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Discussion in 'Electronic Design' started by john, Mar 29, 2006.

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  1. john

    john Guest


    I have to design a voltage source ( +/- 18volts, 13A ) using lead acid
    batteries. My circuit draw is 700mA and requires plus minus 18 volts to
    operate efficiently. I need batteries that can atleast run for 10 to 12
    hours before the voltage drops to +/- 17 volts. I am thinking of adding
    three 6 volts, 13AH ( rated for 20AH ) batteries in series to produce
    +18 volts and adding three 6 volts to generate -18 volts. I choose the
    battery ( BP13-6V ),

    Can anybody advice me that am I doing the right thing that will these
    six batteries last for 10 to 12 hours maintaining +/- 18 volts @

  2. Tim Wescott

    Tim Wescott Guest

    When you say 18V 13A, do you mean 13AH? Or do you mean that your
    circuit usually pulls 700mA but sometimes pulls 13A?

    Everyone rounds cell voltages shamelessly. A '2V' lead-acid battery can
    be expected to have a cell voltage around 1.8V at the end of its useful
    charge -- this would translate to around 16V in your application. The
    only definitive way is to find the discharge curve for the battery and
    check it against your application -- if you can't find a discharge curve
    then you should either distrust that battery, or you should find another
    battery manufacturer that does support their product.

    Switching regulators are so easy to design these days, and 12V
    deep-discharge batteries so easy to obtain, why don't you use a +/- 18V
    switching regulator powered off of 12V? Design (or buy) the regulator
    to work from 11V to 15V to accommodate the full charge to discharge
    characteristic, and just go to town. You'll have to do some calculation
    to get the actual current draw as a function of battery voltage, but you
    may find that your total package is smaller and possibly even less


    Tim Wescott
    Wescott Design Services

    Posting from Google? See
  3. john

    john Guest


    Its 13AH.
  4. Ian Stirling

    Ian Stirling Guest

    Possibly not.
    Have you considered one 12V battery, and a DC-DC converter to generate

    Are you making one or many?
    Is volume/weight/cost most important?
    How many cycles do you want the abtteries to last (cycles).
  5. john

    john Guest


    No, I did not think about DC-DC converter. Would u advice me that how
    will it work. What DC to DC converter I need?

  6. Ian Stirling

    Ian Stirling Guest

    It depends on the following.
    It may be easier especially if you're making many, or using it lots, as
    you have fewer batteries to replace, potentially less often, and the
    charging circuitry is much more simple.
  7. Guest

    Hey, I could use one of dem there DC-DC converters too.

    I've got an ancient laptop, powered by a 12V jumpstart battery, plus
    110VAC inverter, plus the stepdown power supply that comes with the

    I could make this more efficient, with just a DC-DC converter to
    convert the 12V jumpstarter to 18VDC.

    My laptop wants 2.7A at 18VDC. (almost 50W).

    I went to this website:

    and I see there are CUK converters, Boost, Buck Boost, etc.

    My intuition tells me there are trade-offs in efficiency, ease of
    construction, cost, consistency and durability between each of these

    Any suggestions guys?
  8. Robert Baer

    Robert Baer Guest

    Lessee...700mA times 12 hours equals 8.4 Amp-Hours. Multiply that by
    20 (assuming C/20) for a battery rating of 168 AH.
    That rate will allow approximately the voltage you specified near the
    end time period of 12-15 hours.
    The Interstate GC6V200 (6V) is rated at 170AH, their SG-8D (12V) is
    rated at 190AH, their SG-4 (12V) is rated at 164AH, and their SG-4D
    (12V) is rated at 154AH. They have a UPS rated battery, the UPS6-600
    (6V) rated at 180AH.
    Other manufacturers have equivalents; these are heavy duty batteries.
    Be advised that right after charging,the battery will be much higher
    than 18V for the first ten seconds.
    All of this assumes constant current discharge at 25C / 77F.
    Be advised that temperature will be a significant factor in terminal
    voltage and life.
  9. Ross Herbert

    Ross Herbert Guest

    My estimation is that using 20Ah batteries as you suggest (3 x 6V in 2
    banks to give 36V) with a drain of 700mA will give you around 13 - 14
    hours of operation with not less than 80% discharge. So yes, this is
    possible, even if expensive for the initial purchase of 6 batteries.
  10. john wrote...
    You have a 25-watt bipolar-power-supply application, which
    is in the easy sweet spot for a common flyback converter
    with a transformer and bridge-rectifier output. There are
    ICs from LTC, NSC, TI, etc., that make this task easy. I
    generally design and wind my own ferrite-core transformers.
    That might slow you down a bit, but at this power level and
    say 100kHz the transformer will not have very many turns.
  11. Ken Smith

    Ken Smith Guest

    I'd suggest that he look at using 2 DC-DC converters. A very simple
    booster will do the +18V. This would only use one "off the shelf"
    inductor. I'd suggest the LT1270, MUR1660 and a Coilcraft 5022 as a first

    The OP seemed to imply that the -18V load may be different than the +18V
    load. Since we already have the LT1270 etc, I'll suggest a 2 inductor
    "down pumper" Cuk converter for this one.

    I'd put a fuse in the (+) input wire in any case.

    Yes I know the LT1270 is overkill but they are fairly hard to break.
  12. Robert Baer

    Robert Baer Guest

    If you look at the discharge curves given by major battery makers,
    you will find that my answer fits the curve.
  13. john

    john Guest

    thanks for ur reply. But I am using BP13-6V. and it does not have high
    capacity as GC6V200. Its rated as 13AH battery over 20AH. First I do
    not understand this 20AH thing? plus would you think that this battery
    will last that long like 12 hours, if I draw 700 mA from it?

  14. John Fields

    John Fields Guest

    Learn to post properly. From:

    "Summarize what you're following up.

    When you click "Reply" under "show options" to follow up an existing
    article, Google Groups includes the full article in quotes, with the
    cursor at the top of the article. Tempting though it is to just
    typing your message, please STOP and do two things first.
    Look at the quoted text and remove parts that are irrelevant.
    Then, go to the BOTTOM of the article and start typing there.
    Doing this makes it much easier for your readers to get through your
    post. They'll have a reminder of the relevant text before your
    comment, but won't have to re-read the entire article.
    And if your reply appears on a site before the original article
    they'll get the gist of what you're talking about."
  15. john

    john Guest

    thanks for ur reply. But I am using BP13-6V. and it does not have hig
    capacity as GC6V200. Its rated as 13AH battery over 20AH. First I d
    not understand this 20AH thing? plus would you think that this batter
    will last that long like 12 hours, if I draw 700 mA from it

  16. RHRRC

    RHRRC Guest

    The amp-hour capacity you can get from a battery, in practice, depends
    to an extent on how fast it is discharged.
    A high discharge rate (i.e. high current) will give a lower capacity
    than a low discharge rate (i.e. low current).
    To give some standardisation to the amp-hour rating of batteries they
    are rated to the capacity (amp hours) obtained when they are discharged
    to flat in a given time.
    For lead acid batteries of the type you are considering this time is 20
    hours (for NiCads it is 5 hrs for example).
    Your 13Ah battery will give 13Ah if it is discharged at (13/20=) 650mA.
    At a discharge current of 2amps your fully charged 13Ah battery will
    *not* last (13/2=) 6.5hrs - it will be a shorter time than this. You
    will have to look at the manufacturers data for how long it will last.

    'C' is merely a short-hand way to refer to the capacity of a battery at
    its declared (20hr) discharge rate.
  17. John Fields

    John Fields Guest

    On 31 Mar 2006 20:22:30 GMT,
    So you're not posting from Google any more and you think that makes
    it OK to top post?

    And, on top of that, you think it's OK to multipost the same inane
    shit, twice, here, which should never have left seb in the first

    I don't really have a polite and respectful way of saying, "You're a
    fucking moron.", so I guess I won't say anything at all.

    But, if you'll take it back to seb I'll be happy to explain the 20AH
    thing to you and how that relates to the rate of discharge.

    Or, if you'll just learn how to crosspost...

    Oh, what the ****... I'll just do it in the road.

    If you have a lead acid battery with a capacity of 13AH (ampere
    hours), that might lead you to believe that you could take 13
    amperes from the battery for one hour before its voltage decayed to
    what's called the "cutoff voltage"; usually 5.25V for a 6V battery.

    However, for most batteries, in order to attain the full capacity of
    the battery, the rate of discharge is specified as C/10 or C/20,
    which means that if your battery has a capacity of 13AH and is rated
    at C/10, you can only get the full capacity out of it if you
    discharge it at 10% of its capacity. For a 6V 13AH battery rated at
    C/10, that means its output voltage will decay to 5.25V if it's
    discharged at a rate of 1.3 amperes for 10 hours. For a battery
    rated at C/20, the rate of discharge will be for half of that, but
    for twice as long.

    For your particular battery, in order to determine where its voltage
    will be after you've sucked current out of it with time, you'll need
    to get the manufacturer's curves for output voltage VS discharge
    current VS time to determine whether what you think will work will

    John Fields
    Professional Circuit Designer

  18. Robert Baer

    Robert Baer Guest

    There aer published discharge curves that one can look at to see how
    long a battery terminal voltage will stay within a given voltage range.
    The greater the discharge rate, C/n, the sooner that the voltage will
    start dropping "rapidly" below the previous long-term voltage.
    So i looked at the curves to see what discharge rate would give the
    desired "tight" specs mentioned, and C/20 was the highest rate possible.
    Next, i took the 700mA times the 12 hours to get the discharge rate
    of 8.4 amp hours.
    Then multiply by 20 to get the needed rating of the battery.
    It is clear that the battery you chose grossly lacks the capability
    you mentioned, if used directly as a power source.
    You *can* use a switching supply, which would allow the battery to be
    used over a larger terminal voltage range, thereby making more efficent
    use of the total storage capabilities.
    And many others have suggested that approach, which was obvious to me.
    But it seemed that you were somehow tied to using only batteries.
  19. Robert Baer

    Robert Baer Guest

    ------- SNIPped for brevity and sanity ------
    Get off your asinine rants about google groups and posting "rules"
    that do not apply to any of the answers or posting(s) related to this
    Either give a decent answer *without* unrelated griping, or SHUT UP.
  20. budgie

    budgie Guest

    (snip good stuff)
    Sorry, couldn't let that one go through to the keeper.

    'C' refers to the capacity of the battery at the rate that (superficially) is
    the one-hour rate.

    i.e. the 13Ah battery's 'C' is 13A.

    C/20 is the discharge rate of 13/20 for that battery i.e. 650mA.

    and as the O/P's required current is 700mA the redcution in performance would be
    absolutely marginal, giving his target of 12 hours almost spot-on.
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