Driver and COB voltage need clarity

I'm running a hlg-320h-c1050a the voltage range is 152-305. I'm planning on running 5 clu058 1812's their voltage per the data sheet is min 47.8 typ 52 max 56.2. I'm still new to all of this so I want to understand it as much as possible. So if u go by the cob having a 52v I'll be at 260v which falls in that range. Does that mean that the driver is only going to supply only what the cobs needs. I just want to make sure because If u divide the drivers max voltage @305 by 5 that will be 61v which is over the max voltage of the COB. Also say I wanted to run 6 using the drivers max v that will give me 50.8v which is within the COBs range will this work. Driver output is 320.25watts. So 50.8 x 1.05 = 53.375watts x 6 COBs gives me 320.25 watts. So what I want to know is is 5 safe to run as long as it falls in the drivers range and would it be possible to run 6. I just want to make sure I'm looking at this correctly. Appreciate any help Thanx

roll206 said:

Does that mean that the driver is only going to supply only what the cobs needs

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the cob does take only that volts which it needs with this kind of driver.

at 1050ma the cob use around 50 volts (Tc=50C)

Your driver has a Volt pool from 152-305. So you have to connect minimum (152Volt/50Volt=3,04cobs) 3,04cobs, that does mean minimum 4 cobs.

For maximum amount of cobs you calculate: 305volt/50Volt= 6,1cobs . So maximum 6cobs

i hope this helps.

Edit: you are talking about CLU058-1825?my calculations are only for CLU058-1825

No I listed the wrong cob in my post. I'll be running the CLU048-1812'S.

the cob does take only that volts which it needs with this kind of driver
CLU048-1812 at 1050ma use around 52,5-53volt(Tc=50C).

Your driver has a Volt pool from 152-305.
so you to be in this volt range from your driver, above 152volt but under 305volt

calculation to know how much cob you have to use minimum:
152Volt / 53 Volt = 2,86 Cobs so you have to connect minimum 3 cobs


calculation to know how much cob you can use maximum:
305Volt / 53 Volt = 5,75 Cobs so you can use maximum 5 cobs

this calculations are only for hlg-320h-c1050a in combination with CLU048-1812

Thax I appreciate the help.

roll206 said:

Thax I appreciate the help.

Click to expand...

do you know allready, how you wanne mount the cob on the heatsink? you plan to use cob holders or do you wanne screw the cob directly on the heatsink?
I ask because their dont exist cob holders which can handle 2+ cob in serie.

Yes I'm going with the pin fin heatsinks with Ideal holders. I've seen multiple diy fixtures wired in series with holders. Series is the preferred method of wiring cobs and the easiest.

roll206 said:

Yes I'm going with the pin fin heatsinks with Ideal holders. I've seen multiple diy fixtures wired in series with holders. Series is the preferred method of wiring cobs and the easiest.

Click to expand...

Hi,
the problem is that the cob holder can handle maximum 250 volts. That does mean you can only connect 4 cob maximum in series with that ideal holder. 250volt / 53volt = 4,7 so only 4 cobs in serie maximum
https://www.idealind.com/content/pdfs/specsheets/spec_sheet_chip-lok_model_50-22.pdf

Edit: Im correct that the ideal holder can handle maximum 250 volts or?or is it only 150 Volts?
please someone confirm it

It is 250v I was thinking it was 270v for some reason thanx for pointing that out.