Battery Charger

[biasottw]

Hello,
I have a 6V 12Ah SLA battery that is in series with two 5 ohm 1W resisters to make it intrinsically safe. However, this causes a problem for charging. I am not able to use a normal battery charger as it will not detect when the battery voltage has reached the correct point to shut down or switch to the float charge. Do any of you have any ideas that I could try to build my own battery charger. I have attached the circuit I am currently working with, but it is not charging the battery fast enough to begin with. I am only getting about 7.5V (To account for voltage drop across the resisters) and about 300mA. I would like to charge the battery at about 1A until it reaches 6V then reduce down to about the 300mA. Any advice would be greatly appreciated, thanks.

 

[ronv]

Adjust R3 for desired curent. You will need a higher voltage for your final charge as well.

 

[ericgibbs]

Hello,
I have a 6V 12Ah SLA battery that is in series with two 5 ohm 1W resisters to make it intrinsically safe. However, this causes a problem for charging. I am not able to use a normal battery charger as it will not detect when the battery voltage has reached the correct point to shut down or switch to the float charge. Do any of you have any ideas that I could try to build my own battery charger. I have attached the circuit I am currently working with, but it is not charging the battery fast enough to begin with. I am only getting about 7.5V (To account for voltage drop across the resisters) and about 300mA. I would like to charge the battery at about 1A until it reaches 6V then reduce down to about the 300mA. Any advice would be greatly appreciated, thanks.

hi,
You could bypass the two 5R resistors with a suitably rated diode, the diode cathode to the battery terminal end of the 5R's and the anode to the other end of the of the 5R's.
This would reduce the charge voltage drop to approx 0.8/1V or so.

I dont understand why you feel these two resistors are required, is it so you dont short out the SLA.?

 

[MrAl]

Hello,

The wiper arm on the pot is drawn as open. If that is really the case, you need to short that to one end of the pot or this will not work.

Eric, what 5R resistors?

 

[biasottw]

The SLA is part of a pack with all sealed in Epoxy. And yes, the two resistors are in there so the SLA is not shorted out. It is used in potentially explosive situations if there were to be a spark. This is why I need to find/develop a charger for this battery.

No, the pot is adjusted through a turn screw. It is actually not open. That is a mistake in my drawing and thank you for noticing.

 

[MrAl]

Hello again,

Oh ok, i see you wrote that in the post not on the schematic, about the two 5R resistors, no prob with the drawing then, but see below for big prob.

Yes i thought the pot arm was a mistake in the drawing but just in case i though i would mention it anyway.

If you have two 5 ohm resistors in series with the SLA battery then it is very likely that the battery will NEVER charge up all the way, if it charges at all. You can not put that much resistance between the charger and the battery, sorry. That's a total of 10 ohms, and if the battery voltage is 5.9v and the charger is putting out 6v (for example) then the charger current is down to 0.1/10 which is a mere 10ma ! That will never charge an SLA. You have to realize that even a 1 ohm resistor will limit current in that condition to 0.1 amps, which still isnt very good.
Try a fast blow fuse, or a very fast blow fuse instead.

Battery reverse connection protection gets even more interesting

 

[biasottw]

The two 5 ohm resistors are in parallel with each other to create a 2.5 ohm resistor but in series with the battery to prevent a short circuit from ever happening. I am able to charge the battery to the full 6.8V when I attached a DC power supply to the battery. I charged it at constant voltage of 7.5V. As the battery approached completely charged, the current dropped off to only a few dozen milliamps. Now I am trying to get this done with a circuit. I know it is possible to charge a battery in a short time with a large current but eventually that will destroy the battery. I want a full charge in about 8 to 10 hours and be able to be left on the charger almost indefinitely if needed.

 

[MrAl]

Hello again,


Oh ok, that's a little better anyway. If it is fully charging, then what is the problem?

 

[biasottw]

The problem is that I am currently charging it with a DC Power Generator, I can't keep doing it this way. So I am trying to develop a circuit that will do the same thing, but I can't quite get it to work. Hense why I posted on this forum for help.

 

[ericgibbs]

The problem is that I am currently charging it with a DC Power Generator, I can't keep doing it this way. So I am trying to develop a circuit that will do the same thing, but I can't quite get it to work. Hense why I posted on this forum for help.

hi,
In what way does your charger fail to meet your specification.?
Is that the complete circuit in your opening post.?

 

[biasottw]

Yes, that is the complete circuit, minus the battery that has the 2.5 ohms in series with it. The charger works, but just barely. I might get about 6V out of the battery after charging for about 24 hours or more. I need a faster charge. Maybe a circuit that will charge at 1A for a period of time then switch over a constant voltage charge of 7V. But as of right now, I am unsure how or if that circuit is even possible.

 

[ericgibbs]

Yes, that is the complete circuit, minus the battery that has the 2.5 ohms in series with it. The charger works, but just barely. I might get about 6V out of the battery after charging for about 24 hours or more. I need a faster charge. Maybe a circuit that will charge at 1A for a period of time then switch over a constant voltage charge of 7V. But as of right now, I am unsure how or if that circuit is even possible.

I am looking at your circuit with the LTS Sim.
You dont need the 2.5R in the charger path, only the output from the battery.

EDIT:
Two rough simulations of the circuit from the LM317 datasheet for 6V SLA charger and your values.

 

[biasottw]

The battery is sealed in epoxy with the resister so there is no way to remove the 2.5 ohm resister. This is why I am having the problem. This causes normal smart chargers not to work because they can not sense the voltage across the battery. Only what is seen in the diagram can be changed.

 

[ericgibbs]

The battery is sealed in epoxy with the resister so there is no way to remove the 2.5 ohm resister. This is why I am having the problem. This causes normal smart chargers not to work because they can not sense the voltage across the battery. Only what is seen in the diagram can be changed.

I see the resistor are also in the epoxy, you did not make that clear.

EDIT:
this is with the internal 2.5R, using the datasheet version

EDIT2: Note the heating of the LM317

 

[ericgibbs]

hi,
Looking at the requirement I would say the charger has to be smarter than a standard charger.
One way would be limit the charge current as the existing circuit does, but when the charge current falls below a certain level, say 50mA and the charge voltage reaches a certain level, say 6.9V, the charge current is cut to 0mA.

The charger would monitor the voltage, drawing a very low current from the battery and when the battery voltage falls to say 5.4V the recharge starts again.

 

[biasottw]

Thanks for that information. I will try and see if I can't get something like that to work. Again, thanks for your help.

 

[MrAl]

Hi again,

You have to realize that the charging voltage has to be quite a bit higher than the battery voltage in order to charge this thing with a 2.5 ohm resistor in series.
For a 1 amp charge rate and a 6v battery voltage, you have to have 8.5v output from the charger. For 500ma, 7.25v output from the charger, at least.
That means this charger will have to be a little different than most chargers, it will have to know what the output current is too so it can estimate the battery voltage.
The battery voltage is Vout-Iout*2.5, and that requires knowing Vout and Iout too. I am assuming that you dont have access to the internal terminal the resistor is connected to.
Another idea: Buy a new battery without a series resistor in it.