LEAD ACID BATTERY

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 expensive.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Posting from Google?  See http://cfaj.freeshell.org/google/

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

+-18?

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

Hi,

No, I did not th> john wrote:

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.

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 laptop.

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 methods.

Any suggestions guys?

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.

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.

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

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@rahul.net   forging knowledge

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

Hi 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?

Thanks John

--- 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 start 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 does, they'll get the gist of what you're talking about."

-- John Fields Professional Circuit Designer

Hi 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?

Thanks John

times 12 hours equals 8.4 Amp-Hours. Multiply that by

a battery rating of 168 AH.

specified near the

(6V) is rated at 170AH, their SG-8D (12V) is

is rated at 164AH, and their SG-4D

rated battery, the UPS6-600

equivalents; these are heavy duty batteries.

charging,the battery will be much higher

All of this assumes constant current discharge at 25C / 77F.

temperature will be a significant factor in terminal

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.

--- 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 place?

I don't really have a polite and respectful way of saying, "You're a f****ng 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 f*ck... 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 work.

-- John Fields Professional Circuit Designer

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. BUT. 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.

------- 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 message. Either give a decent answer *without* unrelated griping, or SHUT UP.

(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.