OSRAM CSLNM1.TG & CULNM1.TG 1mm², CSLPM1.TG & CULPM1.TG 2mm²

I guess we’re thinking of “intensity” differently then. I think of intensity as a constant, like one unit of brightness. More units take up more space. So, if you have less space, you have less brightness. To get more brightness from the same space (or from less space), you have to increase the intensity itself. Like a building has a specific size footprint. You can get more space by adding more buildings, but you need more land to do that. If your land is limited, the only way to get more space is to make the building taller.

I think there might be some wording confusion. In my quote above, when I said “intensity goes down as you go to the side of the LED” I meant go down in angle, as the angular distribution graph in the datasheet shows.

That graph shows the luminous intensity (cd) as a function of angle. It basically is a graph of the apparent area as a function of angle. The apparent area goes like cos(theta) which is the approximate shape of the graph. The luminance of LED (cd/mm^2) is a constant. Right above the LED (zero degrees) you have the full die area, say 2mm^2, then at an angle of say 60 degrees you have an apparent area of half that, 1mm^2 which gives you half the intensity.

Yes, I understand that too. I was referring to the definition given to me above, that luminous intensity is the product of the surface area and candela. But since the apparent LES will reduce with increased viewing angle from center, then a) if the luminous intensity is constant, and the apparent surface area is reduced, that means the candela increases with the angle, which doesn’t make sense, or b) the luminous intensity decreases as one moves away from the center; because if the candela remained constant, and the surface area became smaller, that has to be. In other words, the intensity of the LED light at a certain angle is directly proportional to that angle it’s being viewed from in relation to head-on. And the reason I bring this up is because I’m trying to understand exactly how_ much_ of the total output of an LED goes into a typical parabolic reflector (using a GT/GT70 as an example), and how much goes into spill. And I postulate that the light that goes into the spill is actually more intense as a percentage than described to me in relation to the light reflected.

That’s what I’m saying all along. If the luminous intensity from 60 degrees to 90 degrees drops off a cliff, then how does 66% of the beam half-angle contain 75% of the lumens as claimed above?

Whoa Bro!

That is is serious reflector thrower… 550-650 kcd?

Luminance is cd/mm^2 (well technically it is per m^2 but we mostly deal with small LEDs here)
Intensity is cd
The luminance of the LED stays the same, but when you look at the LED from off-axis the area is smaller and therefore the intensity (cd) is smaller because luminance * smaller area = less intensity.
That is why a lux meter at 90 degrees from an LED will show low lux even though the LED’s luminance (cd/mm^2) is constant.

Oh, that is another geometric effect. The intensity vs angle graph in the datasheet is not the correct graph to integrate in order to get the total lumens. You have to also integrate as you rotate around the axis normal to the LED surface. Think about how all the light from an LED fills a half hemisphere. There is a lot more area and light at around 45 degrees compared to close to 0 degrees even though it is less intense. See here for discussion and a graph of the resulting light output vs angle.

Bingo! That’s the tidbit of info I overlooked. Thanks…

No prob :slight_smile:

I have it about nailed now. Speaking of full angles, the topmost 60° is (πr/3)² = (π²/9)r² volumetric area, which is just (100(π²/9)/2π)% = ≈17,4533% of the total half sphere volumetric area since a half sphere's area is 2πr². It now becomes clear why ≈25% of the light comes out of the topmost 60° cone.

Uuuugh! Some steam is coming out of my brain right now. :-D

Cheers ^:)

This is what was literally explained in the beginning of this post I linked already.

You need to take into account area because there is more surface area on a hemisphere at larger angles.
A lux meter used to measure the intensity on the datasheet is kept at a constant 1m distance from the LED.
So the total luminous flux output of the LED is the integral of the candela over the whole hemisphere.

When flattened to 2d the result is the lobes graph:

Then you can take the area of one portion of the lobe to determine the fraction of lumens collected or lost.

Here, have this, it makes calculating what % of light you collect easy:

yes 2x26650

I will have for the 1mm ² a CC FET soon,
maybe can drive at 7A with proper board layout and special cooling layout adding 0.5mm copper sheet and fill tons of viases with solder as well

That refers to the CSLPM1.TG (2mm²) emitter, doesn't it?

What host are you speaking of here, Utorch UT02? 0,5mm copper sheet, is that for focusing?

Special cooling?

^:)

sized are planned for 20,22 and 30mm, possibly also 17mm
for high current on CC FET regulation a proper heatsinking is needed, so I am massively enhancing the PCBs thermal properties with solder filled viases and copper sheet,
if there is enough interest and prototypes tested I may produce a 2 layer MCPCB driver (PCB cost for that is very high about 200$ for a bout 100 drivers)
I am not planning on using MOSLED for the FET cooling as this should work in as many hosts as possible

The surface area of said spherical cone comprises about 13.4% of the total surface area of the hemisphere, yet accounts for ~25% of the light. So ~75% of the light comes from the remaining ~86.6%. So if we “weighed” the luminous intensity coming from said 60 deg. cone, which is between .9-1 relative luminous intensity (I’ll guesstimate on average .96 at any point on that surface) and multiply that by .134 I get a result of .1286. Guesstimating the average luminous intensity of the remainder, using a typical chart, between 0 and .9, I’d say about .6. So multiplying .6 by .866 gives me .5196, which is just about 4 times the result I got from the top section. I’m sure the actual equation is a bit more complicated than this (and I hadn’t done calculus or differential equations in about 26 years) but intuitively it seems to jibe now.

It also makes sense why an RLT collar at 33% efficiency would potentially double the luminous intensity of the light coming out its aperture (given a 30 deg. half angle aperture).

While impractical, it also seems clear to me that given a luminous intensity chart for a given LED, there would be an optimal aperture angle for the RLT factoring in the luminous intensity graph, meaning there is an ideal focal length (ratio).

Of course now for a reflector, even a precision electroformed one with the most reflective coating is only approaching 90% efficiency, so that boosts the percentage of light from said 60 deg. cone to about 27% of total light output, and likely more given a typical flashlight reflector. Which makes me ponder that there is an “equilibrium angle” where the light coming from the top cone is equal to that of the lower section with the included efficiency factor. Not clear to me yet what use this would be of, but it should generate the most OTF lumens.

It’s just integration, it’s easy.

2) there is no optimal aperture, the more light that a wavien collar collects the less lumens are output and the more intense the light becomes.
For highest efficiency you use no collar, for highest candela/mm^2 you use a collar that covers almost 100% of the hemisphere.
60 degree opening was just chosen as a good middle ground by wavien.

3) silver coatings are 98, aluminum 90-95, and dielectric coatings can get to 99+

4) the highest OTF lumens are with a mule and no optic. For even highest efficiency, remove the front protection lens too, that will get another 1-3% light output.

1. I should pull out my old textbook… Just don’t have it in me right now.

2. There would be a aperture that gave the most output given a specific emitter. Basically best blend of distance and throw.

3. Yes I’m aware but those coatings are not typically used due to their fragiity. The “standard” coating with a good durability is rhodium which I believe is at around 90%. I’m sure the cast and milled ones we use typically are not as efficient.

4. Yes but we don’t buy White Flats to make mules…

Rhodium has around 75% reflectivity in the visible range.
Dichroic coated glass reflectors have the highest reflectivity. This is what Wavien used.

Yeah, the only people using rhodium are the guys that make short arc flashlights because the UV degrades other coatings.
Protected aluminum is the standard coating for non-lamp applications, that’s what I bought.

Rhodium is also good if you like to clean you reflector with a cloth (for whatever reason) because it’s rather hard. You still need to be careful though.