have i killed my batteries?

[emlclcy]

I've a pair of fairly new 250Ah deep cycle batteries that have been in my shed for 6 months. connecting them up to a variable current variable voltage linear psu and setting the current limit to 4A the voltage rose to 17V. leaving it like this over night hoping the voltage would come down the next day it was still sat at 17v, both respond the same

so have I killed them?

[Nickhlx]

Assuming your voltage measuring device is correct - it sounds like they are somewhat sulphated...

 

I would check with a Hydrometer if they are flooded / wet batteries, and try some of the various "de-sulphating" chargers / tricks, to see if any can do anything, but perhaps budget for new ones...

 

Perhaps time to buy e.g. one ( or two) of those Lidl / Aldi 3 or 4 stage chargers for about £14 each ? I use several to look after boat / car and bike batteries which have kept them in great condition and still going after 6+ years

 

Nick

[Flyboy]

I've a pair of fairly new 250Ah deep cycle batteries that have been in my shed for 6 months. connecting them up to a variable current variable voltage linear psu and setting the current limit to 4A the voltage rose to 17V. leaving it like this over night hoping the voltage would come down the next day it was still sat at 17v, both respond the same

so have I killed them?

Why do you think the voltage should come down? The current should drop as the voltage goes up. I suggest you reduce the voltage to 14 and monitor the current.over a few days.

[Nickhlx]

With the battery connected, and the charging voltage at 17 volts ( current limited at 4 amps) what is the actual current flowing into the battery ? I assumed, perhaps incorrectly) that it was very small, but if it is 4 amps, I would think the battery is warm to hot, and gassing profusely - something that "didn't come across as the situation". If the current is low / negligable, my original comments stand.

 

Nick

[Keeping Up]

Are they 6v batteries connected in series to give a nominal 12v, or 12v batteries connected in parallel to give also a nominal 12v, or 12v batteries connected in series to give a nominal 24v ?

[Iain_S]

When I charge using a power supply, the current drops off if the voltage is turned up too high, due to the power supply restricting its output.

 

As Flyby suggests, try turning the volts down a bit.

 

4A would be below 1% pe hour charge rate, so maybe batteries in reasonable state?

[emlclcy]

Are they 6v batteries connected in series to give a nominal 12v, or 12v batteries connected in parallel to give also a nominal 12v, or 12v batteries connected in series to give a nominal 24v ?

this is the battery http://www.cleversolar.co.uk/shop/deep-cycle-leisure-batteries/elecsol-250ah-agm-battery-1006201.html it looks like 6 x 2v cells

the current flowing into the battery is 4A to achive this requires 17v. if I set the voltage to say 14v I think the current wil be next to nothing but I will check tonight.

maybe they are fine and are fully charged?

It says in the data sheet they can be stored for 6 months and I believe they were fully charged then.

Maybe rigging up say a 100watt load and timing the discharge cycle?

[Paul C]

Why did you decide upon 4A?

[nicknorman]

this is the battery http://www.cleversolar.co.uk/shop/deep-cycle-leisure-batteries/elecsol-250ah-agm-battery-1006201.html it looks like 6 x 2v cells

the current flowing into the battery is 4A to achive this requires 17v. if I set the voltage to say 14v I think the current wil be next to nothing but I will check tonight.

maybe they are fine and are fully charged?

It says in the data sheet they can be stored for 6 months and I believe they were fully charged then.

Maybe rigging up say a 100watt load and timing the discharge cycle?

Yes I would discharge them a bit and see what happens. I've found that batteries left on float for long periods can get a bit "lazy" so perhaps the same applies if they are just left. Did you check what the open circuit voltage was before you started charging?

[emlclcy]

Why did you decide upon 4A?

I have a farnell 0-25v 0-5A linear psu with adjustable current limit. 4A seemed a fair current considering the size of the battery to be left on over night

Yes I would discharge them a bit and see what happens. I've found that batteries left on float for long periods can get a bit "lazy" so perhaps the same applies if they are just left. Did you check what the open circuit voltage was before you started charging?

no I didn't, my multimeter was not with me

[Graham.m]

I've a pair of fairly new 250Ah deep cycle batteries that have been in my shed for 6 months. connecting them up to a variable current variable voltage linear psu and setting the current limit to 4A the voltage rose to 17V. leaving it like this over night hoping the voltage would come down the next day it was still sat at 17v, both respond the same

so have I killed them?

 

this is the battery http://www.cleversolar.co.uk/shop/deep-cycle-leisure-batteries/elecsol-250ah-agm-battery-1006201.html it looks like 6 x 2v cells

the current flowing into the battery is 4A to achive this requires 17v. if I set the voltage to say 14v I think the current wil be next to nothing but I will check tonight.

maybe they are fine and are fully charged?

It says in the data sheet they can be stored for 6 months and I believe they were fully charged then.

Maybe rigging up say a 100watt load and timing the discharge cycle?

 

The batteries that you linked to are AGMs, I don't know what their internal discharge rate is but I am wondering what the battery voltage was when you started. I would suggest that 17V is a bit high for AGMs.

 

Think what I would do is disconnect them from the PSU and leave them for 24 hours and then measure the voltages. Unfortunately it is difficult to tell what damage 17V may have done to them. You could do a test by putting them on a 12 Amp discharge for say 5 hours and see what the voltage is then; that would require a load of 1 Ohm capable of handling 144 Watts without burning out maybe something like this http://www.amazon.com/Watts-High-Power-Ceramic-Resistor/dp/B009PIOT8C. Be careful it will get hot.

Edited December 15, 2015 by Graham.m

[nicknorman]

You could do a test by putting them on a 12 Amp discharge for say 5 hours and see what the voltage is then; that would require a load of 1 Ohm capable of handling 144 Watts without burning out maybe something like this http://www.amazon.com/Watts-High-Power-Ceramic-Resistor/dp/B009PIOT8C. Be careful it will get hot.

It won't just get hot, it will probably burn out. Some basic electrical theory for you graham: power = voltage squared / resistance. Voltage? Well not 12v of course, unless the battery is flat/knackered. More like 12.7 to start with. And the resistor is +-5% so worst case 0.95 ohms. So that gives us 170w into a 150w resistor. Yup, that's some pretty good advice as usual. Anyone got a fire extinguisher?

Edited December 15, 2015 by nicknorman

[ROBDEN]

It won't just get hot, it will probably burn out. Some basic electrical theory for you graham: power = voltage squared / resistance. Voltage? Well not 12v of course, unless the battery is flat/knackered. More like 12.7 to start with. And the resistor is +-5% so worst case 0.95 ohms. So that gives us 170w into a 150w resistor. Yup, that's some pretty good advice as usual. Anyone got a fire extinguisher?

Ouch!!!!!!

[alan_fincher]

It won't just get hot, it will probably burn out. Some basic electrical theory for you graham: power = voltage squared / resistance. Voltage? Well not 12v of course, unless the battery is flat/knackered. More like 12.7 to start with. And the resistor is +-5% so worst case 0.95 ohms. So that gives us 170w into a 150w resistor. Yup, that's some pretty good advice as usual. Anyone got a fire extinguisher?

 

Although not ideal, that works out at only about 13% overloaded assuming the worst case that the resistance is at the lower end of its tolerance, (fairly unlikely for a wire wound resistor of this type I would have thought)?

 

Whilst it may get over-hot, or (unlikely) fail, I would have thought there is little need for a fire extinguisher, because there is unlikely to be anything in them to actually sustain a fire?

 

I personally wouldn't do it, but I think your suggested outcome is unlikely.

[nicknorman]

Although not ideal, that works out at only about 13% overloaded assuming the worst case that the resistance is at the lower end of its tolerance, (fairly unlikely for a wire wound resistor of this type I would have thought)?

 

Whilst it may get over-hot, or (unlikely) fail, I would have thought there is little need for a fire extinguisher, because there is unlikely to be anything in them to actually sustain a fire?

 

I personally wouldn't do it, but I think your suggested outcome is unlikely.

Probably, but my point is that circuits / electrical things should be designed taking into account actual values and tolerances. Personally I would be reluctant to run a resistor at its rated output for a prolonged period without carefully checking the specified conditions its rated at (ambient temperature, air movement etc). I certainly would never recommend to intentionally run a component beyond its rated output. Graham has history in this area. Edited December 15, 2015 by nicknorman

[WotEver]

Just connect up a couple of 55W headlamp bulbs for say 6 hours. That's about 8.5A, so in the region of C17.

 

The added bonus is that the shed will be nice and bright and you can have a tidy up

[blackrose]

I've a pair of fairly new 250Ah deep cycle batteries that have been in my shed for 6 months. connecting them up to a variable current variable voltage linear psu and setting the current limit to 4A the voltage rose to 17V. leaving it like this over night hoping the voltage would come down the next day it was still sat at 17v, both respond the same

so have I killed them?

I have to admit I don't understand any of this. Perhaps down to a lack of electrical knowledge.

 

Firstly, why would you leave big, new (and presumably expensive) batteries uncharged in a shed for 6 months, and why not use an automatic 3 stage charger to keep them properly charged in the first place rather than using what seems like a bit of a dodgy and unreliable setup - at least to my untrained eyes?

 

Perhaps there is a point at which thinking one has a good knowledge about these things actually creates more problems that it solves. My last set of domestic batteries lasted 10 years and all I did was keep them regularly topped up charged with a basic 30 amp 3 stage charger when on mains and a 70 amp alternator or generator running the charger, when not.

 

It really doesn't have to be as complicated as this thread would indicate, but of course it can be if that's your thing.

Edited December 15, 2015 by blackrose

[WotEver]

Firstly, why would you leave big, new (and presumably expensive) batteries uncharged in a shed for 6 months...

 

OP did say in a later post...

 

It says in the data sheet they can be stored for 6 months and I believe they were fully charged then.

[blackrose]

 

OP did say in a later post...

 

I'm sure they can be stored for even longer uncharged, but I doubt they'll ever be good as new despite what it says in the spec sheet. My basic knowledge tells me that batteries need to be kept charged for best performance.

[Graham.m]

It won't just get hot, it will probably burn out. Some basic electrical theory for you graham: power = voltage squared / resistance. Voltage? Well not 12v of course, unless the battery is flat/knackered. More like 12.7 to start with. And the resistor is +-5% so worst case 0.95 ohms. So that gives us 170w into a 150w resistor. Yup, that's some pretty good advice as usual. Anyone got a fire extinguisher?

 

Think it is time for you to get into the real world and away from theory

 

DC watts = V x I simple no squaring

 

The battery voltage at rest will be order of 12.7V the battery voltage under load (10A) will be reduced and reducing.

 

The resistor I have pointed to is capable of taking the power and whilst as I noted it will get hot will not burn or catch fire.

 

 

Just connect up a couple of 55W headlamp bulbs for say 6 hours. That's about 8.5A, so in the region of C17.

 

The added bonus is that the shed will be nice and bright and you can have a tidy up

 

Lamps too fragile and too small the heat from the lamps will be concentrated and there is less area to dissipate the heat so the lamps will get physically hotter, plus they need bases and a way of fastening them so that the lamps make a secure connection. The resistor is wire wound on a ceramic base and is about 10 inches long and a diameter of about 3/4 inch, a biggish area to dissipate the heat and designed for that type of use.

[WotEver]

I'm sure they can be stored for even longer uncharged, but I doubt they'll ever be good as new despite what it says in the spec sheet. My basic knowledge tells me that batteries need to be kept charged for best performance.

 

I don't disagree. Why OP did what he did is unknown, we're simply attempting to determine whether or not the batteries are still healthy.

[WotEver]

Lamps too fragile and... {more boring irrelevancies}

 

Lamps are probably something he has kicking around. A large 1r resistor is unlikely to be. Soldering wires to a lamp base is hardly rocket science. The resistor won't give him lots of added light to clean up by.

[nicknorman]

DC watts = V x I simple no squaring .

So do you refute that a formula for DC power is v squared / R? Yes or no please.

 

The voltage will drop slightly under this modest load but the power will not drop to within the limits of the resistor you specified until it reaches 12.25v. That will take a long time unless the batteries are flat/knackered.

 

... and designed for that type of use.

Designed for use at 150w. Not 170w.

[Graham.m]

 

I don't disagree. Why OP did what he did is unknown, we're simply attempting to determine whether or not the batteries are still healthy.

 

 

And that means doing a load test with knowns

[WotEver]

 

Think it is time for you to get into the real world and away from theory

 

DC watts = V x I simple no squaring

 

The battery voltage at rest will be order of 12.7V the battery voltage under load (10A) will be reduced and reducing.

 

I think it's time you stopped giving poor advice.

 

You cannot load a battery with a resistance and then ignore the resistor value and only talk about current. The current is DETERMINED by the resistance.

 

The power is NOT ExI, it's E²/R, which equates to a shade over 161W @ 12.7V.

Edited December 15, 2015 by WotEver