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Battery question

Discussion in 'Electronic Basics' started by [email protected], Oct 9, 2007.

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  1. Guest

    Hi I have a battery that is 11.1 volt and the device I'm powering will
     
  2. John Fields

    John Fields Guest

    ---
    It depends on the capacity of your battery and the efficiency of
    your conversion.
     
  3. Need to know the capacity of your battery to estimate runtime.
    Battery technology/discharge rate comes into play as well. If the
    drain (2A) exceeds the recommended values you won't have a very happy
    battery.

    Linear regulators, although cheap, are inefficient when dropping
    significant voltage. Using your 10W device in this manner will result
    in ~12W of wasted power. You'd be much better off running a battery
    pack with 50% less capacity (or running two of the devices in series).

    You *could* use a switching regulator, which would be more efficient
    but significantly more expensive than a linear equivalent. Might make
    sense if you have a really large, high-capacity battery.
     
  4. Jamie

    Jamie Guest

    Due to other things not being equal in the equations like
    inefficient regulators etc..., I'll just sling some barn
    yard numbers.
    Get the know the AH, (Amp hour) of the battery.
    basically the constant amount of energy in one
    hour..
    since you're requiring 2 amps, I guess math
    wise, that would give you more than an hour of
    running time. Also, take into account of the
    lowered voltage your circuit is asking for which
    gives you longer time to work with.

    Don't ask me the % of charge decade expected when the
    rated time&load on the battery has been reached. I've seen
    this figure vary all over the place depending on who
    makes the cells.
     
  5. redbelly

    redbelly Guest

    Huh? You're implying the OP's battery has more than 2 AH of charge.
    Where are you getting that from?

    Mark
     
  6. Jamie

    Jamie Guest

    I wasn't implying anything.
     
  7. Eeyore

    Eeyore Guest

    Without knowing the capacity of the battery in Ah it's not possible to even
    begin any calculation.

    Graham
     
  8. Eeyore

    Eeyore Guest

    You're an utter MORON ! You're just pulling numbers out of your ignorant stupid
    ass as usual.

    Graham
     
  9. Eeyore

    Eeyore Guest

    Do please show your calculations for that.

    Graham
     
  10. John Fields

    John Fields Guest

     
  11. Guest

    It is a 20 Ahr battery. with the 78% efficiency of the regulator my
    calculation is about 10 - 12 hrs. I'm going by an old formula. What
    do you think?
     
  12. default

    default Guest

    Switching regulators are specified (more like hyped) in percent
    efficiency.

    A simple linear will just drop the XS voltage (like you were standing
    there with a rheostat continually adjusting the voltage) in your case,
    six volts at whatever current you use would be turned to heat so that
    would be less than 50% efficient. The regulator would use a
    relatively insignificant amount to perform that function.

    A switching regulator might be able to deliver 90%+ efficiency. The
    regulator pulls current from the battery in 11 volt pulses at some
    relatively high current then uses an inductor and capacitor to smooth
    out the voltage to the load.

    With a switcher, you generate a certain amount of noise, so may
    require RFI filtering. IC Switcher specifications generally have a
    graph that shows efficiency as a function of input voltage and supply
    current - you may only get the 97% they claim at one current level and
    90% when using less current. National has a line of "simple
    switchers" that are old technology but relatively efficient and easy
    to apply without a lot of critical parts. (things like inductor and
    cap quality can affect overall efficiency especially at higher switch
    frequencies)

    It isn't all that hard to design a switching regulator from the ground
    up - particularly when you only want to drop voltage.
     
  13. Assuming a "perfect" linear regulator, a constant 11.1V, and 100%
    battery discharge, you'll get 10 hours from your 20 Ah battery. Your
    device consumes 10 W, and the regulator will dissipate 12.2 W as
    heat. Your linear regulator works out to be about 45% efficient.

    An ideal 90% efficient switching regulator will only dissipate 1.1 W,
    and the effective current draw on your battery reduces to 1 A instead
    of 2 A. The result is (idealized) 20 hours of runtime. This would be
    equivalent to using a 5 V, 40 Ah battery pack. At 66% efficiency, the
    runtime decreases to about 15 hours.

    Real-life runtime will not match these idealized conditions. Battery
    chemistry is likely the most significant factor. What type of battery
    are you running? Lithium-ion, NiCd, Lead acid, etc.?

    These are somewhat helpful with respect to regulators (and
    efficiency):
    http://www.maxim-ic.com/appnotes.cfm/appnote_number/751/
    http://www.rason.org/Projects/swregdes/swregdes.htm
     
  14. Guest

    20 amp hour 11.1 volt 78 % efficiency on the conversion.
     
  15. John Fields

    John Fields Guest

    ---
    Your load is dissipating:


    P2 = I2 E2 = 2A * 5V = 10 watts,


    so if your regulator is 78% efficient, then the battery is
    supplying:


    P2
    P1 = ------ = 12.82 watts
    0.78


    into the input of the regulator.

    With a constant 11.1V out of the battery, then, it will have to
    supply:


    P1 12.8W
    I1 = ---- = ------- = 1.153 amperes
    E1 11.1V


    into the input of the regulator.

    Now, since you have a 20AH battery, its time to discharge (to
    terminal voltage) will be:


    20AH
    t = -------- ~ 17.35 hours.
    1.153A


    However, there's a catch...

    The capacity of the battery will depend on the rate of discharge,
    and if that rate is exceeded capacity will decrease.

    For example, some lead-acid batteries are rated at C/10, so if your
    battery was one of those and it was rated for 20AH at C/10, then
    you'd get 20AH out of it if your current draw was 2 amps for 10
    hours. Or more, if the drain was less.

    Since your battery voltage is 11.1V it's probably not lead-acid, but
    no matter what the chemistry, you have to watch the rate of drain in
    order to get C.
     
  16. Guest

    Dang I wish I went to the college you did. You've told me everything
    I need to know, Thanks a million.
     
  17. default

    default Guest

    If you believe that . . . what's the purpose in posting in the first
    place? Do the math.

    In the real world we have to deal with battery chemistry, temperature,
    regulator efficiency (against a whole slew of parameters) and: oh
    yes, the LOAD - where it drops out, how it drops out, etc. etc.

    GVTEK.EDU?????????

    Learn to think. Knowledge is good, but understanding is better,
     
  18. default

    default Guest

    On Wed, 10 Oct 2007 03:31:48 +0100, Eeyore

    snip
    Easy there. This guy has all the hallmarks of a Google Grooper and his
    question is either posted with very limited knowledge of electronics,
    or he's just a chameleon troll.
     
  19. redbelly

    redbelly Guest

    The battery is to run at 2 amps. You're statement that "math wise,
    that would give you more than an hour of running time" implies a 2 Amp-
    hour capacity. If you're not assuming this, how did you come up with
    the 1 hour run time?

    At any rate, it's a moot point now that the OP has told us it's a 20 A-
    hr battery.

    Regards,

    Mark
     
  20. John Fields

    John Fields Guest

     
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