Posted by *noel* on August 14, 2006, 5:47 am

I was trying to figure out how big a deep cycle battery would need to

be to power a printer (300 watts) and a laptop (200watts ?) via an

inverter (true sine), but I'm not getting what I think would be the

right amp/hr results. Say for example in an 8 hour period with the

laptop on all the time and the printer being used a total of 4 hours.

How big should the battery be?

Posted by *Anthony Matonak* on August 14, 2006, 6:49 am

noel wrote:

*> I was trying to figure out how big a deep cycle battery would need to*

*> be to power a printer (300 watts) and a laptop (200watts ?) via an*

*> inverter (true sine), but I'm not getting what I think would be the*

*> right amp/hr results. Say for example in an 8 hour period with the*

*> laptop on all the time and the printer being used a total of 4 hours.*

*> How big should the battery be?*

That would be some laptop that takes 200 Watts. Most range in the

60-80 Watt range. It is possible to get printers that take only 10

or 20 watts. Typically these would be inkjet printers.

That said, the calculation isn't terrible difficult but you would

need to know the efficiency of your inverter. That is, how much

power goes in the inverter compared to how much comes out. It's

never 100%. I would think a 500W true sine inverter could get in

the 80% range. This means 80% of the power you take from the battery

will come out the inverter as AC.

First figure total Watt-hours required.

200W x 8h + 300W x 4h = 2800 Wh AC

Next divide by the efficiency of the inverter to account for its loss.

2800 Wh / .8 = 3500 Wh required from the battery.

Watts = Volts x Amps so to convert Watt-hours to Amp-hours you

need to divide by the Voltage of the battery. Say it's 12 Volts.

3500 Wh / 12V = 292 Ah. You need a 300 Amp-hour battery (roughly).

Of course, if you reduce that to 80W for the laptop and 20W for

the printer then you get (80Wx8h + 20Wx4h)/.8/12V = 75 Amp-hours.

Remember, that you typically don't want to discharge your battery

more than 80% so you'll need to add that much more to the capacity.

Anthony

Posted by *Dale Eastman* on August 14, 2006, 11:44 am

Anthony Matonak wrote:

*> Watts = Volts x Amps so to convert Watt-hours to Amp-hours you*

*> need to divide by the Voltage of the battery. Say it's 12 Volts.*

*> 3500 Wh / 12V = 292 Ah. You need a 300 Amp-hour battery (roughly).*

*> *

*> Of course, if you reduce that to 80W for the laptop and 20W for*

*> the printer then you get (80Wx8h + 20Wx4h)/.8/12V = 75 Amp-hours.*

*> *

*> Remember, that you typically don't want to discharge your battery*

*> more than 80% so you'll need to add that much more to the capacity.*

80% DOD?

I thought what I've read previously in this group was 20% DOD for good

battery life. Or does that depend upon the type of battery?

--

www.synapticsparks.info

Posted by *Harry Chickpea* on August 14, 2006, 12:24 pm

*>Anthony Matonak wrote:*

*>> Watts = Volts x Amps so to convert Watt-hours to Amp-hours you*

*>> need to divide by the Voltage of the battery. Say it's 12 Volts.*

*>> 3500 Wh / 12V = 292 Ah. You need a 300 Amp-hour battery (roughly).*

*>> *

*>> Of course, if you reduce that to 80W for the laptop and 20W for*

*>> the printer then you get (80Wx8h + 20Wx4h)/.8/12V = 75 Amp-hours.*

*>> *

*>> Remember, that you typically don't want to discharge your battery*

*>> more than 80% so you'll need to add that much more to the capacity.*

*>80% DOD?*

*>I thought what I've read previously in this group was 20% DOD for good *

*>battery life. Or does that depend upon the type of battery?*

20% for a lead acid battery is a lot better. 50% would be max

discharge. You also don't want to leave lead acid cells in a

seriously discharged state, so the type of charging and all of that

side of the battery specs start to come in as well.

Now about that printer - 300 watts sounds like a laser printer.

Running a laser printer off batteries sin't the greatest idea in the

world, but to their credit, they often have power saving features, so

the draw isn't a constant draw.

I'm trying to figure out what is driving Noel's question. Why do

this?

Posted by *Neon John* on August 14, 2006, 6:27 pm

On Mon, 14 Aug 2006 11:44:26 GMT, Dale Eastman

*>Anthony Matonak wrote:*

*>> Watts = Volts x Amps so to convert Watt-hours to Amp-hours you*

*>> need to divide by the Voltage of the battery. Say it's 12 Volts.*

*>> 3500 Wh / 12V = 292 Ah. You need a 300 Amp-hour battery (roughly).*

*>> *

*>> Of course, if you reduce that to 80W for the laptop and 20W for*

*>> the printer then you get (80Wx8h + 20Wx4h)/.8/12V = 75 Amp-hours.*

*>> *

*>> Remember, that you typically don't want to discharge your battery*

*>> more than 80% so you'll need to add that much more to the capacity.*

*>80% DOD?*

*>I thought what I've read previously in this group was 20% DOD for good *

*>battery life. Or does that depend upon the type of battery?*

I see people throwing all sorts of BS numbers around. The generic

answer is "it depends". Being an electric vehicle user owning

thousands of dollars worth of batteries, I have more than a casual

interest in this.

Trojan says that their batteries will achieve the specified cycle life

at 80% DOD. Other manufacturers say anything from 50% to 80%. In

large part this is specsmanship. One can quote a MUCH higher cycle

life if it is specified at 50% than at 80%. The slope of the curve of

cycle life between 50% and 80% is fairly gentle so there isn't a huge

amount to be gained.

In my EVs I try to limit DOD to 50% but I don't lose any sleep if I

have to go below that. I put a hard limit (stop and run the gas

generator for awhile) at 80%. The only way to know the DOD is with a

meter that counts amp-hours. All my battery-operated vehicles contain

Link-10s, the cheapest AH meter that I know of that actually works.

In my RV, I routinely discharge to 80%. I'm space limited as to the

amount of capacity I can carry, the batteries are cheap and the

convenience of long cycles is more important to me than cycle life.

Still, I get at least a couple of years of almost weekly use out of

'em. I'm actually into my 4th season on this set of batteries, though

they're getting weak.

John

---

John De Armond

See my website for my current email address

http://www.neon-john.com

Cleveland, Occupied TN

Don't let your schooling interfere with your education-Mark Twain

> I was trying to figure out how big a deep cycle battery would need to> be to power a printer (300 watts) and a laptop (200watts ?) via an> inverter (true sine), but I'm not getting what I think would be the> right amp/hr results. Say for example in an 8 hour period with the> laptop on all the time and the printer being used a total of 4 hours.> How big should the battery be?