800w qbv2

Ok so I just finished building my qb setup.

12x qb288v2 on 3x hlg-320h-c1400b (single fixture for a 4.3' x 4.3')

12 qb288v2's @ 1400ma should be 12*66.78 = 801.36w

I've wired the power cords together, and I've wired the dimming leads together.

I had a 25k ohm pot laying around so I wired it in (knowing that I need 33k ohm pot to have full range dimming for 3 drivers)

Initial startup was ~720 watts. Left it running for a bit and it settled at 700w. heatsinks are barely warmer than ambient temps, drivers are slightly warmer than heatsinks. Turned the knob all the way down, was ~65w before they shut off completely. One thing I noticed was the fixture still pulled ~5w when they were dimmed to off.


I ordered the "cased potentiometer" from rapid led because I asked them through email if they had a 33k ohm version of their cased pot, or if they could do a 30k ohm with a 3k resistor (or anything that would give me full range of my 3 drivers being dimmed with one pot) They said "resistance is variable on the cased pot so one cased pot will work for three drivers"

So yesterday I tried to swap out my 25k ohm pot with the cased one from rapidled, and the fixture doesn't turn on. I tried all 3 of the pot's they sent me, none of them worked. Put the 25k ohm back on and it turned on back to 65w-700w again.

Will get some par readings with the apogee mq-200 probably next week.

700w @ 2.77 board level = 1939 ppf

4 boards per hlg-320h-c1400b is 190.8/229 = 83.3% driver voltage being used.

how do I calculate/estimate the driver loss for this setup in order to determine the umol/J system level?

Pics?

Efficiency graph in the datasheet. It looks like about 92% on 120v and 94% on 240V. https://www.meanwell.com/Upload/PDF/HLG-320H-C/HLG-320H-C-SPEC.PDF

Is the pot from rapid 100k? If so, it should work, but will hit full power when turned up about 1/3 of the way.

What I've found with the pots from Rapid is that leads are polarized(+&-) so polarity matters when connecting it to the dimmer leads. I think that it may have to do with the fact that they use a IC in the case as well as a pot.

banke1 said:

how do I calculate/estimate the driver loss for this setup in order to determine the umol/J system level?

Click to expand...

You just need to know current draw at the wall. Kill-a-watt is your friend there.

HLG's are typically 93% on 115 VAC power and 95% on 230 VAC power.

@GBAUTO hey thanks that did the trick. btw, what is an IC?

@nfhiggs so now at full blast its pulling ~925watts on 120v AC.

is it just me or is that like way higher than it should be?

is there a 240v killawatt?

if I plug the light into the 120v killawatt and dial it down to lets say 750w, then switch it to a 240v power cable and plug it in, will it still pull 750w?

IC- Integrated Circuit

How did you learn that? I wonder if it's 0-10V or PMW?

EDIT: Hmm, it is listed under dimmers, 0-10v accessories, manual dimming.

No real knowledge. Ran into the same issue and just tried that and it worked.

sooooo if the boards are running @ 1400ma and 66.78w per board x 12 boards = 801.36w total at the boards, and the killawatt is reading 920w, does that mean 801.36 / 920 = 87% ?

so then if im getting 66.78w x 12 x 2.77 = 2219 ppf. then divide that by 920w actual draw to get a 2.41 system level? correct maths?

Only thing I'd verify would be the series voltage for your string. Then you would be able to get an accurate power figure.

I don't know how to do that

Simple DC voltmeter-checking voltage on the two output leads from the driver to the led arrays.