I'm trying to find high brightness UV LEDs appropriate for this project:


which includes: "The most critical parts of this project are the UV LEDs, you are looking for 5mm Ultra Violet LED 2000mcd 395nm, 3.4V 20~25mA."

However, it appears that the mcd (luminous intensity) criteria is ambiguous for LEDs and other light sources that emit a peak wavelength that isn't visible to the human eye, such that a directional (solid angle) emitted power (wattage) criteria would be more appropriate. That might explain why it's so difficult to find anyone who can meet those specs. Is there anyone with experience in UV leds who can help? Thanks.

  • 1
    \$\begingroup\$ There is really no reason to waste time on low power 100mW through hole LEDs at 395 nm where 25 or 50 watt COBs are inexpensive. \$\endgroup\$ Commented Jul 21, 2020 at 6:53
  • 1
    \$\begingroup\$ Do you want UVA UVB or UVC? But I agree COB's are really cheap But not so uniform up close since they are almost cylindrical with top emitters too \$\endgroup\$ Commented Jul 21, 2020 at 7:14
  • 1
    \$\begingroup\$ As you correctly state the unit of candela or lumen make no sense in the realm of invisible light. What the author is saying is: buy some cheap 5 mm UV LEDs with doubtful specifications... \$\endgroup\$
    – Arsenal
    Commented Jul 21, 2020 at 9:19
  • \$\begingroup\$ user1850479, here's a COB array that may do the sort of thing the instructable prescribes: amazon.com/Chanzon-Ultraviolet-30V-34V-Components-Lighting/dp/… The package dimensions (4.9 x 4 x 0.3 inches) (as I read it) are for the apparatus (not just the emitting region), but correct me if you think I read that dimension wrong. Since my aim is to develop positive photoresist film for which I'd like good homogeneity of illumination, I just need to hold the board higher over the array (with a corresponding decrease in overall illumination of course). Do you agree? Thanks. \$\endgroup\$ Commented Jul 21, 2020 at 21:52
  • \$\begingroup\$ Tony, according to one source, 320-395 nm is classified as UVC (the UV region that 'kills things'). Since I mentioned 395 nm, I guess I want UVC. And as I mentioned in the previous comment, raising the film higher over the array would appear to assist in providing more uniform illumination. \$\endgroup\$ Commented Jul 21, 2020 at 21:57

1 Answer 1


The datasheets can be a bit all over the place in terms of these units, especially with luminous intensity which is a function of how well the eye perceives the particular wavelength.

Since the application is a box, I would not worry too much about directivity (or the eye), and focus more on getting an LED with a respectable "Radiant power" (total radiated power) to electrical power ratio.

I have used "VLMU1610-365-135" (the datasheet is quite good in my opinion) for a project where it has P_rad=23mW at 20 mA. And it states a forward voltage of 3.5 V at 20 mA. So P_in=70mW, and thus we are around 33 % efficient in radiating light. I might have seen some with 50 % which seems to be around the top.

About this box, remember to be able to dissipate the heat from the LED's through the sides. LED's lose quite a lot of radiant performance when they run hot and the consultant who helped me said that it was the most common error in power LED applications.

  • \$\begingroup\$ Thanks, Bonnevie. I'll be studying that datasheet you included. Do you have an opinion on the link I shared above in terms of the instructable? Any intuitive feel if it's in the same area 'illumination' ballpark as the instructable's discrete LED array? I'll be mindful of heat dissipation considerations. And thanks everyone. Yours are all useful comments. Looks like the COB array can be heat sinked from its underside. \$\endgroup\$ Commented Jul 21, 2020 at 22:06
  • 1
    \$\begingroup\$ Depends on the application. With the COB array linked I think, even at distance, it will apply much more power in the middle than in the sides. It is also not your avg. power supply you will need for it. Personally I would do a PCB with SMD's for this power level. \$\endgroup\$ Commented Jul 22, 2020 at 11:33
  • \$\begingroup\$ The instructable's diode array would apply much more power in the middle than the sides too, so that for truly even illumination, a larger diode spacing at the center decreasing toward the edges would be in order of course. Good point. Since one can buy a pack of 200 LEDs for under $10, that would be the cheapest option. But that $45 COB module would save so much time and trouble that it might be worth it, even without truly even illumination, since, within a certain margin of overexposure (i.e. leak exposing masked regions), once the photoresist film is exposed it can't be further exposed. \$\endgroup\$ Commented Jul 22, 2020 at 18:13
  • \$\begingroup\$ Yes that is almost what I was fishing for. If the application is "It has burned away / it has not burned away" then you are good. I agree that the time spent won't always be worth it. Also the more you try the more you learn, go for COB initially. To note on the LED distribution: It depends on if your input material have a smaller area than the LED panel. I agree that if the input material have the same surface area as the LED panel then the middle will get more power than the sides. But the effect is diminishing with surface ratio. \$\endgroup\$ Commented Jul 23, 2020 at 6:33
  • \$\begingroup\$ Oh there also exists UV LED strips which you can distribute more than that particular COB. \$\endgroup\$ Commented Jul 23, 2020 at 6:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.