LM317 Charger Problem

Hey, folks, I know the topic of LM317 chargers for batteries has been
beaten to death but I've got a problem that I can't solve and my
reading of the groups has yet to yield a similar problem; maybe my
searching capability stinks, but who knows.

Anyway, I have a cordless drill that I want to charge in my vehicle
(13.8 volts when running (12V system)) and I built a charger using a
LM317 that has about 7 to 10 Ohms between the Adj and the Out pins,
runs the output through an LED and finally to my driver (marked as a
2.4v device). The output current is about 75 mA, upped from the prior
value of 50 mA (after I found it wouldn't charge up with the 50 mA
output). It still wouldn't charge up (slow running, no power) so I
checked it *again* against the factory wall wart and it draws, even
after being hooked up for 24 hours, around 110 mA (which explains why
the thing lasts for about 2 days off the charger before, even without
use, the battery goes dead). So, it seems to have some internal
leak-down circuit; I say that because I bought 3 of them (they were
cheap at Home Depot) and my friend has one and all four of them go dead
after a couple of days off the charger. Anyway, I decided to try to
build another charging circuit from scratch and test it on the
workbench before I tear into my vehicle to work on the circuit in it.

What I got is this, and it's messing with my head (which isn't well
tuned for electrical stuff anyway); I built the charger according to
the "Typical Applications" section of the National Semi documentation
to be a constant current charger (having been told two years ago when I
first built the charger by someone here in the groups that these types
of devices are current sensitive not voltage sensitive). It should, I
thought, be really simple: Vin = 13.8, tie Adj to the output, tie Out,
via some resistance, to the output, and BAM, battery charger. Not
quite. I read the notes and the calculation for the Iout should be
1.25/R (1.25 nominal voltage diff between Out and Adj, and R being the
resistance between Out and the load. So, Vin =13.8vdc, Adj goes
straight to the load, and Out goes through R to the load. R=10 Ohms,
so Iout should be 1.25 / 10 = 0.125 or 125 mA, and since the driver
eats (sinks) about 110 mA when fully charged the 125 mA should be
enough to charge the device (slowly) and then provide enough juice to
keep it up.

The only difference between this installation and the one in my vehicle
currently is that I didn't run the final output through a LED before
the load.

Finally, the problem; I measured the current through the device with
two different meters and I get about 460 mA...? I double checked my
pin locations on the TO-220 package, double checked my wiring (it's one
IC, one 10 ohm resistor, 3 pieces of wire or so...it can't be too
difficult to get it right...can it?), cleaned the breadboard of other
projects to make sure I'm not seeing another circuit involved, and yet
it's still drawing the same amount of current. I'm afraid to leave it
running for more than a few seconds because I think it might blow that
battery up. I'm going to try a constant voltage setup to see if that
might not fix it.

But, in case that doesn't work, and even if it does, just so I'll know,
is there something blatantly stupid I'm doing here? I even tried
another LM317 and got the same results (with the way I do electronics I
never buy just one of anything <GRIN&gt.

Hope someone can shed some light on this for me. Thank you for your
time and help.

--HC

 

Hey, Garth, thanks for the reply. Since the cordless driver is
rated/marked at 2.4 volts I would think that the vehicle's power (12v
battery) would be more than enough to handle charging the thing. The
only reason I want the driver is to remove my front license plate which
obscures access to my front-mount receiver on my truck (where I hook up
things like tow hooks and trailer balls), so 1) it won't get used very
often and 2) even when it does, it can take an overnight to charge up,
no problem. The problem is that my current charge setup is only
putting out about 75 mA so it actually will take a fully charged driver
and, over about a week, discharge it to where it barely runs (as
opposed to if I just disconnected the driver from any charge source in
which case it would go dead in about 2 days).

Using an inverter is not a solution I'm willing to entertain because 1)
space (where to mount it?), 2) cost, I don't want to spend any money to
buy an inverter that will only be used to charge a 10 dollar driver,
and 3) it's not an elegant solution like the homemade charger would be.
:-/

I'm kind of obsessive about things. It's hell being me. Thanks
again for your help.

--HC

 

[Lord Garth]

<hboothe@gte.net> wrote in message
news:1110952150.900308.215420@g14g2000cwa.googlegroups.com...

Hey, Garth, thanks for the info, again. I got the thing charging now
with the output at 122 mA, and I'm running it all through a green LED
in parallel with a 22 ohm resistor (to keep the LED from burning too
hot) and it seems to be charging the driver fairly rapidly. The time
limit you mention makes sense (since the current draw is 122 mA (give
or take 1 mA) whether it's the drained driver or the charged driver
that I'm currently testing), but I want to avoid that since I intend to
leave it in my vehicle and I want to be able to use it at a moment's
notice. With the darned thing discharging itself after a day or two it
would be useless to have to leave it unplugged after X number of hours
and then plug it back in for Y number of hours; 1) I have three
vehicles (not braggin', and they're not all new) and I use whichever
suits my needs or tastes on any particular day; the one in question
here is my truck and I try not to use it for short trips (4 miles) to
town to check the mail; so it may be days between when I drive it, and
2) even if I only used/had one vehicle I don't want to have to think,
"gee, did I plug in the charger to the driver? Crap, I did, and that
was yesterday....<run run run> <unplug>". :-/

Maybe this stuff about the drivers draining themselves in a day or two
is Ryobi's cheap way of circumventing complicated (I think the stuff
would be) circuitry to check if the battery is full...just use a load
resistor across the battery so it always eats current, that way you
could never over charge it. Maybe? I'm just not very well versed or
knowledgeable about electronics. :-/

I'm testing the rig now with the LED parallelled with the 22 Ohm
resistor, it's across my meter putting out 122 mA, and it's charging
the deader of the two drivers (the one I've had in my truck on that
system for over a year). I had to put a heatsink on the LM317 (I had
several I robbed off an old VCR that I was trashing) and we'll see how
that's doing shortly.

Thank you again, I appreciate your help and thoughts.

--HC

Very good but what is the charging voltage?

A NiCad can develop an internal short, they do this more easily with
elevated temperatures.
That short can make the unit hot. There are methods to restore NiCads but I
can't vouch
for their effectiveness.