Slashdot Mirror

It actually DOES work in some cases. Many spacecraft presently use triple junction GaAs photovoltaics with ~30% efficiency. Typical single junction Si cells top out at around 12% efficiency. Quad junction cells exist (~43% efficiency), but I'm not aware of any that have flown yet (which doesn't mean they haven't).

It actually DOES work in some cases. Many spacecraft presently use triple junction GaAs photovoltaics with ~30% efficiency. Typical single junction Si cells top out at around 12% efficiency. Quad junction cells exist (~43% efficiency), but I'm not aware of any that have flown yet (which doesn't mean they haven't).

Hi Maury,

* Spectrolab rates their space solar panels for 20 years at GEO: http://www.spectrolab.com/Data.... Since they don't need to withstand weather, they can be much lighter than ground-mounted panels. 13 W/kg for a typical ground panel (not counting mounting and tracker) vs 177 W/kg for the space ones. That has implications for the energy payback time if you manufacture the panels in space.

* Your comparison of operating hours neglects that in space you have 36% higher insolation, because there is no atmospheric absorption. Therefore it takes fewer cells to produce the same output. Also the Nevada desert is an excellent location on Earth. The average location on Earth gets considerably worse hours of sunlight. Since we can't transmit power all over the Earth, cherry-picking a good location is unfair.

Efficiency is not an issue is it. No matter what they do, no solar tech will ever get better than 20% efficient

Fascinating, because just from a quick Google search I find a company selling~30% efficient solar cells today, and that Sharp is at 43.5% in lab cells. I swear I've seen cells in the 30%s being sold commercially (albeit at very high prices), but I forget which company it was.

These cells work at 35% efficiency at 500kW (350 suns) input light:

http://www.spectrolab.com/DataSheets/TerCel/C1MJ_CDO-225-IC.pdf

Therefore the output will be 175kW per square meter. To get 350MW will require 2000 square meters of cells, or a square that's 45m on a side. Thats a bit bigger than one wing of the space station's solar panels. I assume 350MW at the cells, since to deliver 200MW to the utility, you will need to convert the cell output to something, probably microwaves, then convert it back to electricity at the ground.

Now, to get 350 suns focussed on the cells, you will need a big set of mirrors, but those can be just reflectorized plastic sheets. You will also need to cool the cells. The input sunlight is on the same order as the power that modern cpu chips take (50W/cm^2), so that's doable, certainly.

I expect the issues with this system will be (1) can you unfold the very large reflectors needed, and (2) can you keep the heat removal system a reasonable weight.

In economics, let us assume they can sell the power at $0.10/kWh. That comes to $20,000 an hour or about $170 million a year. Assuming something like $5,000/kg launch cost to low orbit, and that it uses ion thrusters to carry itself to high orbit, it takes about 4 years to pay off the launch.

I assume it can use ion thrusters cause its got lots of power onboard, and they are more weight efficient than chemical rockets by a factor of 10.

If you check here, you'll see that due to where they are (north pole of moon on a crater rim) they have almost constant access to sunlight.

If I'm on the north pole of any tidally locked body, and (this is crucial) the normal of the plane of orbit of that body is perpendicular to the sun... then the north and south poles of that body will have almost constant access.

Here's how. If the north pole has none of the rest of the body to shade it, then from the point of view of the observer, the sun will always be halfway on the horizon. In other words, the sun will be half-hidden all of the time, sitting on the horizon and "going around" the body. So if I build a building on the north pole, then the upper floors will see the whole sun. If we add some reflective screens that will rotate and point at the sun, then we've got an increase in how much sunlight will hit our target (garden, photoelectric, or whatever).

Also keep in mind - there's no atmosphere to weaken the amount of sunlight. So even if the sun seems to be on the horizon, the light per square meter (measured with a normal pointing straight at the sun) will be considerably stronger. Instead of a satellite with lots of solar cells, consider a solar concentraing cell (those referred to in the link are already in space) inside an airspace connected to the base. Whatever sunlight isn't converted to electricity is converted to heat. All you have to do is pump cool water past the cell to keep it cool enough, and you can then capture the heat from the water.

So it's an environment with plenty of energy available (once tapped), lots of rock-based nutrients for plants, and a slow speed for landing (relative ground speed for landing is at a minimum at the poles). A perfect place to start hollowing out the inside of the moon for an even more secure moon base. Here's hoping they put one at each pole.

Sorry, but 60% is not the world record. The world-record in efficiency is currently about 42.8%, held by the University of Delaware. Here's their press release.

However, the most efficient cells in production for commercial use are from Spectrolab, a Boeing subsidiary. They claim 40.7% as of December 2006 - which was the world's record until UD broke it 23 July 2007.

According to Spectrolab's web site, the cells they're producing for distribution include their Ultra Triple Junction cells, with a minimum efficiency of 28.3% and a typical terrestrial efficiency of 31% claimed.

In their FAQ, they claim that a concentration of 500 suns is typically optimal. On the earth, you then have to deal with the fact that 2/3 of the energy is not turned into electricity - which means a significant amount of heat to deal with. You would want to cool the cell with something, lest it burn up. Their FAQ mentions that using a 1 cm^2 cell, at 500 suns and 25C will produce about 17.5W - so you'd be "spending" at least 500 cm^2 of real estate to prodcue the 17.5W : 500 cm^2 for a Fresnel lens to focus it down to 1 cm^2 on the cell.

I think they'll sell to anyone as long as you're a U.S. citizen and agree to the export limitations. However, they have a minimum purchase of $5,000 - but you must spend more to get optimal pricing.

Well. My point is this: 60% is not what anyone's achieved. Most companies are just trying to get their $/Watt price as low as possible in order to get widespread acceptance - instead of attempting a new world-record.

I wish that someone had gotten to 60% - it's 2/3 of the way to the Carnot limit of 95% If you're referring to these guys and their "quantum dot cells", from their web site you'll see that it's still all theoretical.

BTW - you can buy a plastic Fresnel lens here, unless they've changed the web page. Be careful and wear a welding helmet (or equivalent) so that the intense concentration of sunlight on something won't be able to cause a light bright enough to burn your retina.

Sorry, but 60% is not the world record. The world-record in efficiency is currently about 42.8%, held by the University of Delaware. Here's their press release.

However, the most efficient cells in production for commercial use are from Spectrolab, a Boeing subsidiary. They claim 40.7% as of December 2006 - which was the world's record until UD broke it 23 July 2007.

According to Spectrolab's web site, the cells they're producing for distribution include their Ultra Triple Junction cells, with a minimum efficiency of 28.3% and a typical terrestrial efficiency of 31% claimed.

In their FAQ, they claim that a concentration of 500 suns is typically optimal. On the earth, you then have to deal with the fact that 2/3 of the energy is not turned into electricity - which means a significant amount of heat to deal with. You would want to cool the cell with something, lest it burn up. Their FAQ mentions that using a 1 cm^2 cell, at 500 suns and 25C will produce about 17.5W - so you'd be "spending" at least 500 cm^2 of real estate to prodcue the 17.5W : 500 cm^2 for a Fresnel lens to focus it down to 1 cm^2 on the cell.

I think they'll sell to anyone as long as you're a U.S. citizen and agree to the export limitations. However, they have a minimum purchase of $5,000 - but you must spend more to get optimal pricing.

Well. My point is this: 60% is not what anyone's achieved. Most companies are just trying to get their $/Watt price as low as possible in order to get widespread acceptance - instead of attempting a new world-record.

I wish that someone had gotten to 60% - it's 2/3 of the way to the Carnot limit of 95% If you're referring to these guys and their "quantum dot cells", from their web site you'll see that it's still all theoretical.

BTW - you can buy a plastic Fresnel lens here, unless they've changed the web page. Be careful and wear a welding helmet (or equivalent) so that the intense concentration of sunlight on something won't be able to cause a light bright enough to burn your retina.

Sorry, but 60% is not the world record. The world-record in efficiency is currently about 42.8%, held by the University of Delaware. Here's their press release.

However, the most efficient cells in production for commercial use are from Spectrolab, a Boeing subsidiary. They claim 40.7% as of December 2006 - which was the world's record until UD broke it 23 July 2007.

According to Spectrolab's web site, the cells they're producing for distribution include their Ultra Triple Junction cells, with a minimum efficiency of 28.3% and a typical terrestrial efficiency of 31% claimed.

In their FAQ, they claim that a concentration of 500 suns is typically optimal. On the earth, you then have to deal with the fact that 2/3 of the energy is not turned into electricity - which means a significant amount of heat to deal with. You would want to cool the cell with something, lest it burn up. Their FAQ mentions that using a 1 cm^2 cell, at 500 suns and 25C will produce about 17.5W - so you'd be "spending" at least 500 cm^2 of real estate to prodcue the 17.5W : 500 cm^2 for a Fresnel lens to focus it down to 1 cm^2 on the cell.

I think they'll sell to anyone as long as you're a U.S. citizen and agree to the export limitations. However, they have a minimum purchase of $5,000 - but you must spend more to get optimal pricing.

Well. My point is this: 60% is not what anyone's achieved. Most companies are just trying to get their $/Watt price as low as possible in order to get widespread acceptance - instead of attempting a new world-record.

I wish that someone had gotten to 60% - it's 2/3 of the way to the Carnot limit of 95% If you're referring to these guys and their "quantum dot cells", from their web site you'll see that it's still all theoretical.

BTW - you can buy a plastic Fresnel lens here, unless they've changed the web page. Be careful and wear a welding helmet (or equivalent) so that the intense concentration of sunlight on something won't be able to cause a light bright enough to burn your retina.

Spectrolab has cells that are over 40% efficient. See here for more details.

JSL

Spectrolab has cells that are over 40% efficient. See here for more details.

JSL

"The concentrator doesn't make the device more efficient at converting solar radiation into electrical power, it just concentrates the light so you don't have to use as large of a device."

Concentrating the sunlight does in fact make the device more efficient in conversion, as long as you keep the temperature (which decreases the efficiency during increase) in check, i.e. near ambient conditions. See the upper right graph on pg2 of this product data sheet from the company that this slashdot article is discussing for an old version of their cells here.

The photovoltaics on the mars rover are likely triple junction GaAs/GaInP2 cells.

Similar products are sold by Spectrolab and Emcore.
Teams that "rayce" solar cars often purchase space-grade reject cells from these two companies.