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Battery charging question

Discussion in 'Electronic Basics' started by steamer, Apr 5, 2005.

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  1. steamer

    steamer Guest

    --I've got a decent battery charger but like most of the ones
    I've seen it's designed to be attached to a car battery. I've tried to
    rig up something that would allow me to run it from a wall socket, but
    I don't have it right. What I've done is add a bridge to the output of
    a Variac, so I've got 120vDC, then I've added a big capacitor to get it
    to something smoother. On a 'scope the output is anything but smooth
    and the charger tends to do that annoying autoatic "trip out" at
    anything even a smidge above what the battery needs; i.e. a trickle
    charge is the best I can do. Can someone suggest a remedy? What I'd
    really like is a ripple-free DC output from the Variac. Can this be
    achieved?
    --TIA,
     
  2. Lord Garth

    Lord Garth Guest

    Firstly, you need 12 Volts DC, not 120VDC. Why don't you use a
    transformer, your bridge rectifier and filter cap and an LM7812 thus
    creating a 12VDC supply. Connect this to the charger.
     
  3. James Beck

    James Beck Guest

    First off, you need to stop trying to kill yourself.
    There are easier ways that won't surprise you as much as what you are
    currently doing.

    You need to drop all that variac crap and to one of the following :

    1) Get a spare car battery and use it as a power source, charging as
    needed.

    2) Get one of the 13.8VDC power supplies designed to do what you are
    doing. A little more information about the battery charger would be
    needed to determine what kind of current you will need.

    Jim
     
  4. John Smith

    John Smith Guest

    Not forgetting to use a power transistor to boost the current capability of
    the poor old LM7812.
     
  5. steamer

    steamer Guest

    --OK here's the problem: I've added a bridge to convert the
    output of my variac to DC and then I used a cap to smooth it a bit, but
    according to the oscilloscope the output is anything but "smooth".
    Anyway I'm just trying to use this to juice my battery charger, which
    is designed to run off of a car battery, but it turns out that the
    variac trick ain't working because the charger keeps tripping out; i.e.
    if I try charging a 12v battery it takes *days* to get it up to snuff;
    I talked to the guy at the local hobby shop and he said this is a
    symptom of insufficient power being sent to the charger. I get better
    results using a 15v wallwart.
    --What I need is to be able to charge a couple of different
    batteries, including motorcycle batteries and my 3 remaining 24-v packs
    of NiMH cells. What would be nice is a benchtop power supply that puts
    out 10 to 20 amps of smooth DC, but I haven't been able to find one.
    --Any suggestions appreciated..
     
  6. John Fields

    John Fields Guest

    ---
    I'm kinda confused here. It sounds like you're saying that your
    charger takes 12V from a car battery and then steps it up to
    whatever's needed to charge other batteries.

    If that's true and you're charging the car battery with rectified AC,
    you shouldn't be able to see any ripple _at all_ on the output of the
    bridge, regardless of whether you have a capacitor across it or not.
    Matter of fact, you don't even need the cap, since the battery will
    smooth the voltage at the outout of the bridge better than any cap
    could.

    So, if you're measuring any ripple at all, I suspect a high-resistance
    connection between the VARIAC and the bridge or between the bridge and
    the battery. Also, miswiring the bridge could do the same thing, and
    remember that a VARIAC is an autotransformer, so if you've got the
    output of the charger connected to mains neutral for some reason,
    OUCH!!!
     
  7. jsmith

    jsmith Guest

    You don't really need "smooth DC" to charge a lead acid battery. The
    pulsating voltage from the rectifier is actually preferable although most
    people don't realize it. Battery manufacturers are probably the least well
    informed in the art of battery charging. And since most electrical engineers
    are stupid and believe anything they read are were taught, they continue to
    revert to ancient concepts regarding battery charging just because some
    moron was able to write a text book with misinformation in it. And by the
    way, a fully charged lead acid battery in prime condition is fully charged
    when it reaches 12.6 to 12.8 volts. Anything above that is "surface charge"
    and is worthless. Once a battery is loaded with "load tester" the voltage
    drops immediately to about 11.5 volts and remains there if the battery has
    not suffered too much from sulfation or grid corrosion. OK you dorks and
    PE's who don't agree, fire away!
     
  8. Rich Grise

    Rich Grise Guest

    John, don't feel pregnant. I can't figure out what the OP is trying to
    accomplish either. I'm guessing that he got his hands on a variac, and is
    trying to use it as a 12VDC supply which would emulate a car 12V supply.

    When he says "a cap", that could mean anything. Steamer, you need a big
    bank of big honking capacitors to make a supply like this have a smooth
    output.

    The charger tripping out, and the output of the variac/diode not being
    smooth indicates that you have nowhere near enough filter capacitance.

    Do you know how to read capacitance values? For what it sounds like you're
    trying to do, you need about 10,000 uF of capacitance - that's about the
    size of a Foster's beer can.

    Do you have a proper multimeter? Or preferably two? (or access to a
    second one - I notice you got your hands on a scope.)

    And, if your "charger" is designed to plug into a lighter socket, you'd
    probably be better off setting your supply voltage to something like
    13.5-14 volts, where they typically sit while the engine is running
    and it's on float charge. I'd say a car battery that is showing a
    terminal voltage of 12V is sorely in need of a proper charge.

    Hope This Helps!
    Rich
     
  9. John Fields

    John Fields Guest

    ---
    Hey, I'm all ears... _Why_ is it preferable?
    ---
    ---
    Really? I guess then, that according to you they just happened to
    luck onto the chemistry and how to do it really inefficiently and you
    know how to do it better. Got some examples?
    ---
    ---
    "are were taught"???
    Yeah, just a trypo...
    That is, you're not _really_ a stupid ****, it just _seems_ that you
    are, so far.
    ---
    ---
    If you've got some novel concepts about recharging and you want to
    share them, I'm sure that there are people here who can make your
    wildest dreams come true.

    But...

    In order to prove that you know what you're talking about, you're
    going to have to show how you can charge a sealed lead-acid battery to
    capacity with less loss than anyone else can and still stay within the
    battery manufacturer's charging limits.

    Or...

    Show how you can, with your methodology, extend the life of the
    battery beyond the battery maufacturer's warranty.
    ---
    ---
    You, obviously, are a neophyte who has wandered far from her area of
    expertise. What you need to be concerned with is if two potatoes (if
    you have aspirations which you think will lead you into management)
    will serve about three customers.

    "D'ya want fries with that" is the catch phrase and, then, if the
    answer is affirmative, and you follow with "Do ya wanna supersize
    that?", you've done your job.

    Congratulations...
     
  10. jsmith

    jsmith Guest

    And do any of your "professional" circuit designs really work?
     
  11. John Fields

    John Fields Guest

    ---
    In an effort try to see where you're coming from, I just went back and
    took a look at your posting history and guess what? The kindest thing
    I can say is that you're a fucking idiot.

    From that first "Ohm's Law" abortion to your last pitiful attempt at
    an insult, above, about the only thing you've done right is learned to
    bottom post. So, there may be hope for you yet. We'll see...
     
  12. steamer

    steamer Guest

    : John, don't feel pregnant. I can't figure out what the OP is trying to
    : accomplish either. I'm guessing that he got his hands on a variac, and is
    : trying to use it as a 12VDC supply which would emulate a car 12V supply.
    --Yes, that's it.

    : And, if your "charger" is designed to plug into a lighter socket, you'd
    : probably be better off setting your supply voltage to something like
    : 13.5-14 volts, where they typically sit while the engine is running
    --Yes, I've done this.
     
  13. steamer

    steamer Guest

    --Meanwhile back at the ranch *I'm* still looking for better
    ways to do what I'm trying to do..
     
  14. jsmith

    jsmith Guest

    A TOP POST. . .
    Now now now, Johnny boy don't get yourself in a dither.
     
  15. IMO, that's largely because your requirement was vaguely specified in
    your original post. It wasn't until 4 days later that you became more
    specific, with:
    "What I need is to be able to charge a couple of different batteries,
    including motorcycle batteries and my 3 remaining 24-v packs of NiMH
    cells. What would be nice is a benchtop power supply that puts out 10
    to 20 amps of smooth DC, but I haven't been able to find one."

    Doesn't that imply that, smoothed or pulsed, a 12V ('120V'!) output
    from your Variac will be inadequate? I'm guessing that you really want
    a heavy duty DC supply capable of delivering about 26V, preferably
    switchable to say 15V for charging your motorcycle batteries if they
    are 12V (you still haven't specified that).
     
  16. steamer

    steamer Guest

    : IMO, that's largely because your requirement was vaguely specified in
    : your original post. It wasn't until 4 days later that you became more
    --Yeah, well, I'm still workin' on the "basics"...

    : Doesn't that imply that, smoothed or pulsed, a 12V ('120V'!) output
    : from your Variac will be inadequate? I'm guessing that you really want
    : a heavy duty DC supply capable of delivering about 26V, preferably
    : switchable to say 15V for charging your motorcycle batteries if they
    : are 12V (you still haven't specified that).
    --Yes. I've been turning the dial on the Variac until the
    output, as measured on a voltmeter, has been around 14 to 15v, then
    I've hooked that up to the battery I'm trying to charge. I've *never*
    cranked the dial past this, for fear of damaging the battery, the
    house, myself, etc.
     
  17. John Fields

    John Fields Guest

    There's nothing wrong with that method, but you neeed to monitor a
    couple of things; the voltage across the battery when it's charging
    and the current going into it while it's charging, like this:


    VARIAC FWB
    +--------+ +-----+
    ACHOT>---O--|C|<--O----|~ +|---[AMMETER]--+------< <---+
    | |O| | | | | |
    | |I| | | | | |
    | |L| | | | [VOLTMETER] [BATTERY]
    | | | | | | |
    ACNEUT>--O---+----O----|~ -|--------------+------< <---+
    +--------+ +-----+


    If you don't have a couple of meters which you can set up to take the
    readings simultaneously, and since you have a VARIAC and can use it to
    adjust the current to whatever you want, you can do the readings
    sequentially (without having to tear down and rebuild anything) by
    substituting a resistor for the ammeter (actually, I prefer this
    method) and using the voltmeter to measure the voltage drop across it
    to determine the current going through it, (and into the battery) like
    this, :


    VARIAC FWB
    +--------+ +-----+
    ACHOT>---O--|C|<--O----|~ +|--+--------------------< <---+
    | |O| | | | | |
    | |I| | | |V O<--O----------+ |+
    | |L| | | | |+ [BATTERY]
    | | | | |I O [VOLTMETER] |
    | | | | | | | |
    ACNEUT>--O---+----O----|~ -|--+--[RESISTOR]--+-----< <---+
    +--------+ +-----+


    To size the resistor, decide what value of current you want the
    voltage indicated by the meter to represent, and the use Ohm's law.

    Let's say that you want the meter to read 1 volt when there's one amp
    flowing through the resistor. Then you would write:

    E 1V
    R = --- = ---- = 1 ohm
    I 1A


    You need to be concerned about the wattage, because the higher the
    current the hotter the resistor will get.

    For typical motorcycle batteries I think the initial xcharging current
    is in the range of 2 to 3 amps, so if you were to use a 1 ohm resistor
    you'd have:


    P = I²R + (3A)² * 1R = 9 watts


    being dissipated by the resistor when the current was first set at 3
    amps. The current would decrease as charging progressed, but it would
    be a good idea to have a hefty resistor in there. A really cheap way
    to get 20 watts worth of resistance would be to get four 1 ohm 5 watt
    "cement" resistors and wire then like this:



    +-----+---->A
    | |
    [1R] [1R]
    | |
    + +
    | |
    [1R] [1R]
    | |
    +-----+---->B

    or like this:

    +-----+---->A
    | |
    [1R] [1R]
    | |
    +-----+
    | |
    [1R] [1R]
    | |
    +-----+---->B

    In either case, the resistance from A to B will be one ohm, and either
    array will dissipate 20 W if the resistors are mounted properly. For
    the cheap "cement" resistors, I'm pretty sure just letting them
    radiate into ambient air will be fine.

    Since your NiMH battery packs may want to be charged from a current
    source, the sizing of the resistor may go differently from that for
    lead acid. What can you tell us about the battery packs?
    Also, what can you tell us about your VARIAC, in terms of how much
    current it's rated to put out?


    Also notice that one of the mains inputs to the VARIAC is labelled
    "ACNEUT". That means "AC NEUTRAL" and you need to be ABSOLUTELY,
    POSITIVELY, SURE that the rig is connected to the mains that way. If
    you don't, you could be killed if you came in contact with what you
    though was "ground" while you were grounded.
     
  18. John Fields

    John Fields Guest

    ---
    P.S. After making sure that mains NEUTRAL is really NEUTRAL, and
    before you plug in the VARIAC, you need to make sure that the rig is
    connected to the VARIAC _EXACTLY_ as shown, that the VARIAC's output
    is set to 0V, that the Variac's ON-OFF switch is set to OFF, and that
    the switch on the rig is set to measure current. ( That it's in the
    "I" position). Once that's done, connect the battery to the rig,
    plug in the VARIAC, and if there's no smoke and the lights are still
    on, place the VARIAC's ON-OFF switch to the ON position. If there's
    still no smoke and the lights are still on, SLOWLY rotate the VARIAC's
    output control while watching the voltmeter which is displaying
    current. If it starts to rise, then things are probably working!
    When you get to whatever the initial charging current is, the VARIAC's
    knob should be somewhere between 10 and 15 volts. Now, change the
    range on the voltmeter to read the battery voltage and flick the
    switch on the rig to the "V" position. You should be reading
    somewhere between 10 and 13V if the battery was fully discharged when
    you put it on the rig. If it is, then just monitor the battery voltage
    (checking current occasionally) until the battery's fully charged and
    you're done.

    There's a little more detail to it than that, depending on whether you
    want to let the batteries float after they've been charged, but the
    best place to get that information is from the battery manufacturer's
    website, since they all use slightly different chemistries and their
    terminal voltages and currents are all a little different.

    Once you've finished charging the battery, turn the VARIACS's knob
    down to 0V, turn it OFF, unplug it, and then (and only then)
    disconnect the battery from the rig.

    Good luck.
     
  19. steamer

    steamer Guest

    --Thanks very much; that's going to help me a lot! :)
     
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