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Some quick questions about cells and batteries...

Discussion in 'Electronic Basics' started by phaeton, Feb 21, 2006.

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  1. phaeton

    phaeton Guest

    Just some quick questions to see if I've got the concepts correctly.
    Any responses, flames or pointers welcome:

    1) With similar types, you can generally expect comparative battery
    life to be proportional to the physical volume of material- i.e. twelve
    D-cells can continuously supply 18V into a load longer than two 9V
    batteries could, right?

    2) Charging a sealed lead-acid battery is about as complicated as
    connecting it to another current source and slowly pulling electrons
    through it? In other words, *could* you safely charge a 12V motorcycle
    or car or similar battery with an AC-to-DC wall adaptor and a simple
    voltage divider circuit?

    3) The same can be said about NiMH or NiCd batteries.

    4) Do connecting cells in series add voltages but keep current the
    same, connecting cells in parallel add current but keep voltages the
    same?

    5) Putting cells in parallel also increases the overall physical volume
    of material. Thus, if you connect 10 ordinary D-cells in parallel you
    still get ~1.5v but it will last roughly 10 times longer than a single
    ordinary D-cell driving the same load. True, False or Maybe?

    Thanks for any and all!
     
  2. phaeton

    phaeton Guest

    Thanks Bob.

    I'm aware of the 'voltage depletion' of batteries, and in fact that's
    what brings me to this. I built a small LM386-based guitar amplifier
    that i've been powering with a 9V battery. All is well if the battery
    is fairly healthy, but after about 30 minutes or so i've 'worn it down'
    to about 7V. That's about where (i assume) it starts affecting
    transistor bias in the chipamp. I get distortion and in a short time
    it will simply bias itself off. (No it's not overheating. it is a
    diminished battery).

    If i turn it off and let it sit for awhile (i assume) the chemical
    reaction has time to bolster the battery back up to about 9V or so and
    I can go at it again, for another 30 minutes. So I was figuring that
    if I were to power this with multiple batteries, i would be able to
    expect more time out of it- and an hour is about as long as I want to
    listen to a 500mW amp. ;)

    (more later, gotta work)
     
  3. Bob Myers

    Bob Myers Guest

    In general, yes, but...where you went wrong in the above is
    that "continuously supply 18V into a load" bit. You're not
    WAY wrong, but I am worried that you might have this
    notion that batteries are truly decent voltage sources (i.e., they
    will maintain their "labeled" voltage until they're out of
    juice, and then just shut off) when in fact they're not all THAT
    great. Batteries/cells will in general show a slow but steady
    decrease in output voltage at a given load, and for that
    matter even the initial voltage of that curve is subject to
    some variation due to their internal resistance. With
    that qualification, though, the above isn't too bad. But for
    example - NiCd cells are supposedly 1.2V each, but that's
    only at full charge; for most of the discharge cycle, you'll
    see closer to 1.1V or under. NiCds have a pretty flat curve,
    but it doesn't just stay at 1.2V for the full cycle and then shut
    down. (This, by the way, brings up a problem with simple
    battery-operated equipment which uses voltage, and ONLY
    voltage, to determine the remaining charge in a battery - a
    problem which contributed to the myth of NiCd "memory."
    The real problem was that NiCd cells suffer from a "voltage
    depression" phenomenon - leave them on the charger a long
    time, and they OVERcharge and the cell voltage actually
    drops off from what it "should" be. All of the expected
    energy or charge is still in there, it's just available at a lower
    voltage - but this lead some equipment to declare these types
    as "out of charge" when in fact most of the stored charge was
    still there!)
    The way most batteries/cells are "supposed" to be charged
    is via a current source limited to a given percentage of the
    specified maximum discharge current (0.1C is a good starting
    point for a lot of technologies), and maintaining that current
    limit UNTIL a given cell voltage is reached, at which point it
    is generally a good idea to switch the current level to a
    "trickle" (0.01C, perhaps, but in both of these it's a really good
    idea to read the specifics for the technology/type in question)
    simply to maintain the thing in a state of "full charge." Too much
    charging current and you run the risk of overheating and/or
    venting (which is generally a Bad Thing); too little isn't
    bad, generally, from a reliability point of view, but most often
    you DO want the thing to charge within some reasonable
    time period. "Quick chargers" pile more charge in at
    once (i.e., higher charging current), but have to very carefully
    monitor the state of the battery to avoid any problems.
    Yes, again to a very rough approximation. Connecting cells
    in series does result in a greater voltage (this is what a simple
    "battery" generally is - e.g., a 9V battery is a stack of 6 1.5V
    cells), but you also, don't forget, are also putting the resistances
    of those cells in series as well. These are NOT little voltage
    sources, quite.
    Yes, this one is pretty close. Of course, putting the cells in
    series ALSO "increases the overall physical volume of material,"
    so you really DO have more energy available in that case as well.
    It's just a matter of how (and how effectively) you can extract
    that energy and do something useful with it.
     
  4. Dan H

    Dan H Guest

    CHARGING BATTERIES IS MORE COMPLICATED. DIFFERENT TYPES REQUIRE
    DIFFERENT METHODS
    DAN H
     
  5. Phil Allison

    Phil Allison Guest

    "phaeton"

    ** Slowly being the operative word.


    ** Again - only very slowly.

    Takes days that way.


    ** Slow as hell, but.




    .......... Phil
     
  6. Yes, adding current capacity (by putting more batteries in parallel, or
    using bigger batteries) will help.

    A concept to understand is "energy capacity", measured in milliamp-hours
    (mAh). Roughly speaking, a battery rated at (say) 600mAh will be able to
    produce 600mA for one hour, or 1mA for 600 hours, before its voltage starts
    falling rapidly. Take a look at a battery manufacturer site, like
    www.duracell.com, to see ratings. But understand that it's a very rough
    measure, because it varies depending on how much current you draw. The
    manufacturer sites give lots of graphs showing how the battery will
    discharge over time with various different loads.

    Most mfr's 9V alkaline batteries are rated 625mAh, so if you're wearing it
    down in 30 minutes, you're drawing roughly 1A. That is counterintuitive,
    because I don't think you can manage to squeeze 1A through an LM386 without
    it burning up. I wonder if there is a problem in your circuit somewhere?

    Also, as was already pointed out, batteries have internal resistance; the
    smaller the battery (physically) the more resistance. When you draw current
    from the battery, that internal resistance causes the voltage at the battery
    terminals to drop (Ohm's Law), and it also causes the battery itself to
    heat. That means that you need to use a battery that is large enough (= has
    small enough internal impedance) to supply the needed current without
    wasting too much energy in internal heating. A 9V battery definitely can't
    supply 1A for very long. That may be part of what's causing the "recovery"
    phenomenon you mention.
     
  7. NoThanks

    NoThanks Guest

    Assuming equivalent construction and equivalent chemistries, I suppose
    thats roughly correct.
    Bad idea. Very bad idea. Every battery chemistry requires its own unique
    manner of charging (and sensing when a charge is finished). Lead acid
    batteries should be typically charged at a constant current until they
    reach a certain voltage, and then charged at constant voltage there-after.
    There's a current limit and a voltage limit when charging lead acid
    batteries, at no time should either be exceeded.

    Two problems with what you describe. First, a wall wart with a resistor
    does not make for a current source, it makes for a voltage source with
    source resistance. Second, the voltage of an unregulated wallwart can
    vary significantly with load. When left open circuit, a 12V wallwart
    might actually be putting out 17V. Lead acid battery resistance increases
    as it becomes charged, so a fully charged battery will appear nearly open
    circuit, and your voltage dropping resistor will cease to work well. This
    will put 17V across the battery which is quite bad for it.
    NiMH and NiCd batteries can probably be safely charged at very low
    currents (1/20th of their capacity?), but if you charge them rapidly you
    better have a much more sophisticated charging scheme or you will risk
    damaging the cells and/or creating a fire.
    Connecting cells in series cause the voltages to increase, but current
    capability will be proportional to the worst of the cells. In theory
    connecting cells in parallel should retain the same voltage but have the
    sum of the current (and charge capacity). There is a big caveat here...
    Putting cells in parallel is not a good idea, because if they do not
    discharge equally, one cell will be trying to charge the other. If one
    cell outright fails (such as a dead short or resistive short -- not
    uncommon) this can lead to very unfortunate consequences.
    In theory, yes, if the cells are perfectly matched, you'd get 10x. In
    reality, it may well be less (or if one cell fails prematurely, all may
    die).
     
  8. phaeton

    phaeton Guest

    4) Do connecting cells in series add voltages but keep current the
    -Connecting cells in series cause the voltages to increase, but current
    -capability will be proportional to the worst of the cells. In theory
    -connecting cells in parallel should retain the same voltage but have
    the
    -sum of the current (and charge capacity). There is a big caveat
    here... -
    -Putting cells in parallel is not a good idea, because if they do not
    -discharge equally, one cell will be trying to charge the other. If
    one
    -cell outright fails (such as a dead short or resistive short -- not
    -uncommon) this can lead to very unfortunate consequences.
    -In theory, yes, if the cells are perfectly matched, you'd get 10x. In
    -reality, it may well be less (or if one cell fails prematurely, all
    may
    -die).

    Would any clever diode networks alleviate you from one dead cell taking
    down the rest?
     
  9. Alan B

    Alan B Guest

    A process called "equalization." I differ here with your idea that the
    current will be constant; rather I think the proper technique is to keep
    the voltage high, while the current will gradually drop off. Again, as you
    say, different vendors may suggest different methods.
    A process called "float."
    Precisely. Much more complicated than "connecting to another current
    source and pulling electrons through." A true statement in an extremely
    general way, but lead and acid are rather harmful substances on their own,
    and when combined create quite a volatile mixture that requires proper care
    at all times.
     
  10. Alan B

    Alan B Guest

    Adding diodes adds voltage drop. With a typical cell (of any chemical
    construction) being <=~2.0VDC, there isn't much room for voltage drop.
    Better to monitor battery voltage and take quick action if it falls out of
    spec.
     
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