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jns 06/09/2020
The four solar cells were about 6.5 inches by 1 7/8 inches each for a calculated total of 48.75 square inches. A square meter is a little under 40 inches by 40 inches. One percent of it is a little under 4 inches by 4 inches or 16 square inches. 48.75 divided by 16 is a bit more than 3. The standard light power for solar cells is the AM 1.5 global solar spectrum sunlight standard = 1,000.4 Watts per square meter. One percent is about 10 Watts and three percent is about 30 Watts so the 30 Watts in the description is the total input wattage at the AM 1.5 standard. 25% efficiency would yield 7.5 Watts, 20% efficiency would yield 6 Watts and 15% efficiency would yield about 4.5 Watts for conversion of solar energy to electrical energy. Those Watts would be converted to a usable voltage and current for a load such as a cellphone. The conversion is not particularly efficient due to the low voltages coming from individual solar cells. Conversion in large solar panels is much more efficient but those cost a lot more. My testing resulted in a maximum of 1.8 Watts to my devices. It seems reasonable to me. I measured a maximum of 93,600 LUX outside whereas the standard is based on a reading of about 109.870 LUX. At the standard, the output should be greater than 2 Watts. Further, I was testing on a 40 degree C day and the black fabric helped to increase the temperature of the cells, I am assuming another 15 degrees C. The folded solar cells are a nice size, about the width of a cellphone, useful for keeping in a glove compartment for an emergency. The ten connectors are: USB Mini-B, 2x Micro-USB, Nokia 2mm, Nokia 3.5mm, PSP 4mm, Apple 30 pin dock connector, LG sync for VX5400 and others, Samsung old 20 pin, Samsung new 20 pin (as best as I can see). I am charging a cellphone and a small tablet with the solar charger. This panel should have similar characteristics to others of a similar size and voltage.
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jns 16/09/2020
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jns Wow,what a difference a properly orienting the solar panel on a clear late winter day makes. Current was 2.67 Amps. It would probably go higher because it was about 90% of the LUX equivalent of AM1.5 and that was an hour earlier. The power was more than the 35W rating of the tester so a 100W 1-Ohm 1% resistor was put in series. The load on the resistor was I^2 x R or simply I^2. The total load came out to 38.29W max and the calculated voltage drop was 14.34V. Because a jumper was used and the current was reasonably high for small solar panels, some additional losses were in the conductors. Adding the losses to the total load and multiplying by 1.1 for the LUX difference and this panel comes reasonably close to the rated power output.

jns 14/02/2021
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jns Iwas able to recondition a sulfated 710 CCA battery from 550 CCA to around 700 CCA. It takes a long time but it works. Make sure you use it before taking a vehicle in for repair if you don't want to possibly have to pay for a new battery when your old one is determined to be sulfated.

jns 13/12/2019
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jns Assuming a 20mm wide top and bottom edge plus a 10mm wide left and right edge, I get a gross solar input of 352.8 W using AM1.5 of 1000W per square meter. It would take a conversion efficiency of greater than 50% to have a 180W output for this panel. At 20% conversion, the output before conductor losses is about 70W. Using that assumption of 20%, if Vmp was 18.0v, Imp would be about 3.89A. So the answer is no, not on the Earth unless you used a solar concentrator which would heat the panel to excessively.

2021-02-26 01:05:02 Helpful (4)
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jns 30W may be the solar power input before conversion using the AM1.5 standard of 1000W per square meter. It is hard to tell since the dimensions are not given.

2021-02-24 11:21:28 Helpful (0)
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