noob about to make a final decision. Please help.

C

Chuff420

Active Member
Danielson999 said:
These are approximates.

CXB3070 x 4 on one HLG-185H-C1400

50w x 4 cobs = 200w dissipation
200w x 50% efficiency = 100 par watts
10% losses from open space, reflectors, walls etc = 90 par watts
90par watts x 4.51 conversion factor = 405.9 PPF
3.5' x 1.8' = 6.3sqft divided by 10.76(sqft per m2) = 0.58m2
405.9PPF divided by 0.58m2 = 699.8 µmols/sec/m2 or PPFD
Click to expand...
So would this be correct for 5 x 3070 AD Bin below?

CXB3070.jpg

CXB3070 x 5 on one HLG-185H-C1400 = 0.7A per COB

50w x 5 cobs = 250w dissipation
250w x 52.37% efficiency = 130.925 par watts
10% losses from open space, reflectors, walls etc = 117.8325 par watts
117.8325par watts x 4.51 conversion factor = 531.424575 PPF
3.5' x 1.8' = 6.3sqft divided by 10.76(sqft per m2) = 0.58m2
531.424575PPF divided by 0.58m2 = 916.25 µmols/sec/m2 or PPFD
 
Danielson999

Danielson999

Well-Known Member
Chuff420 said:
So would this be correct for 5 x 3070 AD Bin below?

View attachment 3615272

CXB3070 x 5 on one HLG-185H-C1400 = 0.7A per COB

50w x 5 cobs = 250w dissipation
250w x 52.37% efficiency = 130.925 par watts
10% losses from open space, reflectors, walls etc = 117.8325 par watts
117.8325par watts x 4.51 conversion factor = 531.424575 PPF
3.5' x 1.8' = 6.3sqft divided by 10.76(sqft per m2) = 0.58m2
531.424575PPF divided by 0.58m2 = 916.25 µmols/sec/m2 or PPFD
Click to expand...
5 cobs will not fit on the HLG-185H-C1400.
5 cobs would fit on the HLG-185H-C1050 and run at 36.5w per cob.

36.5w x 5 cobs = 182.5w dissipation
182.5w x 48.5% efficiency = 88.51 par watts
10% losses = 79.66 par watts
79.66par watts x 4.51 conversion factor = 359.26 PPF
6.3sqft divided by 10.76 = 0.58m2
359.26 divided by 0.58m2 = 619.4PPFD

5 cobs on an HLG-240H-C1400

50w x 5 cobs = 250w dissipation
250w x 45.43% efficiency = 113.57 par watts
10% losses = 102.2 par watts
102 par watts x 4.51 conversion factor = 461 PPF
461PPF divided by 0.58m2 = 794.8PPFD
 
Rahz

Rahz

Well-Known Member
Chuff420 said:
CXB3070 x 5 on one HLG-185H-C1400 = 0.7A per COB
Click to expand...
That datasheet is for AB bin. I'm sure there's a BB bin spreadsheet on the forum somewhere, but regardless, the 185H-C1400 will run cobs at 1.4 amps and each cob draws a specific voltage based on the current(amps). Too many cobs and the driver will attempt to drop the amperage until the voltage is within parameters. Reference the product datasheets to determine the maximum number of cobs on a particular C series driver. #cobs x voltage(at particular current) must be = or < the max voltage of the driver model(particular current).

What you're suggesting will work but with BB bin and 1.4 amps you're looking at about 50-51% efficiency. I do recommend .7 amps, but it takes a C700 driver which I will explain towards the end of this piece as I talk you into spending your money on something other than Tasty.

PAR watts for cobs is just the efficiency x wattage (at a particular temp). The amperage for C series is set (unless adjusted manually) so if we want to know how efficient a cob is at a particular amperage we examine the relationship between the theoretical maximum in lumens (LER) at that Kelvin temp and the actual luemns. The theoretical maximum varies slightly depending on Kelvin temp.

Courtesy of SupraSPL and the other great contributors here:
"CREE CX series:
2700K 80 CRi = 321 (CREE verified LER)
3000K 80 CRi = 325 (CREE verified LER)
3500K 80 CRi = 324 estimated
4000K 70 CRi = 323 (Mr Flux calculated)
5000K 70 CRi = 324 (Mr Flux calculated)

Vero Version 2:
3000K 80 CRi = 323 (Alesh calculated)
3500K 80 CRi = 322 estimated
4000K 80 CRi = 324 (Alesh calculated)
5000K 70 CRi = 334.4 ( Alesh calculated)"

Lumens per watt / LER = efficiency.

The question of lumens per watt at particular currents is answered in the charts in the product datasheets, along with voltage at particular currents. Since you're full of questions I suggest you mystify yourself studying the datasheets since that is where all the information is coming from.

http://www.olfer.com/fuentes/pdf-meanwell/new2006/HLG-240H-C-spec.pdf

But thanks to Supra the spreadsheets do the math for us for the nominal and common currents. .7 amps isn't listed for the 2530 but you can estimate the LPW at 153, efficiency about 46% and voltage about 35.5.

Also, this will interest you: https://www.rollitup.org/t/top-bin-cob-comparison.891010/page-11#post-12179791 particularly the 3070 BB and 3590 CD.

Personally, when it came to building lamps I was conservative on my first build, running Vero 18 at nominal current. The original Vero at nominal is competitive with HID lighting, but not surpassing it by any notable means. Still, for a couple years ago and in an odd shaped space, it was a good deal even though expensive. No optics and an array of softly driven emitters 6 inches above the canopy and there basically are no reflective losses, which BTW I don't suggest you factor into your math in the future. Using good reflectors or getting the cobs close, the losses should be only 1-2 percent or whatever.

The only competitive options then were CXAs and I wasn't as savvy at the time, but the overall expense for CXA -vs- Vero was more pronounced back then so Vero made more sense.

My second build was Vero 29 (still version 1) at .7 amps and I'm still liking that lamp, not just for the output but for the better efficiency. If I were to replace one of them I would want to run 3590s at .7 amps, but might concede to run them at 1.05 amps which is still pretty nice. 60% efficient and I would have about 40% more light than my best lamp for the same wattage/overall heat load. That's worth paying for IMO. People enter this aspect of lighting (DIY) with a budget in mind, but to my mind taking the time and effort to DIY is worth spending a little more money to make something "bad ass". Especially in a smaller box efficiency is important. The more lumens/PAR you can cram in there without affecting the heat load is a good thing. At the very least I would be considering 3590s at 1.4 amps or 3070s at 1.05 amps, and getting the PPFD as close to 1000 as possible.

Ideally, in a 6 foot area, 8 3590s ran on a HLG-185H-C700B. (killing it with 200 watts)

There are cheaper way to go about it, and that's cool (for real) but if you really want my opinion... My business relies on the fact that running these particular COBs at nominal or moderately low current is better than DE HID lighting in several ways, but... for DIY, get dangerous! Spend a couple hundred more and light that box up with minimal wattage. You will be proud of your work still several years from now.
 
C

Chuff420

Active Member
So what determines which driver I use? Why a HLG-185H-C1050 and not a HLG-185H-C1400?

And then once I have selected a suitable driver the following calculations will provide the final solution.
  1. Dissipation (Watts) = [50C diss W] * # of COBs
  2. Gross PAR Watts = Dissipation * 50C%
  3. Net PAR Watts = Gross PAR Watts - (Gross PAR Watts / 10)
  4. PPF = Net PAR Watts x 4.51
  5. Coverage (m2) = Coverage (sqft) / 10.76
  6. PPFD (µmols/sec/m2) = PPF / Coverage (m2)
Correct?

Sorry. Former programmer. It's just the way I think.

Rahz, let me absorb what you've put up and I'm sure I'll have a few more questions. I really need to understand the calculations behind it. An algebraic expression representing the calculation would make perfect sense to me & how I think.

Thanks.
 
Airwalker16

Airwalker16

Well-Known Member
Chuff420 said:
So what determines which driver I use? Why a HLG-185H-C1050 and not a HLG-185H-C1400?

And then once I have selected a suitable driver the following calculations will provide the final solution.
  1. Dissipation (Watts) = [50C diss W] * # of COBs
  2. Gross PAR Watts = Dissipation * 50C%
  3. Net PAR Watts = Gross PAR Watts - (Gross PAR Watts / 10)
  4. PPF = Net PAR Watts x 4.51
  5. Coverage (m2) = Coverage (sqft) / 10.76
  6. PPFD (µmols/sec/m2) = PPF / Coverage (m2)
Correct?

Sorry. Former programmer. It's just the way I think.

Rahz, let me absorb what you've put up and I'm sure I'll have a few more questions. I really need to understand the calculations behind it. An algebraic expression representing the calculation would make perfect sense to me & how I think.

Thanks.
Click to expand...
The driver is dependant on the amperage. Your forward voltage of the cobs is between 33-36% dependent on the current itself (how hard you're driving them). It's a simple equation
(Fv)X anperage = wattage.
So 36 X 1.050= 36watts
36X 1.400= 56watts

Each driver has a maximum output of current, getting less and less as the amperage goes up (500,700,1050,1400,1750,2100)
So to answer your question, you can fit the 5 cobs on a 1050 & NOT a 1400, because it's exceeds the max output. You see?
36x5= 180. Okay ;)
56x5= 280. Not okay :(
The (185) in the driver name relates to the wattage max.
 
Airwalker16

Airwalker16

Well-Known Member
Airwalker16 said:
The driver is dependant on the amperage. Your forward voltage of the cobs is between 33-36% dependent on the current itself (how hard you're driving them). It's a simple equation
(Fv)X anperage = wattage.
So 36 X 1.050= 36watts
36X 1.400= 56watts

Each driver has a maximum output of current, getting less and less as the amperage goes up (500,700,1050,1400,1750,2100)
So to answer your question, you can fit the 5 cobs on a 1050 & NOT a 1400, because it's exceeds the max output. You see?
36x5= 180. Okay ;)
56x5= 280. Not okay :(
The (185) in the driver name relates to the wattage max.
Click to expand...

Watch Growmau5's videos on YouTube. There's a 6 part series on DIY led builds. Watch the first one. You should understand it quite well after that.
 
C

Chuff420

Active Member
So I walked away and educated myself a bit more and discovered that SupraSPL had uploaded a copy of his spreadsheet, and for a guy like me, that's the Holy Grail.

I made some modifications based on what I have learnt here and hopefully discovered all the answers I wanted, only to then find an error in SupraSPL's spreadsheet which impacted directly on my work.

The snippet of the CXB2530 from the COB spreadsheet thread below which I posted previously in this thread is at this stage an unknown.

CXB2530.jpg

Notice the underlined CXB2530 above the spreadsheet is not the same as in the top row in the spreadsheet, which is CXB2350. I've put up a post in the COB spreadsheet thread regarding this discrepancy in order to clarify what the correct labelling should be. You can read my post here for a detailed explanation if you wish. https://www.rollitup.org/t/cob-efficiency-spreadsheets.865238/page-18#post-12360078.

Needless to say, I am keen to find out exactly what COB it represents as will become apparent further down.

So with that in mind, I won't be sure what data is being represented in the samples below were references to CXB2530 are made until SupraSPL or someone else can clarify which COB the data really represents. Regardless I'll press on, as all the other information is correct, assuming my calculations are correct.

I modified the original spreadsheet and added the following information based on the formulas provided:

  • Green area where cabinet dimensions added for later calculations.
  • Red areas indicating the number of COBs to evaluate, as well as there Unit cost & Totals.
  • Blue area:
    • PAR(W)/COB = 50c dissW * 50c %
    • PAR (W) Total = PAR(W)/COB * # COBS
    • PPF = PAR (W) Total * 4.51
    • PPFD = PPF * Area (Metres²)
    • W/Feet² = ((# COBS * 50C %) / Area (Feet²)) * 100
The first thing that became apparent was the area I had to work with. In my first post I noted the chamber size as 860mm wide x 450mm deep and it's 1200mm high (approximately 3.5' x 1.8' & 4' high). My mistake was using "round about" figures for the dimensions in feet, and as such all further calculations were based on those figures equalling about 6 feet². Well, a proper conversion shows that I actually have 4.165668 feet².

Anyway, this is what I have found.

If I give myself a budget of around $150 for COBS this is what I get:

150_Dollars.jpg


If I give myself a budget of around $200 for COBS this is what I get:

200_Dollars.jpg


I haven't costed a driver into the equation, but my budget for COBS & Driver is around $300, so with the 2 spreadsheets above I've allowed enough room to get close or slightly above my $300 limit.

Excluding the CXB2530 figures for now as I'm not sure what the data represents yet, and assuming my formulas are correct, then the 3 * CXB3590 @ 0.7A & 4 * CXB3070 @ 0.35A are getting close to the magic 50W/foot² figure based on the new area of 4.165668 feet² if I spend $200 on COBS. The figures for the CXB2530 look absolutely brilliant, and based on the price would definitely be the way to go, but I'm kind of stuck until I know what COBs that data represents.

Does all this look about right?

Thanks.
 
Rahz

Rahz

Well-Known Member
As far as I know there's no such thing as a 2350 so it's 2530 data you've been looking at.
 
welight

welight

Well-Known Member
Chuff420 said:
So I walked away and educated myself a bit more and discovered that SupraSPL had uploaded a copy of his spreadsheet, and for a guy like me, that's the Holy Grail.

I made some modifications based on what I have learnt here and hopefully discovered all the answers I wanted, only to then find an error in SupraSPL's spreadsheet which impacted directly on my work.

The snippet of the CXB2530 from the COB spreadsheet thread below which I posted previously in this thread is at this stage an unknown.

View attachment 3617064

Notice the underlined CXB2530 above the spreadsheet is not the same as in the top row in the spreadsheet, which is CXB2350. I've put up a post in the COB spreadsheet thread regarding this discrepancy in order to clarify what the correct labelling should be. You can read my post here for a detailed explanation if you wish. https://www.rollitup.org/t/cob-efficiency-spreadsheets.865238/page-18#post-12360078.

Needless to say, I am keen to find out exactly what COB it represents as will become apparent further down.

So with that in mind, I won't be sure what data is being represented in the samples below were references to CXB2530 are made until SupraSPL or someone else can clarify which COB the data really represents. Regardless I'll press on, as all the other information is correct, assuming my calculations are correct.

I modified the original spreadsheet and added the following information based on the formulas provided:

  • Green area where cabinet dimensions added for later calculations.
  • Red areas indicating the number of COBs to evaluate, as well as there Unit cost & Totals.
  • Blue area:
    • PAR(W)/COB = 50c dissW * 50c %
    • PAR (W) Total = PAR(W)/COB * # COBS
    • PPF = PAR (W) Total * 4.51
    • PPFD = PPF * Area (Metres²)
    • W/Feet² = ((# COBS * 50C %) / Area (Feet²)) * 100
The first thing that became apparent was the area I had to work with. In my first post I noted the chamber size as 860mm wide x 450mm deep and it's 1200mm high (approximately 3.5' x 1.8' & 4' high). My mistake was using "round about" figures for the dimensions in feet, and as such all further calculations were based on those figures equalling about 6 feet². Well, a proper conversion shows that I actually have 4.165668 feet².

Anyway, this is what I have found.

If I give myself a budget of around $150 for COBS this is what I get:

View attachment 3617109


If I give myself a budget of around $200 for COBS this is what I get:

View attachment 3617110


I haven't costed a driver into the equation, but my budget for COBS & Driver is around $300, so with the 2 spreadsheets above I've allowed enough room to get close or slightly above my $300 limit.

Excluding the CXB2530 figures for now as I'm not sure what the data represents yet, and assuming my formulas are correct, then the 3 * CXB3590 @ 0.7A & 4 * CXB3070 @ 0.35A are getting close to the magic 50W/foot² figure based on the new area of 4.165668 feet² if I spend $200 on COBS. The figures for the CXB2530 look absolutely brilliant, and based on the price would definitely be the way to go, but I'm kind of stuck until I know what COBs that data represents.

Does all this look about right?

Thanks.
Click to expand...
$72 for CXB3590?? WAY HIGH
Cheers
Mark
 
C

Chuff420

Active Member
Rahz said:
As far as I know there's no such thing as a 2350 so it's 2530 data you've been looking at.
Click to expand...
Well if that's the case, then these are the figures for 4 * 2530 & 5* 2530 respectively which should do the job:

2530_4.jpg

2530_5.jpg


Previously we discussed using 10 * CXB2530 @ 0.7A connected in parallel as 2 rows of 5, and the figures now show that 10 * CXB2530 @ 0.7A will produce approximately 110 W/Foot², which begs the question, can you have too many Watts per Foot²? What's considered the upper limit?

Also, how are the calculations made in order to find a driver for COBs connected in parallel? I assume it's different to having them connected in series.

Thanks.
 
C

Chuff420

Active Member
Airwalker16

Airwalker16

Well-Known Member
You're confusing $72 for the 72V version?@welight sells 3590's for $42. Quit being a smart ass.
SUPRA must have based those numbers off of DIGIKEYS greedy asses.
 
Rahz

Rahz

Well-Known Member
Chuff420 said:
Well if that's the case, then these are the figures for 4 * 2530 & 5* 2530 respectively which should do the job:

View attachment 3617132

View attachment 3617134


Previously we discussed using 10 * CXB2530 @ 0.7A connected in parallel as 2 rows of 5, and the figures now show that 10 * CXB2530 @ 0.7A will produce approximately 110 W/Foot², which begs the question, can you have too many Watts per Foot²? What's considered the upper limit?

Also, how are the calculations made in order to find a driver for COBs connected in parallel? I assume it's different to having them connected in series.

Thanks.
Click to expand...
I'm not following your math leading to 110w/sqft. The 2530 at .7 amps should output about 25 watts right? 10 of them in a 4.16 sq/ft area would be 60 watts per foot. That would produce a very high light intensity towards the upper end of what the plant can tolerate. 110w/ft would certainly be excessive. 10 of them in a 6 foot area is also intense but more reasonable at 42w/ft.

5 2530s in your 4.16 sq/ft area would provide around 700 PPFD, which is not bad but you could certainly push it up a bit.

If you're price for the U2 bins is accurate (?), then I can see how you are finding them to be a better deal. Seems very cheap though, I hadn't mentioned it before because I was thinking in USD.

Two parallel strings would split the current and supply each string with the stated voltage.
 
C

Chuff420

Active Member
Rahz said:
I'm not following your math leading to 110w/sqft. The 2530 at .7 amps should output about 25 watts right? 10 of them in a 4.16 sq/ft area would be 60 watts per foot. That would produce a very high light intensity towards the upper end of what the plant can tolerate. 110w/ft would certainly be excessive. 10 of them in a 6 foot area is also intense but more reasonable at 42w/ft.

5 2530s in your 4.16 sq/ft area would provide around 700 PPFD, which is not bad but you could certainly push it up a bit.

If you're price for the U2 bins is accurate (?), then I can see how you are finding them to be a better deal. Seems very cheap though, I hadn't mentioned it before because I was thinking in USD.

Two parallel strings would split the current and supply each string with the stated voltage.
Click to expand...
Middle group in the spreadsheet below. 0.7A is somewhere between the 0.5A & 0.8A, and the last column shows 117 & 106 respectively. So a guestimate for 0.7A should be about 110.

150_Dollars.jpg

Rahz said:
The 50C% figure is the electrical efficiency. At .7 amps it will be about 46%.
At .7 amps and 35.5 volts the 2350 uses 24.85 watts. .46 x 24.85 = 11.43 PAR watts x 10 cobs = 114 PAR watts.

If you do the same for the 3070 you should find them putting out just over 100 PAR watts, so with 2530s you'd get about 10% more light for 25% more electricity. Not really a bad deal when considering the yield, but for $30 more you could run 5 3070s at 50 watts each... :)
Click to expand...
I simply used the same logic you used in post #38 of this thread, and pretty much got the same figure. You got 114. I got 110.

Hopefully that means we're both right.

Also, something has me confounded. The image below shows the calculations for 4 * CXB2530. At 0.3A it's pumping out 51 W/Feet² which is more than 3 * CXB3590 @ 0.7A which calculates at 46 W/Feet². 4 * CXB2530 @ $58.32 vs 3 * CXB3590 @ $218.10. That's a no brainer.

2530_4.jpg

What am I missing, or is it the way I see it?
 
Last edited:
C

Chuff420

Active Member
Airwalker16 said:
You're confusing $72 for the 72V version?@welight sells 3590's for $42. Quit being a smart ass.
SUPRA must have based those numbers off of DIGIKEYS greedy asses.
Click to expand...
http://www.cutter.com.au/proddetail.php?prod=cut2810

No. The price in the www.cutter.com.au link in the line immediately preceding this one clearly states that the CXB3590-000-000N0HCD35G is $66.09. And since I live in Australia, I have to pay the tax, which brings the total to $72.70.

Could you please provide me with a link to the $42 CXB3590. I'd be very keen to purchase them at that price.

Thanks.
 
Last edited:
Airwalker16

Airwalker16

Well-Known Member
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