Little inverters - 100 to 300 watts. Do they suck batteries???

[MarkJ]

Hi folks

I would like a bit of advice please (Yes, I know I am always afeter advice - blame it on the slack school system when I was a kid!)

I want to charge some camera batteries and maybe power a computer printer (ie for 60 seconds at a go) and am thinking about a small cigarette lighter inverter of 100 watts to 300 watts. Cost about $100

1) Is this a good way to go?
2) does it suck an extraordinary amout of power just to run the inverter?

Gizmomart.com.au - Toys for grown ups

Belkin DC to AC Power Inverter : oo.com.au


Thanky! All great replies get to kiss Nicolle for free! (I am trying to boost the cruise kitty but she needs training)

Mark

[cal40john]

As an example, the Xantrex 175 and 300 specs show 0.15 and 0.2 A draw with no load, and 90% maximum efficiency.

page 25-26
http://www.xantrex.com/web/id/1306/docserve.aspx

John

[MarkJ]

Originally Posted by cal40john

As an example, the Xantrex 175 and 300 specs show 0.15 and 0.2 A draw with no load, and 90% maximum efficiency.

So, that would be good??????????????? <<<<showing signs of niavity <shucks>


Mark

[seafox]

It will be fine. Make sure you buy an inverter that is "pure sine wave" and not just "sinewave" or "modified sinewave".
the pure sinewave ones are most like the power in Australia. They are about twice the price but still cheap. I use a 150w one for my laptop, phone and camera battery for charging. The modified sinewave inverters might stuff up your camera battery. They are around NZ $40. Pure is around $NZ100.

[GMac]

Seeing you mentioned laptops Seafox, may I tap into your knowledge and expend Marks good question (I'm just as dumb as him on this, meant in a loving way obviously Mark )

Can you plug a laptop into the lighter socket with one of them car connection thingys they have or would using a inverter be the better option?

Talking about while in use and for charging purposes. And in relation to total power sucked from battery, as in which is cheaper on amps, if any?

[cal40john]

Originally Posted by MarkJ

So, that would be good??????????????? <<<<showing signs of niavity <shucks>

Mark

If you leave the inverter plugged in with nothing connected to it, it draws 0.2 amps, one incandescant light runs about 1 amp. If for some reason you left it plugged in for 24 hours with it powering nothing, 0.2 x 24 hours is 4.8 amp-hrs in a day, not a huge draw. If you use them like I do, that is plug it in to use it, then unplug it, then you don't care about that number.

90% max efficiency means that at best for every amp drawn out of your battery, 0.9 amps is going to power whatever you have plugged in, and 0.1 amps is being consumed by the inverter, pretty darn good compared to old tech. Where the max efficiency point is I don't exactly know. The inverter will probably not draw less than the 0.2 amps for itself at any power usage level. So if the AC device you plug in only uses 2.4 watts (remember this is an inverter that can handle 175 W), then 2.4 W/12 volts = 0.2 amps, which is the amps coming out of your battery being used to power your AC device, and your inverter uses 0.2 amps, you're down to 50 % efficiency, half the power used from your battery is powering the inverter. Again, drawing 0.4 amps is probably not a big deal, so the efficiency isn't too worrisome at that point. Usually using the inverter closer to the high end of its rating will give better efficiency, but at some power level, this inverter turns on a fan to cool itself, my guess is that the maximum efficiency isn't there either.

After throwing lots of mud on the topic, bottom line, yes, good.

John

[Ex-Calif]

Originally Posted by cal40john

So if the AC device you plug in only uses 2.4 watts (remember this is an inverter that can handle 175 W), then 2.4 W/12 volts = 0.2 amps

Hang on - hang on....

watts = volts X amps. Watts being a unit of power.

My laptop power supply is 65 watts at 110v

65W = 110 X .59a

If I supply 65 watts at 110v from a 12 volt battery

65 = 12 X 5.4 amps

My Camera charger is .065 at 110v.

7W = 110 X .065a

7W = 12 X .58a

So Basically figure out what the unit draws in Amps or Watts at the base voltage and convert.

A 300W inverter will deliver

300W = 12 X 25a
300W = 110 X 2.72a

BTW - Based on last port o' call one could kiss Nicolle for a cheesburger and a beer. Now she wants intellect. Are prices are on the up or jsut too much time on the boat talking to you - LOL....

Where are you now?


New Subject -

GMac - My opinion is that each time you convert you lose:

1) 12V boat ----> 110v Inverter ----> 110v Computer converter ----> XXv dc to laptop

2) 12v boat ----> 12v computer converter ---> XXvdc to laptop

You take the boat inverter out you gain the efficiency of the inverter at least.

[imagine2frolic]

We have used them for the laptop, and tv for years. Never a problem so far.......i2f

[noelex 77]

Dc to Dc converters (they plug into a cig lighter socket then directly into your computer) are more efficient and are available for most laptops. Typically you will save 10% or so in power using one of these for your computer.
There are also DC to DC battery charges available for camera batteries (mine is called a Swallow it will charge almost any rechargeable battery under about 14v). sometime the power savings from this are quite substantial and it will recharge torch batteries etc
For your printer the only option is an inverter unless you find one thats happy on 12V DC or your prepared to build a custom power supply.
If you live aboard your vessel and use a computer a lot these options are worth considering, but for most people a small inverter is the easiest method. Just remember to turn it off when it is not used.
John
PS pleased to see you are enjoying your adventures Mark I catch up with your blog when I have reasonable internet

[cal40john]

I don't understand what you think I did wrong. I was purposely choosing a low wattage device to illustrate that the inverter has low efficiency at low power. So a device that draws 2.4 W, draws 0.02 amps on the 120 volt AC side, and converting with no losses draws 0.2 amps on the 12 volt side. So using your first example of 65 Watts instead of writing

65W = 12 x 5.4,

I wrote

65W/12V = 5.4


Confusedly, John

Originally Posted by Ex-Calif

Hang on - hang on....

watts = volts X amps. Watts being a unit of power.

My laptop power supply is 65 watts at 110v

65W = 110 X .59a

If I supply 65 watts at 110v from a 12 volt battery

65 = 12 X 5.4 amps

My Camera charger is .065 at 110v.

7W = 110 X .065a

7W = 12 X .58a

So Basically figure out what the unit draws in Amps or Watts at the base voltage and convert.

A 300W inverter will deliver

300W = 12 X 25a
300W = 110 X 2.72a

BTW - Based on last port o' call one could kiss Nicolle for a cheesburger and a beer. Now she wants intellect. Are prices are on the up or jsut too much time on the boat talking to you - LOL....

Where are you now?


New Subject -

GMac - My opinion is that each time you convert you lose:

1) 12V boat ----> 110v Inverter ----> 110v Computer converter ----> XXv dc to laptop

2) 12v boat ----> 12v computer converter ---> XXvdc to laptop

You take the boat inverter out you gain the efficiency of the inverter at least.

[denverd0n]

As long as the wattage is up to snuff the cheap inverters work just fine. Yes, they waste some electricity, but not a lot.

I have used a modified sine wave inverter to power battery chargers, a laptop, a small TV, and so on. No problems. Personally, unless I knew for sure that the device I wanted to power was finicky about its electricity I would not pay extra for a true sine wave inverter.

Also, be sure to note the continuous power rating. Especially among the cheapo inverters they will often be advertised with the PEAK current in big, bold letters--600 WATT INVERTER!--and then in the teeny-tiny print it will mention that the continuous rating is only about half that. Buyer beware!

[GMac]

Good stuff. Lighter socket car chargers dodaky thing for me then until I want a printer.

[Ex-Calif]

Originally Posted by cal40john

I don't understand what you think I did wrong. I was purposely choosing a low wattage device to illustrate that the inverter has low efficiency at low power. So a device that draws 2.4 W, draws 0.02 amps on the 120 volt AC side, and converting with no losses draws 0.2 amps on the 12 volt side. So using your first example of 65 Watts instead of writing

65W = 12 x 5.4,

I wrote

65W/12V = 5.4


Confusedly, John

Sorry John - I didn't see things so clearly interspersed with the text.

Also because you were using such a very low wattage I didn't want Mark to think that 0.2a was a typical power draw. He might let Nicolle start using the hair dryer

I have a 300w inverter. If I plug a 300w 110v appliance into it, as I theoretically could do, my house bank would last about 2 1/2 hours to 50% depleted.