a high current switching regulator such as LM2679-ADJ, which is good for 5A and has the current limiting built-in; but be warned: good circuit layout and correct choice of the output resistor and capacitor is critical when building a switching regulator.
Foldback limiting for a battery charger?