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CobKits

Well-Known Member
stupid question but does it dim when you turn it? perhaps its on the wrong poles of the pot

what happens when you disconnect them from the pot and leave them open (separate from one another)?

have you measured the pot with your meter to make sure its really 50K?
 
G

G84

Well-Known Member
CobKits said:
i can do some testing today. i dont think it matters if you put 3 cobs on one wire and 6 on another
Click to expand...
@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

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
 
G

G84

Well-Known Member
@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
 
CobKits

CobKits

Well-Known Member
G84 said:
Well...if I use parallel wiring, I think that I have to split load equally, among driver output,or not?
Click to expand...
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
 
G

G84

Well-Known Member
CobKits said:
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
Click to expand...
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?
 
1

1212ham

Well-Known Member
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.
 
K

kaivorth

Active Member
CobKits said:
stupid question but does it dim when you turn it? perhaps its on the wrong poles of the pot

what happens when you disconnect them from the pot and leave them open (separate from one another)?

have you measured the pot with your meter to make sure its really 50K?
Click to expand...
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.
 
1

1212ham

Well-Known Member
Something has failed or we are missing something. Is that 90w from each driver? Cobs connected in parallel? How many?
 
K

kaivorth

Active Member
1212ham said:
Something has failed or we are missing something. Is that 90w from each driver? Cobs connected in parallel? How many?
Click to expand...
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
 
1

1212ham

Well-Known Member
kaivorth said:
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
Click to expand...
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?
 
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