DIY with Quantum Boards

pop22

pop22

Well-Known Member
54v is a percentage of 120v. 120/54= 0.45 or 45%. The AC side will use 45% of the 2.1 amps. 2.1 x .45 = 0.945a

120v x .945a= 113.4 watts. now an actual reading at the wall will be more because of driver and wiring losses. Running at 50% capacity, the driver is less efficient than it is when ran above 90% capacity. So let's say total efficiency is 90%. that means the draw will be 10% more than the 113.4
113.4 x 1.10+ 124.74 watts. So actual draw at the wall should be aproximately 125 watts for a DC load of 113.4


TWest65 said:
Back in 2017, which is when they were discussing this, they were talking about a HLG-320h-54a driver. If you hook up one qb288v2 to that driver, and set the DC voltage so that you pull 2100ma at the board, the board will use 102.82 watts, according to HLG.
If AC current and DC current were the same then the driver would pull, 120 vac x 2.1amps = 252 watts. We know the driver is using a lot less than 252 watts to output 102.82 watts to the board.
Click to expand...
 
TWest65

TWest65

Well-Known Member
pop22 said:
54v is a percentage of 120v. 120/54= 0.45 or 45%. The AC side will use 45% of the 2.1 amps. 2.1 x .45 = 0.945a

120v x .945a= 113.4 watts. now an actual reading at the wall will be more because of driver and wiring losses. Running at 50% capacity, the driver is less efficient than it is when ran above 90% capacity. So let's say total efficiency is 90%. that means the draw will be 10% more than the 113.4
113.4 x 1.10+ 124.74 watts. So actual draw at the wall should be aproximately 125 watts for a DC load of 113.4
Click to expand...
The easy way to figure it:
wattage at the board divided by the effeciency of the driver.
54v x 2.1a / 90% = 126 watts
The reason our answers are different is because you should have divided by .9 instead of multiplying by 1.1

In reality, the wattage is less because at 2100 ma at the driver, the board voltage on the qb288v2 would be 48.96v, (according to HLG).
48.96v x 2.1a = 102.816 watts at the driver
102.816/.9 = 114.24 watts at the wall
 
pop22

pop22

Well-Known Member
No, multiplying by 1.1 gives the same answer as calculating the amount of current consumed by the efficiency loss, then adding that 10% efficiency loss to the theoretical power draw to give you the actual power draw at the wall. I'm not interested in the Dc side, I'm discussing the AC side, as that's were the loss occurs, in the components converting some of the power to heat. And as to the actual DC power draw,, if we're nit picking, will vary with each set of boards, wiring losses, etc.. the only way to get a true answer to the DC side is to measure both the voltage and the amperage at the end of the string of boards. So no, it's not 54vx2.1/90%, the 10% loss in in the conversion of AC to DC.

TWest65 said:
The easy way to figure it:
wattage at the board divided by the effeciency of the driver.
54v x 2.1a / 90% = 126 watts
The reason our answers are different is because you should have divided by .9 instead of multiplying by 1.1

In reality, the wattage is less because at 2100 ma at the driver, the board voltage on the qb288v2 would be 48.96v, (according to HLG).
48.96v x 2.1a = 102.816 watts at the driver
102.816/.9 = 114.24 watts at the wall
Click to expand...
 
TWest65

TWest65

Well-Known Member
pop22 said:
54v is a percentage of 120v. 120/54= 0.45 or 45%. The AC side will use 45% of the 2.1 amps. 2.1 x .45 = 0.945a

120v x .945a= 113.4 watts. now an actual reading at the wall will be more because of driver and wiring losses. Running at 50% capacity, the driver is less efficient than it is when ran above 90% capacity. So let's say total efficiency is 90%. that means the draw will be 10% more than the 113.4
113.4 x 1.10+ 124.74 watts. So actual draw at the wall should be aproximately 125 watts for a DC load of 113.4
Click to expand...
No, it does not mean that the ac side will be 10% more than the dc side.
It means the dc side will be 10% less than the ac side.

your way:. 113.4 x 110% = 124.74
the correct way: 113.4/90% = 126
 
Budzbuddha

Budzbuddha

Well-Known Member
Flowertime on scorched earth .... decided to throw my last plants under flower during this global problem. Decided these will be the last ones I will grow for awhile, since I managed to stock up on jars already. The KKP S1 are in same medium and had some minor defol for setup. The smallest one is also a KKP S1 but will go 12/12 on it anyways.

A4D18311-1507-45B2-A20A-8FA8D4D4B9CE.jpeg
The Gotham limited were thrown under flower about 3 days ago with the last 2 finishing flower plants ( GSG Sativa lean and GSG bonsai ) .

92635346-B8D1-455A-9C62-003B999FE5DD.jpegC655F7C2-DC9F-4467-BFB7-955AF4F4A61F.jpeg


Well wishes to all .... let’s hope this gets resolved soon . Keep it green. Will update as flowering goes along. Seeds back in fridge
” for now “ ...
 
TWest65

TWest65

Well-Known Member
pop22 said:
No, multiplying by 1.1 gives the same answer as calculating the amount of current consumed by the efficiency loss, then adding that 10% efficiency loss to the theoretical power draw to give you the actual power draw at the wall. I'm not interested in the Dc side, I'm discussing the AC side, as that's were the loss occurs, in the components converting some of the power to heat. And as to the actual DC power draw,, if we're nit picking, will vary with each set of boards, wiring losses, etc.. the only way to get a true answer to the DC side is to measure both the voltage and the amperage at the end of the string of boards. So no, it's not 54vx2.1/90%, the 10% loss in in the conversion of AC to DC.
Click to expand...
Input wattage of driver x efficiency = output wattage
Input wattage of driver = output wattage/efficiency
output wattage/efficiency = Input wattage of driver
113.4 / 90% = 126

For some reason you're using this formula:
output wattage x (2 - efficiency) = input wattage of driver

If your answer is correct then
124.74 x 90% should equal 113.4, but it doesn't!
 
Last edited:
PrometheanLeaf

PrometheanLeaf

Well-Known Member
What is the best spacing and driver options for 8 hlg qb 288 v1 4000k in a 5x5 area? Would two hlg-480-xxxa drivers be a good option as far as wattage?
 
ilovereggae

ilovereggae

Well-Known Member
PrometheanLeaf said:
What is the best spacing and driver options for 8 hlg qb 288 v1 4000k in a 5x5 area?
Click to expand...
I have a 5x4 closet. Currently have 8 v1 boards split b/t 2) hlg240H drivers 4 boards per custom heatsink. Im adding a 3rd w 4 more boards as soon as the heatsink gets here to get to @ 680 w. I run my boards at 60w for greater efficiency, closer hanging height and slightly cooler running temps.

How about 2 boards per slate double heatsink, each on a HLG240? That would be 960W total.
would be killer for that space probably would need to dim it a bit.

If you got one more v1 board you could do 3 slate triple heatsinks w 3 boards each w 320W drivers for same wattage.
 
PrometheanLeaf

PrometheanLeaf

Well-Known Member
I was probably just going to run them on slate 2 single heatsinks with 80/20 connecting them. I was considering running them a bit harder and leaving the center without a board to eliminate that hotspot; but I've been unsure.
 
ilovereggae

ilovereggae

Well-Known Member
PrometheanLeaf said:
I was probably just going to run them on slate 2 single heatsinks with 80/20 connecting them. I was considering running them a bit harder and leaving the center without a board to eliminate that hotspot; but I've been unsure.
Click to expand...
Cool sounds like that would work too. Ill let wiser minds advise on best way to do that setup. How about 2 320w per 3 boards on the edges and 240w driver for the inside boards? That way you can dial in the levels individually? Would still be 880w and 35w/sf.
 
R

RamenToppings

Member
4 x 288, which driver do I need? I don't see a 4 board kit on hlg to tell me what driver I need :(. Not sure where to buy the wiring necessary either.

Thank you
 
R

RamenToppings

Member
I can't find the appropriate foot print tent for 4 x 288qbs, vivosun tents with bigger dimensions are not square... I don't know anything. What kind of tent or dimensions would optimize the lights?
 
You must log in or register to reply here.
Top