Problems charging a lead-acid battery

[warfreak131]

Hello guys, I have a problem charging a lead-acid battery.

I have a 0-30V 0-3A power supply. If I hook up the power supply directly to the battery (with a resistor and/or diode) and set the voltage to an appropriate level, the LA battery will only draw about 150 mA. After a few minutes though, the battery starts to draw a huge amount of current to the point where it peaks out the power supply. And the power supply drops in voltage trying to deliver so much current. Trying to deliver 3A to a battery definitely isn't safe so I immediately unplug it.

Then I tried to make a more complicated circuit where I placed an appropriate voltage across a resistor. I then placed a unity gain amplifier (through an LM741) across the resistor, and hooked the output to the battery. I placed a diode biased against the positive terminal of the battery to stop current from flowing from the battery to the op-amp. Although, when I take the voltage across the battery while it's connected, it never changes. If it happened to be 6V when I started, it'll never change. However, when I disconnect the battery and verify the voltage from the op-amp, it's 12V.

Can anyone suggest how to fix this?

 

[skeptic2]

Hello guys, I have a problem charging a lead-acid battery.

I have a 0-30V 0-3A power supply. If I hook up the power supply directly to the battery (with a resistor and/or diode) and set the voltage to an appropriate level, the LA battery will only draw about 150 mA. After a few minutes though, the battery starts to draw a huge amount of current to the point where it peaks out the power supply. And the power supply drops in voltage trying to deliver so much current. Trying to deliver 3A to a battery definitely isn't safe so I immediately unplug it.

Is this a car battery, a GelCel or what?
What is its voltage and amp hour rating?
What voltage do you set your power supply to?
What is the resistance and wattage of the resistor?
Does your power supply have a current limit setting? If so, how is it set?

Then I tried to make a more complicated circuit where I placed an appropriate voltage across a resistor. I then placed a unity gain amplifier (through an LM741) across the resistor, and hooked the output to the battery. I placed a diode biased against the positive terminal of the battery to stop current from flowing from the battery to the op-amp. Although, when I take the voltage across the battery while it's connected, it never changes. If it happened to be 6V when I started, it'll never change. However, when I disconnect the battery and verify the voltage from the op-amp, it's 12V.

Can anyone suggest how to fix this?

Can you post a schematic of your circuit?

 

[AlephZero]

If your battery measures 6V with no load, then either it's a 6V battery, or it is beyond repair anyway so don't bother trying to charge it.

What you need is a charger designed for a 6V lead-acid battery. Messing about with a 30V power supply and trying to design your own circuit (and your OP doesn't give much confidence you know how to do that properly) probably won't achieve anything except wrecking the battery and/or the power supply, and/or starting a fire, or even causing an explosion, if you accidentally electrolyze enough hydrogen and oxygen gas in a confined space.

Simple battery chargers only cost 10 or 20 dollars. It's hardly worth trying to make your own.

 

[tfr000]

Op amps really aren't designed for high current stuff - why are you using one?
If you really want to get into charging lead-acid batteries, search around on the web. There are some good technical reports about how to do it properly. Although, like the other posters said, battery chargers are cheap - why bother? It pretty easy to either fry the battery or drain it down so low that it gets damaged.

 

[warfreak131]

Is this a car battery, a GelCel or what?
What is its voltage and amp hour rating?
What voltage do you set your power supply to?
What is the resistance and wattage of the resistor?
Does your power supply have a current limit setting? If so, how is it set?


Can you post a schematic of your circuit?

I don't have the battery or resistor in front of me, so I don't know it, but I can get back to you on that. But the resistor was several kOhms and 1 Watt. But it is a small LA battery probably used to start a motorcycle. On the battery, it lists the charging voltage as anywhere from 13.8-14.5 volts (it is a 12V battery), so I erred on the side of caution and went with 13.8 V. Whenever I used a diode, I added 0.7V to that number.

The power supply does have a current limit, it's set via a knob. My supply is a BK precision 1670A. I have attached a schematic.

Op amps really aren't designed for high current stuff - why are you using one?
If you really want to get into charging lead-acid batteries, search around on the web. There are some good technical reports about how to do it properly. Although, like the other posters said, battery chargers are cheap - why bother? It pretty easy to either fry the battery or drain it down so low that it gets damaged.

I'm just doing this as a learning experience, just to see how I could tackle a problem.

 

[MalcolmB]

http://batteryuniversity.com/learn/article/charging_the_lead_acid_battery

This might help.

 

[tfr000]

Well, a 741 has an output short circuit current of about 25 mA, so that's all you will ever be able to drive into the battery with that circuit. Opamps are generally used to amplify small signals, like audio or RF, usually in a relatively high-impedance environment. Definitely not what you want for charging batteries.
So the power supply has a knob-controlled current limit? Perfect... Turn it off, short the output with a jumper, turn it on and watch the current meter. Set whatever current you want with the knob. Remove the jumper and connect the power supply directly across the battery. Set it to about 14.0V. Turn it on. The current limit will hold the current at whatever you set until the battery is charged, at which time you should see the current drop off. That is, unless the battery is damaged - then it will do one of two things - 1) it won't accept any current 2) it will suck up current forever and never charge up.

 

[warfreak131]

Well, a 741 has an output short circuit current of about 25 mA, so that's all you will ever be able to drive into the battery with that circuit. Opamps are generally used to amplify small signals, like audio or RF, usually in a relatively high-impedance environment. Definitely not what you want for charging batteries.
So the power supply has a knob-controlled current limit? Perfect... Turn it off, short the output with a jumper, turn it on and watch the current meter. Set whatever current you want with the knob. Remove the jumper and connect the power supply directly across the battery. Set it to about 14.0V. Turn it on. The current limit will hold the current at whatever you set until the battery is charged, at which time you should see the current drop off. That is, unless the battery is damaged - then it will do one of two things - 1) it won't accept any current 2) it will suck up current forever and never charge up.

Thank you sir, I appreciate it.

 

[skeptic2]

It sounds to me like you can forget about the resistor, diode and opamp and just connect the power supply, set to 13.8 V and 3 A, directly to the battery (with correct polarity of course). If the battery is still good, it may take a long time to charge, at least overnight but you should be able to see an increase in voltage after a few hours. If you don't, you should have the battery tested.

 

[warfreak131]

I was wondering if there would be any harm in leaving it running at 3A, but when I first tried it out, after about 10 seconds, I could hear some unnerving sounds coming from the battery, so I switched it off.

 

[davenn]

I was wondering if there would be any harm in leaving it running at 3A, but when I first tried it out, after about 10 seconds, I could hear some unnerving sounds coming from the battery, so I switched it off.

did you connect it correctly ?
positive of PSU to positive of battery and negative of PSU to negative of battery

you still haven't told us exactly what sort of battery it is and what its A/hr rating is
its really difficult for anyone to give you accurate help when you can't give info asked for

unless its a very low A/Hr rating, charging it at 13.8V and 3A for a couple of hours is quite ok

I work with sealed lead/acid batteries quite a lot, used for emergency alarm systems on public address amplifiers 2 x 12V 26 A/Hr batteries in series to provide the PA with 24V

These batteries are replaced after 2 years regardless of it they need it or not, the good ones go home for my ham radio backup supply.
SLA batteries want a constant voltage and as they charge up their current drawn will slowly decrease from ~ 3 to 4A down to less than an Amp

cheers
Dave

 

[davenn]

It sounds to me like you can forget about the resistor, diode and opamp and just connect the power supply, set to 13.8 V and 3 A, directly to the battery (with correct polarity of course). If the battery is still good, it may take a long time to charge, at least overnight but you should be able to see an increase in voltage after a few hours. If you don't, you should have the battery tested.

agreed

and as commented earlier by some one if your 12V battery is now only 6V, chances are it's had it
anything below ~ 8-10 V and a 12V SLA battery has pretty much had its life. You may bring it up close to 12V but it won't have any staying power ie. it will loose charge really quickly


cheers
Dave

 

[warfreak131]

did you connect it correctly ?
positive of PSU to positive of battery and negative of PSU to negative of battery

you still haven't told us exactly what sort of battery it is and what its A/hr rating is
its really difficult for anyone to give you accurate help when you can't give info asked for

unless its a very low A/Hr rating, charging it at 13.8V and 3A for a couple of hours is quite ok

I work with sealed lead/acid batteries quite a lot, used for emergency alarm systems on public address amplifiers 2 x 12V 26 A/Hr batteries in series to provide the PA with 24V

These batteries are replaced after 2 years regardless of it they need it or not, the good ones go home for my ham radio backup supply.
SLA batteries want a constant voltage and as they charge up their current drawn will slowly decrease from ~ 3 to 4A down to less than an Amp

cheers
Dave

Hello Dave, sorry for the late reply. I checked the battery. All that it says, aside from things like "Do not dispose of in fire", etc, is 12 VDC Sealed lead-acid battery 5 Ah. It also mentions that during charging, the amount of initial current should not exceed 1.5 A.

 

[hisham.i]

In your connection the resistor is useless, in order to limit the current the resistor should be in series with the battery.
Place a variable resistor with correct power in series with ammeter and the battery, then change the resistance to adjust the charging current.
Also remove the op amp the opamp is not designed for these currents.
Or even you can directly connect the power supply to the terminals of the battery & it will charge.
BUT the best way is to get LA battery charger get cheap one it will work fine.

 

[davenn]

In your connection the resistor is useless, in order to limit the current the resistor should be in series with the battery.
Place a variable resistor with correct power in series with ammeter and the battery, then change the resistance to adjust the charging current.Also remove the op amp the opamp is not designed for these currents.
Or even you can directly connect the power supply to the terminals of the battery & it will charge.
BUT the best way is to get LA battery charger get cheap one it will work fine.

please read ALL the posts, he doesn't need a resistor

he has a variable voltage and current PSU, he can set the current to whatever he likes to limit the current

Hello Dave, sorry for the late reply. I checked the battery. All that it says, aside from things like "Do not dispose of in fire", etc, is 12 VDC Sealed lead-acid battery 5 Ah. It also mentions that during charging, the amount of initial current should not exceed 1.5 A.

OK good
so set the current to 1.5A so that isn't exceeded and set the voltage to ~ 14V ...
the SLA I have i front of me states a cycle voltage of 14.4 to 15V and a standby voltage of 13.5 to 13.8V

you didnt confirm if the battery was only reading 6V ... see my and other earlier posts regarding this. If it is, the battery may well be "stuffed" and any charging you try to do to it is a waste of time

cheers
Dave

 

[warfreak131]

Thank you for the prompt response Dave. I had tried what tfr000 had suggested, where I set the voltage to 14.5, then shorted the outputs and set the current limit to 1.5A. However when I turned it on it did what I mentioned in my first post. It starts out drawing only about 150mA, and it will hover there for a while, go up a bit, go down a bit, then skyrocket and max out the supply. Then it must hit some maximum power output curve, and the voltage will drop. And even if I drop the current limit to decrease the power, the voltage drops along with it. What do you think is happening here? I have another power supply I can try it with though.

The battery seems to more or less retain whatever voltage it was previously at. So just for example, if it is at 6 volts, and I charge it long enough to 7 volts, it will stay at 7 volts for quite a while before any appreciable self-discharge sets in.

 

[tfr000]

The power supply might reduce voltage in order to hold the current at 1.5A. If it does anything more than that (hold the output current at 1.5A at some reduced voltage) then something is wrong - either the power supply isn't working right, or the battery is fried and is acting like a dead short, or something.
If you are indeed measuring 6V out of a 12V lead-acid battery, open circuit, it is fried. You usually can't drain them down to less than 9-10V without damaging them.

 

[warfreak131]

Yeah I think I've come to terms with this stopping in it's tracks at this points. Thanks to everyone and their help, I appreciate it.