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LM317 Charger Problem

Discussion in 'Electronic Basics' started by [email protected], Mar 16, 2005.

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  1. Guest

    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>).

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

    --HC
     
  2. Lord Garth

    Lord Garth Guest

    You need a constant current at more than 12 volts to charge the battery.
    The
    LM317 will lose a couple of volts across itself and that alone puts you
    below
    what is needed.

    You need to boost the voltage first and then regulate the current. There
    are
    many IC's that can boost the applied voltage but their output current will
    be limited.
    I doubt you'll achieve a rapid charge though you will probably be able to
    trickle charge.
    That could take about 10 hours which I'm sure is unacceptable.

    How about going the easy way and get an inverter...this will create 120 VAC
    and you
    can use your home charger to do the job. This is not efficient so watch the
    drain on
    your vehicle battery.
     
  3. Guest

    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
     
  4. Guest

    Um, Hello, my name is HC and I'm an Idiot. <sigh>

    If you hook up the LM317 just like the technotes document from National
    says to, running Vin to 13.8 volts, and Out to the load, but you
    MISPLACE the wire on the breadboard to the Adj pin, so that the
    connection that is supposed to exist between the Adj pin and the load
    does NOT exist, it runs basically straight 13.8 volts (Vin voltage) to
    the load. :-/ I went to start on my constant voltage regulator and
    was pulling wires off the breadboard and saw that I had missed placing
    the Adj wire connection in the right row, it was one row off...and I
    had not seen it. I hooked it up right and it's putting out 120 mA like
    it should. Now the question is whether or not that'll burn that
    battery out because 120 mA is more than it draws at idle...but at least
    I'm now in the right range.

    Sorry folks, sometimes I just screw stuff up. :)

    --HC
     
  5. Guest

    Now that I've (hopefully) eradicated sheer stupidity from my project I
    have it charging the driver but the problem is that it's putting 120 mA
    across the driver even though the driver is fully charged (the same
    driver on the factory wall wart only draws 110 mA). I'm afraid that
    difference may mean that I'm going to boil/burn/ruin the battery.
    Plus, I have had to put a heatsink on the LM317 because it's puttin'
    off some heat (enough that it's painful to touch it). Can anyone shed
    some light on what I'm doing? Am I doing (yet again) something stupid?

    The constant voltage regulator tried to push 260 mA over the same
    driver and got hotter than all git out, too. :-/

    Thanks, all.

    --HC
     
  6. Lord Garth

    Lord Garth Guest

    For some reason I got the idea that this was a 12V driver. Sorry!

    Is this driver using sealed in rechargeable C cells? Perhaps they are
    sub C cells. In any case, you still need a greater voltage than the cells
    in order to charge them. Based on the voltage, these are NiCad
    batteries. The voltage needed is related to temperature but something
    close to 1.4 volts per cell at about 68 degrees F. Your two cell screw
    driver is going to need 2.8 volts to do the job. the required current
    should be one tenth the cell capacity for no more than 16 hours at a
    time. More than this will cause the cells to heat and reduce their
    capacity.
    The lesson is to charge it and then don't charge again until it is empty.

    A high capacity sub C cell might be rated at 2Ahr but this is a guess.
    1.6AHr might be more typical. One tenth of this 160mA to 200mA.
    The easiest thing I can think of is to get a mobile cell phone charger
    and modify it. I got my Nokia mobile charger with a new phone body
    for $5 at the T-Mobile store...can't go wrong at that price!

    The NiMH battery in the phone is a higher voltage than you need but
    a few series diodes will reduce that to a suitable value or better, use
    a low dropout regulator to precisly set 2.8 volts. Measure the current with
    your discharged screwdriver connected to the modified phone charger.
    If it falls within the range above, happy days! if it is too high, control
    it
    with a resistor ( but watch for the voltage drop ) or shorten the recharge
    time.
    Just remember that a hot battery isn't a happy battery.
     
  7. Lord Garth

    Lord Garth Guest

    Good, now knock the voltage down to 2.8V and you are good to go!
    Yeah, I have seen 1200mAH batteries too so 120mA is good .1C rate
    to charge your batteries. Just keep that time limit in mind, 16 hrs max
    or stop when the cell warms up.
     
  8. Guest

    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
     
  9. Lord Garth

    Lord Garth Guest

    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.
     
  10. Guest

    Hey, Garth, the voltage across the positive and negative leads of the
    charger (while actually charging) is 2.6v (which I think means that I'm
    sinking all available voltage EXCEPT 2.6 volts. The voltage across the
    leads of the device with the driver NOT attached, so it's open circuit
    potential is 13.6 (approx.) volts. I ran the unit for a while last
    night without the housing of the driver getting hot or even warm, but
    the LM317 sure gets hot. I set one device up as a constant voltage
    device, as I think I mentioned in one of my posts and it got hot, too.
    I still don't know if I should be setting this thing up as a constant
    voltage or a constant current device, or if I should be looking for
    some different technology to do this. If it's constant current, won't
    it do just that, remain constant so, say the driver's battery is
    getting fully charged so it's current tries to drop (it's drawing less
    current), won't the setup of the LM317 as a constant current device
    just increase voltage to still manage the same amount of current
    output? With the constant voltage device it would seem like it
    couldn't increase the current because it can't increase the voltage, so
    that might be better, but it was getting hot, too.

    The heat tells me that maybe the problem is inherent to this device
    trying to dissipate/sink the difference between a 13.8v car system and
    a 2.4v driver. It would help if I knew a lot more about electronics,
    and it's frustrating not knowing this stuff any better than I do; I
    fear that my ingnorance may be preventing me from seeing or
    understanding some simple thing that would make this clear.

    Anyway, thanks for your help and time.

    --HC
     
  11. Lord Garth

    Lord Garth Guest

    The chargers output voltage is not 'stiff' so you need to measure
    close to 2.4 volts open circuit. That means without anything attached
    to the output of the voltage regulator. Given an input of 13.8V and an
    output of 2.4V, that means the 317 is going to loose 11.4 volts across
    itself. It will need a heatsink since that is quit a bit of difference.
    Make it
    reasonably large.

    Now that you have created the proper voltage, you can regulate the current.
    The second regulator will loose some voltage across itself as well so be
    prepared to go back and adjust the voltage regulator to make up for the loss
    on the current regulator. Ultimately, you should read the output of the
    open circuit current regulator to be close to 2.4 volts and if connected to
    a battery, the current should be limited to 120mA for a .1C assuming 1200mAH
    sub C cells.

    You need both conditions to do this properly.

    Too high of a voltage applied across the battery will damage that battery.
    The way one sets up bench power supply is to adjust the open circuit voltage
    and then set the current limit to zero, short the supply and SLOWLY turn up
    the
    current limit until the desired level is reached.
     
  12. HC

    HC Guest

    Hey, Garth, what I've got right now is the circuit as I've described it
    with decent-sized heatsink on it and it's working well. I'm leaving it
    on the bench setup running almost 24x7 (well, as much 24x7 as there is
    in between yesterday sometime and today <grin>) with the only exception
    of that I shut it off when I leave the house (since I work out of the
    house that's not much down time). So far it's charged the battery
    quickly and yet the battery/housing doesn't seem to be getting warm
    (which it would, I think, if it was hurting it). The LM317 get's warm,
    but with the larger heatsink it's not unpleasant to touch (like it was
    by itself).

    If I understand your most recent post, though, you're suggesting that I
    run two LM317's in series? One as a constant voltage supply and one as
    a constant current supply?

    Thank you again for your time and help.

    --HC
     
  13. Lord Garth

    Lord Garth Guest

    Sorry, I'm running today...

    Yes, that is what I'm suggesting...
     
  14. HC

    HC Guest

    Hmm, okay, I'll try that. I won't be able to do so for a couple of
    days (other problems/projects), but I'll let you know when I do.

    Thanks again for your time and help.

    --HC
     
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