Minimalistic multichip DIY LED build

Power LED´s shall be connected in series - its easier that way. If you use the driver I mentioned - yes you can take the 5 whites in series to one of the drivers and the 5 blues in the other chain.

The heatsink i got from Ebay (somewhere) The other construction (arround the heatsink) I found at my job, it is two covers from a cable duct. Use your imagination :)

Sincerely Lasse

can i use this driver in series my led ?

http://www.ebay.com/itm/DC30-36V-Di...ing_Parts_and_Accessories&hash=item564a8c5baa

or other driver you think more cheap.HLG-40H-54B probably too expensive for me :D
cz i'm only a highscool student,so it's little bit hard to buy a expensive gadged..would you help me please ?
 
It will nor work with the normal 10 watt (the driver gives 1500 mA) but probably work for three 20 watts in a serie (it is just at the lower edge for the FV so i do not know for sure). If it is enough with manual dim this and this should work for 5*10 watts in a serie

Sincerely Lasse
 
It will nor work with the normal 10 watt (the driver gives 1500 mA) but probably work for three 20 watts in a serie (it is just at the lower edge for the FV so i do not know for sure). If it is enough with manual dim this and this should work for 5*10 watts in a serie

Sincerely Lasse

wow..thanks soo much lasse..so i must build my led like this picture ?

5x10W_series_connection.jpg


both in the blue and the white one ?

once again thank you so much :)
 
Yes - five RB connected in series to drive 1, 5 white connected in series to drive 2.

Remember that you can not use a computer to dim this driver. You have to manually dim it with the potentiometer that comes with it. I have experience with a similar drive (100 watts), and there is no smooth dimming - but of course it is possible to dim it.

Sincerely Lasse
 
Yes - five RB connected in series to drive 1, 5 white connected in series to drive 2.

Remember that you can not use a computer to dim this driver. You have to manually dim it with the potentiometer that comes with it. I have experience with a similar drive (100 watts), and there is no smooth dimming - but of course it is possible to dim it.

Sincerely Lasse

wow.. quick repond from you Mr.Lasse.. thanks so much..

you said in few page in back that 10W 20K is look like a more "yellow" than a 10W 16K.then with my combination,did you think that the light will be more white or more blue ?

thanks for helping a dummies like me :)
 
The bigger point is power output flexibility.

The cost savings also includes a lower electricity bill due to less power for the same amount of light. Also less heat requires less cooling, less evaporation and therefore less top off water.

The added benefit is the temperature stability you can gain.

My tank used 550 watts of metal halide requiring fans on for atleast 6 hours to keep it under 81 or so. And that's from starting at 78. Now my temp swings from 78 to 78.5 max and 2 (92mm) fans run for the lighting duration instead of the 4 (two 160mm and two 120mm fans) I used to use with the halides. I also top-off 1/2 the amount I used too. For me I also run my ventilation fan much less (controlled by a hygrometer) to keep the humidity under 60. Average humidity now is about 50-53 depending on the house humidity levels.
 
wow.. quick repond from you Mr.Lasse.. thanks so much..

you said in few page in back that 10W 20K is look like a more "yellow" than a 10W 16K.then with my combination,did you think that the light will be more white or more blue ?

thanks for helping a dummies like me :)

This observation was made ​​in relation to between this chip of 16 000 K and the corresponding chip (from the same supplier) of 20 000 K. How it is with chips from other vendors I do not know.

If you combine 5 white (16 000 K) and 5 RB and has the ability to separately dim them, you will be able to get the lighting temperature you want. Note that this observation applies only to the chip I purchased from AC-RC. If it's the same with chips from other vendors I do not know.

Sincerely Lasse
 
Thanks everyone for all of the information on this thread. I have a standard 125 just like RiddleEagle18 and my MH fixture just lost a ballast forcing me to look at lighting. I have been very interested in LED-DIY for a while now so I am looking at this as an opportunity :)

It's a bit of a disadvantage not being able to see one of these multi chip LED's in person. What is the color difference between these three LED's?

100W 20000K High Power 10000LM LED Panel

EPISTAR 100W Super Actinic Blue Hybrid Led Panel

EPISTAR 60W Actinic Blue Hybrid Led Panel

I am leaning toward three of the 60 Watt hybrids for my setup.

Will this mean well dimming power supply work for the 60W Panel? Power supplies seem to be the biggest mystery to me:) Can someone suggest a better fit for the 60W?
 
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Here is a google sketch of a standard 125 with LED's hung at 27" above tank base or 6" above water line.

ledlights.jpg


The larger cones are 90 degree optics and the smaller cones are 60 degree optics.

How concerned should I be about the light only having a 12 inch diameter at the water surface? Even smaller with the 60 degree optics.

You should be fine. If you can tilt the outside multichips towards the middle it would be even better, but as it looks right now.... you're good!
 
The cost savings also includes a lower electricity bill due to less power for the same amount of light. Also less heat requires less cooling, less evaporation and therefore less top off water.
The original comment was on the idea of $100(ish) saved on a DIY LED vs a Kessil LED... not LED vs MH.
 
The use of Lens do not drasticly chage the equasions for the distribution of light per distance. The formula still remains that if double the distance you will reduce the light by 1/4 th. What lenses do however is concentrate the light in a smaller area.

Sorry, this just isn't the case. If what you were saying was true, then using a reflector/lens on any light would gain you nothing. If you concentrate the same amount of light, ie number of photons per area per second, on a smaller area you have increased intensity, and this is precisely what reflector/lens do, albeit with some loss. One way to think of it is to consider the size of the circle of light that's projected. If you have a 60 degree lens and place it so you have a projected circle of say 18" diameter, say 18" distance, and then took a 30 degree lens and made the same size circle though from a greater distance, you will have the same intensity of light in that same area, less transmission losses (virtually zero through air) and any difference in lens/reflector efficiency. However, if you measured 6" closer to the lights from that 18" plane, the narrower beam will gain less intensity because the size of the circle will be larger than with a wider cone angle.

Basically, a tighter lens will net you less difference in PAR from one level in the tank to the next, but you will have to run the lights higher to achieve this. It won't gain you more PAR over a wider angle lens at a given spread, but can make for much more even lighting.

With a sufficiently tight beam, there would be no difference between the intensity at the emitter and at the destination. An excellent example of this effect is coherent light (IE LASER beams). There is (ideally) no cone angle to the beam, and so the only losses are in transmission, something blocking or absorbing the beam.

edit: Also something else to consider is refraction/reflection of light off the surface increases the farther you get from straight on, though this probably doesn't account for all that much except the extreme edges of very wide optics.
 
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Sorry, this just isn't the case. If what you were saying was true, then using a reflector/lens on any light would gain you nothing. If you concentrate the same amount of light, ie number of photons per area per second, on a smaller area you have increased intensity, and this is precisely what reflector/lens do, albeit with some loss. One way to think of it is to consider the size of the circle of light that's projected. If you have a 60 degree lens and place it so you have a projected circle of say 18" diameter, say 18" distance, and then took a 30 degree lens and made the same size circle though from a greater distance, you will have the same intensity of light in that same area, less transmission losses (virtually zero through air) and any difference in lens/reflector efficiency. However, if you measured 6" closer to the lights from that 18" plane, the narrower beam will gain less intensity because the size of the circle will be larger than with a wider cone angle.

Basically, a tighter lens will net you less difference in PAR from one level in the tank to the next, but you will have to run the lights higher to achieve this. It won't gain you more PAR over a wider angle lens at a given spread, but can make for much more even lighting.

With a sufficiently tight beam, there would be no difference between the intensity at the emitter and at the destination. An excellent example of this effect is coherent light (IE LASER beams). There is (ideally) no cone angle to the beam, and so the only losses are in transmission, something blocking or absorbing the beam.

The inverse square law holds true. We're not talking about lasers here.
 
Perfect lasers do not exist, all light will diverge at the rate of proportional to 1/r². Adding lenses simply changes the intensity over a smaller area, the cone that is being projected will have a surface area ~ r² at any distance r from the source, however you need to take into account the distance after then lens and not before it.

The air-water interface is where things get tricky, if there was a perfectly flat surface with no reflection, the water would bend light rays towards the surface normal, in effect causing a lensing effect and reset the whole distance equation with intensity. However taking into account reflections, and the fact the water is a bit turbulent you get all sorts of brain numbing effects to take into account, so to the best of our ability we'll simply say 1/r² type behavior ;)
 
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The air-water interface is where things get tricky, if there was a perfectly flat surface with no reflection, the water would bend light rays towards the surface normal, in effect causing a lensing effect and reset the whole distance equation with intensity. However taking into account reflections, and the fact the water is a bit turbulent you get all sorts of brain numbing effects to take into account, so to the best of our ability we'll simply say 1/r² type behavior ;)



My experience is that just this will happen. I have on several occasions compared PAR value for a distance only in the air with the same distance but in combined air-water media. The combination of air - water has always given a higher PAR value than the same distance in air. And then I have not a smooth water surface. A small difference but still fully measurable

Sincerely Lasse
 
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