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Registered Member #71
Joined: Thu Feb 09 2006, 08:23AM
Location:
Posts: 63
Hi all,
I have a question about LEDs and photodiodes/LDRs.
The dominant wavelength of an LED is a "mixed colour" output of the LED which appears monochromatic to the human eye. Thus, the dominant wavelength is a function of the human eye spectral response.
Now imagine there is an LDR or photodiode which does not have a spectral response similar to that of the human eye. Its perceived "one-colour wavelength" of the LED will be different to the dominant wavelength.
So I am wondering how one would calculate the "dominant wavelength" as appears to the LDR or photodiode? Is it a matter of multiplying the LED output spectrum with the sensitivity of the LDR/photodiode and then performing some kind of calculation?
Registered Member #3215
Joined: Sun Sept 19 2010, 08:42PM
Location:
Posts: 780
for what I know, you need to multiply the emission spectrum of the LED by the sensitivity spectrum of the LDR or photodiode
what you'll get is a somewhat arbitrary unit but it should give an accurate idea
what do you plan to do? maybe I could be of some help...
if you are looking to distinguish precise wavelengths of light there are some dichroic mirrors or interference filters which are a valuable tool, and I know a very nice ebay seller who gives spectrums for each piece of filter he sells
Registered Member #71
Joined: Thu Feb 09 2006, 08:23AM
Location:
Posts: 63
Multiplying my two functions together, the peak is at around 470nm for my blue LED. Take this to be the dominant "LDR perceived" wavelength.
In the RGB colour space this wavelength is approx. R=0,G=153,B=255 (using 0=min, 255=max). This is assuming we are concerned only about chromaticity and that luminosity is at a unit level.
So, when I shine this blue LED onto a purely red surface (ie R=255,G=0,B=0), the reading from the LDR should be the same as if the surface was totally black? This is my understanding?
I have tried this and the reading is not the same, in fact there is a large difference between a totally black surface and the red surface. Is there something wrong with my logic here, and is it even viable to use the RGB model in this sense?
Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
To get some kind of a meaningful answer, you have to take the spectrum you got by multiplying the two spectra together as discussed above, and multiply it by the reflectance spectrum of the red surface. (Even though it looks red, it probably reflects a tiny amount of blue.)
The area under the resulting curve will (AFAICS) be a measure of the output from the LDR. For instance, if it is half what the area was under the previous curve, the output will be halved.
(Disclaimer: up to linearity considerations of the LDR, and the absolute attenuation of the reflectance)
The RGB colour space is too limiting for physical problems. It isn't even quite adequate for describing the response of the human eye: there are quite a few colours we can perceive that are slightly outside of it, as anyone who has dabbled in digital photography knows.
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Dominant wavelengths and other jazz!
