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

1. johnGuest

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

2. Tim WescottGuest

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

hi,

Its 13AH.
Regards
jonn

4. Ian StirlingGuest

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).

5. johnGuest

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?

Thanks
John

6. Ian StirlingGuest

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
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:
http://www.powerdesigners.com/InfoWeb/design_center/articles/DC-DC/converter.shtm

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?

8. Robert BaerGuest

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 HerbertGuest

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. Winfield HillGuest

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 SmithGuest

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.

12. Robert BaerGuest

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

13. johnGuest

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

14. John FieldsGuest

---
Learn to post properly. From:

"Summarize what you're following up.

When you click "Reply" under "show options" to follow up an existing
cursor at the top of the article. Tempting though it is to just
start
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."

15. johnGuest

H
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

Thank
Joh

16. RHRRCGuest

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 FieldsGuest

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
place?

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
work.

--
John Fields
Professional Circuit Designer

18. Robert BaerGuest

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.

19. Robert BaerGuest

------- SNIPped for brevity and sanity ------
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.

20. budgieGuest

(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.