New Cree J Series 3030

nfhiggs

Well-Known Member
See, initially you said it was "2% in favour of the Samsung". But by the time you had made up your figures - after Greengenes (not me) called you on them - it was now 1.5% in favour of the Cree. Or 1.9%. Whatever.
Since those factors are divisors and not multipliers, those numbers do indeed favor the Samsung. In this instance lower is better.
 

alesh

Well-Known Member
Indeed. Good work :)

:edit: Never mind, that's clearly the 400-700 figures.
Is that only 400-700 or all wavelengths? "True" PAR is only 400-700, so if you want to compare umol values then you should only fill that part of the Excel sheet and leave the rest on 0.

I personally like to just go for the LER/QER lumen to par conversion factor. So I can simply divide lumen by that factor. I never saw any need for the individual LER and QER values.

I see people calculating efficiency using LER and then use QER to go to umol. Why not just go from lumen to umol directly?

BTW I just tried, but the sheet doesn't care about the scale of the SPD either. Or rather, I just did the 3000K 80 CRI SPD and I forgot to divide the percentages by 100 and it didn't make any difference. Makes sense, because it's all relative anyway. The distribution doesn't change by changing the vertical scale.
It would be better to have complete SPD and then only use 400-700nm values to calculate QER. Otherwise a small error is introduced as lumens are defined over a wider spectrum.

I used LER and QER separately mainly to explain all the steps needed to calculate all these values. Just blurting out the correct answer is poor teaching. Also useful to be able to calcucate radiometric output (and indirectly quantity of heat being generated).
 

wietefras

Well-Known Member
It would be better to have complete SPD and then only use 400-700nm values to calculate QER. Otherwise a small error is introduced as lumens are defined over a wider spectrum.
So your saying the LER should be the same for full range SPD as the 400-700 range?

Otherwise there would be no difference between a selection and zeroing out the "out of range" SPD values.

I used LER and QER separately mainly to explain all the steps needed to calculate all these values. Just blurting out the correct answer is poor teaching. Also useful to be able to calculate radiometric output (and indirectly quantity of heat being generated).
Yes, in understanding how the calculations are done it helped, but in general use at some point it stops being useful to make the distinction. Especially since people seem to be wasting time and are often making calculation errors on the two step conversion.

If they even understand at all what needs to happen with those two numbers. There is so much misunderstanding in what those two numbers mean. For instance, greengenes claimed that efficiency numbers calculated using the LER alone are all you need.

Simply dividing lumen/lux values by a number is much more fool proof and faster.
 

alesh

Well-Known Member
So your saying the LER should be the same for full range SPD as the 400-700 range?
Yes, if you want to be accurate. LER should be calculated over the whole range of wavelengths with output >0 or over the whole range of luminosity function.
Actually I just compared the difference between "correct" method and zeroing out the "out of range" SPD values and it's surprisingly huge.
For the 1750K Vero Decor spectrum correct 400-700 nm QER is about 4.59 µmol/J while zeroed SPD gives 5.13 µmol/J - 12% error.
 
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Prawn Connery

Well-Known Member
Try this alesh. There are graphs for the 4000K 80CRI Cree 380-780 and LM561C 400-780.

While you're here, did you ever end up creating a file to plot SPD/energy per mole against the McCree Curve?
 

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alesh

Well-Known Member
Try this alesh. There are graphs for the 4000K 80CRI Cree 380-780 and LM561C 400-780.

While you're here, did you ever end up creating a file to plot SPD/energy per mole against the McCree Curve?
I digitized both SPDs too and came up very close to your numbers. The spectra are very similar, almost identical I'd say.

There are such spreadsheets flowing but I don't believe I ever made one public. McCree's curve is very speculative, while my spreadsheets are just for converting well defined units of measurement.
 

Prawn Connery

Well-Known Member
Since those factors are divisors and not multipliers, those numbers do indeed favor the Samsung. In this instance lower is better.
You're right. But PAR watts still don't give the full picture - because it's simply a measure of energy over a range of wavelengths without taking into account how efficiently each of those wavelengths is used by the plant at different stages of growth. For example, if a DE HPS puts out the same PAR watts as, say, a generic 6500K LED (real-world example), which one's going to be better for flowering?

And isn't that what we're trying to achieve here?

My whole argument - bullshit aside - is that if you open those files you will see that the Samsung clearly puts out a lot more of its energy in the green spectrum compared to the Cree. So even though the LER and QER numbers are similar, which one do you think will be more efficient in terms of photosynthesis plotted against the McCree Curve?

The simple fact luminous flux is weighted around the green spectra where the human eye is most sensitive - and where photosynthesis is arguably least reliant (for flowering purposes at least) - means a "super efficient" LED in terms of LPW that puts out a significant amount of its radiant energy in the 500-600nm wavelengths may not be all that efficient at all when it comes to horticulture.
^ That was post #17.

I maintain my argument: to my eye, the Cree is the more efficient LED in terms of photosynthesis. I believe plotting it against the McCree Curve will prove this. And I believe it will be more than a 2% difference.

That's my nest mission when I have a bit more time. I'm off to bed again - it's late. :razz:
 

Prawn Connery

Well-Known Member
I digitized both SPDs too and came up very close to your numbers. The spectra are very similar, almost identical I'd say.

There are such spreadsheets flowing but I don't believe I ever made one public. McCree's curve is very speculative, while my spreadsheets are just for converting well defined units of measurement.
I understand the argument for the McCree curve being speculative. One could also argue the SPD graphs (certainly by the time they've been digitised) may not be that accurate either . . .

But my question (as above) would be, do you not think there is an appreciable difference between the two LEDs considering the Samsung puts out a lot more of its energy in the green (500-600nm) spectra than the red/blue compared to the Cree?

You say the spectra are almost identical, except the numbers relative to each wavelength are clearly quite different. That's why I'm asking the question.

EDIT: I really have to go to bed now, so will read/respond later :oops:
 
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wietefras

Well-Known Member
You're right. But PAR watts still don't give the full picture
PAR watts is what you get after dividing by the LER. When you divide by that lumen to PAR conversion (LER/QER) or multiply PAR Watts by QER, you end up with umol/s and not PAR watts.

I maintain my argument: to my eye, the Cree is the more efficient LED in terms of photosynthesis. I believe plotting it against the McCree Curve will prove this. And I believe it will be more than a 2% difference.
Even though the thinner coating will make the Cree COB/SMD itself slightly more efficient, for equal lumen/lux output those of the Samsung produce a higher umol/s PAR value.

The Cree spectrum contains a slightly higher ratio of blue. Blue photons take more energy, but umol and plants only take into account the number of photons. So equal PAR watts on blue and red will produce more photons on the red side and less on the blue side. That extra bit of blue in the Cree spectrum actually produces less photons (umol) than the extra green/red in the Samsung.

Also, again, check out CobKit's comparisons where Cree COBs tend to come out lower on output than expected. So somewhere in that 7% margin they all put in their small print, Cree tends to be a bit on the low side of that margin. Tiny spectral differences don't have anywhere near that effect

Perhaps you are thinking of the McCree quantum yield chart though? That's where green is supposedly less efficient, but then so is 450nm blue light. You can also correct for that chart and calculate the "YPF". I'm pretty sure in YPF, the Samsung SPD is also more efficient.

if a DE HPS puts out the same PAR watts as, say, a generic 6500K LED (real-world example), which one's going to be better for flowering?
Why would you flower under 6500K led? How is that a comparison to a 2000K HPS?

The problem with HPS is that the PAR watt or rather actual umol/s output, is not directly comparable to that of leds. HPS radiates all around the bulb. So you need a reflector and then you lose around 18% of the light already on the reflector. With COBs and SMD's you don't have to waste light on reflectors. The light goes in the correct direction already.
 

alesh

Well-Known Member
I understand the argument for the McCree curve being speculative. One could also argue the SPD graphs (certainly by the time they've been digitised) may not be that accurate either . . .

But my question (as above) would be, do you not think there is an appreciable difference between the two LEDs considering the Samsung puts out a lot more of its energy in the green (500-600nm) spectra than the red/blue compared to the Cree?

You say the spectra are almost identical, except the numbers relative to each wavelength are clearly quite different. That's why I'm asking the question.

EDIT: I really have to go to bed now, so will read/respond later :oops:
There's certainly some error that comes with digitizing but it's not that big and the margin of error could be calculated if I cared. The McCree's curve is speculative as a concept.

These relatives graphs throw you off.
If we move from relative to absolute it looks like this (both are hypothetical 100W_radiometric sources with respective SPDs). Cree has slightly more blue that's all. Remember, you have to compare area under the graph, not a y peak values.
cree_x_samsung.png
 

Prawn Connery

Well-Known Member
Area under the curve I understand. And a few basic calculations show the above graphs have the same relationship (relativity) as the original SPD charts (as they should). It's not the relative vs absolute figures I'm questioning.

Going by the numbers, the Samsung is clearly the more efficient diode in terms of overall photometric output from 400-700nm (4.73 vs 4.70, or thereabouts). But by converting SPD to LER to QER and ultimately PPF/W, it doesn't appear there is a step further to calculate yield photon flux.

If there is, I don't see it.
 

Prawn Connery

Well-Known Member
PAR watts is what you get after dividing by the LER. When you divide by that lumen to PAR conversion (LER/QER) or multiply PAR Watts by QER, you end up with umol/s and not PAR watts.
PAR watts/PPF whatever - you're missing the point if you're not going that extra step to calculate YPF. Whether you believe in spectrum-weighted photosynthetic efficiency or not (which you must do to put any credence in PAR in the first place - though that's another argument).

wietefras said:
Even though the thinner coating will make the Cree COB/SMD itself slightly more efficient, for equal lumen/lux output those of the Samsung produce a higher umol/s PAR value.
Over a very wide range of 400-700nm (which still isn't as wide as it arguably should be). But we know the value of each of those wavelengths in terms of photosynthetic absorption is not equal, don't we?

Or do you know something to the contrary? As I said, perhaps that's another debate . . . But my argument all along has been "photons ain't photons".

wietefras said:
The Cree spectrum contains a slightly higher ratio of blue. Blue photons take more energy, but umol and plants only take into account the number of photons. So equal PAR watts on blue and red will produce more photons on the red side and less on the blue side. That extra bit of blue in the Cree spectrum actually produces less photons (umol) than the extra green/red in the Samsung.
You're not following, are you?

Those "green" photons are arguably worth less than the blue if we are to base photosynthetic absorption on established chlorophyll pigment curves.

"Photons ain't photons".

wietefras said:
Also, again, check out CobKit's comparisons where Cree COBs tend to come out lower on output than expected. So somewhere in that 7% margin they all put in their small print, Cree tends to be a bit on the low side of that margin. Tiny spectral differences don't have anywhere near that effect
Red herring. At least for the sake of the original argument. Though it is a valid point in absolute terms, the original argument is based purely on the SPD graphs - not on caveats or disclaimers.

wietefras said:
Perhaps you are thinking of the McCree quantum yield chart though? That's where green is supposedly less efficient, but then so is 450nm blue light. You can also correct for that chart and calculate the "YPF". I'm pretty sure in YPF, the Samsung SPD is also more efficient.
Now we're getting somewhere - it only took you 6 pages of obfuscation!

wietefras said:
Why would you flower under 6500K led? How is that a comparison to a 2000K HPS?
You wouldn't - and that is my whole point.

But while we're on the subject, some old-skool indoor growers still flower under 5000+K MH, even though 2200+K HPS gives consistently better yields - and not just because HPS bulbs are more efficient than equivalent MH.

BTW, doesn't the Foton Phantom add 6500K LEDs to his Fusion flowering boards? Just sayin'.

wietefras said:
The problem with HPS is that the PAR watt or rather actual umol/s output, is not directly comparable to that of leds. HPS radiates all around the bulb. So you need a reflector and then you lose around 18% of the light already on the reflector. With COBs and SMD's you don't have to waste light on reflectors. The light goes in the correct direction already.
Yeah, that's crap. I haven't used a reflector for almost 20 years:
HazeHarvestSideCloser.jpg

But I'll forgive you, because I did mention DE HPS and not SE HPS. DE bulbs don't really lend themselves to vertical growing . . .
 

Prawn Connery

Well-Known Member
I don't want to sound like an ungrateful sod, so thanks alesh for taking the time to explain things to a novice.

And thank you, too, wietafras: I've learned shit-loads! I now know what the fuck I'm talking about! :bigjoint:

But I'll still fight you to the death :fire:
 

wietefras

Well-Known Member
PAR watts/PPF whatever - you're missing the point if you're not going that extra step to calculate YPF. Whether you believe in spectrum-weighted photosynthetic efficiency or not (which you must do to put any credence in PAR in the first place - though that's another argument).
How am I missing the point? I told you that's what you need to do if you want to correct for photosynthetic efficiency. Get for instance the McCree RQE chart data and adjust the SPD for that chart.

I do "know" something to the contrary though. Those photosynthetic response charts are created by putting a leaf in a sensor and measuring it's response to the light. When applied to real life plants, the research results show over and over that spectrum hardly makes any difference to yield (within reason). Less at least than McCree or any of the other photosynthetic response charts based on leaf measurements would indicate.

BTW Check out the McCree chart. There isn't much difference between blue photons and green photons.
McCree vs Cree 3000K.png

Besides, the real point still is though that the SPD's are almost identical. So whichever way you look at it, there won't be much of a difference.
 

Prawn Connery

Well-Known Member
That they're both 4000K emitters based on the same technology, they should be similar. But the curves are not identical. There's also the fact we don't have data for the Samsung below 400nm. And realistically, who knows how accurate those curves are to begin with - let alone what margins of error are introduced after digitising them?

So nit-picking aside, there's probably not a huge difference. Agreed.

The problem with the above McCree curve is it's an average of quantum photosynthetic response of light energy absorbed - not light that reaches the plant - based on 22 food crops.

Yes, photosynthetic response (energy conversion) to green light is more efficient. But only once it's absorbed. There is also the matter of how much of each spectra is absorbed in the first place to convert it energy.

The absorptance of 450nm acoss all 22 plant species was 24% higher than 550nm. So even though 550nm is more efficient, the average plant absorbs 24% less of it than 450nm.

The relative action of 550nm is 13% higher than 450nm. So the efficiency of absorbed green light is higher than that of absorbed blue light.

That still leaves a disparity in favour of 450nm in terms of actual light that hits the plant.

Quantum yields were not the same across all species of plant, either. Relative Quantum Yield of pigweed was 10% higher at 400nm than it was at 550nm, according to McCree's study.

I'm sure we don't have all day and night to dissect this, but suffice it to say, most plants absorb more blue light than green light, but use the green light they absorb more efficiently than the blue.

http://www.inda-gro.com/IG/sites/default/files/pdf/ACTION-SPECTRUM-KJMCCREE.pdf

In any case, I think this thread has served its purpose. At least it has for me.
 
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