LEAD ACID BATTERY
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From: Ian Stirling on 29 Mar 2006 16:59 john <conphiloso(a)hotmail.com> wrote: > Hi, > > 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? 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. >> Are you making one or many? >> Is volume/weight/cost most important? >> How many cycles do you want the abtteries to last (cycles).
From: Robert Baer on 30 Mar 2006 04:49 john wrote: > Hi, > > 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 ), http://www.zbattery.com/zbattery/ub13-6.html. > > 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 @ > 600mA. > > Thanks > Regards > John > 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.
From: Winfield Hill on 30 Mar 2006 07:35 john wrote... > > 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? 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. -- Thanks, - Win
From: Ken Smith on 30 Mar 2006 10:03 In article <e0gjb501a3b(a)drn.newsguy.com>, Winfield Hill <Winfield_member(a)newsguy.com> wrote: >john wrote... >> >> 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? > > 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. 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 guess. 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. -- -- kensmith(a)rahul.net forging knowledge
From: Robert Baer on 31 Mar 2006 05:27
Ross Herbert wrote: > On 29 Mar 2006 10:49:27 -0800, "john" <conphiloso(a)hotmail.com> wrote: > > >>Hi, >> >>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 ), http://www.zbattery.com/zbattery/ub13-6.html. >> >>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 @ >>600mA. >> >>Thanks >>Regards >>John > > > 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. If you look at the discharge curves given by major battery makers, you will find that my answer fits the curve. |