Calculating Watts for laptops and scanners

I need to calculate power consumption for three devices:

- Laptop (with a Digital Camera) = 50 Watts
- Scanner Fingerprint = 35 Watts
- Scanner Signature = 35 Watts

I've read that a laptop needs 50 Watts and a standard scanner needs 35 Watts. So, I assume that I would need 120 Watts in total. Say 155 Watts, just in case.
I'd like to supply energy through a solar panel. This solar panel has to supply 155 Watts.

Could anybody tell me if I am correct in my calculations?
And could you recommend me a brand of solar panel for the devices that I need?

Thank you!
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I think we would need to know how you intend to connect the devices to the solar panel - are you attempting to tailor the setup to feed the devices directly with DC, or are you going via an inverter > AC and then back to DC using the standard power bricks that come with laptops etc?

I looked at a couple of laptops and thery are rated 60 - 70 W max, which probably means that they'll take something like that when running AND charging their main batteries. Depending on your funding of this setup, you may take into account that the laptops will take less power if run without the batttery inside; solar panels aren't cheap. Running without batteries in a laptop will of course have some drawbacks... a brownout will probably be pretty inconvenient.

What kind of scanners are the ones you mention - fingerprint scanner?
Rosa2003Author Commented:
Thank you Rid,
Sorry for my bad english. Yes, I meant fingerprint scanner and signature scanner. I suppose that they will take same power consumption as standard scanners.
I thought that it would be enough to feed the devices with solar panels.
What would be the difference between feeding devices with DC and via inverter > AC ?
Rosa2003Author Commented:
Do you think that using a 155-Watt UPS for my devices would be cheaper than using solar panels? I have to add the cost of using a mini-generator or a car lighter for the UPS.
I'm in a Caribbean island, so that's why I'm thinking using solar panels.
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In the first case (direct DC feed) the solar panel output voltage must match the required input voltage of the devices. May be tricky if they are not the same.
In the second case (AC feed) you need to calculate for losses in the inverter (solar panel DC to AC) and the "power bricks" (the "transformers" that change the AC back into DC to feed the laptop and scanners).

Any voltage conversion will induce losses.

My guess (a guess, mind you) is that a fingerprint scanner takes very little power as these are built into some laptops. An extra 35 W seems a bit heavy for a laptop...
Very generally, it looks easier to make a system with a solar panel array which feeds a standard accumulator (like a car battery) and then use an inverter to produce 220 V (or 110 V) AC and then run your machines off this AC. Some losses will be there, but you can use standard equipment and the battery can absorb any short periods of heavier load. A smallish petrol engine/generator would be useful for charging the batteries as well.

If you're into tinkering you could get a used UPS unit and use it to build your own inverter (or rather use the inverter part inside it), but that requires some knowledge of electronics.

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>- Scanner Fingerprint = 35 Watts
>- Scanner Signature = 35 Watts

This is completely unrealistic.  3.5W for each would be closer to reality. Use maybe 5W for your estimate.

There are laptops that are 40 watts and ones that are 120 watts..
Look at the adapter..
That would be the max...

You can calculate the scanners the same way from the adapter or label plate if they are AC.

If they are AC you will need an inverter.
If you need an inverter account for about 27% losses in wattage between DC-in and AC-out.. (Lost in the inverter circuitry and dissipated as heat.)
(You can make the same assumption about any electrical conversion that doesn't use high effieciency (specialty) equipment.)

Also you need to take into account that the actual output of solar panels is seldom the full rating.. You'll probably be safe assuming a 50% efficiency..
(I otherwords at least double what your calculations say you need...)

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