Lux Calculations?

Hello Matt,

I suppose that you are talking about white light.

You may expect about 40 lm/W (40 lumen per watt electrical input). Assuming 10W electrical input, you will get 400 lm output.

When you want to have an illumination of 10k lux (lm/m2), your spot can have a maximum size of 400lm/10k lm/m2 = 0.04m2. That is equivalent to a circle of 22cm (8.7"). Because of the non-uniformity of the radiated field and stray radiation, you will probably not be able to produce 10k lm/m2 on a disc with D=22cm.

When you want a distance from e light source to the object of about

So probably you will use several LED's with really narrow angle, so you can form your own spot. Using more LED's may reduce die temperature, enhancing life span and efficiency. With LED's virtually all heat goes to the heat sink (less direct heat radiation as with other light sources).

When you like green light, you can have far more lumen output with same electrical input because green is in the middle of the eye sensitivity curve.

Hope this will help you a bit.

Wim PA3DJS

How did you come up with 12.5 degrees? Would this be for a single LED at 400 lm ?

The LED's that I've been looking at have had an angle of 120-140 degrees, and about 100-140 luminous flux, could you help me make a calculation with the higher angle and reduced luminous flux?

Here's the spec sheet that I'm getting my numbers from. Maybe I'm looking at the wrong thing?

The 10k lux is for phototherepy, to reduce the effects of winter-time blues.

If I was to use a wider angle, how many LED's would I need to use.

Is it as simple as calculating the lux of one at this wider angle, and adding them together to make 10k lux?

Yes, Green light is the color that I'll be using for this project, just for that reason too.

Helped a bunch, Thanks Matt

TazaTek snipped-for-privacy@tazatek.com posted to sci.electronics.design:

Nearly, from the datasheet the patterns are about 40 degrees at 70% but when you add them up you have to take position into account as well, then overlay the patterns and add the pattern correct contours up.

Hi matt,

The 12.5 degrees It was just an example, when you want to illuminate a circle with D=0.22m, from 1m distance, your light source should have a beam width = invtan(0.22/1).

In general, for green light (555nm), 683lm = 1W radiation power. Lux = lm/m2.

Determining the required optical power is the first step. This can be done by determining the area to be illuminated (A, m2). Multiply this with the minimum required illumination (Iv) in lm/m2. The result is lm (optical power).

Now you know the minimum required optical power in lm. You should take into account stray light, so probably you want more.

The second step is to find a light source with the required optical output and suitable radiation pattern (maybe optics required).

For a certain convenience, you probably want some distance between your book and the light source. The area to be illuminated and distance from source to area determines the beam witdth. By using more LEDs with narrow beam witdth, you have some control over the light distribution. When you use a diffuse wide angle LED, many optical power is lost.

A good source on light is the "light handbook" You can download it for free.

When you want to illuminate a very large area with indirect light (gives better comfort), you also should take into account the reflectivity of wall cover.

Hope this will help a bit.

Best regards,

Wim PA3DJS

In article , TazaTek wrote in part:

A lot of high power LEDs have a nominally lambertian radiation pattern. In the lambertian pattern, the "nominal beam width" (edge of which has half the central intensity) is 120 degrees, and candela is lumens divided by pi. Lux is candela divided by square of distance in meters.

You probably want something with optics to concentrate the light into a narrower beam.

- Don Klipstein ( snipped-for-privacy@misty.com)

In article , Wimpie wrote in part:

Since visible LEDs of color other than "royal blue" generally have output specification in photometric terms rather than radiometric terms, I would skip everything having to do with watts of optical power.

Also keep in mind, some LED manufacturers have their best white LEDs giving more photometric output than their best green ones. This means that the white ones have greater radiometric output by an even larger margin.

- Don Klipstein ( snipped-for-privacy@misty.com)