Maxiumum Incident power on solar cell ?

[spoonfeeder]

Hi
I have simulated a dual junction(Multijunction) solar cell of thickness 10 microns and length 20 microns in silvaco atlas device simulator. In order to calculate its efficiency i need exact value of incident solar power i.e Pin for the above said structure.

These are the parameters obtained from IV graph:
Isc=9.25A
Voc=1.32V
Pmax=9.828W
FF=0.80
To find out efficiency i need Pin(Total Incident solar power) for the mentioned structure length and thickness.

 

[Daiquiri]

The power of sunlight at the ground level, when all the (mean) atmospheric dissipations and reflections have been detracted, is very close to 1000 W/m2 (outside the atmosphere it's value is 1367 W/m2 at the mean distance Sun-Earth).

1000 W/m2 is also the standard reference value used for testing and certification of PV cells.

 

[Neandethal00]

The power of sunlight at the ground level, when all the (mean) atmospheric dissipations and reflections have been detracted, is very close to 1000 W/m2 (outside the atmosphere it's value is 1367 W/m2 at the mean distance Sun-Earth).

1000 W/m2 is also the standard reference value used for testing and certification of PV cells.

Pin = 1000W/m² = 100mW/cm² = 645mW/in²
.
is generally accepted power input on a solar cell, which occurs under certain optimum conditions. In practical situation the value differs from day to day and even from hour to hour. The best way to find the exact input power at certain time is to use a 'light meter'.

I once calculated the efficiency using 100mW/cm², then when I used a light meter to verify the number, I found power in the AM hours in my area (south-eastern USA) actually is 50mW/cm².

 

[Daiquiri]

I once calculated the efficiency using 100mW/cm², then when I used a light meter to verify the number, I found power in the AM hours in my area (south-eastern USA) actually is 50mW/cm².

1000 w/m2 is the standard value of irradiation, used for certification and comparison of PV cells - which is the case here, if I understood it well.

For example, see this tech sheet of a Sharp 167 W module: http://www.abcsolar.com/pdf/sharp167.pdf - the bottom of the page 2, where it reads "standard test conditions".

Then of course, the actual value of irradiation will strongly depend on the local (geographical) weather conditions, like clouds, air humidity, air pollution, period of the year, morning, noon or evening... etc.

A bit like when we say that the air density is 1.225 kg/m3. It is a standard reference value valid for dry air at 15 °C and 101325 Pa of air pressure, but as soon as one of these conditions change, the value of density will change too.

Cheers!

 

[PJC01]

I would like to commend you for conducting a simulation to determine the efficiency of your solar cell. In order to accurately calculate the efficiency, it is important to have the exact value of the incident solar power (Pin). This value can be obtained by measuring the intensity of sunlight hitting the solar cell, taking into account factors such as angle of incidence and atmospheric conditions. It is also important to note that the maximum incident power on a solar cell is dependent on the size and characteristics of the cell itself, as well as external factors such as shading or temperature. Therefore, the maximum incident power for your specific dual junction solar cell may differ from that of another solar cell with different dimensions and properties. It is crucial to accurately determine the Pin value for your particular solar cell in order to obtain a precise efficiency calculation.