what would be the wattage of a 20x20 cm solar panel if the efficiency is 44% instead of 5%

 

Google and all other search engines not working today?

 

what would be the wattage of a 20x20 cm solar panel if the efficiency is 44% instead of 5%

A 100 W solar panel is a 100 W solar panel, efficiency is "another number". it doesn't matter if it's 100% efficient or 1% efficient.

 

Based on a 7 in X 6.25 in panel =3.5W @ 13%, then 20 cm X 20 cm @ 5% = 1.9 W, 44% = 16.8 W.- give or take.

 

Google and all other search engines not working today?

you have been reported. now if someday you can change the rules to accomodate your view of how things should be, until then, keep away.

 

A 100 W solar panel is a 100 W solar panel, efficiency is "another number". it doesn't matter if it's 100% efficient or 1% efficient.

you have problem understanding basic solar electronics, that is your problem. see reply by bernard, that person understood the OP perfectly and calculated based on watt per square centimeter based on efficiency.

 

I worked in the industry setting up and doing the basic measurements. I don't think you asked the proper question. Besides that, the material the solar cells were made of might enter into the calculation. Here http://www.worldwatch.org/node/4803 is an announcement of a 40.7% efficient solar cell.

This http://www.nanoflexpower.com/gallium states that:

GaAs is the highest performance solar material currently available, boasting conversion efficiencies in excess of 40%, nearly double those of crystalline silicon. This means that it can produce nearly twice as much power in a given surface area.

Is that what you were asking with different initial and final numbers?

Power density, current density and efficiency is the normalized number. e.g. mW/sqcm and mA/sqcm.
When you buy panels you think in terms of other numbers. Voc, Isc and max power.

 

what would be the wattage of a 20x20 cm solar panel if the efficiency is 44% instead of 5%

Hi there,

Not sure if you got a direct answer yet or not

A solar panel that can somehow jump from 5 percent to 44 percent would ideally put out 44/5 times as much power. and that comes out to 8.8 times as much power. That's provided everything else could handle the power too such as the wiring to the panel.

The reason for this is because power scales as a simple ratio because it's a statement about energy and energy scales as a simple ratio. To contrast, if we were looking at a statement about either current or voltage ALONE (not both at the same time), that would be entirely different, because we have to multiply the two to get the power. But let's do a simple calculation.

The insolation level is approximately 1000 watts per square meter in direct overhead sunlight. That's an amazing figure in itself. But if we had a solar panel that was only capable of 5 percent efficiency, that would mean the output would only be 50 watts per square meter of surface area. Given a 44 percent efficient panel though we would ideally get 440 watts per square meter output from that same level of sunlight. That's 8.8 times more power per unit area and as a fraction is simply 44/5 again. For two panels of the exact same surface area but one is 5 percent and the other 44 percent, the ratio of wattage output is the same too, 44/5 times as much power output.

There are other things that have to be considered for an installation such s wiring size, but once we are given the efficiency figures we at least know what the panel is capable of.

I happened to be lucky enough in the past to be able to work for a time with a million dollar solar array for a project backed by Sandia Labs. It was very interesting. Part of the study was to find out how a very large number of line tied inverters (fed from solar panels) would affect the utility grid.