Different colored light on solar panels?

AI Thread Summary
Different wavelengths of light affect solar panel performance, with blue light generally producing the most energy due to its higher frequency, while red light generates less energy as it requires more photons to move electrons. Ultraviolet light can also be absorbed, but excess energy is often lost as heat. Solar cells have a cut-off frequency; below this threshold, no electricity is generated, and light above it can create unwanted heat. Composite solar cells can optimize energy conversion by using two materials to absorb different light frequencies effectively. Overall, both the color and brightness of light are crucial for maximizing energy output from solar panels.
algilaera
Messages
2
Reaction score
0
would there be any difference if different wavelengths of light is shined on a solar panel? like red, blue and green light. would blue produce the must power becuse the frequency is the heighest? also would ultrviolet light work on solar panels? would it produce more energy?
 
Physics news on Phys.org
It depends upon the solar panel. Different materials are sensitive to different frequencies.
 
Solar cells only use multiples of there absorbtion frequency, all other energy is dumped as heat.

Some types of solar cells absorb blue light so,
if you gave it red light, it takes a lot of potons to move the electron, so it generates verry little energy/light used
if you gave it UV light, it would take the energy it needed, then 'burn' off the rest as heat.
if you gave it blue light, it would thank you, and give you a lot of energy.
 
Single material solar cells have a cut-off frequency. Below the cut-off, no electricity is generated. The light passes through the active layer, and is lost. Light at the cut-off generates one electroon for each photon. Light above the cut-off generates one electron for each photon, but also creates unwanted heat.

There are composite solar cells that use 2 materials. The top layer let's low frequency light through, and converts high energy photons to electricity. The bottom layer converts the low frequency light, and is shielded from the more energetic photons that would cause heating.

Recent discovery of the band-gap of InN allows for the creation of a graded material. It would have a continuously varying band-gap that would allow for the design of a theoretically optimized solar cell, extracting the maximum conceivable electricity from sunlight.

Njorl
 
so you guys are saying different colored lights wount matter? and it wount matter if its a dim light or a bright light becasue if the lights strong enough to be above the "cut-off" point, it would produce the same amount of voltage?
 
Originally posted by algilaera
so you guys are saying different colored lights wount matter? and it wount matter if its a dim light or a bright light becasue if the lights strong enough to be above the "cut-off" point, it would produce the same amount of voltage?

Not quite.

The voltage produced is dependent on the structure of the cell.

The current produced depends on the number of photons absorbed.

The color of the light does matter. It needs to be of at least the cut-off frequency, or it won't power the cell. It is an all-or-nothing effect.

The brightness does matter. It is essentially a measure of the number of photons hitting the cell. This is a continuous effect - more brightness means more current. It is possible to saturate a cell though, in which case adding more photons will have diminishing returns.

Njorl
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top