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There's a white and blue wire on each. I combined the white wires together, then the blue wires together, and then hooked them up to the pot, one middle, one outside if that makes sense
50k
@CobKits
Are you replying to myour question?
"hlg-480h has output split in 2 different pair (+and-) of dc.So it has 2 + wire and 2 - wire.I want to use it to drive 3 bar with 4 cob each, so I have to split 2 output in 3.This way I have 2 option:
use each pair from hlg to drive 6 cob
or (that what I am asking for) use a 5 pole wago, connect 2 + or - output from driver on it, and plug 3 + or - from bar(1 from each bar) to split output from driver equally to each bar wiring in parallel"
CobKits
Well-Known Member
sorry i thought you had 9 cobs not 12. but yeah i think you can do 8 and 4 on the outputs
i gotta test it, but i dont think its sensitive to the loads being matched on each pair. im pretty sure they come from the same circuit and are just multiple leads as the current gets pretty high on those.and HLG 600-12 is north of 40A output, i think theyd rather have a pair of leads than a single #8 wire which would be more difficult for people to wire to. even though the 480 and 600 at higher voltage are still under 20A, manufacturing probably dictates a standard wire size design for the whole series
i gotta test it, but i dont think its sensitive to the loads being matched on each pair. im pretty sure they come from the same circuit and are just multiple leads as the current gets pretty high on those.and HLG 600-12 is north of 40A output, i think theyd rather have a pair of leads than a single #8 wire which would be more difficult for people to wire to. even though the 480 and 600 at higher voltage are still under 20A, manufacturing probably dictates a standard wire size design for the whole series
@CobKits
Well...if I use parallel wiring, I think that I have to split load equally, among driver output,or not?So if I put the 2 + output from driver, and 3 + wire from bars on same wago 5 pole, than I am splitting driver output in 3 output ,1 for each bar.Than I can connecT 4 cob in parallel to 1 of that 3 wire,using another 5 pole wago.It could be easier if I draw it?
Edit: I am just reading again your reply and it is more clear(sorry English is not my lenguage ).So even if driver has 2 different output pairs,load is from same circuit and I do not have to be balanced. ..anyway, I will wait for your testing. Thanks for your help
Well...if I use parallel wiring, I think that I have to split load equally, among driver output,or not?So if I put the 2 + output from driver, and 3 + wire from bars on same wago 5 pole, than I am splitting driver output in 3 output ,1 for each bar.Than I can connecT 4 cob in parallel to 1 of that 3 wire,using another 5 pole wago.It could be easier if I draw it?
Edit: I am just reading again your reply and it is more clear(sorry English is not my lenguage ).So even if driver has 2 different output pairs,load is from same circuit and I do not have to be balanced. ..anyway, I will wait for your testing. Thanks for your help
CobKits
Well-Known Member
yes you have to be in parallel, no i dont think it needs to be equal load per pair of leads. in fact it might be bad to gang both + leads together, no reason i can see to do that
id do it like this if you want to balance (might be cumbersome if you dont want wiring between bars but youd need to find 6-slot wagos for that (or use marine busbars or the 8-slot non reusable ideal push-in terminals):
wago#1
1 + from driver output#1
2 to cob A +
3 to cob B +
4 to cob C+
5 to wago #2
wago #2
1 + from wago #1
2 to cob D +
3 to cob E +
4 to cob F+
5 open slot for voltage measurement
wago#3
1 - from driver output#1
2 to cob A -
3 to cob B -
4 to cob C-
5 to wago #4
wago #4
1 - from wago #3
2 to cob D -
3 to cob E -
4 to cob F-
5 open slot for voltage measurement
wago#5
1 + from driver output#2
2 to cob G +
3 to cob H +
4 to cob I+
5 to wago #6
wago #6
1 + from wago #5
2 to cob J +
3 to cob K +
4 to cob L+
5 open slot for voltage measurement
wago#7
1 - from driver output#2
2 to cob G -
3 to cob H -
4 to cob I-
5 to wago #8
wago #8
1 - from wago #7
2 to cob J -
3 to cob K -
4 to cob L-
5 open slot for voltage measurement
It will get very complicated and with too many wire through fixture.
Wago#1
1 + from driver output#1
2 + from driver output#2
3 wire to bar A as +(call it a)
4 wire to bar B as +(call it b)
5 wire to bar C as +(call it c)
Wago#2
1 + wire from wago#1 (wire a)
2 + to cob 1a(first cob of bar A)
3 + to cob 2a(second cob of bar A)
4 + to cob 3a(third cob of bar A)
5 + to cob 4a(fourth cob of bar A)
Wago#3
1 + from wago#1 (wire b)
2 + to cob 1b (first cob of bar B)
3 + to cob 2b (second cob of bar B)
4 + to cob 3b (third cob of bar B)
5 + to cob 4b ( fourth cob of bar B)
Wago#4
1 + from wago#1(wire c)
2 + to cob 1c (first cob of bar C)
3 + to cob 2c (second cob of bar C)
4 + to cob 3c (third cob of bar C)
5 + to cob 4c (fourth cob at bar C)
Do the same for -
Using 3 junction box (1 for each bar).
Central bar will have junction box with:
wago#1(to split + from driver in 3) wago#2 (to split + between its 4 cob)
Wago 1bis (configured as wago#1 but for -)
Wago 2bis (configured as wago#2 for -)
Than each pair from wago#1 and wago#1bis will go to Junction box of bar B and bar C.It should be more clean configuration, with 3 box and just few wire connecting junction box between them.Is this ok?
Instead of multiple connections, connecting all the wires at one point will give equal voltage/current to each LED. I would use a wire nut to connect together both positive wires from one driver and 6 wires, one to each LED. Do the same for negative. Duplicate for the other 6 LED and the other Driver. 4 wire nuts and you are done.
From the left, it's dim, then I turn it and it gets brighter the more you turn it. However, about 1/4 throw it just suddenly stops getting brighter.
Confirmed the 50k resistance of the pot
Disconnected the pot, left the leads open, and I'm still getting 90w from the wall. What the hell!
Correction, these are the HLG-185-48B drivers, not 54.
Update.
Unplugged one driver, so I'm only testing one driver.
Pulling half of what I was pulling. 44w.
These are wired in parallel. 3 CXM22 to each HLG-185-48 driver. This isn't making any sense at all
JCave
Well-Known Member
Would
https://www.radioshack.com/products/radioshack-100k-ohm-linear-taper-potentiometer
and
https://www.radioshack.com/products/radioshack-10-ohm-1-2w-5-carbon-film-resistor-5-pack
work?
I have a MeanWell HLG-185H-C1400B, I read that its a good idea to to put a 10k ohm resister on negative also?
https://www.radioshack.com/products/radioshack-100k-ohm-linear-taper-potentiometer
and
https://www.radioshack.com/products/radioshack-10-ohm-1-2w-5-carbon-film-resistor-5-pack
work?
I have a MeanWell HLG-185H-C1400B, I read that its a good idea to to put a 10k ohm resister on negative also?
Strange.... does it seem like 44W? Three CMX22 should draw about 150 watts at the max driver voltage of 51.7. Can you check voltage at the cob?
guy_incognito251
Well-Known Member
HLG-185-48B output voltage is 48V which is too low for the cxm22 - the driver will be running in constant voltage mode and limiting the output current (only HLG-185-48A has voltage adjustment).
Anyway to work with what I have, or do I have to order more drivers bleh.