need a good COB

Thorhax

Well-Known Member
I have 3 of these dual output drivers that I want to use. I was wondering which chip would be best for the drivers output. Here is a pic of itIMG_8035.jpg
I was thinking of hooking up two Vero18s to each even though they will run at 30watts. Would be cool if there was a better array I could use though
 

bicit

Well-Known Member
You could also use the vero 29 orthe cxa 3070 for a boost in power ddisplacement and efficiency. But it will be more expensive.
 

churchhaze

Well-Known Member
Vero 18 or anything with similar V-A characteristic is probably the best for that supply.

The remaining voltage could be dropped using monos at 650mA in series if you wanted.
 

Thorhax

Well-Known Member
Vero 18 or anything with similar V-A characteristic is probably the best for that supply.

The remaining voltage could be dropped using monos at 650mA in series if you wanted.
so is it safe to hook up 2 veto 18s to each?
 

churchhaze

Well-Known Member
Actually, the vero 18 might not work because of the 30V minimum. At 650mA, they'd probably drop more like 29V.

Maybe there's another cob between the voltage range.

Edit: You could use monos to drop a few extra volts dropped!! problem solved!

If you add just 1 XP-E photo red to the series string, it will drop enough volts to operate within the voltage range!
 

AquariusPanta

Well-Known Member
29V * 0.65A = 18.85W

For per channel so 113.1W electric dissipation for 6.

At 38% efficient, 43W of radiant power is put out from the lights.
Huh? I get the first line of math but the rest leaves me scratching my head. Care to share some brotherly insight Churchy?

Edited: After doing some personal math, I can decode your english.

You were trying to say 2x channels per driver, 3x drivers, which equals 6x connections for 6x Vero 18's, running @~650mA, with voltage around 30 or so. The math comes out to around 115W in some cases.

(For clarity): [Vero(29.5V * .650mA) * (2x channels * 3x drivers)] = ~115W

The efficiency though... still trying to decode the mysterious algorithm you used.

:joint:
 
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AquariusPanta

Well-Known Member
"At 38% efficient, 43W of radiant power is put out from the lights."

38% is a given because i said "at 38%".
Right, but why not 100% or 420%? Why choose 38% arbitrarily? Obviously your withholding some information that you feel fit to introduce but not fully explain, something like that of your math.
 

churchhaze

Well-Known Member
The efficiency of your every day vero 18 3500k 80cri at 650mA is closer to 38% than 100% or 420%... lets just call it a hunch that efficiency is not 100% or 420% if you catch my drift. Then you'd be breaking the laws (of thermo).

If your efficiency is 39%, plug in 39%.
 

AquariusPanta

Well-Known Member
I have a feeilng the efficiency of your every day vero 18 3500k 80cri at 650mA is closer to 38% than 100% or 420%... lets just call it a hunch that efficiency is not 100% or 420% if you catch my drift. Then you'd be breaking the laws (of thermo).

If your efficiency is 39%, plug in 39%.
I would like to think you have more widespread knowledge and maybe even experience over these matters but I could be kidding myself. Only your participation, which has been gradually picking up, and demonstrations will overcome my doubt.

I sort of catch your drift but not entirely. So you are saying that, in short, even though a given Bridgelux Vero 18 may show ~30W actual on a meter, that, which matters most, the radiancy of the light, meaning the spectra of UVC to Infrared, may only be ~%40 while being ran at ~650mA, while the rest is quite possibly heat or some other form of energy? Or is this not the case?
 

churchhaze

Well-Known Member
Yes, exactly. If the light is 40% efficient and it takes 100W of electric power, it will give off 40W of radiant power and 60W of direct heat..

I sort of catch your drift but not entirely. So you are saying that, in short, even though a given Bridgelux Vero 18 may show ~30W actual on a meter, that, which matters most, the radiancy of the light, meaning the spectra of UVC to Infrared, may only be ~%40 while being ran at ~650mA, while the rest is quite possibly heat or some other form of energy? Or is this not the case?
 

AquariusPanta

Well-Known Member
Yes, exactly. If the light is 40% efficient and it takes 100W of electric power, it will give off 40W of radiant power and 60W of direct heat..
Do you believe the efficiency variable, 0-100%, varies due to the current that is discharged from the driver or do you think it's based off the materials used to build the light/COB/star and how well it was constructed?

For example, most legitimate drivers are advertised with efficiencies around 80 to 90 %. Also to include for consideration is the different BINs that some of the popular Cree COBs offer, with AB, I believe, being the best BIN type. It would seem that the quality of the BIN, such as AB, correlates with the potential radiancy efficiency of a given COB. Is this so?
 
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