I found a schematic on the LM317 datasheet (attatched). How does Rs limit current? I'd like to use this to charge my motorcycle batteries (7AH and 12AH), not at the same time. Will it work alright? It'd be powered by an 18v 680ma DC output walwart.

Right now I charge the battery with a 12v 100ma DC output walwart, it takes a few days but it does charge it. But I'd like a little voltage protection (the 100ma 12v is 15.51v with no load, so I'm sure eventually it could get the battery there), and also 100ma just takes forever to charge a battery. 680ma would get it charged in less than 24 hours.

Yes, that will do it. Rs makes the charger look like it is driving the charger through a resistor (2.2Ω) that's bigger than the one in the circuit (.2Ω) by the formula at the bottom. With those values it should put out about 13.75V, so you have voltage and a certain amount of current limiting.

How discharged does your battery typically get?

How much would that limit the current then? I'm not really worried about limiting the current as it can't go beyond 680ma anyway. Is it possible to damage a 7A 12v battery by holding it at 13.7v @ 680ma after it is fully charged?

Normally the battery doesn't need a whole lot of charging, just top it off mid winter and then before the season starts to keep it fresh. It probably gets down to 80% before I charge it. Although there was once or twice I've needed to charge it up from 0% charge...once was an electrical short, the other was when I was trouble shooting and I kept cranking on the electric starter. After an afternoon of the bike not starting, the battery wouldn't even crank the engine at all.

Originally Posted by duffy

The current will be zero when it hits 13.75V, no problem there. This is a very good circuit for what you want. If it was deep discharged, there would be some issue with it taxing the wall transformer.

Why will the current be zero when it hits 13.75v? Is that due to Rs? What if it hits 13.75v before the battery is done charging?

Also, my 100ma transformer charged it fine...it took half a week, but it did it. So could I just leave out Rs then since it doesn't need any current regulation? Also, what wattage resistors should I use?

>>Why will the current be zero when it hits 13.75v?

When the battery's internal voltage hits 13.75V there will be no differential between the charger and the battery, so no current. If you check the unloaded float voltage of that lead acid battery, fully charged and disconnected from anything, I believe you will find it's over 13V. 13.75 is an excellent choice for the charge point voltage.

>>What if it hits 13.75v before the battery is done charging?

It's normal to see the charger at full voltage across a battery that hasn't REALLY finished charging. To do that, you need to measure the current.

>>Is that due to Rs?
>>Also, what wattage resistors should I use?

Rs just makes the LM317 appear to have a 2.2ohm power resistor in series with it. Incredibly, with a maximum output of .68A, all those resistors can be quarter-watt. But use a 1 watt for Rs, a really drained battery might just pull more more current.

>>So could I just leave out Rs then since it doesn't need any current regulation?

Probably. The battery can easily handle more current, Rs is mainly in there to reduce the load on the LM317 when it's dealing with a heavily discharged battery. It DOES have some internal protection against over temperature (high wattage). Help it out and give it a good heatsink, this is a series-pass regulator. The wallwart will almost certainly put out more current than 680ma if the voltage is loaded to under 12V. Try hooking an ammeter (on a high setting) in series with the wallwart and a heavy load (a headlight or something). See how much it can REALLY put out. THat's what the regulator is going to have to deal with

So if I put a large heatsink on the chip, it should be fine without Rs? And I can use standard 1/4w resistors then for the R1 and R2.

Sounds good to me.

If you want to make it safe (including overcharge protection) you're better off using a UC3906 lead-acid-battery charger IC.

You can kill any battery easily overcharging it.

Boncuk

Originally Posted by Boncuk

If you want to make it safe (including overcharge protection) you're better off using a UC3906 lead-acid-battery charger IC.

You can kill any battery easily overcharging it.

Boncuk

I'm not worried about overcharging. The charge will happen slow enough that it won't just randomly explode, and I'm normally impatient and check all the time to see how it's coming along.

Also, limiting voltage to below gassing voltage (which is ~14.5v at room temp), will keep it from being dangerous. And as soon as the battery reaches the charger's voltage of 13.75v, it will no longer flow any current, keeping it from overcharging. That said, it could be slightly overcharged by the time it might hit 13.75v, but I doubt it and I'm not worried about it.

If I was wanting to limit it to 500ma output, what value would I use for Rs? Is that the Zout equation? If so, I get 0.22 with the stock values. Would that mean with 0.2 ohms at Rs it would put out 0.22 amps (220ma)?

The reason I'm looking for 500ma output is because I have a cheapy analog ($5-10 from walmart 8 years ago) multimeter that reads 500ma max, and I'd like to use it for this charger project. Would using the circuit as is with my 680ma wall mount transformer break my multimeter?

Originally Posted by duffy

It might. You can use a pass resistor to change the scale of a cheap meter - look at the voltage developed across a .1 ohm resistor and the voltage will be 10x the current. In the circuit above you can simply read the voltage off the .2 ohm resistor.

That circuit doesn't really have current limiting, it just has a large apparent series resistance.

So if there is 1.25v across a 0.1 ohm resistor, there will be 12.5ma passing through it or 12.5 amps?