Do-It-Yourself Solar-Powered PC: Hardware

Anaximander

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Sep 13, 2007
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Excellent article so far. I especially liked the monitor wiring job! The main thing I would like to see would be a comparison to show if theirs any difference between a WinXP and Vista Setup (maybe even a Linux distro). And possibly if theirs anything that could be done to either OS (background services come to mind) to lessen their overall impact on the wattage scale.
 

wsbsteven

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May 23, 2006
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Much better than the previous article.

"We also took into account the fact that transporting energy over a distance as short as 50 feet (15 meters) requires incredibly thick power cables to avoid excessive power losses."

That answers my question from the first article on why 1-2 gauge wire was used.
 

warezme

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Dec 18, 2006
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WOW! I love this stuff :love:

When your through with this, and I can't wait to see the hook ups to the panels and stuff but how much thought have you guys put into like a larger array of panels that would maybe power a higher end PC? Is this a point of diminishing returns due to cost and complexity?

 

joex444

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Feb 16, 2006
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I found a GE 170W panel for $760. It should end up in a diminishing return situation -- you spend about $1500 for panels to power a high end PC. So, you'd essentially need about 380W grid power to make 340W of solar power. I think on average, 9 cent/kWh is about right. So, there is 5 years of 24/7 up time at max load you've already purchased in the panels. BTW, a low power system like this, if the panel is about $750, would mean that THG's also just bought 5 years of grid power for that low power PC to be on 24/7. (And we haven't priced in batteries ;)

Also, found this in the article: "The reason is that the voltage levels are low, which means the current levels are high, and loss is proportional to current."

V=IR (we'll as-sume (you really need to stop filtering that) no internal resistance and all that good stuff)
So, if V is low then I is low; if V is high, then I is high. You pretty much implied that the resistance of your device depends on what it's doing, which I find wrong.

Loss, is in terms of power. P = IV = I(IR) = I^2*R. So, loss is proportional to current SQUARED.

Perhaps you know something I don't in regards to this (after all, I've only had introductory E&M, and we didn't really do alternating currents, though it looks like you were talking about DC).
 

mboekerster

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Sep 14, 2007
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Hey guys, very interesting article!

A couple of things I noticed though: Near the beginning, you have a diagram showing the inverter and the PSU going from DC to AC back to DC. Then, the article explains how you guys got around that, but the image is the same one! Basically, the 'wrong' and 'right' way have an identical illustration.

I was also wondering why you only chose WD hard drives.. I know Samsung has some very well performing and efficient drives, as does Seagate. To add to that, what about a SSD? I know it would jack up the price, but it would be interesting to see how solid state (or even hybrid) drives compare under those precise measurements.

Additionally, the small tables you have throughout the article that show a summary of power usage, the total is always 61.23 (idle) and 115.6 (load). I thought it was strange when I first saw it that the numbers did not add up, then I noticed in later versions that the list of components was increasing and figured that the last one would then add up correctly.

Again, very interesting article! Thanks TomsHardware.

-Martin
 

adobo76

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I noticed a few places where you could save on power here.
- Use SSD hard drives instead of a Rotating magnetic disk drive.
- I saw 2 fans used in this project. I wonder if the same job could
be accomplished with 2 really big heat sinks.
- Use an LED backlit monitor instead of one that uses a florescent bulb.
 

Kari

Splendid

But in this case you need to transfer certain amount of power to the PC, and as we know P=UI, so if your voltage is low you'll need higher current to transfer the needed power. To transfer 115W you'll need 9,6 amps with 12V and only 1,05 amps if you're using 110V.

edit:typos
 

spireneusz

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Sep 16, 2007
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I just want point out a little error made by Voltech PM3000A.
As you see on this picture:

power is 61,23W, but is it true? Lets count it manually: 11,854*5,342 = 63,324 so from the 61,23 came out?
 

Erik74

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Sep 17, 2007
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Great Article.

I am just curious where you got the power figures from.

I once wanted to update my old G4 imac with a efficient hard drive. And it was a nightmare to get the power figures.

best regards,

Erik
 

ebattleon

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Sep 19, 2007
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Great article thus far. There are a few questions but Iwant to reserve those until next article as I want to see the approach you will take in setting up the solar panels and battery bank.

 

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