New Setup LED Reccomendations

Trippyness

Well-Known Member
I got it , the only thing im confuzed on now is how to select drivers. Here is what I got.
Ill need 2 x LM561c Reels of 2500.
At 171 Lm/w at 2.59 umol/j which is 934watts aprox. With a PPFD
in a 25 Sq/f space is about 1,040 PPFD.

So my question is this, how do you calculate drivers for these? Im aware on the QB at 304 Diodes, has a current of 1050 and a Fv min and max of
108.3 112.5. Any help on what drivers I would need as well as how I should wire the strips. Built COBS before but this is slightly confuzing even after reading a few threads.
 

Aolelon

Well-Known Member
It's fairly simple. Find out if you're going to be wiring in parallel or series. If parallel, CV driver most likely. If in series, a CC driver most likely.
The 304 you will need a CC driver. If in series add your Voltages up and make sure they are within range of the Constant Current mode of the driver.
With the 304, chaining a couple drivers is hard because the connectors are only rated at 300v.
So putting more than 2 on a driver isnt recommended. As an example.
2 304 have a min max of 105.2-109v ea. At 700 mA so I'll take the max and add it. Which gives me 218v. Take 218 x .7 = 152.6w
So I'll need a CC driver of at least 153w @ 700mA
The meanwell hlg-185h-c700b fits that. With a CC region of 143-286v.
But you'll only have a max of 700 mA so I'll look at the next one up.
The hlg-240h-c1050b has a max voltage of 240w and puts out up to 1050mA
CC region is 119-238 within our range of 218-224(112@1050)
So that's the driver I'd pick.

Wiring strips is easier with a CV driver. If its 46v strip, pick a 48v driver and wire them up in parallel.
22vstrips get a 24v driver.
Just have to make sure you're not going over the Amperage.
5 strips at 22v is still 22v but each driven at 700ma is 3500mA or 3.5A

If that's confusing there are calculators on Ledgardener site. While it doesnt have stripa it has the qb boards and most cobs,
If it comes to you needing more help I'm sure there are tons of people, including myself willing to help.
http://ledgardener.com/cob-led-driver-selection-tool/
 

Aolelon

Well-Known Member
I'm not sure how they figure out the vf and current for the strips. @Stephenj37826 might be able to help. It's how they are wired in seriesxparallel on the board or strip which determines each diodes current and Vf. So I wish I could be of more help figuring that out.
With the strips that are already there it's easy cause they give you the values so if you're just going to be remaking those.
Here's how the strips are wired in a circut.
Which is gibberish to me.
 

Attachments

Trippyness

Well-Known Member
It's fairly simple. Find out if you're going to be wiring in parallel or series. If parallel, CV driver most likely. If in series, a CC driver most likely.
The 304 you will need a CC driver. If in series add your Voltages up and make sure they are within range of the Constant Current mode of the driver.
With the 304, chaining a couple drivers is hard because the connectors are only rated at 300v.
So putting more than 2 on a driver isnt recommended. As an example.
2 304 have a min max of 105.2-109v ea. At 700 mA so I'll take the max and add it. Which gives me 218v. Take 218 x .7 = 152.6w
So I'll need a CC driver of at least 153w @ 700mA
The meanwell hlg-185h-c700b fits that. With a CC region of 143-286v.
But you'll only have a max of 700 mA so I'll look at the next one up.
The hlg-240h-c1050b has a max voltage of 240w and puts out up to 1050mA
CC region is 119-238 within our range of 218-224(112@1050)
So that's the driver I'd pick.

Wiring strips is easier with a CV driver. If its 46v strip, pick a 48v driver and wire them up in parallel.
22vstrips get a 24v driver.
Just have to make sure you're not going over the Amperage.
5 strips at 22v is still 22v but each driven at 700ma is 3500mA or 3.5A

If that's confusing there are calculators on Ledgardener site. While it doesnt have stripa it has the qb boards and most cobs,
If it comes to you needing more help I'm sure there are tons of people, including myself willing to help.
http://ledgardener.com/cob-led-driver-selection-tool/
Gotcha, however I will be using the 2500 SMD strips
https://www.digikey.com/product-detail/en/samsung-semiconductor-inc/SPMWHT541ML5XAVMS0/1510-1286-2-ND/5766179

I went of the 304 based on the QB diode amount. Should have specified. Ill be using an entire reel per light.
I am going off the spec sheet here: https://horticulturelightinggroup.com/collections/quantum-boards/products/quantum-boards

Looking at this section:
108.3 112.5 1050 113.71 118.1 177.5 171.2 2.59

Want to use 1 reel per light. Some help there would be great :)
Sorry for teh confusion, not using the rigid strips but the tape.
@Stephenj37826 Any help there?
 

Randomblame

Well-Known Member
Gotcha, however I will be using the 2500 SMD strips
https://www.digikey.com/product-detail/en/samsung-semiconductor-inc/SPMWHT541ML5XAVMS0/1510-1286-2-ND/5766179

I went of the 304 based on the QB diode amount. Should have specified. Ill be using an entire reel per light.
I am going off the spec sheet here: https://horticulturelightinggroup.com/collections/quantum-boards/products/quantum-boards

Looking at this section:
108.3 112.5 1050 113.71 118.1 177.5 171.2 2.59

Want to use 1 reel per light. Some help there would be great :)
Sorry for teh confusion, not using the rigid strips but the tape.
@Stephenj37826 Any help there?

That's not an LED strip for lightning pourposes it simply a roll of diskret diodes. You would need to solder them to an flex PCB to create strips you can use. I would recommend to use 4ft. single row F-strips(46V/1,12A nom.) or single row EBgen.2 strips. The more you take the cheaper they get.

This are the data for F- and EBgen2. strips:

Single row 4ft. F-Series Gen3 3000K(46v/1,12A, 125mA per diode)
QER=4.86 µmol/J
LER=321.60 lm/W
= 2.54 µmol/J
51,52W * 2.54 µmol/J / 28,97$ = 4.52 µmol/$
https://www.digikey.de/product-detail/de/samsung-semiconductor-inc/SI-B8V521B20WW/1510-2225-ND/6624009

4ft. EB Series Gen2 3000K(39v/700mA, 43,75mA per diode)
QER=4.84 µmol/J
LER=331.66 lm/W
= 2.55 µmol/J
27,3w x 2,55μmol/J / 12,37$ = 5,63 μmol/$
https://www.digikey.com/products/en?keywords=BXEB-L1120Z-35E4000-C-B3

F-strips are in stock most of the time but with EBgen.2 it's a gamble to get them.
For 4ft. F-strips use a HLG-240H-48A for parallel wiring and for 4ft. EBgen2. take the HLG-240H-42A. The HLG-240 is the MW driver with the best price/performance rating and properly used you will get ~260w/280w at the wall.
For a 5x 5' area three of them would provide around 780w net./840w at the wall.
If you take 5 4ft.single row strips per driver they would run with ~1040mA, which means each diode gets only 115mA. Effiency would increase by a certain margin so ~2,6μMol/J maybe.
So 3x HLG-240H-48A(~65$, mousser.com) & 15 4ft. F-strips (29$, digikey) =630$ US.

Take some 1" aluminum c-channel to mount the strips on. It's easy to make a frame with them and they keep the strips nice and cool.(~50°C max.). You can screw them or simply use the cheap blue double sided thermal tape(25m rolls, 20mm wide, 0,2mm thickness, 7,50$, am4zon/e3ay)
 
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Aolelon

Well-Known Member
Wouldn't the double row be more efficient and less cost? 28$ for ea single and 52 for each double, he could use 7 of the double row.
 

Randomblame

Well-Known Member
Figured that was the case. So you would actually have to use those to send to a pcb maker to.make your own board.
Yepp!
He could create his own PCB's the only thing needed would be a gerber file with dimensions, circuits and other specs. But there are several online services or chinese manufactures offering different LED PCB's.
The diodes he could place himself after he has applied solder paste on the solder joints. Then bake it for 5 around minutes in the oven and finished are your own boards.

But the whole thing probably costs more than a certain number of finished F-strips, not to mention the extra effort. Producing Gerber files is not that easy, but you can also use Gerber files from the manufacturers.
 

Randomblame

Well-Known Member
Wouldn't the double row be more efficient and less cost? 28$ for ea single and 52 for each double, he could use 7 of the double row.
Sure, he could use the double row strips to save a few bucks but single row strips offer better coverage. Effiency is the same(same diodes) unless you drive them lower but than you need even more strips. Also 7 means he would need a HLG-240 (3 strips) and a HLG-320 (4 strips) to get nearly the same brightness from the strips.
 

Dave455

Well-Known Member
Gotcha, however I will be using the 2500 SMD strips
https://www.digikey.com/product-detail/en/samsung-semiconductor-inc/SPMWHT541ML5XAVMS0/1510-1286-2-ND/5766179

I went of the 304 based on the QB diode amount. Should have specified. Ill be using an entire reel per light.
I am going off the spec sheet here: https://horticulturelightinggroup.com/collections/quantum-boards/products/quantum-boards

Looking at this section:
108.3 112.5 1050 113.71 118.1 177.5 171.2 2.59

Want to use 1 reel per light. Some help there would be great :)
Sorry for teh confusion, not using the rigid strips but the tape.
@Stephenj37826 Any help there?
Make it easy...get some quantum boards from HLG......
 

Aolelon

Well-Known Member
That's not an LED strip for lightning pourposes it simply a roll of diskret diodes. You would need to solder them to an flex PCB to create strips you can use. I would recommend to use 4ft. single row F-strips(46V/1,12A nom.) or single row EBgen.2 strips. The more you take the cheaper they get.

This are the data for F- and EBgen2. strips:

Single row 4ft. F-Series Gen3 3000K(46v/1,12A, 125mA per diode)
QER=4.86 µmol/J
LER=321.60 lm/W
= 2.54 µmol/J
51,52W * 2.54 µmol/J / 28,97$ = 4.52 µmol/$
https://www.digikey.de/product-detail/de/samsung-semiconductor-inc/SI-B8V521B20WW/1510-2225-ND/6624009

4ft. EB Series Gen2 3000K(39v/700mA, 43,75mA per diode)
QER=4.84 µmol/J
LER=331.66 lm/W
= 2.55 µmol/J
27,3w x 2,55μmol/J / 12,37$ = 5,63 μmol/$
https://www.digikey.com/products/en?keywords=BXEB-L1120Z-35E4000-C-B3

F-strips are in stock most of the time but with EBgen.2 it's a gamble to get them.
For 4ft. F-strips use a HLG-240H-48A for parallel wiring and for 4ft. EBgen2. take the HLG-240H-42A. The HLG-240 is the MW driver with the best price/performance rating and properly used you will get ~260w/280w at the wall.
For a 5x 5' area three of them would provide around 780w net./840w at the wall.
If you take 5 4ft.single row strips per driver they would run with ~1040mA, which means each diode gets only 115mA. Effiency would increase by a certain margin so ~2,6μMol/J maybe.
So 3x HLG-240H-48A(~65$, mousser.com) & 15 4ft. F-strips (29$, digikey) =630$ US.

Take some 1" aluminum c-channel to mount the strips on. It's easy to make a frame with them and they keep the strips nice and cool.(~50°C max.). You can screw them or simply use the cheap blue double sided thermal tape(25m rolls, 20mm wide, 0,2mm thickness, 7,50$, am4zon/e3ay)
How do you calculate how many volts and mA each diode gets? I am thinking of designing my own PCB in the future and want to know how to calculate all that.
 

Randomblame

Well-Known Member
How do you calculate how many volts and mA each diode gets? I am thinking of designing my own PCB in the future and want to know how to calculate all that.

From F-strip datasheet! For instance, a 2ft. single row strip has 72 diodes and the internal circuit is 8s9p or 8 diodes in series 9 times parallel. This means at 1120mA nominal current each string gets 1120:9=124,45mA. And with 23v each diodes would get 2,875v.
The 4ft single row is 16s 9p and 46v/1,12A and the 4ft double row are 16s18p at 46v/2,24A. The current each diode gets it always the same with this layouts.
EBgen2 2ft. strips has 112 diodes in an 7s16p layout and run with 700mA nom. current, so 700mA:16=43,75mA per diode.
 

Randomblame

Well-Known Member
The EB strip diodes are far less efficient but run at a third of the LM561c current. You could squeeze out up to 185lm/w and run them ultra low(~20mA). LM561c is a lot better because @65mA you can get up to 210lm/w from a 4k diode. Ultra low it would be probably ~220lm/w.
But the prices for LM561c and LM301b differ only a little if you buy them in rolls and the LM301b would be again 10-20lm/w more efficient.
 
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