I concur on all counts.Quote:
Originally Posted by DreadedHermie
Good correction DH. :)
"You must spread, yadayada"
Aloha y'all
Weeze
Printable View
I concur on all counts.Quote:
Originally Posted by DreadedHermie
Good correction DH. :)
"You must spread, yadayada"
Aloha y'all
Weeze
i was waiting for you guys to chime in...rep to all:thumbsup:
i es no tekneecal Weezard...but i do's know how to grow:D
Was wondering when the others would come in to weigh in. :)
Hi guys! :hippy:
Seems people here are totally missing my point:Quote:
Originally Posted by khyberkitsune
The watt per watt LED grow match the HPS grow with 80% deep red and no yellow while the HPS has 80% yellow/orange and only maybe 5% deep red.
The current science tells us HPS should be an epic fail in comparison - and yet it doesn't.
Is no one else truly curious? Methinks there is something critical here that we are overlooking that could help in future LED designs.
you have answered your own question.....this is why LED's are more efficient with more than 80% of the light used by the plant and HPS only about 20 - 30 % of the light absorbed is actually used...this is why less wattage can match production of HPS...LED's hit the main lighting wavelenghts with little energy wasted:smokin:
not to mention the energy for cooling costs...thats why it takes 1000w HPS to get good results!:thumbsup:
Not even close. If the science as presented was complete and accurate, and 80% of HPS light was wasted then LEDs would be outperforming HPS by a huge factor.Quote:
Originally Posted by stra8outtaWeed
Not happening.
I am not challenging you guys as you seem to think, but there is definitely a piece of knowledge missing.
as i stated i do not need science to tell me my meds are better....i have incredible pain from SCI and when i can match production and have more potent meds...then LED's are blowing HPS out as i said earlier I WILL NEVER GO BACK TO GROWING WITH HPS....i am producing outdoor quality product with indoor controls basically!
There is nothing missing, the math hasn't been expressed and there's still one important unit that hasn't been discussed.Quote:
Originally Posted by RackitMan
Take a look at the light spectral chart - 80% is emitted in yellow/green, 20% in the blue/red.
Out of a 400W HPS, 320 watts of light are from yellow/green. 80 watts blue/red. This is why 90w UFO panels have the equivalent (even as stated on my panels) "Photosynthetic power" - only the honest panel makers refuse to say "Get the same yield as" because that is misleading.
The DIFFERENCE is that the HPS emits a higher photon flux density even for just those 80 watts of actual blue/red versus typical 1W led panels. Also, the angle of emission of LEDs directly determines the photon flux density - the wider the beam angle the less dense the flux of photons. You want the penetration power of a 400w HPS out of a panel of 1W diodes? Put the diodes down to a 30 degree beam to increase the photon flux and thus the penetration power.
But even watt-for-watt 400w LED in a tight panel will smash 400w HPS for yield. 400w 100% usable plant power versus 400w 20% usable plant power. Add in the fact they last longer and the entire thing is a long-term win situation.
I want to show you an e-mail I got recently.
Growth is directly proportional to the rate of photosynthesis. The rate of photosynthesis varies based on what wavelength of light the plant is exposed to. Anyway, theres some math for you. The spectral output for an HPS can be approximated and calculated for every 10 nm or so, such that you get the ammount of CO2 reduced per whatever-you-want-to-measure-your-light-output-in (I suggest a spreadsheet or some such thing) then compare to deep red LED.Quote:
I just went back to look at Fig. 7 of the PDF in:
The Photosynthetic Action Spectrum of the Bean Plant -- Balegh and Biddulph 46 (1): 1 -- PLANT PHYSIOLOGY
The action curve mostly ranges from 72 to 113 molecules of CO2 utilized
per 1,000 photons.
Typical red LED light with peak wavelength of 635 nm appears to me,
according to that curve, to achieve utilization of 108 molecules of CO2
per 1,000 photons incident upon the leaves.
These photons have average eergy close enough to 1239.7/635 electron-
volts each, which is 1.952 electron-volts.
Multiply 1.952 by 1,000/108, and so far it looks like 18.075
electron-volts of energy in the form of red LED light is required to
convert one CO2 molecule and one water molecule into one O2 molecule and
carbohydrate.
To convert this to joules per mole, multiply by Faraday's Number, which
is 96,485 coulombs per mole. This indicates that 1.744 megajoules of red
light from LEDs having peak wavelength of 635 nm is required to remove one
mole (44 grams) of CO2 from the air.
Good LEDs of this peak wavelength are around 25% efficient. This means
about 7 megajoules, or about 1.94 kilowatt-hours, of electric energy must
be delivered to such red LEDs to remove 1 mole (44 grams) of CO2 from the
air.
Keep in mind that these red LEDs usually have their dominant wavelength
mentioned more than their peak wavelength. The dominant wavelength is a
color specification and is in the mid 620's of nm for these red LEDs.
This usually appears as a slightly orangish shade of red.
====================================
I am repeating these calculations for two other LED wavelengths used for
growing plants:
660 nm: My eyeball estimate is that that one's spectrum achieves 109
molecules CO2 utilized per 1,000 incident photons, rather than 108. A 660
nm photon has 1.878 electron-volts of energy rather than 1.952. So light
energy requirement to remove 1 mole of CO2 from the air is 1.744
megajoules * (108/109) * (1.878/1.952), which is 1.663 megajoules.
I know of one manufacturer of such LEDs achieving about 25% efficiency
(LEDEngin). That indicates about 6.7 megajoules (about 1.85
kilowatt-hours) of electrical energy must be delivered to these 660 nm
LEDs to remove 1 mole (44 grams) of CO2 from the air.
Since plants have some requirement for blue light, I am repeating these
calculations again for 450 nm, from a Philips-Lumileds "Luxeon Rebel" LED
of "Royal Blue" color. My "eyeball estimate" of photosynthetic action by
that LED's spectrum is 92 molecules of CO2 per 1,000 photons. The average
energy per photon here is 2.75 electron-volts.
So, 1.744 * (108/92) * (1.952 / 2.75) means about 2.88 megajoules of
such blue LED light are required to remove 1 mole (44 grams) of CO2 from
the air. It appears to me nowadays that a top rank royal blue Luxeon
Rebel is about 30% efficient at a conservative current of 350 milliamps.
That means 9.6 megajoules, or 2.67 KWH, of electrical energy must be
delivered to these LEDs to remove 1 mole of CO2 from the air.
Red LEDs need conservative amounts of current and excellent heatsinking
to achieve 25% efficiency. LEDEngin 660 nm red LEDs probably achieve 25%
efficiency at 350 milliamps, possibly at 700 milliamps. The shorter
wavelength orangish red ones usually achieve 25% efficiency at 500 mA at
the most, preferably 350 mA.
Interesting. So if I am understanding you or the article properly, it is not so much that a plant specifically needs 660nm light, it is just that that wavelength gives the greatest photon->CO2 conversion rate.
Hypothetical: So if a monochromatic LED could convert 50% of its electrical energy to yellow light, it would outperform a monochromatic LED that could only convert 25% of its electrical energy to deep red light. Is this (more or less) correct?
Or does our fave plant require a SPECIFIC trigger at 660nm that cannot be compensated for by other spectra?