DiY LEDs - How to Power Them

stardustsailor

Well-Known Member
(...) Well, I certainly was a noob (not sure how it's improved). I was just desperately searching for a 4.7K resistor (didn't find it) and came across remnants of my first LED grow light. 3W of brutal power :) (...)

:P

My first high power led grow light :


(Cheapo 1W LEDS : NW 4000°K - WW 3000°K - RED 620 nm -RED 660 nm - Violet 395 nm ,Year 2011-2012 )
My second attempt ,based on Guod's modular design ...


( Oslons, 2700°K 90Ra + Hyper RED 660 nm,home- made MCPCB,LEDs reflowed at kitchen ,
Recom DC/DC buck drivers +MW AC/DC PSU ,Year 2013 ,I think )
3rd light ,the "monster "...



( 4x CXA3070 AB 3000°K ,MW HLP-80H-42,Arduino TC monitor and overheat protection,adjustable fan speed )
............
And the story continues ,up to the latest V series Twin and V series Tetras ...



Showin' off :P ....

Cheers.
:peace:
 
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dionysus4

Well-Known Member
Again ...And again ....And again ...
FGS ,read the fuckin' report ,for the HLG 185H-C-1400B ,at MeanWells site ....

When nothing is connected at Dim wires (no pot ,no resistor ...just nothin' ) ,
Most -if not ALL- the CC drivers featuring 3in1 dimming ,output more current than the rated .
View attachment 3594808


Your driver is not outputting 1400 mA ,but ~1570 mA .
Plus the losses from the driver ( as it has a max efficiency of 94% ) .
That's why you get 220 W power dissipation from the wall ,instead of 185 or 200 W .


So ,the calculation goes like this :

At 1570 mA ,and with a Tc of ~40°C ,each CXB3590 will have a Vf of ~ 34,7 V ( roughly )
You use 4 of them ..
4 * 34,7 = 138.8 VDC
138.8 * 1,57 = 217,9 W ( total power dissipation at the DC side )

P ac * eff = P dc => P ac = P dc / eff

217,9/0.94 = 231,8 W ( total power dissipation at the AC side )...
That's the figure you should be getting ,from your power meter ...
( consider also the tolerance / accuracy of your power meter )
Is it clear to u now ?
Nothin' to worry about ...
(:

Cheers.
:peace:
so actually the 1400mA drivers put out 1570mA?

once i rig a 100k pot to the dim wires it will reduce to a max of 1400?

thanks as always
 

modular

Member
Again ...And again ....And again ...
FGS ,read the fuckin' report ,for the HLG 185H-C-1400B ,at MeanWells site ....

When nothing is connected at Dim wires (no pot ,no resistor ...just nothin' ) ,
Most -if not ALL- the CC drivers featuring 3in1 dimming ,output more current than the rated .
View attachment 3594808


Your driver is not outputting 1400 mA ,but ~1570 mA .
Plus the losses from the driver ( as it has a max efficiency of 94% ) .
That's why you get 220 W power dissipation from the wall ,instead of 185 or 200 W .


So ,the calculation goes like this :

At 1570 mA ,and with a Tc of ~40°C ,each CXB3590 will have a Vf of ~ 34,7 V ( roughly )
You use 4 of them ..
4 * 34,7 = 138.8 VDC
138.8 * 1,57 = 217,9 W ( total power dissipation at the DC side )

P ac * eff = P dc => P ac = P dc / eff

217,9/0.94 = 231,8 W ( total power dissipation at the AC side )...
That's the figure you should be getting ,from your power meter ...
( consider also the tolerance / accuracy of your power meter )
Is it clear to u now ?
Nothin' to worry about ...
(:

Cheers.
:peace:
Cheers man. Thanks so much for the info, apologies if the question had been asked million times before. I tried searching before posting but was struggling to find any relevant posts.

Makes a lot more sense now Hope yours and @alesh 's super informative responses help others as much it help me!

:peace:
 

stardustsailor

Well-Known Member
so actually the 1400mA drivers put out 1570mA?

once i rig a 100k pot to the dim wires it will reduce to a max of 1400?

thanks as always
As @alesh mentioned : Most common 100K pots have a tolerance of 10% if not more
and almost always they are under 100K ( 80-90 K ) .
That's another reason * you will have to connect a 10 K ( <= min value ),
"limiter " resistor in series with the pot.
So, their resistance will add up and reach ~100K .
Otherwise ,the max output ain't gonna be 1400mA ,but less ..

(* the main reason for placing a 10K limiter resistor ,is to limit the lower end of output current .
3in1 drivers should not be dimmed down less than 10% of their max output ,when using resistive -aka analog-dimming )

Of course ,you can find (multiturn / high precision ) pots with a tolerance of 1% or even less ...
But do not expect prices lower than $30-$40 per unit ...
 
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stardustsailor

Well-Known Member
Cheers man. Thanks so much for the info, apologies if the question had been asked million times before. I tried searching before posting but was struggling to find any relevant posts.

Makes a lot more sense now Hope yours and @alesh 's super informative responses help others as much it help me!

:peace:
https://www.rollitup.org/t/meanwell-led-drivers-3-in-1-dimming-function.838760/

https://www.rollitup.org/t/3in1-dimming-part-ii-a-more-advanced-approach.855193/

Cheers.
 

uzerneims

Well-Known Member
Hey, help a brother out!

What is the best, efficient way to drive 10x V29?
Like - imo all ten by single driver - 1.4A, but maybe there are way to run them little bit lower and on less driver count...
Any ideas?
 

kony brado

Well-Known Member
So I bought the drivers from KB. Turns out that with a small tax fraud they were cheaper than TME (including shipping and DHL fee).
i also got my package of 10X elg-150-c1050A from jerry today,the invoice on the package said 28$ :clap::clap::clap:,jerry is the man:bigjoint:. the only problem with the elg A is that the built in dimer is on the bottom of the driver,where the wall wold be . (i don't know what they where thinking >:() i guess its up to me to be flexible and redesign .

alesh,is there any problem mounting the drivers upside down on the wall (heat dissipation ? ) in order to make the dimmer easy to reach ?
:peace: thanks :peace:
 

stardustsailor

Well-Known Member
Hey, help a brother out!

What is the best, efficient way to drive 10x V29?
Like - imo all ten by single driver - 1.4A, but maybe there are way to run them little bit lower and on less driver count...
Any ideas?
I don't think that you can find easily a 580-600 W CC driver ( 400 V ,1.4 A output )
You can use two of these though ...(200 W )

Harvard Engineering CLH200-1000A-UNI-B
At 1000mA ...200V output

http://www.harvardeng.com/coolLED%2FCLH200W.pdf
 
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bggrass

Well-Known Member
I saw someone posted a meanwell hlg 300w driver picture at some point. Anyone knows when are those expected to be on the market?
 

nevergoodenuf

Well-Known Member
@nogod_ They are actually CC / CV drivers with a cap at set voltage, but yes you do need to run in parallel. So far the 300 watt COB has needed to be ran at 150 watts until the girls get used to the intensity.
 

bggrass

Well-Known Member
They are not CC drivers so you will have to wire in parallel or just use one 300w cob like @nevergoodenuf
I think it was a cc driver. I know about the ones that can be run in parallel, but I believe someone posted a picture of a mw 300w driver that was coming at somepoint. I believe it was the kind you can use to wire in series not in parallel. Same like let's say hlg 240 c1050b, but it was a 300w. I could be wrong though....
 

kony brado

Well-Known Member
I think it was a cc driver. I know about the ones that can be run in parallel, but I believe someone posted a picture of a mw 300w driver that was coming at somepoint. I believe it was the kind you can use to wire in series not in parallel. Same like let's say hlg 240 c1050b, but it was a 300w. I could be wrong though....
yes,meanwell hlg-320h-c,still under development.
320 watt cc driver.
http://www.meanwell.com/catalog/led/#p=35
 

uzerneims

Well-Known Member
It not necessary to run on one, it would be cool to drive 14x cobs with few drivers instead of buying for every cob one, because imho it would be more expensive.
 

OLD MOTHER SATIVA

Well-Known Member
i know so little but i may be able to answer my own question

i think the 24 on the driver means a max of 24 v can be used so ..if that is true

then i would need a different, higher voltage driver to run one 3590 @ 50 w

24 would not even power avero 29 at 28% according to what i[think i] have learned here

i am learning..hahaha..but a little knowledge can be a dangerous and/orexpensive thing

if our dollar was not so darned low i would have bought a light from one of the vendors on here
 
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