Subject: Re: A question on solar cells (probably a FAQ)

From: [email-masked] (Anmar Caves)
Date: Fri, 10 Sep 1993 14:21:47 GMT

In article <[email-masked]> [email-masked] (Matthew DeBell) writes:
> Just a quick question about solar cells:
>
>What is the theoretical maximum amount of electric power that can 
>be generated using a solar cell of a given size (say, 1 m^2)? 
>

Current cost effective efficiencies are in the 14-17% range.  The
maximum achieved using multi-layer thin film has been 29%.  I think
the theoretical maximum is ~60%, but that's from memory and is
probably wrong.

Now, you ask about how much power that translates into, that will
depend on where you are, time of day, and orientation in respect to
the sun.  Around here in Southern Indiana we average 4.5 peak hours
solar radiation daily.  Dept. of Geography here at IU maintains a
weather station that measures solar radiation, and around here the
peak amount during the clearest days of fall (the peak time) is around
830 watts per square meter.  Which corresponds with my measured usable
output at short circuit of my panels at home, when the efficiencies
are calculated in.

Now, if you are out in space, solar radiation is going to be higher, I
think it's 1.3 kW per square meter, so your cell will put out a higher
amount of power.

What does this mean?  Ok, some real world statistics based on my
measurements over a year period.

A 48 watt panel, 36 cells, spec'd at 15.9 volts, 3.02A (a Siemans
M75), running a constant load source at 12.9 volts, put out 3.1A at
peak. The panel was flat mounted on my roof with some eastern
obstruction.  Over the year the daily average current output was 21.3A.
The highest peak recorded was during a partially cloudy day and that
was at 3.6A.  The reason for that comes from rereflected light from
the undersides of clouds.  This effect was first noted during a
Solar-Powered car race in Australia I believe, but is a transient
effect and the partially cloudy day lowers overall power output.

Other effects noted.

During cloudy days the panel would put out anywhere from 10-30% of
rated power depending on how overcast.  Even on the most overcast days
the panel would be putting out a couple hundred milliamps.  The bulk
of this seemed to be UV, as a glass cover placed over the panel
dropped output significantly.  Clouds don't attentuate UV as much it
seems. 

When I plotted it out I noted an odd anomaly, there were three periods
over the year where for about a week there would be power measured at
night.  I checked, and those three periods corresponded with the
phases of the moon.  It seems I was getting about a couple hundred mA
from the light reflected from the moon.  Those three periods also
corresponded to times of clear weather, and higher daily averages.

I also couldn't zero the plot.  It seems the streetlights were giving
me a few mA as well.

-- 
Anmar Mirza   #Chief of Tranquility  # I don't need to carry a weapon,
EMT-D         #Base, Lawrence Co. IN # People talk to me long enough they
N9ISY (tech)  #Somewhere out on the  # then go and hurt themselves.
Networks Tech.#Mirza Ranch.C'mon over# ---Patti Cummings

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From: [email-masked] (Steve Frysinger of Blue Feather Farm)
Date: 10 Sep 93 14:00:52 GMT

>From article <[email-masked]>, by [email-masked] (Matthew DeBell):

Maximum solar energy available on the surface of the earth is 1360 watts/m^2.
(Straightforward calculation based upon the Stefan-Boltzman Law E=kAT^4).
 
The average, accounting for day/night variation is 1/4 of this (340 W/m^2).
(This ignores other losses, of course.))

Steve Frysinger

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From: [email-masked] (Gary Coffman)
Date: Fri, 10 Sep 1993 16:52:41 GMT

>be generated using a solar cell of a given size (say, 1 m^2)? 

Well, you can start with the solar constant which gives a total
radiant flux of 1,388 watts/sq meter in orbit. Then reduce that
by the amount that is energetic enough to activate the photoelectric
effect in the cell material, basically the visible and UV portions
of the spectrum, and factor that by the absorption of the atmosphere
at those wavelengths. That gives you the amount available at Earth's
surface. That leaves you with about 450 watts/meter^2 the cell could
use if the cell were 100% efficient at it's active wavelengths. Cells 
aren't 100% efficient, so it is somewhat less.

Gary

-- 
Gary Coffman KE4ZV          |"If 10% is good enough | gatech!wa4mei!ke4zv!gary
Destructive Testing Systems | for Jesus, it's good  | uunet!rsiatl!ke4zv!gary
534 Shannon Way             | enough for Uncle Sam."| emory!kd4nc!ke4zv!gary 
Lawrenceville, GA 30244     | -Ray Stevens          |

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From: [email-masked] (Dan Eldred)
Date: 10 Sep 1993 11:44:11 -0700

>aren't 100% efficient, so it is somewhat less.
Let me add to the previous post.  Gallium arsenide solar cells are
about 18% efficient and silicon is about 12% efficient, for spacecraft
grade solar cells.  Commercial grade solar cells will be less efficient.
Plus, the gallium arsenide cells are relatively new and not generally
available.  So, assuming 10% efficiciency in silicon cells you end
up with about 140 watts/m^2.  That's in outer space.  From this you
have to subtract for atmospheric absorption plus account for the fact
that the cells may not be pointed directly at the sun (multiply by the
cosine of the angle).  There is also a "packing factor" which is the
ratio of surface area on the solar cell array occupied by cells vs. wasted
space.  The packing ratio for round solar cells is less than 78%.  There
is also dirt on the cover plate.  You can get down to 5% efficiency (70 W)
pretty quickly.

	- Dan

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From: [email-masked] (Gary Coffman)
Date: Fri, 10 Sep 1993 16:52:41 GMT

Gary