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Overunity Machines Forum



2 Watts from an old 2N3055 as a solarcell ?

Started by hartiberlin, June 04, 2005, 01:34:25 PM

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prometheus_effect

Hi Guys,

Some solar data:

Irridation of 1,000w/m^2 (Max surface level). Probably better to use 900w/m^2.

6" dia lens or 150mm = 0.01767m^2 area * 900w/m^2 = 15.9 watts input to transistor (not including lens transmission losses of about 10%)

Irridation efficiency:

0.5v @ 3a = 1.5w =? ?9.4%
0.5v @ 4a = 2.0w = 12.6%
0.5v @ 5a = 2.5w = 15.7%

Power loss for silicon is about 0.5% per degree C temp rise so good heat sinking is important. Heat sink to about twice the lens area. Suggest you use at least 4mm thick aluminum and use a good heat sink paste between the transistor and the heat sink.

So heat load = 15.9w (solar input) * 0.9 (lens efficiency) * (1 - .126 (efficiency @ 4a)) = 12.5 watts to heat sink at no more than 10C above ambient for good cell efficiency.

For the lens I suggest you use a rigid acrylic Fresnel lense as they have lower losses (about 92 - 93% efficiency).

To get the best efficiency adjust the distance between the transistor and the lens until the bright image of the sun just coveres the silicon of the transistor. There is no need for the image to be focused but just to fully illuminate the silicon surface.

Then you need to track.

Commercial solar concentrator cells are available with efficiencies in the 25 - 37% efficiency range. Checkout:
http://www.sunpowercorp.com/html/Products/SPG/conc.html

All the best,
Greg
Now its just engineering effort, time and money,
Prometheus_Effect

prometheus_effect

Guys,

64 x 2n3055 @ $0.39us each:

http://store.americanmicrosemiconductor.com/2n3055.html?referrer=google&adlink=2N3055_order

At 2.5 w thats $0.156us per peak watt.

With a good 2 axis tracker you can get a conversion ratio of peak watts to annual kWh of about 2,500. So we should produce about 2.5Wp * 2,500 = 6.25 kWh annually. As most homes use around 5 mWh annually you would need about 800 of these transistors tracking the sun.

Using 6" square Fresnel lens we would get about 28% more energy or about 8 kWh annually per unit. Packing the lens and transistors in a 1 yard square panel we would get 6 * 6 = 36 lens / transistors = 290 kwh annually. Then our 5 mWh energy usage would require 17 such panels and we are off the grid.

All the best,
Greg



Now its just engineering effort, time and money,
Prometheus_Effect

prometheus_effect

Guys,

Just so we know what we are talking about here is a photo of a 2n3055 minus the case. Note the die size is about 2.8mm square or about 7.9mm^2.

Now a 150mm (6") square Fresnel lens has a surface area of about 22,500mm^2. The sun (concentration) factor is then 22,500 / 7.9 = 2,848 suns. That is an amazing figure as most leading edge concentrators work at around 500 suns, while commercial ones work closer to 20 - 50 suns.

That you guys are seeing 2 to 3 watts from such a simple setup is truly amazing.

Walter can you please repost your readings, load resistor value, lens data (diameter, focal length, etc), 2n3055 die size, manufacturer, etc.

Thanks,
Greg
Now its just engineering effort, time and money,
Prometheus_Effect

Walter Hofmann

Quote from: prometheus_effect on June 13, 2005, 10:29:05 PM
Guys,

64 x 2n3055 @ $0.39us each:

http://store.americanmicrosemiconductor.com/2n3055.html?referrer=google&adlink=2N3055_order

At 2.5 w thats $0.156us per peak watt.

With a good 2 axis tracker you can get a conversion ratio of peak watts to annual kWh of about 2,500. So we should produce about 2.5Wp * 2,500 = 6.25 kWh annually. As most homes use around 5 mWh annually you would need about 800 of these transistors tracking the sun.

Using 6" square Fresnel lens we would get about 28% more energy or about 8 kWh annually per unit. Packing the lens and transistors in a 1 yard square panel we would get 6 * 6 = 36 lens / transistors = 290 kwh annually. Then our 5 mWh energy usage would require 17 such panels and we are off the grid.

All the best,
Greg




Hi greg,
I dont know where you get the average home usage , but I need at my house down in florida with A/C 2,400 SF, two well and 1 pool pump each 1 1/2Hp, septic pump,1,000SF workshop,washer and dryer etc. 18 to 20 KW anually.
greetings
walt

Walter Hofmann

Hi stefan,
I honest dont know which I have connected I will check this again because I try both basis - collector and basis - emitter and both did bring out somethings one more then the other.
I also dont know what manufacturer it just sayd 2N3055 on top. I just took of the top (grinding around with cooling)then took off the plastic cover from the die.Th diameter of the camera lens is around 3/4inch dont know and doesnt say the anything about focul lenght,
The current readings are short circuit with a radio shack DVM set to 20 Amp.
Down here you can not use acrylic lenses because they whould not last more then a few hours in the sun then they start to deform due to the heat (up to 200 degr F)
I try 2 and 4 inch magnifier and they whould not bring more but the distence to focus is to far to make something stable, I aslo try one of this cheap reading glasse with 2.75 and at a distance of about 16 inches it brought allready 0.8 Amp short circuit.I do have a few of this trans laying around and if I find the time I will put a smal array together to see how much it can provide.
But overall in my opinion something like this cannot replace the power for the hole house it only can substitute portions, maybe up to 1kW, for light or so.
greeting
walt