Math behind

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speedyganga

Well-Known Member
Thank you alesh,
I understood the principle but I am not sure to understand how to do it...
should I normalize the Mc cree curve and apply the coefficient to each wavelength of the Led spectrum curve ?

I am interested in this because I wonder if 4000K/3500K + red and blue would be more efficient than 3000K alone.

Anyway it is not only about efficiency but mainly about plant reactions and triggering hormones.
 
qwerkus

qwerkus

Well-Known Member
Thank you so much - really useful thread. Has any of you cross-checked the results with support from cree or bridgelux? They surely have all that data...
 
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alesh

alesh

Well-Known Member
speedyganga said:
Thank you alesh,
I understood the principle but I am not sure to understand how to do it...
should I normalize the Mc cree curve and apply the coefficient to each wavelength of the Led spectrum curve ?

I am interested in this because I wonder if 4000K/3500K + red and blue would be more efficient than 3000K alone.

Anyway it is not only about efficiency but mainly about plant reactions and triggering hormones.
Click to expand...
You'd have to digitize McCree's curve. Then you'd have to calculate 4000K+RB (or whatever) SPD. Then multiply desired SPDs by McCree's curve and integrate the result. Higher is better. Probably.
 
SupraSPL

SupraSPL

Well-Known Member
It means the Plantastar makes 2.3% more photons/PAR W than the 3000K Cree, but that is only a small part of the story:

-The Cree can be set up to make 100% more PAR W per dissipation W versus HPS (twice as efficient)
-The Cree can deliver its photons to the canopy more efficiently than the HPS (reflector losses)
-The Cree can spread the photons in the canopy more uniformly than the HPS
-HPS cost much less/PAR W than the Cree
 
S

speedyganga

Well-Known Member
Hence when comparing to light I must refer to this spreadsheet and the electrical efficiency ?
It means for exemple HPS plantastar (let say 36% efficient) will be better than a 39% efficient cree ?
Then I could apply same method for monos ?!! (Yes I know this is another thread haha)
 
SupraSPL

SupraSPL

Well-Known Member
Because of reflector losses, even if the HPS efficiency was higher, the LED would still put more photons into the canopy because HIDs emit in 360 degrees and rely heavily on the reflector. 20+% of the light emitted from the HPS converts to heat before it even leaves the reflector, another 10% of the remainder is lost if there is a sheet of glass for a cooled hood. Then another 5%+ of what is left is lost bouncing off the walls.

The LED can lose as little as 6% according to the Cree TEMPO test of the spectrum king LED which uses a single large glass lens. Another small % of the lateral strays may be lost bouncing off the walls before reaching the canopy
 
bggrass

bggrass

Well-Known Member
I found in another thread how to calculate par and mmol/s/W

PAR = efficiency x dissipated watts@50C
mmol/s/W = PAR x conversion factor

Is that right? Where do I find the conversion factor?
 
bggrass

bggrass

Well-Known Member
alesh said:
It is the quality that I'm referring to as QER.
Click to expand...
So for the PAR values if efficiency = 0.5 and dissipatedWatts@50*C = 100

PAR = 0.5 x 100 = 50 PAR ?

the dissipatedWatts@50*C, is that the total watts, i.e. voltage x amps = watts?


The QER might take me some time to take in.
 
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