DIY with Quantum Boards

Frank Nitty

Well-Known Member
@Frank Nitty here is how i would wire up the QB120 fixture in series using your current setup to keep the wiring simple. You just need to make sure the + / - are in the proper positions so u might need to flip the 2 'bottom' boards so they are setup like in this diagram.

red is driver +
black is driver -
gray is wires connecting the boards together in series

View attachment 5000029
So what is the proper driver
 

Frank Nitty

Well-Known Member
That's because you are set up to run your 54v boards in parallel, and that top driver is the correct voltage for that setup. The other drivers run higher output voltages, so you would need to reconfigure your boards in series for them to work with those other drivers.

When in doubt, check the specs:

View attachment 5000024
These are the things that I don't understand, and that is why I come to you all...
 

ilovereggae

Well-Known Member
So what is the proper driver
that is the recommended driver for 4 QB120s in series.

if you want to run the 2 QB288s in parallel then get this driver:

HLG-240H-54A

That will give you constant voltage, and then split the current between the 2 boards.

Personally I would get the A/B model instead so that you can also add a dimmer:

and then order this dimmer
 

ilovereggae

Well-Known Member
What do you mean only 150 watts?

Sorry I don't have it totally figured out yet. What's the math again?
All good. This stuff took me a long time to get a grasp of. A few years ago I couldnt tell you confidently if the power from a wall plug was AC or DC lol. Big ups to @Rocket Soul and @The Loafter and everyone else I kept bothering in PMs with questions.

If you check the QB288 board guide you can see the info there on what the boards pull at various currents and voltages. (No clue why these pages arent linked on the site anymore but if you google u can find them). *once i did the math some of my original guestimates were off.

Screen Shot 2021-10-01 at 8.24.41 PM.png

Screen Shot 2021-10-01 at 8.25.05 PM.png

Anyway with Constant Current drivers, you are sending each module the same current. This is actually the way preferred by the manufacturers and what they recommend. This means when you connect them in series, each led module gets the same current, and then you add up the voltage used by each module.

For the 288s, lets start with the C2100 first. You can see from the HLG page that at 2100ma (or 2.1A) they will draw 48.96V each.
48.96 x 2.1 x 2 = 205.6 Watts.

For the C1750 you dont see that exact current on their list but you can figure it out by averaging the values at 1400 and 2100. So lets call it 48.1V (someone feel free to correct me but its been a long day and dont have that brainpower rn) at 1750ma which is same as 1.75A.
48.1 x 1.75 x 2 = 168 Watts.

for the QB120 at 2100ma they will pull 22.32V each.
22.32 x 2.1 x 4 = 188 Watts.

If you look at the datasheets for the drivers you will see that the

C1750 outputs 71 ~ 143V
C2100 outputs 59 ~ 119V

Now the one thing that makes this not 100% is that these drivers are actually CC/CV. Not sure how this works exactly but the simple explanation is it will output slightly more current to make up the difference. I know with my 240H-C1050 drivers if I have them turned all the way up they output 240W at the wall no matter what so I think that they are either pushing more current, or since there is more voltage available to each module it causes them to pull more current. Someone more experienced please clarify.
 

PadawanWarrior

Well-Known Member
All good. This stuff took me a long time to get a grasp of. A few years ago I couldnt tell you confidently if the power from a wall plug was AC or DC lol. Big ups to @Rocket Soul and @The Loafter and everyone else I kept bothering in PMs with questions.

If you check the QB288 board guide you can see the info there on what the boards pull at various currents and voltages. (No clue why these pages arent linked on the site anymore but if you google u can find them). *once i did the math some of my original guestimates were off.

View attachment 5000367

View attachment 5000368

Anyway with Constant Current drivers, you are sending each module the same current. This is actually the way preferred by the manufacturers and what they recommend. This means when you connect them in series, each led module gets the same current, and then you add up the voltage used by each module.

For the 288s, lets start with the C2100 first. You can see from the HLG page that at 2100ma (or 2.1A) they will draw 48.96V each.
48.96 x 2.1 x 2 = 205.6 Watts.

For the C1750 you dont see that exact current on their list but you can figure it out by averaging the values at 1400 and 2100. So lets call it 48.1V (someone feel free to correct me but its been a long day and dont have that brainpower rn) at 1750ma which is same as 1.75A.
48.1 x 1.75 x 2 = 168 Watts.

for the QB120 at 2100ma they will pull 22.32V each.
22.32 x 2.1 x 4 = 188 Watts.

If you look at the datasheets for the drivers you will see that the

C1750 outputs 71 ~ 143V
C2100 outputs 59 ~ 119V

Now the one thing that makes this not 100% is that these drivers are actually CC/CV. Not sure how this works exactly but the simple explanation is it will output slightly more current to make up the difference. I know with my 240H-C1050 drivers if I have them turned all the way up they output 240W at the wall no matter what so I think that they are either pushing more current, or since there is more voltage available to each module it causes them to pull more current. Someone more experienced please clarify.
I have a math degree. I kinda get it, but I don't, lol.
 

ilovereggae

Well-Known Member
I get that. I just am having a hard time with the numbers. I have the 1750's on my 2 board Rspecs, and I can turn it up to like 260 watts. According to the math @ilovereggae gave it should only be 168 watts. That's my confusion.
yeah i dont get it either tbh. I have my 240H-c1050B connected to 4 QB288s. according to the datasheets that should mean there is like 52V left over and i should only get 200W from it. however if u look at the datasheets there is a chart that shows the curve where it switches b/t CC and CV.

Screen Shot 2021-10-01 at 10.16.12 PM.png

The only thing I am legit wondering - is the driver pulling its max power when turned up to full blast, and there are more volts being pulled than used? I havent taken time to get out a volt meter to see. Meaning maybe I am only getting 200W to my boards? And you are only actually getting 168W.

Can someone answer that? In a CC driver, is the full voltage "pushed" to the modules so they will split whatever is available? Or will they each only pull what they need and the extra is just going "unused"?
 

pop22

Well-Known Member
The voltage will peak at the voltage drawn by the board with the highest draw. The boards are not exact matches ( although they are pretty close ) so if one board draws .2 more voltage that's what will flow through the chain.

Either of those drivers will only run 2 boards in series. Using as an example 51v per board you'll get about 90 watts per board draw with the 1750ma driver and about 108-110 watts with the 2100ma driver. The "left over" voltage is meaningless as the 2 boards have already used the majority of the available amperage.
Ameprage is like fuel in your gas tank, and voltage is the fuel pump, pushing the "fuel" to your boards. You may have the pressure to push more fuel but your injectors are already using the max they are capable of. The rest is just reserve.

yeah i dont get it either tbh. I have my 240H-c1050B connected to 4 QB288s. according to the datasheets that should mean there is like 52V left over and i should only get 200W from it. however if u look at the datasheets there is a chart that shows the curve where it switches b/t CC and CV.

View attachment 5000422

The only thing I am legit wondering - is the driver pulling its max power when turned up to full blast, and there are more volts being pulled than used? I havent taken time to get out a volt meter to see. Meaning maybe I am only getting 200W to my boards? And you are only actually getting 168W.

Can someone answer that? In a CC driver, is the full voltage "pushed" to the modules so they will split whatever is available? Or will they each only pull what they need and the extra is just going "unused"?
 

ilovereggae

Well-Known Member
The voltage will peak at the voltage drawn by the board with the highest draw. The boards are not exact matches ( although they are pretty close ) so if one board draws .2 more voltage that's what will flow through the chain.

Either of those drivers will only run 2 boards in series. Using as an example 51v per board you'll get about 90 watts per board draw with the 1750ma driver and about 108-110 watts with the 2100ma driver. The "left over" voltage is meaningless as the 2 boards have already used the majority of the available amperage.
Ameprage is like fuel in your gas tank, and voltage is the fuel pump, pushing the "fuel" to your boards. You may have the pressure to push more fuel but your injectors are already using the max they are capable of. The rest is just reserve.
good morning.. thanks Pop that fuel analogy was the best explanation I have heard.

What is confusing me is that according to the datasheet on the QBs at lower currents like 1750 they are only pulling closer to 48V not 51. Since the C1750 can output 143V it seems pretty close that u could get a 3rd board on there.

I'm still a little confused in your example. Is the driver compensating by outputting more current than it's rated for in order to get to that wattage? I knew that was true for the regular CV drivers but didn't know the CC ones worked the same way.
 
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