ChiefRunningPhist
Well-Known Member
Good stuff 
Base pump, WV, μmol/J, phosphor CE, LED effeciency
- Thread starter ChiefRunningPhist
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alesh
Well-Known Member
Hey. I haven't been very active lately. Haven't been respoding to anyone not just particularly you.I hadn't seen that particular thread but have tried to reach out to @alesh and the supra guy. It seems they just do the labor of finding every lm output per nm for a specific SPD and then use excel to multiply by the CIE factor.
I was trying to use the actual equations of the curves, the Luminosity curve, and the SPD curve, multiply them and integrate from bottom WV emission to top WV emission. I was using techniques I learned in school to approximate curve equations from a visual representation, (asymptotes, parabolas, normal curves, ect), but these proved harder to approximate than anticipated (or remember how lol) and I found a tool that seemed to do what I needed to find the conversion without the calculus, or me personally digitizing (someone probably already had done it).
I've been wanting to bounce shit off of them for awhile but they don't seem to be around anymore.This technique was one of those ideas.
Mathematical approximation of the curves in not the way to go imo. Neither luminosity function nor monochromatic LEDs' SPDs are normal distribution functions or any other standard functions. And of course, phosphor converted LEDs' SPDs are not even wildly similar to any standard function.
The easiest somewhat accurate way is to digitize the SPD and use excel as you describe. Basically to do a (graphical) approximation of function multiplication and integration.
I haven't been able to read the whole thread but looking at the first post - it's not correct at all. You can't use peak wvl for something as not-monochromatic as warm white spectrum. It's actually unsuitable for mono LEDs too, but to a lesser degree. Also stated µmol/J should probably be at the LED or even fixture level - that is, with phosphor losses counted in.
Feel free to @me with specific questions.
ChiefRunningPhist
Well-Known Member
The luminosity function is described as being "normalized" about 555nm. And looks like a normal distribution. Albeit you'd have to tweak the equation to model it more or less but it looks like a normal distribution, no?
A mono LED can be modeled relatively easy I think?? At least if its symmetrical? Though I hear ya, if I were to find an easily modeled equation of the Luminosity function, adding and subtracting known functions to create the SPD function is a huge pain in the ass lol
I guess I'd ask for more clarification as to why I can't use peak base pump, (caveat of assuming the base pump SPD is symmetrical)?
If the phosphor CE is at best 1, then the qty that is produced at base = total qty. The resulting energy difference in photon downgrading is expressed as heat on the phosphor film, it would mean low CCT chips were warmer than high CCT chips, I think this is observed in reality. When you measure μmol/s in a sphere, you're counting the total emmision. If it's understood that there is no loss in qty of photons, only loss in energy of photons, then you should be able to approximate base pump chip effeciency based on the base pump peak WV and total μmol/s measured. I care less so about the additional losses on the phosphor film, more so about raw performance, Ie what I put in in terms of watts to what I get out in terms of photons.
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ChiefRunningPhist
Well-Known Member
ChiefRunningPhist
Well-Known Member
@alesh have you digitized the photoreceptor absorption & action spectra? That's something I'm trying to do, then I can compare lamp SPDs and their output to the total % utilized by the plant.. Create a new metric to judge grow lights from.. Like a UVI but weighted to photoreceptor absorption peak WVs instead of just 295nm like UVI.
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ChiefRunningPhist
Well-Known Member
ChiefRunningPhist
Well-Known Member
In my estimation, it seems ~22.5% of base pump radiance is converted to heat via phosphor film for this luxeon 5700k, even though photon qty is essentially maintained. I need to correct and use a different term than electrical effeciency. Effeciency is energy out over energy in, so the effeciency metric I've been describing is not exactly total effeciency because it's not accounting for phosphor film energy losses, it's only describing the electrical effeciency per base pump photons created.
I used the luxeon color line to come up with the 22% figure.
According to the data sheet a cool white L1SPCW90002000000 pushes 1.57μmol/J at typical/nominal.
According to the data sheet the deep red L1SPDRD0002000000 pushes 2.65μmol/J at typical/nominal.
Looks like peak base WV are ~455nm for the cool white, & ~664nm for the deep red.
455nm/119.6 = 3.80μmol/J
664nm/119.6 = 5.55μmol/J
455nm cool white:
(1.57μmol/J) ÷ (3.8μmol/J)
=
0.413; or 41.3%
663nm deep red:
(2.65μmol/J) ÷ (5.55μmol/J)
=
0.477; or 47.7%
When comparing these effeciencies to a different and trusted source, I get..
"
455nm cool white = 32.0% effecient
664nm deep red = 47.9% effecient "
It seems to me that the 0.4% difference (47.7% ÷ 47.9% = 99.6%) on the 664 nm deep red was due to the SPD of the 664nm not being symmetrical (amoung other roundings etc), and the 445nm cool white had a 22.5% reduction in effeciency (32.0% ÷ 41.3% = 77.5%), or a 22.5% reduction in [(mW of total radiant output) vs (mW/μmol of base pump)], and due to the losses on the phosphor film. For this 5700k example it looks like ~22.5% reduction in total effeciency due to phosphor film losses.
This could explain the discrepancy I've been having with Samsung saying 3.0+ μmol/J ( I calculate 79% effecient in base pump of 455nm) despite the effeciency from the trusted source giving a much lower effeciency (~70% effecient). Aha! I thought they were just lying through their teeth..
Maybe, phosphor film losses scale linearly with CCT & CRI, but idk, would be useful info, most SPDs of like CCTs & CRIs are pretty similar company to company.
In conclusion, it seems you can calculate a monochromatic LED chips total electrical effeciency, via its measured μmol/s output and its peak WV, but, using this technique for white light requires an additional phosphor loss be added to the the electrical effeciency per base pump photon created (I'm dubbing it the "base pump EE") to finally arrive at a "total electrical effeciency".
@alesh I couldn't message you, says you're locked up like fort Knox lol, hmu sometime..
I used the luxeon color line to come up with the 22% figure.
According to the data sheet a cool white L1SPCW90002000000 pushes 1.57μmol/J at typical/nominal.
According to the data sheet the deep red L1SPDRD0002000000 pushes 2.65μmol/J at typical/nominal.
Looks like peak base WV are ~455nm for the cool white, & ~664nm for the deep red.
455nm/119.6 = 3.80μmol/J
664nm/119.6 = 5.55μmol/J
455nm cool white:
(1.57μmol/J) ÷ (3.8μmol/J)
=
0.413; or 41.3%
663nm deep red:
(2.65μmol/J) ÷ (5.55μmol/J)
=
0.477; or 47.7%
When comparing these effeciencies to a different and trusted source, I get..
"
455nm cool white = 32.0% effecient
664nm deep red = 47.9% effecient "
It seems to me that the 0.4% difference (47.7% ÷ 47.9% = 99.6%) on the 664 nm deep red was due to the SPD of the 664nm not being symmetrical (amoung other roundings etc), and the 445nm cool white had a 22.5% reduction in effeciency (32.0% ÷ 41.3% = 77.5%), or a 22.5% reduction in [(mW of total radiant output) vs (mW/μmol of base pump)], and due to the losses on the phosphor film. For this 5700k example it looks like ~22.5% reduction in total effeciency due to phosphor film losses.
This could explain the discrepancy I've been having with Samsung saying 3.0+ μmol/J ( I calculate 79% effecient in base pump of 455nm) despite the effeciency from the trusted source giving a much lower effeciency (~70% effecient). Aha! I thought they were just lying through their teeth..
Maybe, phosphor film losses scale linearly with CCT & CRI, but idk, would be useful info, most SPDs of like CCTs & CRIs are pretty similar company to company.
In conclusion, it seems you can calculate a monochromatic LED chips total electrical effeciency, via its measured μmol/s output and its peak WV, but, using this technique for white light requires an additional phosphor loss be added to the the electrical effeciency per base pump photon created (I'm dubbing it the "base pump EE") to finally arrive at a "total electrical effeciency".
@alesh I couldn't message you, says you're locked up like fort Knox lol, hmu sometime..
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eatled
Active Member
If you want to look into this further you may want to look at Cree XP3G Royal Blue.
The Cree Royal Blue is spec'd in both radiant watts and µmol/J.
The royal blue is the same as a white without any phosphor.
So if @alesh has a spreadsheet for a white XP3G then you can compare the radiant watts of the white with the royal blue spec and you will know the phosphor loss.
I'm not sure but the Luxeon Color C line may be the same as their Sun Plus red and blue. If so you can get the radiant watts from the Color C spec.
When comparing these Luxeons with other manufactures keep in mind the Luxeon are spec'd at 85° C where most others are spec'd at 25° C.
eatled
Active Member
I have a PHP routine that does that. But it's easy enough to just look at the LED's spectrum and see where the red and blue peaks and their widths to compare.
McCree showed us which spectrum to look at.
I you want a better analysis use an @aleash spreadsheet and sum the blue and red wavelengths.
When looking at the spectrum for these phosphor pumped LEDs, sum the deep red from 625 nm to 675 nm and the blue 425 nm to 475 nm.
Generally what is important is the ratio between deep red and deep (royal) blue.
Research has shown that the BR ratio is particular by species. Such research for cannabis does not exist. That I know of.
Have you seen the Luxeon 2835 Horticulture White? It's a Luxeon 2835 mid power LED with a very good looking spectrum.
Also in the LUXEON 2835 with FreshFocus are some good looking spectrum. Red Meat, Marbled Meat, and Produce.
ChiefRunningPhist
Well-Known Member
I'm using some of the luxeon 2835 color line, but not many. The 474nm, & their 545nm lime.
Theres actually been a significant amount of research that points to green light being just as useful as R:B. Lots of people write about LED deficiency, and many say the symptoms these growers experience is due to VPD, but if you look at the research, it looks like its more SPD to me. I can post research paper links if you'd like. I post them all the time but have begun to get the feeling nobody really reads them and so it's just a waste of time, but there have been 2 or 3 recent papers dealing with SPD and plant growth, some cannabis specific.
eatled
Active Member
McCree was done with antiquated 1960's equipment and kitchen sponges. Forget McCree. There is newer more accurate reports since.
Read the attached PDF, Chapter 7 of the plant physiology Bible of today, in this century.
The relative quantum yield is the difference between what was absorbed and what increased oxygen output. Useless information IMHO.
It confuses the shit out of most people and creates too many arguments on this site for what it's worth.
All RQE shows is that there is very little difference between what is absorbed and how much CO2 is converted to oxygen.
All we need to know is 450 nm and 650 nm is what plants like the best. The question is in what BR ratio for any given species is optimal.
All we care about is Action. In order for there to be Action there must have been Absorbance. I only want Action. Plants want 450 and 650 nm for photosynthesis.
For photosynthesis. That opens another can of worms. What is important is the effect of spectrum on secondary metabolites e.g. THC, flavor, aroma, and taste.
For example it has been found that the yellow spectrum affects the aroma of basil. No such research for cannabis.
Who gives a fuck about the difference between absorption and action? Not me.
The difference between action and absorbance is used by grow fixture vendors for purposes of deception. If you see relative quantum yield mentioned by a grow light vendor, they are a charlatan.
This image is from the attached PDF.
RQE is the difference between these two curves. So what? All it is saying is if a photon is absorbed, it is very likely to be used.
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eatled
Active Member
Whoa big guy.
The SPDs are typically radiometric, not photometric (luminosity).
555 nm does not play a role here.
What normalized means is given any output level this is the distribution. If the point of incidence is at this angle and this distance, this is the intensity. Not related to spectrum.
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eatled
Active Member
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Renfro
Well-Known Member
Green light does do something http://ursalighting.com/effect-green-light-plants/
ChiefRunningPhist
Well-Known Member
Checked the the XPG royal blue numbers with this base pump peak WV technique, 99.7% accuracy.
Idk, I'm not counting mcree out, I just disagree with the significance of relative quantum yield. Ill check out the .pdf.
They measured what percent of CO2 was being evolved into O2 under different WVs and intensities. O2 evolves when CO2 is converted into sugar, ie photosythensis. So imo it's very significant. It tells which WVs stimulated the greatest level of photosythensis. Maybe you can help me see your POV.
I could be mistaken but I'm under the impression that action and absorbance are different things. Absorbance is how many photons are absorbed out of how many there are, and action is how effecient the actual absorbed photons are utilized. They compound each other. If a chrollophyll molecule absorbs 80% of photons emitted at 455nm, and also has a 75% rate of action utilizing 455nm photons that were absorbed, then the chlorophyll molecule will utilize 60% of the total energy supplied at 455nm (0.8 × 0.75 = 0.6). Ect ect..
Here's a yield photon flux & mW flux chart I pulled off wiki..
Heres all the photoreceptor absorption rates from a different site (I can't remember)...
Here's the 2 overlapped...
The yield photon flux graph should actually be taking action and absorbance into account I'm pretty sure, but interesting the YPF in green seems much greater than anticipated... Hmmm...
I've attached a paper showing that past ~450μmol/s of overhead illumination, green is actually stimulating photosythensis more effeciently than red.
I def think R:B is important and most important, but just focusing on R:B ratio seems antiquated given all the research on SPD and morphology. Ill check out that .pdf, lol, I've been known to change my mind
eatled
Active Member
How do your eyes interpret a leaf as green? A green photon must be reflected off the leaf and be received by your eye's sensor.
Unless green photons positively affect the secondary metabolites, for the sake of photosynthesis green is very ineffective/inefficient. And I do not care about how deeply green is absorbed.
See Chapter 7.
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eatled
Active Member
Look at this graph. RQE is the difference between these two curves. Not relevant.
eatled
Active Member
In photosynthesis photons are converted to electrons. After that, it depends upon the electron transport chain and the photo chemical reactions.
eatled
Active Member
No.
In order for a photon to be utilized it must first be absorbed. What McCree is saying is, if a photon is absorbed, is it very likely to be utilized.
So what is actually important is what is absorbed rather than of the absorbed what is utilized because nearly every absorbed photon is utilized.
The charlatan grow light vendors are trying to confuse you. And they did.
Forget about the absorption spectrum all that is important is the action spectrum.
BTW McCree did not consider reflection, only absorption and transmittance.
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ChiefRunningPhist
Well-Known Member
Well I'm a bit rusty on my calculus, but was thinking it does matter if you're trying to convert the lm measurement they give to a chip effeciency.
Take the f(spd) × f(Luminosity) and integrate resultant from top WV to bottom WV and then divide recorded lm output to the best case scenario figure you just calculated (integration solution).