# The Effect of Optics on LED light



## Hoppy (Dec 24, 2005)

We have been talking about whether or not optics, lenses, are needed on our LED light fixtures, and which ones to use, if they are. I've been thinking about this for several days now, and I think I have it figured out, at least theoretically. Do you see any mistakes in this?










When I finally get the LEDs for the light I'm making now, I will be able to do enough testing to see if this is close enough for our purposes, and hopefully figure out what the constants involved are, so we can more accurately calculate the PAR any array of LEDs will produce.


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## Mxx (Dec 29, 2010)

If we're to ignore half the lumens put out by an LED then that's quite a substantial portion to ignore. Has this ever been borne out by practical testing or is this a theoretical margin of error which we're using?

I suppose it depends as well on whether you have suspended lights which might happen to have a considerable amount of 'spill', or whether they're directly on top of the tank, as I personally prefer to use them. 

For LEDs directly on top of the tank for example, even with a 120 cone of light all of the light is still focused into the tank. You might lose a bit from surface reflection of course, but doesn't most light which hits the sides of the tank tend to still be reflected back into the tank? Which is I believe roughly the same principle as a fibre optic cable relies upon, that so long as the light is below the angle of incidence which is I believe 30 degrees, then the light will be reflected back into the fibre, (or vessel as is the case for us)? 

If that is the case, and the vast majority of light does happen to be contained within and bounced around the tank until its absorbed by plants or something else, then couldn't we basically simplify the equation down to the following? -
Total Lumens/tank area in square feet = Lux?
(I'm not sure if it's as simple as that, but it would be beneficial to find out through actual testing of a tank containing nothing but water). 

And as soon as you or someone else manages to measure and determine a Lux/PAR conversion for the three different colours of white LEDs then we'd be able to determine what we're each working with? 

Of course, if you have a suspended fixture with or without optics then that's rather a different situation if there is a lot of light spill, and that might require practical measurements to ascertain.


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## jcgd (Feb 18, 2004)

With no optics and the light at the surface, the light would strike the glass at 60 degrees at the edge of the beam. So you would get 30 degrees pa spill it it starts reflecting back in a 30 degrees. 

A led with 40 degrees would therefore have no spill, as it would hit the glass at a maximum of 30 degrees. 

I think this is right, correct me if I'm wrong. 

From what I can gather, it is silly to both.have the fixture on the glass (generally inconvenient) and have no optics. It wastes light and doesn't give you as much light as you can acheive with optics. 

This is just my opinion though, and I may be wrong.


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## Hoppy (Dec 24, 2005)

Since lux or PAR are light per unit area, and since the area in the outer half of the cone of light is 3X the area of the inner half, for a 40 degree optic, 3.6X for a 60 degree optic, and 7.9X for a 120 degree optic, it is not a bad idea to ignore the outer half. We aren't seeking exact numbers here, just good approximations. With the approximations we can extrapolate data with some confidence. You can't just divide total lumens by tank footprint, because that ignores the drop in intensity with distance, and because the reflected light off the glass isn't nearly 100%. That light has losses at the water to glass interface, at the glass to air interface, where the reflection occurs, and at the glass to water interface, plus the loss through the glass itself. The reflected light is more likely to be 50% than it is to be 90%. And, from the point of the reflection, the intensity drops still further before it reaches the substrate in that roundabout way.

Also, just because we develop the relationships based on the inner half of the cone of light doesn't keep us from determining the constants using real life measurements. The relationships remain nearly the same, but the constants change.

Justincgdick is right that with 40 degree optics very little light is lost due to spillover, but that requires raising the lights to get some light in the upper corners of the tank. This raising of the light has the benefit of reducing the difference between the intensity near the top of the tank and near the bottom of the tank.


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## Neya (Nov 21, 2009)

One thing I noticed off the bat is I do not see a specific led mentioned. Different LEDs has different base optics/spread, not all are 120 degrees. Even within the same manufacturer. 

Ie: bare(no optics) XP-G give off less par then xR-e driven at the same mA... XR-e have a tighter base optic lens.


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## Hoppy (Dec 24, 2005)

Again, the goal was to determine the relationship between PAR and lenses with different cone angles, with LED spacing, and with total lumen output of the LEDs. Specifics have to await having something to test. The reason for doing something like this is to find out what the plot of the data should look like, as well as to get a good feel for what data are needed. Since the relationships are simple ones, it should be possible to determine the multiplier that gives you PAR if you know the lumen output for one of the LEDs you are using, how they are spaced, and at what distance from the LEDs you want to know the PAR.

A complication that can't be picked up with simple testing is the amount of added PAR you get with water in the tank. From what I have already done I expect it to be around 10-20% in the middle of the tank, and more near the glass sides. But, I won't be able to determine that with the light fixture I'm working on, because I don't have a tank anywhere near big enough, even if I had the desire to have that tank sitting on my floor full of water.


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## evilc66 (Feb 28, 2008)

The theory is right. The complication comes in when you look at the efficiencies of different optics. For most of the "name brand" optics that are readily available, it's not that great of a concern, providing that the lens in question doesn't have any odd distribution issues. Where is starts becoming problematic is when you start getting into the many cheap Chinese lens options that are out there. You wouldn't think that you could lose 20% of your output through a poor lens, but you can.


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## Hoppy (Dec 24, 2005)

evilc66 said:


> The theory is right. The complication comes in when you look at the efficiencies of different optics. For most of the "name brand" optics that are readily available, it's not that great of a concern, providing that the lens in question doesn't have any odd distribution issues. Where is starts becoming problematic is when you start getting into the many cheap Chinese lens options that are out there. You wouldn't think that you could lose 20% of your output through a poor lens, but you can.


I've always wondered just how much loss there is with a typical LED lens. Considering that they are anything but expensive, precision optics, I can see it being as high as 20%. But, that doesn't greatly concern me, since I have enough trouble just guessing the PAR with a +/- 50% accuracy. It is probably always going to be a good idea to over design, then use dimmable drivers to drop the PAR to what we want. The trick will be determining what is an over design. There are so many variables to contend with it makes your head spin.


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## cggorman (May 9, 2009)

Most of the L'edil optics are in the 90% range. Not sure about the other manufacturers as they don't make optics for my LEDs.

I know it adds another layer of complexity, but don't forget the intensity distribution within the light cone. inside a 60° (TIA) cone, 75% of the total light energy may be confined to the interior 40°, as an example. It may bear research. Again, I haven't done a tremendous amount of looking at the Cree product, but it should be similar to what I have looked at.


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## Hoppy (Dec 24, 2005)

cggorman said:


> Most of the L'edil optics are in the 90% range. Not sure about the other manufacturers as they don't make optics for my LEDs.
> 
> I know it adds another layer of complexity, but don't forget the intensity distribution within the light cone. inside a 60° (TIA) cone, 75% of the total light energy may be confined to the interior 40°, as an example. It may bear research. Again, I haven't done a tremendous amount of looking at the Cree product, but it should be similar to what I have looked at.


That's why I ignored the part of the cone of light in the outer half of the cone angle. I'm hoping it will be possible to end up with an equation of something like PAR at a given distance = a constant times the lumen output of one LED divided by the distance between LEDs squared, for a panel of LEDs, or divided by the distance between LEDs, for a single row or two closely spaced rows of LEDs. Then to account for different distances from the LEDs, change that to PAR at distance H equals a different constant times the lumen output of one LED divided by the distance H squared divided by the spacing between LEDs or the spacing squared (for a panel of LEDs). This for each cone angle optic, or with another variable, probably the tangent of half the cone angle. And, of course, another variable to include the effect of the phase of the moon :icon_mrgr


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## Mxx (Dec 29, 2010)

My guess is that with LEDs, that we're perhaps overstating the importance of distance here, with or without optics. And it's not as if we're dealing with an inverse square falloff, which would only be applicable to a single bare bulb. 

LEDs are directional enough that all of the light is at least pointed at the tank, so much if not the majority of it will make it into the tank. And from there, the further the measuring point is away from the LEDs, the less the closest LEDs will contribute to the intensity, but you will however have more LEDs contributing intensity at that point which to some degree should more or less make up for that. And optics would mean that the intensity from the closest LEDs would be higher, but fewer LEDs would be contributing at any point so the results may work out to be similar even. (Disclosure: I'm rooting against optics as I do happen to like the look of light filling my tank well, capturing the colours of my fish, and avoiding excessively harsh shadows underneath plants and other decor). 

We lose light at a few points, spill around the tank if the lights are raised, reflection off the surface, and then some light spilling out the sides of the tank (though I suspect that's relatively minimal), and perhaps a certain portion is lost passing through the water itself. I don't know how much these factors contribute to lost light with a directional light source, but I very much do look forward to finding out what the measured results would be. And optics would mean less light is lost to 'spill', but I haven't the slightest idea to what extent that sort of spillage impacts the intensity at the target points.


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## xmas_one (Feb 5, 2010)

Here's something relevant...

http://en.m.wikipedia.org/wiki/Snell's_law


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## jcgd (Feb 18, 2004)

Mxx said:


> My guess is that with LEDs, that we're perhaps overstating the importance of distance here, with or without optics. And it's not as if we're dealing with an inverse square falloff, which would only be applicable to a single bare bulb.


Why would it only apply to a single bulb? The inverse square falloff would completely apply. The resulting par at any one spot would be a product of the par supplied by each LED. The par of each LED would have fallen off at the inverse square rate.



Mxx said:


> LEDs are directional enough that all of the light is at least pointed at the tank, so much if not the majority of it will make it into the tank. And from there, the further the measuring point is away from the LEDs, the less the closest LEDs will contribute to the intensity, but you will however have more LEDs contributing intensity at that point which to some degree should more or less make up for that. And optics would mean that the intensity from the closest LEDs would be higher, but fewer LEDs would be contributing at any point so the results may work out to be similar even. (Disclosure: I'm rooting against optics as I do happen to like the look of light filling my tank well, capturing the colours of my fish, and avoiding excessively harsh shadows underneath plants and other decor).


You are missing one point though. Hoppy has stated a few times that when you use the optics, you must raise the fixture. This causes the spot lighting to diminish. Your point is valid if you left the fixture at the water level, but you would simply not do this using optics. Expecially the 40 degree. You raise until until the coverage is even.

The intesity will be nearly the same, or possibly even greater still, and as a bonus, due to the par dropping off at the inverse square, the par levels from the surface to the bottom will vary less than without optics and the fixture being raised up. It's a win, win.


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## Hoppy (Dec 24, 2005)

I have been playing with this "theory" a bit. I have a LED light, made from Ebay 3 watt LEDs, for which I have quite a bit of PAR data. So, I assumed the seller was being honest in giving the lumens each LED will emit, and used that single data point to make a chart that lets me determine the PAR produced, for any LED arrangement, with any LED that I have lumens data for. Then I used the chart to predict what my newest LED light, if I ever get the LEDs, will produce: 










Based on just the one very unreliable data point, my new light will give about 350 mms of PAR at full current, with the 40 degree optics installed, about 30 mms of PAR without the optics. If I use the optics and want only 40 mms of PAR, the LED current will be about 200 mAmps.

It is very unlikely that this is accurate, but with some good data it could become extremely useful.

EDIT: I spent more time on this, and included data from my LED light with Cree XR-E LEDs, making sure all of the data was taken in the air, no water. The chart looks better now, and the relationship appears to be valid. If anyone has more data from Cree LEDs I would appreciate getting it - PAR at what distance, with how many LEDs per square foot of the fixture, and what model LED at what current.


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## evilc66 (Feb 28, 2008)

Could be useful, but you would have to characterize every LED in every color temperature to get good results. Every manufacturer has slightly different spectral distributions based on their proprietary phosphor mix, and as a result, will change PAR and lumen values for the same relative color temperature. Because lumen measurements are weighted in the green range, increases in blue and red output (that we care about and will affect PAR readings) will have little effect on lumen output, but will affect PAR output considerably.


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## Hoppy (Dec 24, 2005)

evilc66 said:


> Could be useful, but you would have to characterize every LED in every color temperature to get good results. Every manufacturer has slightly different spectral distributions based on their proprietary phosphor mix, and as a result, will change PAR and lumen values for the same relative color temperature. Because lumen measurements are weighted in the green range, increases in blue and red output (that we care about and will affect PAR readings) will have little effect on lumen output, but will affect PAR output considerably.


This is only good for white LEDs, cool white, warm white, neutral white, etc. where the differences in spectra are not extreme. Knowing the Cree model number, you can refer to the published Cree data and get good enough values for lumen output. Then using the output vs current curve in the same data, you can get the lumens at the current you will be using. 

This lets you determine about what layout of LEDs will give you the PAR you want/need for your tank. But, always, to be sure, you need to use dimmable drivers, and aim for at least a third more PAR than you want, expecting to adjust the LED current downwards to get the right PAR.

But, more data is needed to be sure this really works. I think it looks promising. If more data still fits the chart, it can be reduced to a simple equation, which is a lot easier to apply than the chart.

Edit:

The simple equation for the chart as it is now is:


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## jcgd (Feb 18, 2004)

Hoppy, it seems most of the reefers who have been switching to LED lighting in the last year or so have been using around double the LEDs most of us are planning on using for our planted tank fixtures. The main difference being that they have used XPGs while we are starting to use XML. I believe the XMLs are about 40% more powerful than the XPGs so I'm thinking having around 10 XML per foot will provide very high light possibilities.

What kind of light levels are you guesstimating you will get with these fixtures, mainly yours? These reefers seen to have been replacing 250-400 watt MH intensityies with 20 or so XPG LEDs per square foot, even often bleaching corals.

Thoughts?


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## Hoppy (Dec 24, 2005)

You can fill in the numbers in that equation and solve it for PAR, for whatever combination of LEDs and optics you wish. Sometime today I plan to do that for one of the currently being designed lights in the DIY section. It is pretty risky to use an equation like that and rely on it right now. Until I get more data it is mostly just an idea.

One thing I'm now sure of is that we can get just about any amount of PAR we want, for any size tank, by selecting the right optics and arrangement of LEDs. And, I am almost sure that an equation of that form will work very well for determining which of many LED configurations will work best, be most economical, and do what we want it to do. It is the aquarium lighting option of the future.


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## cggorman (May 9, 2009)

I was SOO excited when lighting-class diodes finally became affordable enough to consider designing my own fixture. Lighting a wedge-shaped tank is such a pain with long glowy tubes of glass.


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## jcgd (Feb 18, 2004)

Hoppy said:


> The simple equation for the chart as it is now is:


Okay Hoppy, I don't know how to make the equation work, aha. My calc. isn't liking what I'm inputting and I'm not sure of a few things.

H^2 tan^2 ...... do you mean square the tan of H squared? So for 48" height, I square 48, take the tan of that number and then square it?

For N, say you are doing pendants like me. 4 12"x6" pendants over a 72"x24" tank. What would I use for N?

Thanks.


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## Hoppy (Dec 24, 2005)

Typing math equations on a standard keyboard isn't easy. That equation includes the distance from the LEDs squared times the square of the tangent of one quarter of the lens cone angle.

My latest iteration of that equation is somewhat different. But, I'm close to having a lot more better data, so I'm waiting on that to do any further adjusting of that equation (or dumping it entirely.)


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## redfishsc (Aug 29, 2010)

Hoppy said:


> This is only good for white LEDs, cool white, warm white, neutral white, etc. where the differences in spectra are not extreme.



FWIW, royal blues give very similar PAR numbers to whites. I've seen them higher and lower than the same type white, same current, same optic angle. 


This is the reading given by the PAR meter, so most likely it's actually under-reading the blue and the PAR is going to be even higher.


Similar situation with 460nm blue T5's like the ATI Blue+


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## Hoppy (Dec 24, 2005)

Oddly enough, the blue LEDs I'm using on the LED light I'm working on now look like white light to me. I can't see a difference from the cool white. Perhaps I will see it when it is reflected light instead of direct light. So, I can easily believe that they produce comparable PAR to the whites. (When I tested warm white vs cool white, the warm whites gave significantly less PAR)


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