LED Spotlight

[stasikdz]

Hello,
I want to ask for help - advice because I don't understand anything from optics.
The situation is as follows: it is necessary to create a LED spotlight that will illuminate the object about 50x50Cm in size, the object from the source is located at a distance of 20m, led matrix size 40x40mm.
I tried to assemble a complex from fresnel lenses but nothing works, as I realized I bought the wrong lenses or frenel doesn't want to work in a complex.
I ordered these lenses : https://www.aliexpress.com/item/328...c99f1987703ba16f7-1594225931579-05524-_sSETun
with focal length 40mm; 60mm, 80mm.
Can anyone suggest anything, I don't need higher math calculations, it would be enough to tell how many lenses need to be in the complex to achieve the desired result, what lenses to use and where to put them.
Thanks.

 

[DIWdiver]

That's a pretty tight beam you are looking for. To achieve that with any sort of optical efficiency, you'll need a lens that's much larger than your emitter array. I'm not an optics expert by any means, but what little experience I do have suggests that the lens should be at least 10 times the diameter of your emitter.

There was some discussion of this years ago on CPF, and some optics experts did get involved. I suspect it's in the Homemade and Modified forum.

 

[JustAnOldFashionedLEDGuy]

Is this for commercial use? A one off art project?

When you say "illuminate", how bright is your LED array, and how bright do you need that 50*50cm object to be?

Why would you start with a 40*40mm array? That is huge to illuminate such a small area. That sounds more like that one would do with 1 high powered LED.

50cm * 50cm = at 20 meter = 2 degrees. To keep most of the light on a square that size with a 40mm * 40mm source would require an optic several meters in diameter. Obviously that is not practical.

Let's take a step back and figure out what you are trying to accomplish.

Hello,
I want to ask for help - advice because I don't understand anything from optics.
The situation is as follows: it is necessary to create a LED spotlight that will illuminate the object about 50x50Cm in size, the object from the source is located at a distance of 20m, led matrix size 40x40mm.
I tried to assemble a complex from fresnel lenses but nothing works, as I realized I bought the wrong lenses or frenel doesn't want to work in a complex.
I ordered these lenses : https://www.aliexpress.com/item/328...c99f1987703ba16f7-1594225931579-05524-_sSETun
with focal length 40mm; 60mm, 80mm.
Can anyone suggest anything, I don't need higher math calculations, it would be enough to tell how many lenses need to be in the complex to achieve the desired result, what lenses to use and where to put them.
Thanks.

 

[stasikdz]

Hello,
This is not a commercial project, I would like to build 2 lights for myself.
I should illuminate a spherical traffic mirror, on the mirror should be not less than 50W of light, it is necessary to illuminate the area (sized as a parking lot for 10 cars), from the opposite side it is impossible to impose "active" light sources.
I wanted to use an LED area with a power of 100W, but if you say that the mirror must be HUGE then I will have to change the strategy. I just got new 50W LED modules with an area of 25x25mm, but I understand I need an even smaller source.
I could order such LEDs 90W or better 150W, or is it possible for a person who does not understand anything from optics to build something like that?:thinking:

 

[JustAnOldFashionedLEDGuy]

Hello,
This is not a commercial project, I would like to build 2 lights for myself.
I should illuminate a spherical traffic mirror, on the mirror should be not less than 50W of light, it is necessary to illuminate the area (sized as a parking lot for 10 cars), from the opposite side it is impossible to impose "active" light sources.
I wanted to use an LED area with a power of 100W, but if you say that the mirror must be HUGE then I will have to change the strategy. I just got new 50W LED modules with an area of 25x25mm, but I understand I need an even smaller source.
I could order such LEDs 90W or better 150W, or is it possible for a person who does not understand anything from optics to build something like that?:thinking:

What you are attempting is not so much as impossible, as pretty much totally impractical. It will be much easier to find a way to run wires, even if it seems difficult at this time. As opposed to running 120/230/277VAC (not sure where you are), you could likely run 24 VDC if you prefer but then you will need an LED driver that accepts 24VDC. They exist. This isn't something you are going to accomplish for $100.

 

[stasikdz]

I know that it is impractical but there are no other options, the cable must not be pulled on the air (the cable will damage the landscape), while the asphalt must not be cut.
There are no problems with electricity, electronics, from a mechanical point of view there should also be no problems (housing construction, etc.)
The only problem is the optics I haven't worked with.
I will order everything I need to do, I just need to understand what to order and what to do ....

 

[arek98]

There is no such optic, you looking for less than 1.5 degree beam angle. Even LEP has about 3 degree and is nowhere close to power you want.

EDIT: Maybe you can try Ledil FCP13895_SEANNA-A and smallest LED that will give you enough power, overdrive LED.

 

[stasikdz]

Thanks for the suggestion, I'll order 1 piece and see what it looks like.
Maybe someone else has an idea?

 

[stasikdz]

I probably won't buy this lens because a fairly small LED is provided.
What do you think of such an option? https://youtu.be/AxTGS5i2AHY

 

[RetroTechie]

There is no such optic, you looking for less than 1.5 degree beam angle. Even LEP has about 3 degree and is nowhere close to power you want.

Nonsense. Question this thread started with, is the kind of job every "thrower" style flashlight is designed for. Or perhaps some fixtures as used to illuminate the outside of buildings. Just have a look at those things, and you'll find that for a tight beam (read: parallel = non-diverting rays of light), you will need:

That's a pretty tight beam you are looking for. To achieve that with any sort of optical efficiency, you'll need a lens that's much larger than your emitter array.

That's the general rule: lens or reflector much bigger than the light source. So that as seen 'from' the lens (or mirror), light source can be approximated as a point source. How well that approximation is, depends on that size ratio, how accurate the lens / reflector is shaped, and whether light source sits in the correct position. IIRC for reflectors that's usually parabolic shape. And for lenses: Fresnel lenses are just a shortcut to making thinner / lighter lenses, not better ones.

At the light source you can do this by making it smaller. That is: a higher-intensity light source = same amount of light coming from a smaller surface area. Higher lumens output or more powerful (W) light source is only one part of getting there.

Unless you want a big, heavy lens, I'd say go the reflector route. Will probably also capture a higher % of the light coming from your LED. So you're looking for a big reflector, whose shape is designed to produce a tight beam from a 'point' source sitting at the base of it. That is: not a reflector that's designed around a bulb, since that is placed somewhat inside the reflector. Although maybe one could find such a reflector that works reasonably well for this purpose, or fiddle a bit with positioning the LED (but cooling it might be a problem then).

Maybe a stupid question, but why not try & find some of the biggest reflectors around, that are sold as DIY parts for LED flashlight hosts? Or DIY "light up the sky" types of projects. Then find the most powerful LED you can fit in that reflector's base? I'll predict right here that will be a smaller LED than 40x40mm. Unless you want ~1m reflector... :laughing:

 

[JustAnOldFashionedLEDGuy]

No offence to Retrotechie, but what he wrote is all pretty much wrong. To maintain most of the light in a 2 degree total angle, you need both a small LED (low lumens) and a big reflector. To get a lot of lumens, you need a bunch of these, or as discussed before one super huge optic. The closest off the shelf optic is about a 6" round compound optic (LEDIL I believe) that matches with an XHP35. This may give you 300-500 lumens hitting that reflector ... may!


I don't know how wide your parking lot is, but for 10 cars, you don't cut the asphalt. You could of course go around, but there also machines for boring underneath for conduit for just such situations.

 

[stasikdz]

one thing is clear is that this task is not as simple as it seemed at first.
I'll do the following: I ordered a 180W LED, and a 300mm (almost 12 inch) fresnel lens with a 120mm focal length, let's see what comes together. If the result is not satisfactory then I will have to talk to the owner from whom I can take the electricity, put the electricity meter, pull the cable (~200m) and put the Led light on the remote.
Thanks for the suggestions.

 

[arek98]

Nonsense. Question this thread started with, is the kind of job every "thrower" style flashlight is designed for. Or perhaps some fixtures as used to illuminate the outside of buildings. Just have a look at those things, and you'll find that for a tight beam (read: parallel = non-diverting rays of light), you will need:

I guess you didn't get to the point where OP said 50W of light in that 1.5 degree beam (he wrote 50x50cm from 20m but this is what it is). Point me to a thrower that can do this and is smaller than satellite antenna, please, I would like to buy one.

Edit: Assuming 5000lm output (that's about 50W on LED assuming good efficiency - don't know how much on target) it means about 9,275,000cd!

light source can be approximated as a point source

No, it cannot be. LED is far to big to be approximated as point source when considering smallish reflectors, maybe if you plan using few meters in diameter reflector.