Science Buddies: "Ask an Expert"

I am measuring the output of a solar cell while changing the position of the solar cell. I am not sure if I am using the right measurements? I started with 0, 30 and 60 degrees, then added 90 degrees. Am I using the right angles? I am doing this outside AT 4:15 PM each day. My output averages 6.72 to 133.32.

Hello and welcome to the forum:

In an experiment, it 's hard to know what the "right" anything is If you could give us some more information, we could help more. You might have to modify your experiment based on the data you're collecting, in other words the angles that give you the best data comparisons might change after you look at the data.

Just a tip about measuring solar cell output. Students often measure either voltage or current from the solar cell as a measure of its output. Neither are good measurements alone. You really want to know the power output instead. You could use a standard resistor across the solar cell and measure the voltage or current and calculate the power - that's a much better way to go.

Best,
Keith

Erick+ - The optimum angle for a solar cell to receive maximum radiation is to be absolutely perpendicular to the direction of the sun or light source. Any change from perfectly perpendicular will limit the radiation per unit area of the solar cell to convert the solar energy. As a theoretical, assume the cell is perfectly perpendicular to the sun or light source. Call that 0⁰. This will give you maximum energy conversion. If you move the cell 90⁰ from that position, direct energy from the source will be zero. Any angle in between will give you a varying degree of energy generation. It will not be a linear relationship due to several other factors, but it might be close. For example, a position of 45⁰ may measure close to 50% compared to the perpendicular measurement. Keith correctly notes that a simple voltage measurement or current measurement with a multimeter is misleading. You should calculate power generated in a load, a resistor that approximates the internal resistance of the solar cell for example, to make the most accurate assessment of energy generated. You can get an approximate calculation of that internal resistance by measuring the near short circuit current as measured by a low impedence multimeter (try the 200 mA range), then measure open circuit (only the voltmeter as load) voltage in the same illumination. This represents a load line for the cell. Divide the voltage by the current to get a rough approximation of the internal resistance of the cell. For example, you measure 200 mA at zero voltage, and 6 volts at zero current. Divide 6 volts by 0.2 Amps and the resistance is 30Ω. Use that approximation for your load to measure current and voltage for your power calculation.

Rick Marz

Thank you so much Mr Rick and Mr Keith !!! That is what I needed.
Erick