0
$\begingroup$

I am building an LED-light and am trying to figure out how many LEDs I need to be on it. I know how much illuminance I need as well as the distance to the subject and the luminous flux of one of the LEDs. The problem is that the LED's data sheet doesn't provide a clear viewing angle but instead a directivity diagram / radiation pattern (page 11).

Question: How do I compute the illuminance of one LED (26lm) at a given distance (1m) based on the directivity diagram / radiation pattern?

Edit: More specifically, I would like be able to compute the average illuminance on a surface of a given size at a given distance, for example the average illuminance on a circle with a radius of 10cm at a distance of 1m that's center point is straight in front of the LED and that's not tilted in any direction.

$\endgroup$
1
  • 2
    $\begingroup$ Hi and welcome to Physics SE! I think this is a good question; however, it might be more appropriate for Engineering SE (engineering.stackexchange.com), as it seems more design-based than physics-based. I will propose that it be transferred. $\endgroup$
    – Time4Tea
    Jul 30 '18 at 18:48
0
$\begingroup$

I assume this is not a homework question.

First, you need to calculate the angle subtended by the circle.

For numbers in your example, the angle will be $2\times arctan(0.1m/1m)=11.4^{\circ}$.

Then, using the radiation diagram, you need to determine (roughly) the percentage of the total luminous flux of the device within that angle. It will be the same in 2D and 3D, so you can calculate it for 2D.

Then you can determine the average illuminance of the circle as a ratio between that flux and the area of the circle.

$\endgroup$
4
  • $\begingroup$ This is no homework question, I wish we did something like this in school. Just to make sure I understand correctly: at 5.7° the relative illuminance is still roughly 1 meaning the average illuminance equals 26lm / ((0.1m)^2 * pi) = 828lx. This number seems very high. On another note, wouldn't this mean that if I had an LED that had the same relative illuminance along all angles, the illuminance would change based on how big the area I consider is? $\endgroup$
    – user202748
    Jul 30 '18 at 21:19
  • $\begingroup$ @john Sure. But in your calculation, $26lm$ is the total luminous flux of the LED. You have to take a percentage of that flux corresponding to $11.4^{\circ}$ angle, which would be on the order of $10$%. $\endgroup$
    – V.F.
    Jul 30 '18 at 22:04
  • $\begingroup$ Okay, thank you. How do I take that percentage? Is it the ratio of area within the angle that is covered by the circle and the full circle's area? $\endgroup$
    – user202748
    Jul 30 '18 at 22:46
  • $\begingroup$ @john I believe so. $\endgroup$
    – V.F.
    Jul 30 '18 at 22:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy