TWest65
Well-Known Member
I can't decide what to use for four qb135 boards. (fV = 36v, Max current per board = 2000mA)
hlg-240h-c1750b (looks 1 volt short at 71-143v. 250mA short of max current, is that sort of limiting?)
hlg-320h-c2100b (76-162v can handle 4 boards. 100mA over max current, is that a bad idea)
hlg-320h-c1750b (91-183v can handle 4 boards, or 5. But if 5 boards used would 'extra'/leftover voltage (over the 144v needed for the four boards) harm anything? Also, would run boards softer than the 320h-c2100b.)
hlg-320h-36b (8.9A *available*, right? I.e.: 4 boards could be hooked up and they'll just draw the current they "want", is that right?
hlg-240h-36b (I'm also considering this)
Gonna cover a 3x9ish area with qb135s, looks like 8 boards just because running odd numbers (like 5 boards per driver, times 2), or six and four, seems awkward. Though maybe 5 and 5, just a weird fixure or wiring arrangement - I don't usually put more than 2 per fixture, don't wanna.
Consider it a 3x10; 30sq foot, veg area. I'd like to put both driver dimmer leads on a single dimmer (rapidled pot) for even levels on both halves of canopy (and have the option to dim each individually still), that's why I want both/all same drivers.
What would you do?
@TWest65 @Barristan Whitebeard @Stephenj37826 Whoever.
Thanks.
I'm going to assume you meant QB 132's.
The HLG-xxxH-36B, will work fine but, if you every plan to repurpose the driver, a constant current driver would have more options.
If you think there is a chance you might run 5 boards in the future get the hlg-320h-c1750b. If budget is tight or your pretty sure this driver will stay married to these boards, save a little money and get the hlg-240h-c1750b. My personal preference would be the hlg-320h-c2100b, because of the possible repurposing options.
If you plan on running your boards much over 1750mA, then the hlg-320h-c2100b is your only option.
Without using a controller like an arduino, there is no simple way to control the dimming of the lights as you described. You can have one potentiometer control both drivers, or two potentiometers, each controlling one driver.
As far as "left over" voltage or current, there is none per se. It's more a case of unused potential.
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