I need to select LED/LEDs for a light source in my project. The requirement is that the light source should approximately be able to project light beam of 1.5m diameter on a surface, from a distance of 1.5m. The incident light should have lux equal to 1500. How do I go about selecting the LEDs for this application? What factors should be considered?Diagram attached to illustrate the scenario

  • \$\begingroup\$ Figure out how many watts of light output you need, and find an LED that's rated for that. If you need the beam to have that specific a shape, you're probably going to need to add lenses yourself. \$\endgroup\$ – Hearth Mar 14 at 17:27
  • \$\begingroup\$ Buy a suitable flashlight. \$\endgroup\$ – Chris Stratton Mar 14 at 18:02
  • \$\begingroup\$ LEDs typically don't just cast a beam out of the box, there's surrounding peripherals to gather all the light (usually in a pretty wide cone, maybe 170degrees) and focus it. In particular, I'd look at Total Internal Reflection lenses (TIR) in conjunction with the other advice given here. \$\endgroup\$ – Orotavia Mar 14 at 20:08
  • \$\begingroup\$ lux is a unit, you want an illuminance of 15000 lux. You should consider whether the illuminance should be homogeneous in the drawn circle of 1.5 meter (quite hard to achieve i think) Or is illuminance allowed to be not homogeneous and do you want just 15000 lux at the centre or at least 15000 lux at the circumference of said drawn circle? If it should be homogeneous, dont forget to take the spatial distribution of the LED into account. \$\endgroup\$ – Huisman Mar 14 at 22:51
  • \$\begingroup\$ 15000 lux should be the OP's 1500 lux of course... \$\endgroup\$ – Huisman Mar 14 at 23:02

This site has a lumens to lux calculator, given beam angle and distance.


Looks like ~2300 lumen would yield ~1500 lux at 45 degrees and 1.5 meter.

Might look at "tactical flashlights" like these and see what kind of LED & lenses they use to achieve 1300 lumen, go from there, use 2 of them kind of thing.

  • \$\begingroup\$ I think the beam angle should be 2*arctan(1/2) = 53°. \$\endgroup\$ – Huisman Mar 14 at 23:31

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