12v Lead Acid Battery Charger aka I need to stop killing my

[Kevin Walton]

Hi

I have killed a 12v lead acid battery, and would like some help
understanding why, and how to stop it:

I currently have a 'plug in the wall' style 12v lead acid battery
charger that I use to recharge a sealed 7ah 12v gel cell.

The cell is generally used one afternoon a week to start model
aircraft engines (i.e. high discharge rates for short periods) and is
left in a cold garage the rest of the time, usually on charge. After
less than a years use the capacity of the cell is less than half that
of new.

The charger I have just researched a little more, no load it outputs
12v, loaded (ie charging the battery) the voltage is 15v. The current
flow is a constant 400ma (all measured using a electric flight watt
meter) and does not change between a discharged battery and a fully
charged battery. I opened up the wall charger and it is simply a
transformer and a rectifier - nothing else.

I don't understand the theory of above, how come a current is flowing
when attaching the battery 'appears' to raise the voltage across the
charger and the rest voltage of the cell is higher than the rest
voltage of the charger?

So, the reason I am killing the battery is that I routinely over
charge it for long period of time. Once I understand the theory above
I hope to add something to the output of the existing charger to stop
me killing the battery - be that turning it into a float charger by
limiting the voltage (so all I would need to do here is to add a 13.x
v regulator to the output?) or by something more sophisticated.
Options welcomed.

Thanks in advance for the help

Cheers
Kev

 

[Externet]

Hi Kevin.
Lead acid batteries are to be charged with constant voltage. If you are
using that constant current charger, they are being killed.
A 7812 regulator with a couple of 1N4004 diodes in series to its ground
reference pin will raise the regulation to about 13.3 V, good enough.
Miguel

 

[Eric R Snow]

On 14 Mar 2005 06:49:39 -0800, kevin@unseen.org (Kevin Walton) wrote:

Hi

I have killed a 12v lead acid battery, and would like some help
understanding why, and how to stop it:

I currently have a 'plug in the wall' style 12v lead acid battery
charger that I use to recharge a sealed 7ah 12v gel cell.

The cell is generally used one afternoon a week to start model
aircraft engines (i.e. high discharge rates for short periods) and is
left in a cold garage the rest of the time, usually on charge. After
less than a years use the capacity of the cell is less than half that
of new.

The charger I have just researched a little more, no load it outputs
12v, loaded (ie charging the battery) the voltage is 15v. The current
flow is a constant 400ma (all measured using a electric flight watt
meter) and does not change between a discharged battery and a fully
charged battery. I opened up the wall charger and it is simply a
transformer and a rectifier - nothing else.

I don't understand the theory of above, how come a current is flowing
when attaching the battery 'appears' to raise the voltage across the
charger and the rest voltage of the cell is higher than the rest
voltage of the charger?

So, the reason I am killing the battery is that I routinely over
charge it for long period of time. Once I understand the theory above
I hope to add something to the output of the existing charger to stop
me killing the battery - be that turning it into a float charger by
limiting the voltage (so all I would need to do here is to add a 13.x
v regulator to the output?) or by something more sophisticated.
Options welcomed.

Thanks in advance for the help

Cheers
Kev
Greetings Kevin,

Though I'm no great shakes in electronics I think I can answer your
questions. First, since there is no filter on the charger it is
delivering pulsed DC. When measured with your meter it will show the
wrong voltage. Connecting to the battery puts a filter in the circuit
and your voltage reading is now closer to correct. Second, use a
resistor to limit the current. 15 volts should be OK for the battery.
It's the high current that kills the battery. Does the battery maker
have a website or a phone number? If so, maybe you can ask them about
the voltage and current requirements to keep the battery charged
without overcharging.
Cheers,
Eric

 

[John Fields]

On 14 Mar 2005 06:49:39 -0800, kevin@unseen.org (Kevin Walton) wrote:

Hi

I have killed a 12v lead acid battery, and would like some help
understanding why, and how to stop it:

I currently have a 'plug in the wall' style 12v lead acid battery
charger that I use to recharge a sealed 7ah 12v gel cell.

The cell is generally used one afternoon a week to start model
aircraft engines (i.e. high discharge rates for short periods) and is
left in a cold garage the rest of the time, usually on charge. After
less than a years use the capacity of the cell is less than half that
of new.

The charger I have just researched a little more, no load it outputs
12v, loaded (ie charging the battery) the voltage is 15v. The current
flow is a constant 400ma (all measured using a electric flight watt
meter) and does not change between a discharged battery and a fully
charged battery. I opened up the wall charger and it is simply a
transformer and a rectifier - nothing else.

I don't understand the theory of above, how come a current is flowing
when attaching the battery 'appears' to raise the voltage across the
charger and the rest voltage of the cell is higher than the rest
voltage of the charger?

---
If your meter is average-reading and is reading 12VDC out of the
charger when the charger isn't connected to the battery, then the peak
voltage out of the charger will be about 19V even though your meter
only shows 12V. However, as soon as you hook it up to the battery the
battery will soak up the peaks and your meter will then be reading the
steady DC voltage across the battery terminals which, when you
measured it, was 15VDC. If you were to let the battery discharge to
the point where it had about 11V across it, under load, and then you
removed the load and started to charge the battery, the voltage would
jump up a little, maybe to 12V or so at first, but then it would take
much longer to get to its fully charged voltage if it was a good
battery. If you have a discharged battery and and it reads 15V as
soon as you put the charger across it, then its internal resistance
has risen to the point where much of its capacity will be lost.
---

So, the reason I am killing the battery is that I routinely over
charge it for long period of time.

---
Yes.
---

Once I understand the theory above
I hope to add something to the output of the existing charger to stop
me killing the battery - be that turning it into a float charger by
limiting the voltage (so all I would need to do here is to add a 13.x
v regulator to the output?) or by something more sophisticated.
Options welcomed.

one of the easiest ways to do this is shown in the middle illustration
on page 19 at:

http://cache.national.com/ds/LM/LM117.pdf

You should contact the battery manufacturer to find out what the
terminal voltage should be if you're going to float charge the battery
for a long time or if you want to cycle-charge it when you're using it
frequently. They'll be different voltages, and they'll vary a little
between manufacturers and with temperature, so you may want to use a
pot to select the voltage out of the LM317 to suit your situation.

Also, it would be a good idea to heat sink the LM317.
--
John Fields

 

[Kevin Walton]

"Externet" <externet@inorbit.com> wrote in message news:<1110815860.279714.277210@z14g2000cwz.googlegroups.com>...

Hi Kevin.

Hi, Thanks for the reply

Lead acid batteries are to be charged with constant voltage. If you are
using that constant current charger, they are being killed.

I think you missunderstood my posting a little, it is not a 'constant
current charger', it is a charger designed for 12v lead acid
batteries, just that measuring the current flow from the charger
during the charge cycle it is a consistant 400ma.

A 7812 regulator with a couple of 1N4004 diodes in series to its ground
reference pin will raise the regulation to about 13.3 V, good enough.

Sounds like a nice simple option to improve the wall charger that will
also physically fit inside the wall charger.

Thanks
Kev

 

[Kevin Walton]

kevin@unseen.org (Kevin Walton) wrote in message news:<c97617a9.0503140649.4889853d@posting.google.com>...

After
less than a years use the capacity of the cell is less than half that
of new.

Actually, after just doing a discharge test it barely has 10% of its
original capacity, so alhtough this thread is relevant as I need to
not kill the next one, I dont think I will be geting much use out of
this one again unless I can rejuvinate it
(http://www.shaka.com/~kalepa/lowpower.htm)?

Cheers
Kev

 

[Larry Brasfield]

"Kevin Walton" <kevin@unseen.org> wrote in message
news:c97617a9.0503141215.44b92b19@posting.google.com...

"Externet" <externet@inorbit.com> wrote in message
news:<1110815860.279714.277210@z14g2000cwz.googlegroups.com>...
....
Lead acid batteries are to be charged with constant voltage. If you are
using that constant current charger, they are being killed.

I think you missunderstood my posting a little, it is not a 'constant
current charger', it is a charger designed for 12v lead acid
batteries, just that measuring the current flow from the charger
during the charge cycle it is a consistant 400ma.

The current is pulsatile, but its average value
is approximately constant under the conditions
that exist while it is destroying your battery(s).
See the post by Mr. Fields to get a better idea
why that current is approximately constant.

A 7812 regulator with a couple of 1N4004 diodes in series to its ground
reference pin will raise the regulation to about 13.3 V, good enough.

Sounds like a nice simple option to improve the wall charger that will
also physically fit inside the wall charger.

It is a decent solution, but it will dissipate
more power than the diodes and tranformer
presently do, so you need to be sure that
the circuit does not overheat in that location.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[John Fields]

On 14 Mar 2005 12:15:43 -0800, kevin@unseen.org (Kevin Walton) wrote:

"Externet" <externet@inorbit.com> wrote in message news:<1110815860.279714.277210@z14g2000cwz.googlegroups.com>...
Hi Kevin.

Hi, Thanks for the reply

Lead acid batteries are to be charged with constant voltage. If you are
using that constant current charger, they are being killed.

I think you missunderstood my posting a little, it is not a 'constant
current charger', it is a charger designed for 12v lead acid
batteries, just that measuring the current flow from the charger
during the charge cycle it is a consistant 400ma.

---
Yes but, unfortunately, if it's putting 400 mA into the battery when
it's discharged and also putting 400 mA into the battery when the
battery is fully charged then it is, in fact, a constant current
supply!

Lead-acid batteries generally want to be charged using a
constant-voltage supply because as the battery voltage rises when it's
charging, the current being put into the battery tapers off with time
because the difference between the battery voltage and the charger
voltage gets to be smaller and smaller, so the current being forced
into the battery gets lower and lower as the battery voltage goes
higher and higher.

However, I suspect that what's happening in your case is that as the
battery voltage increases when it charges, the diminished load on the
transformer causes its output voltage to rise as well, forcing the
voltage difference between the battery and the transformer to remain
more or less constant, forcing the current to remain more or less
constant. Also, I believe that as the battery charges its internal
resistance decreases somewhat, which would aggravate the effect. With
a regulated (or voltage-limited) supply, though, the change in the
battery's internal resistance would be more than compensated for by
the rise in battery voltage, so eventually the system would come to
rest with only the trickle charge current flowing into the battery.

--
John Fields

 

[Kevin Walton]

John Fields <jfields@austininstruments.com> wrote in message news:<ljlb31pq526n8gkgblfk7d1c5v0bbrvife@4ax.com>...

On 14 Mar 2005 06:49:39 -0800, kevin@unseen.org (Kevin Walton) wrote:

The charger I have just researched a little more, no load it outputs
12v, loaded (ie charging the battery) the voltage is 15v. The current
flow is a constant 400ma (all measured using a electric flight watt
meter) and does not change between a discharged battery and a fully
charged battery. I opened up the wall charger and it is simply a
transformer and a rectifier - nothing else.

I don't understand the theory of above, how come a current is flowing
when attaching the battery 'appears' to raise the voltage across the
charger and the rest voltage of the cell is higher than the rest
voltage of the charger?

If your meter is average-reading and is reading 12VDC out of the
charger when the charger isn't connected to the battery, then the peak
voltage out of the charger will be about 19V even though your meter
only shows 12V. However, as soon as you hook it up to the battery the
battery will soak up the peaks and your meter will then be reading the
steady DC voltage across the battery terminals which, when you
measured it, was 15VDC.

Ahhh, ofcourse, RMS and all that.

If 12v = average, then peak = 12/0.637 = 18.8v, rms = 1.11 x 12 =
13.32v

Not quite the 15v seen but then without putting a scope on it, it does
explain it.

If you were to let the battery discharge to
the point where it had about 11V across it, under load, and then you
removed the load and started to charge the battery, the voltage would
jump up a little, maybe to 12V or so at first, but then it would take
much longer to get to its fully charged voltage if it was a good
battery. If you have a discharged battery and and it reads 15V as
soon as you put the charger across it, then its internal resistance
has risen to the point where much of its capacity will be lost.

Yup, I have now realised that this battery is pretty much dead and
have been experimenting with a battery that does have 50% of it's
capacity left. Charging the second cell (12ah) started with a lower
voltage and a current of around 600mah, the voltage slowly rose
through the charge to 15v and then sat at 400mah current.

Once I understand the theory above
I hope to add something to the output of the existing charger to stop
me killing the battery - be that turning it into a float charger by
limiting the voltage (so all I would need to do here is to add a 13.x
v regulator to the output?) or by something more sophisticated.
Options welcomed.

one of the easiest ways to do this is shown in the middle illustration
on page 19 at:

http://cache.national.com/ds/LM/LM117.pdf

What part does the Rs play in that circuit? I appretiate it sets the
output impedance, which has an effect on the current flowing. If I
know the internal resistance in a battery limits the initial charging
rate, and the voltage is only set at the appropriate float charging
voltage, when the battery is fully charged the voltages will equalise
and the current flow will become zero?

Cheers
Kev

 

[John Fields]

On 15 Mar 2005 04:57:33 -0800, kevin@unseen.org (Kevin Walton) wrote:

John Fields <jfields@austininstruments.com> wrote in message news:<ljlb31pq526n8gkgblfk7d1c5v0bbrvife@4ax.com>...
On 14 Mar 2005 06:49:39 -0800, kevin@unseen.org (Kevin Walton) wrote:

The charger I have just researched a little more, no load it outputs
12v, loaded (ie charging the battery) the voltage is 15v. The current
flow is a constant 400ma (all measured using a electric flight watt
meter) and does not change between a discharged battery and a fully
charged battery. I opened up the wall charger and it is simply a
transformer and a rectifier - nothing else.

I don't understand the theory of above, how come a current is flowing
when attaching the battery 'appears' to raise the voltage across the
charger and the rest voltage of the cell is higher than the rest
voltage of the charger?

If your meter is average-reading and is reading 12VDC out of the
charger when the charger isn't connected to the battery, then the peak
voltage out of the charger will be about 19V even though your meter
only shows 12V. However, as soon as you hook it up to the battery the
battery will soak up the peaks and your meter will then be reading the
steady DC voltage across the battery terminals which, when you
measured it, was 15VDC.

Ahhh, ofcourse, RMS and all that.

If 12v = average, then peak = 12/0.637 = 18.8v, rms = 1.11 x 12 =
13.32v

Not quite the 15v seen but then without putting a scope on it, it does
explain it.

---
What's your point? The 15 volts you saw, you said, was when you read
the DC battery voltage with the charger hooked up to the battery.
That means that was the voltage the battery was limiting the 19V peak
input DC pulses to.
---

If you were to let the battery discharge to
the point where it had about 11V across it, under load, and then you
removed the load and started to charge the battery, the voltage would
jump up a little, maybe to 12V or so at first, but then it would take
much longer to get to its fully charged voltage if it was a good
battery. If you have a discharged battery and and it reads 15V as
soon as you put the charger across it, then its internal resistance
has risen to the point where much of its capacity will be lost.

Yup, I have now realised that this battery is pretty much dead and
have been experimenting with a battery that does have 50% of it's
capacity left. Charging the second cell (12ah) started with a lower
voltage and a current of around 600mah, the voltage slowly rose
through the charge to 15v and then sat at 400mah current.

Once I understand the theory above
I hope to add something to the output of the existing charger to stop
me killing the battery - be that turning it into a float charger by
limiting the voltage (so all I would need to do here is to add a 13.x
v regulator to the output?) or by something more sophisticated.
Options welcomed.

one of the easiest ways to do this is shown in the middle illustration
on page 19 at:

http://cache.national.com/ds/LM/LM117.pdf

What part does the Rs play in that circuit? I appretiate it sets the
output impedance, which has an effect on the current flowing. If I
know the internal resistance in a battery limits the initial charging
rate, and the voltage is only set at the appropriate float charging
voltage, when the battery is fully charged the voltages will equalise
and the current flow will become zero?

---
I don't think it'll ever get to zero, but it'll be low.

It'll also limit the initial current into a discharged battery.

--
John Fields

 

[Lawrence Oravetz]

Kevin Walton formulated on Monday :

Hi

I have killed a 12v lead acid battery, and would like some help
understanding why, and how to stop it:

I currently have a 'plug in the wall' style 12v lead acid battery
charger that I use to recharge a sealed 7ah 12v gel cell.

The cell is generally used one afternoon a week to start model
aircraft engines (i.e. high discharge rates for short periods) and is
left in a cold garage the rest of the time, usually on charge. After
less than a years use the capacity of the cell is less than half that
of new.

The charger I have just researched a little more, no load it outputs
12v, loaded (ie charging the battery) the voltage is 15v. The current
flow is a constant 400ma (all measured using a electric flight watt
meter) and does not change between a discharged battery and a fully
charged battery. I opened up the wall charger and it is simply a
transformer and a rectifier - nothing else.

I don't understand the theory of above, how come a current is flowing
when attaching the battery 'appears' to raise the voltage across the
charger and the rest voltage of the cell is higher than the rest
voltage of the charger?

So, the reason I am killing the battery is that I routinely over
charge it for long period of time. Once I understand the theory above
I hope to add something to the output of the existing charger to stop
me killing the battery - be that turning it into a float charger by
limiting the voltage (so all I would need to do here is to add a 13.x
v regulator to the output?) or by something more sophisticated.
Options welcomed.

Thanks in advance for the help

Cheers
Kev

I keep an automatic motoecycle battery charger in my flite box. never
killed a battery that way

--
This is an automatic signature of MesNews.
Site : http://www.mesnews.net