Minimalistic multichip DIY LED build

[hillscp]

I have been saying that I'd post pictures of my multichip build, and while it's not done I do have some in progress pics to share..

The light body is a 4" DWV ABS coupler, they are slightly larger than SCH40 fittings, and as a coupler it's solid ABS, no foam core stuff. The heatsink, shown without fan (but the fan fits I just didn't take it all back apart to put that on again) is a Zalman CNPS7000Cu, tested at 27C over ambient @ 100W heat input.

This will be housing a 60w hybrid (10k/445), and I'm hoping it won't be too blue after the comments above. The plan is to drive it all the way to 150w if so desired.

If it's not too blue already, I have plans to add a ring of 6 420nm emitters at the bottom around the main reflector.

Mounting options could be hanging or I will be tapping a 1/2 NPT hole in the back, so a gooseneck could be made using metal or plastic conduit and the appropriate fitting. It's a bit too heavy for a kessil gooseneck.

Nice job. I have been thinking of a way to enclose mine. Right now they look like FrankenLights.

I have three of those 60w hybrid (10k/445) panels and I don't think you will find them too blue. I have three 150W 14K Phoenix double ended metal halides right now and the LED's are not as blue. I used to run 14K XM's and they were more white than the Phoenix. I would rate the hybrid panel as looking more like the XM than the phoenix (crisp white with a slight blue tinge). I am thinking about adding some 20W Royal blues with dimmers to so I can adjust the color.

BTW, what kind of connector are you looking at for the pendants? I am thinking about using a 4 pin Molex power supply extension cut in half so I can connect the power supply and fan in one quick click.

 

[Lassef]

Also, just so I'm clear, when I add LEDs in series this will reduce the forward voltage from the driver. I ask because the driver puts out 48v but the LEDs are rated at 1-10v. Having 4 of them in series will lower the vf to 10v?

Thanks,
Tony

No the driver puts out 1.7 A and adapt the voltage to the load - its a constant current device - it works thus contrary to the common DC sources do (constant voltage and adjusting the current to the load). In this case the voltage it will put out is the sum of each VF in the daisy chain. In your case around 40 V. If you look at the spec. of the driver on page 1 you can see something named Constan Current Region. For your driver it is 28.8 to 48 V. Your total FV in a daisy chain must fit into this range. It means that with your driver you can have 3 - 4 chip (FV each 10-11 V) in a daisy chain (daisy chain = connected in serie) Not more than 4 - not less than 3.

Rated 1-10 V? can you give me the link to your chip´s ?

Sincerely Lasse

 

[Lassef]

@ tomservo: Nice build - one question you blow in a lot of air - where will it go out? I do not see any outlets for air. It looks like you have a plastic cover between the LED and the cooler.

A remark:

Many people are afraid of letting the warm air going down to the surface of the water because it is a belive that this warm air will heat up the water. My experience is contrary. Because the warmed air has less humidity, it will instead increase the evaporation of the aquarium. This takes more energy (read heat) than you get in the water by blowing the warm air over the surface. The resulting sum is normally increasing cooling of the tank instead of heating (within certain limits of course) !


Sincerely Lasse

 

[Lassef]

I have my HLG-80H-36B diming with a 10V wall wart and a 10K Pot. It works great except the light comes on rather suddenly. It does dim completely (off) and (as far as I can tell) as high as when the dimmer circuit is open (all the way on).

I am interested in dimming it using the POT as a rheostat and varying the resistence between the two dimmer wires. I'm a noob so I initially tried to do it with the 10K pot but it would not brighten all the way. Well duh, the PDF says you need to vary the resistance to 100K. And if (like in my case) you are trying to dim three drivers with a single POT you will need a 33K POT. Just pointing that out so nobody makes the same rooky mistake I did

So later tonite I'll wire up a 100K POT and give that a try on a single light.

I think that the Meanwell drivers not are true PWM drivers - ie they can take a PWM signal but they do not send out a PWM signal to the LED. They dim the LED through taking the current down to a certain level. This mean that you never can dim to 0

The other thing is that the spec (page 5) of some reason says 102 - 108 % if DIM+ and DIM - are open so I think that your are right in your conclusion that you do not get the same brightness as you have with the wires open.

Keep posting - I have no experiences to dim with pot´s

Sincerely Lasse

 

[tronyony]

Rated 1-10 V? can you give me the link to your chip´s ?

Sincerely Lasse

Lasse, forgive me. The Budlight got the best of me last night. The Vf is 10-11v not 1-10v for the chips. They are from ac-rc.


Thanks again,
Tony

 

[tronyony]

Ok, so after googling how to wire the pot to the driver I'm a bit confused. Some indicate that you only need to connect the blue and white dimmer wire to the pot while other say that a 10v source also needs to be connected to the pot. Can anybody point me in the right direction. My driver is the Meanwell HLG-80H-48B


Thanks,
Tony

 

[bemenaker]

tronyony, to make it dim, you have to supply a voltage there on one of them. Based on Lassef's post, it sounds like if they are not connected at all, the driver will run at full power. With the two wires not connected (the blue and white dimming wires), it will either run at 100% or 0%. To make it dim, you connect an external power source maximum 10V+ to one of the wires, (blue I believe, but you'd have to look at the data sheet, I am not looking at it currently) and the other wire (white) would go to the negative side of the 10V supply. You would put a variable resistor inline with the 10V+ to give you your dimming effect.

 

[Lassef]

tronyony, to make it dim, you have to supply a voltage there on one of them. Based on Lassef's post, it sounds like if they are not connected at all, the driver will run at full power. With the two wires not connected (the blue and white dimming wires), it will either run at 100% or 0%. To make it dim, you connect an external power source maximum 10V+ to one of the wires, (blue I believe, but you'd have to look at the data sheet, I am not looking at it currently) and the other wire (white) would go to the negative side of the 10V supply. You would put a variable resistor inline with the 10V+ to give you your dimming effect.

Yea but I think it works with a real resitance also as the spec shows a table with resistances from 10 K Ohm to 100 K Ohm - there 10 gives 10 % and 100 K gives 100 %. Ie. you can do both ways with this driver.

Look at Hillscp:s post 1938

I´ll ask my electronic guru and comeback later

Sincerely Lasse

 

[janne68]

Ok, so after googling how to wire the pot to the driver I'm a bit confused. Some indicate that you only need to connect the blue and white dimmer wire to the pot while other say that a 10v source also needs to be connected to the pot. Can anybody point me in the right direction. My driver is the Meanwell HLG-80H-48B


Thanks,
Tony

Hi Tony

If you want to controll the MeanWell HLG type of driver with a potentiometer then you do NOT need any power supply. Just connect the potentiometer between the two dimming connector, one to the center and one to one of the outer connections. If it dimming upp/down goes in the wrong direction, then swap which outer pin you are using.

This type of driver can also be driven with a Open Collector type of PWM signal (or a 0/10Volt PWM) or you can even use a 1-10Volt signal overpowering the "build in supply".

So in your case, no extra power is needed.

For one driver you use 100K potentiometer, for two driver use 50K and for three use 33K.

Regards,
Janne

 

[tronyony]

Yea but I think it works with a real resitance also as the spec shows a table with resistances from 10 K Ohm to 100 K Ohm - there 10 gives 10 % and 100 K gives 100 %. Eg. you can do both ways with this driver.

Look at Hillscp:s post 1938

I´ll ask my electronic guru and comeback later

Sincerely Lasse

That's what I saw too. Resistor(pot), 0-10v or 10v pwm. To me this means that I can connect the blue to the center post on the pot and the white to either side depending on if I want to in crease resistance by turning clockwise or counter clockwise. Yes?

@Bemenaker, that's what I've been reading and that's why I'm a bit confused. That's what a lot of people seem to say it needs to be wired up but when I reed the data sheet it looks like it can be accomplished in three different ways.

 

[Psionicdragon]

The multichip in the photo is a three colour channel multichip.

Wouldn't multichip have different voltage requirements that could fry the chip?

 

[tronyony]

Hi Tony

If you want to controll the MeanWell HGL type of driver with a potentiometer then you do NOT need any power supply. Just connect the potentiometer between the two dimming connector, one to the center and one to one of the outer connections. If it dimming upp/down goes in the wrong direction, then swap which outer pin you are using.

This type of driver can also be driven with a Open Collector type of PWM signal (or a 0/10Volt PWM) or you can even use a 1-10Volt signal overpowering the "build in supply".

So in your case, no extra power is needed.

For one driver you use 100K potentiometer, for two driver use 50K and for three use 33K.

Regards,
Janne

Janne,

Perfect! This is what I thought but didn't want to burn up my driver. This is how I read the data sheet.

Thanks,
Tony

 

[Lassef]

Hi Tony

If you want to controll the MeanWell HLG type of driver with a potentiometer then you do NOT need any power supply. Just connect the potentiometer between the two dimming connector, one to the center and one to one of the outer connections. If it dimming upp/down goes in the wrong direction, then swap which outer pin you are using.

This type of driver can also be driven with a Open Collector type of PWM signal (or a 0/10Volt PWM) or you can even use a 1-10Volt signal overpowering the "build in supply".

So in your case, no extra power is needed.

For one driver you use 100K potentiometer, for two driver use 50K and for three use 33K.

Regards,
Janne

Thank you my Guru :beer:

Sincerely Lasse

 

[Landsailor]

. Because the warmed air has less humidity, it will instead increase the evaporation of the aquarium. This takes more energy (read heat) than you get in the water by blowing the warm air over the surface. The resulting sum is normally increasing cooling of the tank instead of heating (within certain limits of course) !

This is absolutely true.

I live in Las Vegas. My pool has a solar heater...in the hot desert. Why?

So much more heat is lost by evaporation than any other method. How do you cool a hot cup of coffee? You blow across the surface. Any anticipated heat transfer from the warm air from the LEDs will be offset by the evaporation. Metal halides are an entirely different story since they are so hot in the first place.

Exhausting the hot air from your LED will keep your tank warmer.

 

[Lassef]

Exhausting the hot air from your LED will keep your tank warmer.

You mean cooler I think.

Why MH put in heat in the tank is because there is a lot of IR wavelenghts in the spectrum from a MH.

You do not have IR in the blue LED - you have some in the white LED - but rather little.

Sincerely Lasse

 

[hillscp]

Hi Tony

If you want to controll the MeanWell HLG type of driver with a potentiometer then you do NOT need any power supply. Just connect the potentiometer between the two dimming connector, one to the center and one to one of the outer connections. If it dimming upp/down goes in the wrong direction, then swap which outer pin you are using.

This type of driver can also be driven with a Open Collector type of PWM signal (or a 0/10Volt PWM) or you can even use a 1-10Volt signal overpowering the "build in supply".

So in your case, no extra power is needed.

For one driver you use 100K potentiometer, for two driver use 50K and for three use 33K.

Regards,
Janne

I hooked it up two ways. With the 10k powered 10v potentiometer and the rheostat (100k hooked to two legs). Both work the same. Since I will ultimately be dimming with my controller I will ultimately be using the 0 - 10v option but it is nice to have options. I don't have a pwm signal to try

 

[ambryatim]

Was wondering if I could run 4 of the 10watters on this driver?? Meanwell LPF-40d

 

[BeanAnimal]

This is absolutely true.

Sorry, gotta step in here because the "advice" is starting to teeter on misinformation...

Absolutely True? Not really to any extent. The "warmth" of the air and its relative humidity is not really driving the evaporation, it is simply the rapid exchange of air at the water surface. Warm, or cold the effect is going to be about the same. To that end, the moisture content of the "warm" air coming from the fixture is going to be pretty much the same as the rest of the air on the room...

Exhausting the hot air from your LED will keep your tank warmer.

So no...


In any case, evaporation (phase change) is extremely efficient at moving energy. 1 gallon of evaporation moves over 8000 BTUs of heat. It takes 1 BTU to raise (or lower) 1 pound of water 1 degree F.

Enjoy

 

[Lassef]

Chip spec FV = 9-10V; Current 900 mA. 4 in a daisy chain -> FV (tot) 38-40V current still 900 mA

Drivers sp LPF-40D - 48. Constant current region = 28,8-48 V. Constant current 840 mA

It should work FV(tot) is in the Constant current region of the driver and the drivers current is below 900 mA.

If you want a lower current LPF-40D-54 works also and give you 760 mA.

First cas you will have around 9 Watts consumtion from the chip - at the lowe current - its aroun 7.5 W

Sincerely Lasse

 

[ambryatim]

So, the 900mA current is the max the led will draw, but will work below that too??

Cause I was wondering about it being to low on the amps and not being able to supply enough voltage to the leds and causing the driver to shutoff due to overtemp or overcurrent.

Thanks

Chip spec FV = 9-10V; Current 900 mA. 4 in a daisy chain -> FV (tot) 38-40V current still 900 mA

Drivers sp LPF-40D - 48. Constant current region = 28,8-48 V. Constant current 840 mA

It should work FV(tot) is in the Constant current region of the driver and the drivers current is below 900 mA.

If you want a lower current LPF-40D-54 works also and give you 760 mA.

First cas you will have around 9 Watts consumtion from the chip - at the lowe current - its aroun 7.5 W

Sincerely Lasse