Hello,

Here is my problem.

I want to compute the irradiance over a flat area at distance d from the light source (LS).

The light source is a led. From all the Cree, Lumileds, Osram or Led Engin datasheets, I can extract these values:

Radiant Flux express in Ф_{e}(W)

FWHM or viewing angle express in °

Typical radiation pattern chart.

With these value, I try to compute the radiant intensity I_{e}(W/sr) and after the irradiance E_{e}(W/m˛) of a surface at d from the LS and view with an angle θ theta relative to the normal to the surface.

So, some leds have a radiation pattern which is close to a lambertian pattern I_{θ}= I_{0}.cos(θ) but other leds and lenses have dissimilar patterns.

All these radiation patterns are relative which means that the radiant intensity vary from 1 to 0 for 0< θ < 90°.

As I know the Radiant Flux how can I compute the radiant intensity I_{0}for θ = 0° ?

I try different methods, the last one:

I assume that the radiant flux ФSo, what is the good ?_{E}is contain in a 2π sr angle which is an hemisphere.

I divide Ф_{E }by 2π = W/sr and look at this value as the radiant intensity I_{0}for θ = 0°.

But it is not the good method.

I compute the area under the radiation pattern chart. This area is equal to 100% of the radiant flux Ф_{e}. but with this method at θ = 0° the radiant flux = 0 and of course, θ = 0° means solid angle equal to 0 !

Perhaps but….

So I take a relative radiation pattern, extract values with “Datathief” soft and work on the value.

Using Excel, I compute the area under the line. Assuming that at the half angle θ_{1/2}value the radiant intensity is half the peak intensity, I compute the relative part in % of the area from 0° to θ_{1/2}value and from θ_{1/2}to the include angle.

So for example using the datasheet of the Osram LH-CP7P, I have :Ф_{e}= 0.728W

θ_{1/2}± 40°

Values from the relative radiation pattern => Compute array

From 0 to θ_{1/2}, S_{0 to θ1/2}= 79.29 % of the total surface

From θ_{1/2}to 90°, S_{θ1/2 to 90°}= 20.71 % of the total surface

So, If I think properly, 79,29% of Ф_{e}is contained in the solid angle describeb by the θ_{1/2}angle.

As θ_{1/2}=> 2π . (1-cos(θ_{1/2})) sr , 40° => 1.46999 sr

Therefore, as I have 79.29% of Ф_{e}in 1.46999 sr , can I compute the radiant intensity at 40° as I_{40°}= 0.728 * 0.7929 / 1.46999 = 0.393 W/sr ????

As I_{40°}= 0.5 I_{0°}=> I_{0°}= 2 . I_{40°}= 2 . 0.393 = 0.785 W/sr ???????

Does any of the forum member do this sort of computation ?

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