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You are right that the use of "Candela" (as mcd etc) is somewhat dubious for UV LEDs.Candela is the actual output power weighted (according to wavelength) by the eye-sensitivity. This means that an X-candela UV led at (for example) 390nm may be pumping out far less photons than an X-candela UV LED at 360nm (where X is the same) ... as the eye is much less sensitive to 360nm than 390nm. The Candela weighting function is the CIE luminosity function, sometimes known as Ybar. Ybar on my webpage ...sorry, you cannot view external links. To see them, please
REGISTER or LOGINThere isn't a well-defined cutoff when visible becomes UV - there's a continuum where the eye becomes progressively less sensitive, as you'll see from the Ybar function (on my webpage or linked tables).It's probably dodgey/dubious to use any photopic measurement unit (Candelas or lumens) for wavelengths shorter than about 420nm. You'll notice that Philips' LumiLEDs are quoted in watts even for the deep blue rather than Candela/lumens for this reason.(Similar arguments arise with deep red LEDs - 625nm (an orangey-red) tend to have higher brightnesses (mcd) than 645nm (deep red) again because the eye is less sensitive to the longer wavelength so the Candela rating for 645nm is lower for the same radiometric (light photons or light watts)).The second thing to be aware of is that the manufacturers' measure of mcd (or Cd) is _in the main beam_... therefore if you have the same LED-chip in two LED-packages running at the same power and efficiency... the one with the optics giving the narrowest beam angle has the highest mcd-rating (even though the total number of photons given out summed over all directions is identical).If you compared the solid-angle of the beamwidth you could roughly compensate for this effect in the specification.Another question you need to ask yourself is what you want the UV for. Is there something specific you are trying to make fluoresce which needs a specific wavelength?Conventional mercury UV "blacklight" tubes (fluorescent lamps without the fluorescent material) are normally filtered to emit primarily the 365nm wavelengh which is _relatively_ eye-safe. Shorter wavelengths (mercury tubes can also be filtered to pass ~305nm or 254nm) are significantly more dangerous for a given (watts) power.Regardng eye-safety you need to be a bit careful with anything deep-blue ("Royal blue") or further into the UV at sufficient powers. The LumiLEDs/Luxeon Royal Blue if driven at a watt or so (eg 350mA) will easily give you temporary retinal burn (spots for a minute or two) if looked at directly even for a few seconds. 5mm UV LEDs driven at a few 10's milliamps are less concerning, but if you've got UV LEDs where you're passing 100~200mA or more then you really should be taking some care - certainly not looking directly at the LEDs ... even more so as the wavelengths get shorter than 390nm.Wikipedia reckons that "LED efficiency at 365 nm is about 5-8%, whereas efficiency at 395 nm is closer to 20%"... that efficiency _ought_ to be quoted in watts of light out vs watts of power in. I don't know the accuracy of the figures, but they sound plausible. Thus a 395nm LED of a given input power will give out perhaps around 3x as much UV light power (in watts or photons) as a 365nm LED. But the 395 will be more visible as a dull violet than the 365 which will be more invisible on non-fluorescent surfaces - also the 365 will (for the same amount of light Watts, or photons) probably excite stronger fluorescence effects in general - if that's what you're looking for.So..... not a simple answer, but hopefully you'll have a better understanding of the problem now :-)
Based on the materials I can use at home, I have thought the next experiment to "measure" (even relatively) the UV outputs:* to open one tube edge of one CFL that got out of service.* to put the LED inside the CFL tube pointing the beam outside the glass wall.* to light the LED on.* to take a picture of the illuminated point (from outside) in a dark room. I plan to use a DSLR camera with a tripod.* to repeat with the other LED in the same conditions.* to compare both pictures in a Photoshop type software.What do you think? Will it work? At least, will I be able to check that the LEDs are yielding UV light? Should I diffract the light with a CD/DVD before take the photos?How can I avoid the violet (visible) light to get to the lamp (to be sure that I only photograph the white light produced by the UV light)?.
I am reminded of an April front cover on Electronic Design News (EDN), probably 1983, where they showcased a brilliant new technology, complete with colour photographs - the FED - Flame Emitting Diode.I wish I had kept the front cover. If anyone still has it, please let me know.
(...) Try woos glass.
* the first LED has reduced its luminosity (or whatever is the name) to about 50% after a few minutes of use. (It can be due to a shortcut in the circuit that has killed most of the visible light, perhaps also some of the UV component. It can also be a defective LED). I had already take several photos. As the LED was able to produce luminescence on a 50 note, I thought it was able to still shed UV.
* the camera "see" the violet light as blue. Only when I have set up the white balance to 10000K it has begun to display the light as purplish, with a blue (not real) halo and white in the centre of the LED emitter.
That doesn't sound good. Perhaps you're over-running the LED and the heat is damaging it. Make sure you don't exceed the rated current, and if it's designed to be affixed onto a heatsink, then do so (using a suitable thermal paste)!
You are right. In spite of I can't notice any heat, I guess I've been overrunning the LED. I thought 4.5V would be fine (since blue and deeper need higher voltage than classic red and yellow LEDs), but today I've seen (in a page with similar LEDs) that they require only 3.2 to 3.6V. I still have another one to test (and the other 2 are untouched).And I have to add a resistor.