Charging A Lead Acid Battery
Dave.H
Jaycar Electronics sells a 5 amp hour 6 volt SLA battery for use in
flashlight lanterns, but as far as I can tell they don't sell the
charger.
Battery
http://www.jaycar.com.au/productResults.asp?FORM=KEYWORD
(CAT. NO. SB2498)
Thanks
Phil Allison
"Dave.H"
** N19309
...... Phil
Ecnerwal
<135de534-4250-4e24...@n20g2000hsh.googlegroups.com>,
"Dave.H" <the1...@googlemail.com> wrote:
> How would I go about building a charger for a lead acid battery.
> Jaycar Electronics sells a 5 amp hour 6 volt SLA battery for use in
> flashlight lanterns, but as far as I can tell they don't sell the
They have a truly irritating number of ads
(site-loads-like-sightless-burrowing-rodent-posteriors), but EDN has an
extensive archive of "design ideas" that can be useful to browse through.
Uses a dedicated IC, and is a 4 cell rather than 3 cell charger:
http://www.edn.com/contents/images/121103di.pdf
Here's one from a bicycle enthusiast - not terribly sophisticated, but should
work:
http://www.myra-simon.com/bike/charger.html
Here's a more complex one from a Ham Radio enthusiast:
http://www.geocities.com/vk3em/sla-charger/PDFs/SLA-ChargerMkIIManualV1-3.PDF
--
Cats, coffee, chocolate...vices to live by
ehsjr
Very simple charger:
-----
+9 ------Vin|LM317|Vout---+
----- |
Adj [2.5R]
| | 1N400x
+----------+---->|---+
| +
[Batt]
|
Gnd --------------------------------+
You need a 9 or 12 volt wall wart capable of
at least 500 mA, an LM317 chip, a heat sink for
the chip, and 2 5 ohm, 1 watt (at least) resistors
in parallel to make the 2.5 ohm resistance, and
a 1N400x diode. Charge for 12-14 hours.
Ed
Ross Herbert
While that might work, it is a constant current source more appropriate to
charging Ni-Cd or Ni-Mh cells, not a lead acid battery.
LA battery is best charged with a constant voltage source or better still, a
regulated 2 or 3 stage charger.
Read Jaycar's own reference sheet on SLA batteries.
http://www.jaycar.com.au/images_uploaded/slabatts.pdf
Ross Herbert
wrote:
:How would I go about building a charger for a lead acid battery.
There are several suitable Jaycar 6V SLA chargers.
MB3516, MB3525 for example.
Dave.H
> There are several suitable Jaycar 6V SLA chargers.
> MB3516, MB3525 for example.
I noticed that, right after I sent the original post. I think the
reason I missed it is because I was only looking in the "Battery
Charger" section. I will use the 6V 500 mA (MB3516) one.
ehsjr
Read Jaycar's own description for the specific battery.
It states: "Charge current 500mA for 10- 14 hrs"
There is no "might" about it (as in your statement
"While that might work"). CC charging *does* work on SLAs.
Note that this is not comparing CC charging to other
methods. It is correcting the "might" to "does".
If you want to talk about "best" chargers, don't snipe
at my post which addressed a "Very simple charger"
It's simple, it works, and it matches Jaycar's description.
Ed
Phil Allison
"ehsjr"
Ross Herbert wrote:
>>
>> While that might work, it is a constant current source more appropriate
>> to
>> charging Ni-Cd or Ni-Mh cells, not a lead acid battery.
>>
>> LA battery is best charged with a constant voltage source or better
>> still, a
>> regulated 2 or 3 stage charger.
>>
>> Read Jaycar's own reference sheet on SLA batteries.
>> http://www.jaycar.com.au/images_uploaded/slabatts.pdf
>
>
> Read Jaycar's own description for the specific battery.
> It states: "Charge current 500mA for 10- 14 hrs"
** Jaycar are a mob of dumb parts grocers - they are experts on NOTHING
!!
> There is no "might" about it (as in your statement
> "While that might work").
** No - it will not work as a good charger should.
It will destroy the battery.
> CC charging *does* work on SLAs.
** It destroys them by overcharging - you ASS.
> If you want to talk about "best" chargers, don't snipe
> at my post which addressed a "Very simple charger"
> It's simple, it works, and it matches Jaycar's description.
** YOU are a good match for the creeps at Jaycar.
In the KNOW NOTHING stakes.
...... Phil
Bill Bowden
>
> - Show quoted text -
Why is your simple charger so complicated? Why not use a 12 volt DC
wall transformer and 13 ohm resistor (5 watt)? You get 540mA when the
battery is low at 5 volts, and about 400mA as the battery voltage
rises to 7 volts.
-Bill
redbelly
If you forget to turn off a charger like that, it will seriously
overcharge a 6V battery! On the other hand Ed's charger will not do
that.
Mark
James Beck
@y5g2000hsf.googlegroups.com>, redbe...@yahoo.com says...
>
> > Why is your simple charger so complicated? Why not use a 12 volt DC
> > wall transformer and 13 ohm resistor (5 watt)? You get 540mA when the
> > battery is low at 5 volts, and about 400mA as the battery voltage
> > rises to 7 volts.
> >
> > -Bill
>
> If you forget to turn off a charger like that, it will seriously
> overcharge a 6V battery! On the other hand Ed's charger will not do
> that
>
It should continue to charge the battery up to the wall wart voltage,
which is still too high if you leave it plugged in too long.
I would, and do, just use a CV source that is set to the float voltage
of the battery. Pick a regulator that has over temp and current
limiting and let it float.
Jim
redbelly
> In article <6194982a-5fc6-406e-b08d-dd15013ba985
>
> > > Why is your simple charger so complicated? Why not use a 12 volt DC
> > > wall transformer and 13 ohm resistor (5 watt)? You get 540mA when the
> > > battery is low at 5 volts, and about 400mA as the battery voltage
> > > rises to 7 volts.
>
> > > -Bill
>
> > If you forget to turn off a charger like that, it will seriously
> > overcharge a 6V battery! On the other hand Ed's charger will not do
> > that
>
> It won't?
> It should continue to charge the battery up to the wall wart voltage,
> which is still too high if you leave it plugged in too long.
> I would, and do, just use a CV source that is set to the float voltage
> of the battery. Pick a regulator that has over temp and current
> limiting and let it float.
>
> Jim
What about all the voltage drops between the wall wart and battery:
1 to 1.5V drop-out voltage of regulator
1.2-1.3V between regulator "out" and "adj" pins (across 2.5R resistor)
0.6-0.7V diode drop across 1N400x
Mark
Bob Monsen
news:135de534-4250-4e24...@n20g2000hsh.googlegroups.com...
If you use a constant current source, make sure it can only rise to the
trickle voltage.
Here is a charger that should work for you:
(view courier)
12V ----o---------------o------.
| | |
.-. | |
| |1k | |
| | | |
'-' 2N2222 | |
| |c |
o------o-----b| |
| | |e |
| | | |c
| | '---b| 2N3055 (may need heat sink)
| | |e
| | |
| | |
| | 2N2222 |
| c| |
z |b-----. --- Your battery
A e| | -
8.1V Zener | | |
| | | |
| | ___ | |
GND------o------o--|___|-o------'
1R 5W
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)
This will supply a constant current of 700mA or so until the battery gets up
to about 6.8V, at which time it'll keep the voltage about constant to
trickle charge the battery.
You can affect the current by choosing a different sense resistor. The 1R
resistor is convienient to obtain, but may not be what you want. Typically,
you charge one of these SLA batteries with 500mA, so a 1.2R resistor might
be better. Also, the zener diode is only required if your power supply isn't
regulated. If it is regulated, you can use a pot and adjust the voltage to
about 8V at the wiper instead.
Hang around while you are charging the battery for the first time. If it is
getting hot, use a smaller current. You don't want those SLA batteries to
start outgassing.
Regards,
Bob Monsen
ehsjr
The circuit you describe would require *regulated* 12 volts,
making it _more_ complicated than the "very simple" circuit.
The very simple circuit is designed to match the charging
requirements in Jaycar's description. Can't do that with the
12 volt DC wall wart and 13 ohm resistor.
Ed
Phil Allison
"ehsjr"
> The very simple circuit is designed to match the charging
> requirements in Jaycar's description.
** Jaycar are a mob of dumb parts grocers -
they are experts on NOTHING !!
> CC charging *does* work on SLAs.
** It destroys them by overcharging - you ASS.
> If you want to talk about "best" chargers, don't snipe
> at my post which addressed a "Very simple charger"
> It's simple, it works, and it matches Jaycar's description.
** YOU are a good match for the creeps at Jaycar.
In the KNOW NOTHING stakes.
...... Phil
Can't do that with the
Ross Herbert
I said "might work" for the following reasons.
Provided that the OP does in fact stick to a "wall wart" transformer limited to
not very much more than 500mA capability, it will probably work ok.
Unfortunately, I don't know if the OP will not use a 12V "wall wart" or indeed
any other higher VA rated transformer which is capable of 1A or more, and this
is where the problem can arise.
Remember, your circuit is an add-on and CC circuits attempt to do do just that,
ie. source a constant current irrespective of load conditions.
The problem with such a simple circuit is that it depends to a large extent on
the specification of the DC input source.
If using a higher VA rated transformer (than a 500mA wall wart) the actual
output voltage of the BR and the output current rating of the transformer will
definitely have a say in how well it will work and whether the battery will be
damaged if left connected too long.
If we accept that the CC circuit you proposed is set at 500mA with such a
transformer, then it will pump this much current through the battery no matter
whether the battery is fully charged or not.
Consider the case where the transformer is capable of putting out 12V at say 2A
from the BR. The CC circuit will successfully limit the charging current to
500mA and the majority of the source voltage will be dropped across the BR and
CC circuits. When the battery is fully charged to say 6.8V or thereabouts, there
is still a surplus of 5.2V available from the BR which will continue to pump
500mA through the battery. It will destroy the battery if left connected.
You might remember those cheap and nasty Arlec chargers
http://cgi.ebay.com.au/ARLEC-BATTERY-CHARGER-6V-AND-12-VOLT-CAR-BATTERY-CHARGE_W0QQitemZ320217446445QQihZ011QQcategoryZ79253QQcmdZViewItem
They had nothing more than a transformer, a rectifier and a thermal current
limiting switch to control the charge rate. They depended upon the fact that
when the battery was fully charged the DC voltage from the rectifier and the
battery voltage were almost equal and thus very little additional current was
pumped into the battery. Unfortunately, depending on the local AC supply voltage
the transformer output voltage was usually several volts higher than the fully
charged battery voltage, so it continued to pump quite a hefty current into the
battery thus destroying it. Your CC circuit is an electronic equivalent of the
thermal current switch in those cheap chargers.
ehsjr
Right! The "very simple charger" does not include
circuitry for protection against overcharge.
Ed
James Beck
@e6g2000prf.googlegroups.com>, redbe...@yahoo.com says...
Build it and try.
As the current drops I'll bet you don't get the drops you think,
especially across the resistor used for the current sense. You are also
assuming that the 9V wall wart is regulated. A cheap unregulated wall
wart that is a "nominal" 9V under X% of load will usually be quite a bit
higher than you expect as the load drops. A CV float charger could be
used with any wall wart that is the V drop of the regulator or higher.
Phil Allison
"ehsjr"
> Right! The "very simple charger" does not include
> circuitry for protection against overcharge.
>
> Ed
>
** What a posturing, fucking arrogant turd you are - Ed.
A very simple charger circuit WILL do just that.
Its called a voltage regulator with current limit.
........ Phil
Bill Bowden
No, it doesn't require a regulated input.
If the transformer is rated at 500mA and 12 volts, the starting
current will be 500mA. As the battery voltage rises, the load
decreases, causing the transformer voltage to rise and the current to
remain fairly constant, or more constant than using a regulated input.
Therefore, an unregulated input is better than the alternative you
suggest.
-Bill
redbelly
> In article <4b78e703-68e3-4822-b318-a1cd9fb8ff04
> @e6g2000prf.googlegroups.com>, redbell...@yahoo.com says...
You make good points, I had been thinking simplistically about the
voltage drops at 500 mA.
Mark
ehsjr
Yes, it does, to get the 540 mA you specified,
and the "about 400 mA" (depending on what you mean
by "about" 400 mA) you specified.
>
> If the transformer is rated at 500mA and 12 volts, the starting
> current will be 500mA.
But you did not specify any current rating.
> As the battery voltage rises, the load
> decreases, causing the transformer voltage to rise and the current to
> remain fairly constant, or more constant than using a regulated input.
> Therefore, an unregulated input is better than the alternative you
> suggest.
You are trying to argue a point not being debated.
There are many chargers that are *better*. No debate
on that.
My objection to your post was that what you claimed was
not correct. You *cannot* specify that the current will
be some particular number without regulation. That
regulation adds complexity to your circuit.
By the way, I assume your question "Why is your simple
charger so complicated?" was facetious, and you were
just busting balls. 6 simple parts does not make a
complicated charger! Your 2 part charger is even
simpler - no debate - and it may be better.
Ed
>
> -Bill
James Beck
3f4365...@i7g2000prf.googlegroups.com>, redbe...@yahoo.com says...
If you rely on the regulator's current and thermal limit to hold the
current under a certain level it will get HOT. I also had a National
part that the info in the datasheet didn't match real life and we blew
out a few wall warts before I caught it.
Jim
ehsjr
I'm amazed at this thread. The "very simple charger"
I diagrammed *stinks* if _used_ as a float charger, which
is what you were de facto discussing when you talked about
forgetting and leaving the batteries on it too long.
If used that way, it *might* prevent damage to the
batteries by the cumulative voltage drop which you and
Mark have discussed, but it is the *wrong* tool for
that job, and the wrong usage of the tool.
I am glad to see that you are taking the discussion to a
float charger. I'm responding below because you mentioned
relying on the chip's thermasl & current limits and that
some wall warts blew.
It would be poor practice to design a float charger that
relied on the regulator's current and thermal limit to
hold the current under a certain level, assuming by
"the regulator's" you mean the IC chip. Those things - the
current and thermal limit - only indirectly hold the
current under a certain level. They are design maximums
for the chip, not for whatever load the chip is feeding.
The circuit design needs to keep the current under the
maximum rating of the chip under worst case conditions.
The designer specifies a heat sink and/or a design that
keeps the chip temperature below the maximum spec. He/she
needs to ensure that any other limitations (eg Vin-Vout
rating) for the chip are adhered to.
A float charger may not need current limiting for normal
conditions, but it does for worst case: a shorted battery.
That's where a float charger without current limiting fails.
Under normal conditions, the battery will limit the current
drawn as the battery voltage increases, and additional
limiting may not be required. Still, you need to consider
the whole circuit. You mentioned that you had some blown wall
warts. If the batteries require more than the wall wart can
deliver, that may be a specification rather than circuit
problem. (ie use a bigger wall wart) If they blew because the
circuit relied on the chip to shut down when it got too hot,
that's a design issue. You indicated an error in the National
datasheet was the cause - do you still have the details? It
could be helpful to know which part and what spec was wrong.
Ed
James Beck
When the datasheet and app note show the part being used in this very
way, I would assume the manufacturer has rated the part as such.
Wouldn't you?
>
> The circuit design needs to keep the current under the
> maximum rating of the chip under worst case conditions.
> The designer specifies a heat sink and/or a design that
> keeps the chip temperature below the maximum spec. He/she
> needs to ensure that any other limitations (eg Vin-Vout
> rating) for the chip are adhered to.
The IC had a built in current limit and, like I mentioned above,
National had several design examples that used over current protection
as an integral part of the design. The problem was that the limit was
wrong on the datasheet and in the app note. 1 - 1.2A was suppose to be
min to max range, it turned out to be 1.9 - 2.2A or some such.
>
> A float charger may not need current limiting for normal
> conditions, but it does for worst case: a shorted battery.
> That's where a float charger without current limiting fails.
> Under normal conditions, the battery will limit the current
> drawn as the battery voltage increases, and additional
> limiting may not be required. Still, you need to consider
> the whole circuit. You mentioned that you had some blown wall
> warts. If the batteries require more than the wall wart can
> deliver, that may be a specification rather than circuit
> problem. (ie use a bigger wall wart) If they blew because the
> circuit relied on the chip to shut down when it got too hot,
> that's a design issue. You indicated an error in the National
> datasheet was the cause - do you still have the details? It
> could be helpful to know which part and what spec was wrong.
>
> Ed
>
The part was a National LM2941CT and the wall warts blew because the
current limit of the IC was wrong on the datasheet. I didn't want to go
to a bigger wall wart, we didn't want to charge at a greater rate
anyway. I'm sure (hope) they have corrected the datasheets by now, but
the funny thing was one of the applications engineers sent me a "fix"
for this problem that entailed a small sense resistor and a transistor
to implement a half assed current regulator and they had that wrong too.
It was set up to pull the feedback pin lower as the current increased,
that would increase the voltage out as the current demand increased
causing the system to slam hard against the + rail. I think they were
relying on the internal current limiting to do its' thing at the level
the datasheet had listed erroneously. I had already fixed the problem
myself. The point was I didn't want to add complexity to the design,
that was why I chose the part in the first place, but I ended up adding
parts anyway.
As a side note, they have changed the datasheet, but there is still one
hold over. In the application notes they still show a "1A Coil Driver"
or some such, that is suppose to limit the current to 1A using the
internal current regulation of the part. Well, it won't do it.
Jim
ehsjr
That would depend on the datasheet, but in general, yes, I would
probably trust it.
Thanks, that's good information to know. Also as a side note,
I've read comments by others who are skeptical, or suspicious
or uncertain (searching for the right term, not sure what
it is) of LDO's in general because they've been burned
by them, too.
Ed
>
>
>