DiY LEDs - How to Power Them

alesh

alesh

Well-Known Member
I feel pretty dumb asking this but this is just a cover of the Iadj pot and is supposed to be removed in order to regulate the current, right? Never had any A type MW driver and this one is for a friend so I wouldn't like to damage it.
a_driver.jpg

edit: Nevermind I should have read the data sheet.
2015-10-29 10_35_50-Mean Well Switching Power Supply - Opera.png
 
alesh

alesh

Well-Known Member
D

Devildenis69

Well-Known Member
Hey guys, is there a curve "Effeciency vs Out_Current" for the hlg185h-c anywhere ?
I feel a bit like that question has been asked several times, but can't find the info.

by the way, can't see why you're saying that this driver would be pulling 2A through the 3590, everything seems fine to me, maybe I'm not quiet awake yet bongsmilie:sleep:
 
alesh

alesh

Well-Known Member
Devildenis69 said:
Hey guys, is there a curve "Effeciency vs Out_Current" for the hlg185h-c anywhere ?
I feel a bit like that question has been asked several times, but can't find the info.

by the way, can't see why you're saying that this driver would be pulling 2A through the 3590, everything seems fine to me, maybe I'm not quiet awake yet bongsmilie:sleep:
Click to expand...
Try coffee and the 3590 data sheet - section Electrical characteristics, voltage x current chart.
 
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Devildenis69

Well-Known Member
are you referring to the fact that the extra volt from the driver would be "converted" to an increase in current ?
like when you're exceeding the voltage output of a driver, its current output may decrease
I'm running right now a poorly chosen driver/cob combo (LPC-60-1400/ V18 ) and I can assure you that my cob is running at 1.4A
or maybe just Coffe/Hangover is bad combo and I m totally missing your point.

would you have an idea of effeciency lost while running a 185h-c700 at 500mA ?
 
alesh

alesh

Well-Known Member
Devildenis69 said:
are you referring to the fact that the extra volt from the driver would be "converted" to an increase in current ?
like when you're exceeding the voltage output of a driver, its current output may decrease
I'm running right now a poorly chosen driver/cob combo (LPC-60-1400/ V18 ) and I can assure you that my cob is running at 1.4A
or maybe just Coffe/Hangover is bad combo and I m totally missing your point.

would you have an idea of effeciency lost while running a 185h-c700 at 500mA ?
Click to expand...
You're missing that it's a constant voltage driver.

The efficiency loss of the HLG depends on the voltage needed to achieve the desired current. The closer to max voltage the smaller the loss is.
 
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Devildenis69

Well-Known Member
Soooorry man, makes sense now

and I'm speaking about losses created by dimming the current, like I said a 185h-c700 ran at 500mA
are thoses losses negligibles ?
 
S

snooplover5

Active Member
Devildenis69 said:
Soooorry man, makes sense now

and I'm speaking about losses created by dimming the current, like I said a 185h-c700 ran at 500mA
are thoses losses negligibles ?
Click to expand...
Hi, long time lurker of this thread and by no means a pro. But I think the losses can be anticipated by looking at the load vs. Efficiency curve of the driver's datasheet

Edit: Attempt clarify
 
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alesh

alesh

Well-Known Member
Devildenis69 said:
Soooorry man, makes sense now

and I'm speaking about losses created by dimming the current, like I said a 185h-c700 ran at 500mA
are thoses losses negligibles ?
Click to expand...
snooplover5 said:
Hi, long time lurker of this thread and by no means a pro. But I think the losses can be anticipated by crossing the load vs. Efficiency curve of the driver's datasheet
Click to expand...
Depends on the voltage. You can have a string which needs 285V to pull 500mA through and you can have a string which needs only 150V. The 285V is going to be more efficient.
 
D

Devildenis69

Well-Known Member
Yes of course I know that, but I m speaking about reducing current by dimming, so both voltage are equal.
285*0.5 = 142W, 285*0.7 = 200W
maybe we could transpose it on the load curve, as power proportional to current in the datasheet case.
so running this driver at 500mA is equel to run it at 70% load, I think it's kinda what snooplover was saying ...
 
Greengenes707

Greengenes707

Well-Known Member
Read the data sheet.
If after you have looked at ever picture in it(you don't actually have to read anything), and still don't understand...then I do to know what to tell you
 
bicit

bicit

Well-Known Member
alesh said:
I feel pretty dumb asking this but this is just a cover of the Iadj pot and is supposed to be removed in order to regulate the current, right? Never had any A type MW driver and this one is for a friend so I wouldn't like to damage it.
View attachment 3531137

edit: Nevermind I should have read the data sheet.
View attachment 3531138
Click to expand...
It's a rubber stopper, you can just pinch and pull it out. Took me a minute to figure out as well :P
 
smokehog

smokehog

Well-Known Member
Hi Alesh, I have read the 3590 data sheet - section Electrical characteristics, voltage x current chart.

Please tell me if i have got this right.

On the cxb3590 chart for 36v it says it should be 1400mA so it needs a 1400mA dimmable driver HLG185-1400b?

The driver i asked about is a Meanwell LPV-series 60-36, so this is no good because it would be 2600mA which would make the driver's overload protection kick in. Which means the driver is no good? Is the maximum for the cxb3590, 3800mA?

Also, is it because it cant be dimmed from 1.67mA so stays constant?

I have no electrical experience apart from changing a plug or light bulb so are sorry for the silly questions.

Trying to understand it all before i construct my light, just waiting on 6 100mm glass lens arriving from Jerry before i attempt my build.
 
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Z

ziggydb

Active Member
Devildenis69 said:
Yes of course I know that, but I m speaking about reducing current by dimming, so both voltage are equal.
285*0.5 = 142W, 285*0.7 = 200W
maybe we could transpose it on the load curve, as power proportional to current in the datasheet case.
so running this driver at 500mA is equel to run it at 70% load, I think it's kinda what snooplover was saying ...
Click to expand...
Screenshot_2015-10-30-00-03-36.png This should help.
 
alesh

alesh

Well-Known Member
smokehog said:
Hi Alesh, I have read the 3590 data sheet - section Electrical characteristics, voltage x current chart.

Please tell me if i have got this right.

On the cxb3590 chart for 36v it says it should be 1400mA so it needs a 1400mA dimmable driver HLG185-1400b?

The driver i asked about is a Meanwell LPV-series 60-36, so this is no good because it would be 2600mA which would make the driver's overload protection kick in. Which means the driver is no good? Is the maximum for the cxb3590, 3800mA?

Also, is it because it cant be dimmed from 1.67mA so stays constant?

I have no electrical experience apart from changing a plug or light bulb so are sorry for the silly questions.

Trying to understand it all before i construct my light, just waiting on 6 100mm glass lens arriving from Jerry before i attempt my build.
Click to expand...
Yeah LPV-60-36 is no good for CXB3590 but HLG-60H-36 can power single CXB3590 up to 1.7A (and is dimmable).
 
alesh

alesh

Well-Known Member
Devildenis69 said:
Yes of course I know that, but I m speaking about reducing current by dimming, so both voltage are equal.
285*0.5 = 142W, 285*0.7 = 200W
maybe we could transpose it on the load curve, as power proportional to current in the datasheet case.
so running this driver at 500mA is equel to run it at 70% load, I think it's kinda what snooplover was saying ...
Click to expand...
Good news for you then, the losses will be minimal. On the other hand if you ran full current at 70% voltage (translating into 70% load) the losses would be higher.
 
OrganicCanuck

OrganicCanuck

Well-Known Member
Does any one know how far a meanwell hlg-185h-1400c driver can be placed from 3590 36v Cobs? 4 cobs/ driver

I am looking to have the drivers placed outside of the grow space, possibly 10-15 feet away, wired with 18/10 soow cable, I used a voltage drop calculator to come up with a % of drop. And for 15 feet of 18awg wire 36v @ 1 amp , the drop is .053% . Is this acceptable ?
 
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