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Things worked better than I expected. So here's an announcement. I figured out how to build photovoltaic panels with a 100% efficiency. I know some PhD said it couldn't be done and if I would've stuck with convention he would've been right. Well convention is a restriction we don't necessarily have to abide. I didn't and I achieved stunning results.

The next step is testing. Basically set up analog and record amperage and voltage for months. For those of you they can't wait I know how to build panel but not according to California standards which I've tried to find. So if you want were panel or by a licensed build panels from me give me a call at 614-706-1179.\par

Here are readings and calculations from a demonstration panel of 19 inches by 16 inches by 6 inches.

Cloudy day in April, near noon.

1) 26.24 fc is roughly 262.4 lumens (approximately 25%) of max daylight value.

2) (36" * 0.0254cm/in)"2=0.836sq ydl sq m 0.836 * 120 wattsl sq m = 1 00. 32watts/sq yd

3) (16" * 19")/36""2=0.2346 23% of a square yard

4) 5.27 v * 1.34 a =7.0618 Watts

5) 7.06 * 0.23"-1 = 30.7 Watts

6) If the panel was 100% efficient it would have generated 26.24 watts but the reading and math indicate the design generates 30.7 watts which is more than 116% efficient.

Clear day in April, near noon.

7) 5.5v * 4.73a = 26.015 Watts

8) 26.15 w x (0.2346"-1) = 110.89 w

9) 0.9144"2 is 0.836 x 120 W/m"2 is 100.3 w/sq. yd. So the panel has n efficiency of 110.5 % in direct sunlight but 110.89 projected watts 11 OO.s watts Isq yd=11 0.6% efficient

1. I like to use lumens when talking about sunlight because it provides a greater resolution than other measurement systems. I read 26.24 foot candles on the meter I had available and footcandles is roughly 1/10 of a lumen therefore 262.4 1m would have been delighted available through the clouds and approximately 25% of maximum sunlight. Note if a panel were 100% efficient and operated off of that efficiency then its wattage output would have been roughly 25 Watts.

Line 2 convert inches to meter and then converts to watts per square yard. Line 3.converts panel measurements in inches and gives the decimal equivalent of a square yard roughly 23%. Line 4 lists voltage and amperage readings taken in those light conditions and the wattage from multiplying the readings together. Line 5 converts the wattage from a fractional panel to a full square yard panel and its following line demonstrates the panel output is in excess of 100% for a lower light condition. 30 watts per square yard in a 26% light available from the sun.

Line 7 readings were taken on a clear day in direct sunlight to observe panel efficiency in those conditions. The readings wattage is the numerator and the denominator is calculated wattage of a fractional panel which indicates panel efficiency in excess of 100%.

Works better than I thought.

Now we are preparing for testing an about one square yard panel.