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I'm trying to design an LED light, but I'm not sure how many LED's I need for a given light level.

Specifically, the Luxeon LXR7-RW57 1000 lumen white LED (data sheet)

Q: How many lux should I see at 1 meter with just one of these LED's? (Assuming no reflector, and a radiation pattern as in the data sheet page 16)

Flash
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    I'm voting to close this question as off-topic because it is about the engineering/design of a lighting system, not about the creation, editing, processing, or presentation of a still photograph. – scottbb Jul 30 '18 at 17:25
  • It appears to be ~1200 lux according to the calculator at http://www.ledstuff.co.nz/data_calculators.php. I chose a beam angle of 60° to preserve 90% of the luminary strength – Stan Aug 02 '18 at 15:39

2 Answers2

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The polar radiation pattern of Fig. 15 (p. 16 of the datasheet) for the white LEDs looks very close to a circle. It is thus reasonable to assume that these LEDs have approximately a lambertian pattern, which is confirmed by the fact that the intensity falls to 1/2 the maximum at 60° from the optical axis (cos 60° = 1/2). Based on this, you can deduce that the on-axis luminous intensity is

I = Φ / π = (1000 lm) / (π sr) = 318 cd.

At 1 m distance, and assuming the plane you are lighting is facing the LED, the illuminance is

E = I / (1 m)² = 318 lx

But this is only straight under the LED. If you are lighting an extended plane, the illuminance will fall-off with a cos⁴ law as you move out of the center of the light spot.

Edit: I am adding some rigorous derivations to support my sayings. You can skip them if you are afraid by math or you just trust me with the integrals.

Computing the on-axis intensity

Let’s assume that the luminous intensity I has an axially-symmetric distribution, i.e. it depends only on the angle θ between the direction of measurement and the axis of the LED. Then, the total luminous flux emitted by the LED is the intensity integrated over all directions of space:

Φ = ∫ I(θ) dΩ = ∫ I(θ) 2π sin(θ) dθ,

where dΩ = 2π sin(θ) dθ is the element of solid angle. Judging from Figs. 14 and 15 of the datasheet, it appears that I(θ) quite closely follows Lambert’s cosine law:

I(θ) ≈ I(0) cos(θ) for θ < π/2, zero otherwise

(the relevant curves are the ones labeled “White”, the “Royal Blue” has a different radiation pattern). Then, the total flux is

Φ = 2π I(0) ∫ cos(θ) sin(θ) dθ

The integral is for θ in [0, π/2], and it evaluates to 1/2. See Wikipedia on Lambert's cosine law for the derivation. Thus we have

I(0) = Φ / π = 318 cd.

It is worth noting that the same result can be achieved by a very crude approximation: that I(θ) is equal to I(0) inside the 120° cone, and zero otherwise. Then

Φ = ∫ I(0) 2π sin(θ) dθ for θ in [0, π/3]

By a mere coincidence, this crude approximation happens to give the very same result as the cosine law. On the other hand, if we really need more precision, we could digitize the curve from the datasheet and compute the integral numerically. I leave this as an exercise to the reader. ;-)

Computing the illuminance

Let’s assume that we have a flat surface at a distance z = 1 m, directly facing the LED, i.e. normal to the LED’s optical axis. We then have a light spot which is brighter in the center (on-axis with the LED) and fades away progressively as one moves out of the center. Let dS be an elementary surface at the center of the spot. This surface captures the light emitted over the elementary solid angle

dΩ = dS / z²

and thus the flux

dΦ = I(0) dΩ = I(0) dS / z²

The received illuminance is then

E(0) = dΦ / dS = I(0) / z² = 318 lx

This calculation can be extended to a point lying at a distance r from the center, for which the light rays arrive at an angle θ from the LED’s axis. We get:

dΩ = dS cos(θ) / (z² + r²)
dΦ = I(θ) dΩ = I(0) dS cos²(θ) / (z² + r²)
E(r) = dΦ / dS = I(0) cos²(θ) / (z² + r²)

but since cos(θ) = z / √(z² + r²), and I(0) = E(0) z²,

E(r) = E(0) cos⁴(θ) = E(0) z⁴ / (z² + r²)²

which leads to the following iluminance pattern, as a function of the distance r to the center of the spot:

r (m)   E (lx)
 0       318
 0.5     204
 1        80
 1.5      30
 2        13
 2.5       6
Edgar Bonet
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  • The angle is not because it is lambertian (which implies diffuse and this is hte opposite). You can buy these with different angles and it is the lens on top that controls it. it is chosen by design. It makes no sense to talk about X lux at a point in front, unless you are talking about a laser, as 1 lux is defined as 1 lumen at 1 m2 , and this is not assuming a 100% to 0% hard barrier. – Michael Nielsen Oct 07 '13 at 21:57
  • @Michael Nielsen: By “lambertian” I mean it follows the Lambert’s cosine law. Whether it’s diffuse or not is irrelevant: only the angular distribution matters for computing the ratio of flux to intensity. I know you can buy them in different angles, but this particular item (i.e. 120° full angle at half maximum) just happens to have a lambertian angular distribution. Or very close to it: look at the datasheet. – Edgar Bonet Oct 08 '13 at 06:43
  • @Michael Nielsen: No, 1 lux is not defined as 1 lm at 1 m², it’s defined at 1 lm per m². In practical terms it’s the flux collected by your luxmeter (in lm) divided by it’s integrating area (in m²), which is always much less than 1 m². – Edgar Bonet Oct 08 '13 at 06:45
  • Comment only: Not a true lambertian pattern - according to my eye-brain after looking at 'a few' LED radiation patterns in recent years BUT will no doubt be close enough for the purposes. – Russell McMahon Oct 08 '13 at 14:14
  • @Russell McMahon: I am not talking about LEDs in general, but very specifically about the Luxeon LXR7-RW57. Look at the datasheet: if it’s not perfectly lambertian (which it’s probably not), it’s nevertheless pretty darn close. – Edgar Bonet Oct 08 '13 at 15:08
  • @EdgarBonet yes thats what I said. 1 lux is 1 lm spread on 1 m2. and if you get more m2 you have to divide by them. you cannot just choose to look at 0.00001m2 from the same source and decide it has 1324235lux. – Michael Nielsen Oct 08 '13 at 15:36
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    @Michael Nielsen: Of course you can look at an area as small as you wish! It would make no sense to arbitrarily impose a minimum integrating area. Just an example: if you were repairing a watch (under loupe) you would need good illuminance, but only over a few cm². Look up the definition of “illuminance” in a good physics book: it’s defined for an infinitesimal area (as dΦ/dS). Or search for “luxmeter” in Google Images: you can easily see that most of them integrate over an area far smaller than 1 m², the order of magnitude being more like 0.001 m². – Edgar Bonet Oct 08 '13 at 16:15
  • but it doesnt change that the 1000 lumens get spread over 9.42m2. this is a constant you cannot change. if you look at 0.001m2 it is not 1000 lumens you are looking at anymore, only part of the light. and I didnt say it doesnt make sense to integrate les than 1m2, but less than the full m2 that is covered. if I want light on a 1mm*1mm watch part I will get a light with a narrower angle. your number is 3times the real number and you claim that it is because my number is average, but the light changes only a factor 2 within my area, so something is going wrong. – Michael Nielsen Oct 08 '13 at 16:24
  • you also dont account for that it is lens design that controls the variation within the area, not cosine laws. – Michael Nielsen Oct 08 '13 at 16:25
  • What's going wrong is that you based your calculation on very rough approximations. That's why you are off by a factor 3, which is OK for this kind of back-of-the-envelope order-of-magnitude estimate you are doing. The fact that the angular distribution is controlled by a lens is perfectly irrelevant: only the actual distribution matters, and for this particular LED it just happens to be a cosine (look at the curves in the datasheet please!). – Edgar Bonet Oct 08 '13 at 16:38
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You need to divide the Lumens by the area it covers at that distance. You can find that area using the polar radiation pattern on page 16.

The beam angle is defined as: "beam angle is the point at which the Intensity of a source drops to 50% of maximum" makes your LED a 120 degree light due to the lens on it.

So if you make a triangle with one side a=1m b = half the width of the light cone at 1 m distance angle = 120/2 degrees at the top of the light. Then you need tan(angle) = b/a. but it is b you need to find. b=tan(angle)*a = 1.73m

This is the radius of the circle of light you are getting. so Area = b^2*pi and your lux is then Lumens/Area = 1000/9.425 = 106lx.

Michael Nielsen
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  • Rushing - Area calc may be right - = projected area / 4Pi. Also: Note that in fig 14 the White LED (green curve) falls off from the ventre to be only 1/2 as bright at 60 degrees. So while calculating lux = incident_lumen/area you will get a 1:2 reduction across the area. Total energy radiated at any given angle can be calculated by summing squares in fig 14 under curve at that angle and comparing to total squares under curve. Royal blue has 90% of light at 140 degrees (see note 4 page 6) and white output curve is less steep sided so 90% angle is lower - maybe 100-110 degrees range? – Russell McMahon Oct 06 '13 at 03:16
  • It is normal when describing starts in natural things than you count the span where it is 50-100% ( the -3/6 db crossing). they call it a 120degree light themselves (2x60) and probably accounted for the 50% fallout out to that in the 1000lumens. – Michael Nielsen Oct 06 '13 at 16:50
  • as for area vs projected area, no matter if you model it as a sphere or a disc with zero angle, it comes out at A_proj = pi*r^2. Also note that 50% light is to us a slight change, barely discernible, like a linear volume button that barely changes anything from max to half volume. – Michael Nielsen Oct 06 '13 at 17:01
  • You typically need about +100%/-50% difference in amplitude to perceive level differences when you can see only one source at a time. If the two sources are visible but not majorly overlapping about a +50% change is visible. BUT for "wall washing" type applications where multiple sources illuminate adjacent areas people can typically discern +10% to +20% variations in intensity (brighter to dimmer). In an LED light used as a "torch" or RoomLight or lantern all patterns may be nearly wholly overlapped. But If his LED light is a eg strip fluro replacement it may form adjacent areas. – Russell McMahon Oct 07 '13 at 06:55
  • FWIW: I design LED lighting for portable applications. That doesn't guarantee that I know what I'm talking about - but I hope I do :-). The above comments are approximations and there is of course more to it than that but that's probably a good starting point. – Russell McMahon Oct 07 '13 at 06:56
  • I know that the detectability depends on the context,you will see the falloff because it is part of a ramp and you have the following falloff to 0%, and the comment about this perception was besides the point. but my point was that in technical specs you always define teh boundaries from the 50% crossing, which is why it is safe to assume that they defined the lumens and the light angle from the area covered from -60 to +60 degrees, making the area 9.42m2 and lux 106. – Michael Nielsen Oct 07 '13 at 12:04
  • Let me point out that by dividing the luminous flux by the area covered you are calculating the average illumination over said area. My answer, on the other hand, is about the illumination in the center of the light spot. – Edgar Bonet Oct 07 '13 at 16:13
  • Two issues with this answer: first, you are assuming that the specified luminous flux is only the flux inside the viewing angle. This is dubious. The Philips’ LED glossary defines the “Lumen Output” as “The total lumens emitted of a light source, system, or solution.” (emphasis mine). Thus you only have ≈ 750 lm inside the viewing angle, and the average illuminance over your 1.73 m-radius disc is 80 lx. – Edgar Bonet Oct 09 '13 at 07:58
  • Second: you are computing an average over a very large disc, with highly non-uniform illuminance: from 318 lx at the center down to only 20 lx at the edge (it’s a 4-stops variation!). Averaging over such a wide range yields a number that is hardly representative of anything. – Edgar Bonet Oct 09 '13 at 07:59