# Narrowband photodiode vs LED

I am looking for a photodiode with a peak reception in the 630nm range. I really want high rejection of the 450nm band. However, this doesn't appear to be easily found, for sure not for low cost.

So my options are:

1. Use long-pass filters on more standard photodiodes.
2. Use an LED for a photodiode.

So the real question is: What are the drawbacks/advantages of using a standard LED for a photodiode? Speed? Reliability? Bandpass filtering? Linearity over temperature and input light intensity?

This approach will work if you are making a transmissive measurement like absorbance or a reflectance measurement, since these require both a light source and a detector. It will not work if you are trying to measure an emissive source, like fluorescence or irradiance. If you are making an emissive measurement, you are going to need to use the longpass approach with a real photodetector. For low quantities, try Anchor Optics, Edmund Optics, Thorlabs, or Newport. Here's a 500nm longpass filter for $21.50. http://www.edmundoptics.com/optics/optical-filters/longpass-edge-filters/longpass-glass-color-filters/66039 To answer your question about the drawbacks of using an LED as a photodiode specifically... I think the major thing to consider for your application would be that if you want to use an LED as a photodetector, you can't assume that the spectral distribution its response is equivalent to the spectral distribution of its output. For instance, just because the output of an LED has a center wavelength of 600nm with a 25nm bandwidth may not mean that it is not sensitive to light at 450nm. I am not positive about this, but I would be absolutely shocked if that were the case. • Without knowing the source of the unwanted 430 nm signal, I don't see how you can conclude that OP's problem is solved by changing the light soruce. – The Photon May 29 '14 at 17:15 • You need to read the entire response. The second paragraph of the response specifically addresses this issue. If the OP is trying to make an emissive measurement then you are right - he cannot solve his problem by conditioning the stimulus. – FL_Engineer May 29 '14 at 17:19 • Why do you think his current setup has 630 nm and 450 but not any other wavelengths present? Is there a vapor lamp source or something that would have these characteristics? – The Photon May 29 '14 at 17:30 • To your point, my guess is that yes, he is probably measuring fluorescence. But I see this exact question asked all the time ... even by people doing reflection measurements. The simplest approach in this case is to condition the stimulus but most people who don't work in this industry just don't think like that. You and I agree, he is probably going to need to purchase a filter. But maybe he's lucky and he can just condition the light source. – FL_Engineer May 29 '14 at 17:40 • Yes, its fluorescence, so the real issue is to filter out the 450nm source. – Erik Friesen May 29 '14 at 17:46 Low cost optical filters that will fill your needs are readily available. What you want are Wratten optical filters, in this case a 12 or 15 will do the job. These will give you better than 1000:1 response differences between 630 and 450 nm. See, for instance http://motion.kodak.com/motion/uploadedfiles/Kodak/motion/Products/Lab_And_Post_Production/Kodak_Filters/W2-12.pdf for the response curve of a 12. As an example source, Edmund optics will sell you a 3" x 3" filter for$50, and you can cut this down to size as you like. Cutting down to 1/2 x 1/2 (a decent size window for any but a monster photodiode) will give you a per filter cost of less than \$2. The filtering is intrinsic to the glass, so you don't have to worry about babying a dichroic coating. Transmission is a little less than you can get with higher tech, but not outrageously bad.