Base pump, WV, μmol/J, phosphor CE, LED effeciency

[alesh]

No this is incorrect. The energy per photon per nm scales linearly. Its why a constant can be used as a conversion factor, ie 119.6.

Nope. Energy per photon can be calculated as:
E = (h*c) / λ
where h is Planck's constant, c is speed of light a λ is wavelength. E is obviously inversely proportional to wvl.

In my experience, using this technique with unphosphored chips results within a max of ~3% deviation. Most most times less than 1% deviation. I understand the principle behind the concern but due to the narrow emmision range and due to the small difference in energy per photon/nm the differences are neglible as you say.

How many examples does your experience include? And how did you actually calculate 3% deviation?
In my experience, the deviation is highly dependent on a specific model of LED.

No, I'm pretty sure it's correct. Please re-read.

I'm describing base pump EE, not total chip effeciency. If there is a phosphor loss then this phosphor loss would further reduce base pump EE. Base pump EE = the total chip effeciency when no phosphor is applied. When phosphor is applied then chip effeciency = base pump EE - (phosphor loss). Like @eatled said, we only care about total photon output, so the excess energy left on the phosphor film is reducing the total chip effeciency but when CE =1 does nothing to reduce base pump EE or total photon output. Base pump EE is what I care about when it comes to effeciency of plant utilization.

The original calculation is incorrect. If you want to calculate how many photons per unit of energy of cool white spectrum are there you have to take into account the whole SPD not just the peak.
Base pump efficiency (why tho?) of a white LED with its output specified in µmol/J cannot be easily and accurately determined from a data sheet. You can find the efficiency of the whole package - that is LED chip + phosphor losses + lens losses. But you can't determine the ratio of these.
You could perhaps approximate it by looking at the specs of the royal blue version of said LED, but there's no guarantee that both of those would use the exact same chip.
Also...you want to include phosphor losses in the designing process of a LED fixture.

 

[ChiefRunningPhist]

Nope. Energy per photon can be calculated as:
E = (h*c) / λ
where h is Planck's constant, c is speed of light a λ is wavelength. E is obviously inversely proportional to wvl.

Well you are most definitely right. 1/x is not a linear function, I stand corrected. Apologies.

How many examples does your experience include? And how did you actually calculate 3%

Ive been looking through about every manufacturer of mono LEDs for the past year for the most effecient, I can really only verify the ones that give the intensity rating in μmol/J, so mainly the reds, though like I said all have been within ~3% max. Ill calculate based on the effeciency % and then work back to find μmol/J based on the peak WV technique, or visa versa and derive the effeciency based on the μmol/J they give. I think I posted an example but I'll have to go back and look.

The original calculation is incorrect. If you want to calculate how many photons per unit of energy of cool white spectrum are there you have to take into account the whole SPD not just the peak.

I don't want to calculate output per unit of energy output, I want to calculate output per unit of energy input. That's what base pump EE is. Photon out per unit energy in.

Base pump efficiency (why tho?) of a white LED with its output specified in µmol/J cannot be easily and accurately determined from a data sheet.

I guess I'm stlill waiting for an explanation why you can't estimate base pump EE with peak base pump WV and a radiometric intensity measurement. When Samsung claims 3.0+ on their lm301b, are you saying that I can't equate the 3.0 to 3.0 of base pump? With a CE of 1 I'm pretty sure that logic follows?

Also...you want to include phosphor losses in the designing process of a LED fixture.

Ya, you'd want to know what they are. You'd want to know how much energy you're wasting leaving on the phosphor (at least I do). I don't think you can calculate that without a base pump EE.

When I meant linear, I meant it scales at a constant rate. 119.6.

Trying to say that WV energies are proportional to their nm And can be found by multiplying by a constant. That's not the same as linear and I was wrong.

(650nm)/(450nm) = 1.44
^^^ same as vvv
(650/119.6) = 5.43μmol/J; X
(450/119.6) = 3.76μmol/J; Y
X/Y = 1.44


I stand by my first calculation. I didn't specify effeciency as "base pump EE", but that is what I'm referring to, and that is what I'm solving for. Knowing base pump EE, will allow you to determine phosphor losses. Also, was pointing out what the difference in base pump peak WV means initially for chip performance.

Just think about it. If you look at mono SPDs, they are narrow, maybe only 50-75nm wide. This equates to approximately only a ~10% difference in energy between the top of the WV range to the bottom. So even if I used an x-intercept point as my peak WV, it's still within ~5% of measured. The % deviation when dealing with monos and the base pump peak WV is really small. I'm sure you have your ways of getting chip effeciency estimates from suppliers, or from when you digitize, ect, so try it for yourself, compare the effeciency you get/calc for the power input specified on the datasheet, and then do the peak WV calc and see how far off you are. Idk how much rounding the data sheets do in the bullet points, but here's a quick example anyways...

I used to think it would be an issue, but as you can see this SPD leans pretty hard to the right and is still pretty close to accurate. If I used 675nm as my peak WV point to back-calculate effeciency from, I'd get 58.8% effecient, a proportional 5% difference in the figure that was stated (56%).

 

[ANC]

I'm going to have to read this thread again, when I'm more stoned or more sober. Right now it is feeling like 12th-grade science class.
Fuck I even married someone who has probably never heard or read the name, Avogadro.

 

[alesh]

Ive been looking through about every manufacturer of mono LEDs for the past year for the most effecient, I can really only verify the ones that give the intensity rating in μmol/J, so mainly the reds, though like I said all have been within ~3% max. Ill calculate based on the effeciency % and then work back to find μmol/J based on the peak WV technique, or visa versa and derive the effeciency based on the μmol/J they give. I think I posted an example but I'll have to go back and look.

Works reasonably well for LEDs with a narrow FWHM and symmetric SPD. Works not so well for LEDs with a broader FWHM not so symmetric SPD. Introduces an uknown inaccuracy in all cases.
Doesn't work at all for something with as broad SPD as white phosphor converted LEDs.

I don't want to calculate output per unit of energy output, I want to calculate output per unit of energy input. That's what base pump EE is. Photon out per unit energy in.

I guess I'm stlill waiting for an explanation why you can't estimate base pump EE with peak base pump WV and a radiometric intensity measurement. When Samsung claims 3.0+ on their lm301b, are you saying that I can't equate the 3.0 to 3.0 of base pump? With a CE of 1 I'm pretty sure that logic follows?

Well if you want to convert output of an LED specified as photon flux (in µmol/J) to radiant power (in mW) you need to calculate how many photons there are in each joule of the output. Then of course you divide the LED's output in µmol/J by this number to get radiant power output of the LED. Then efficiency is simply Pout/Pin. This of course includes phosphor losses as the original output specified as photon flux (in µmol/J) is for the whole LED with phosphor and lens (if there's one) and everything.
Now there's no way to determine how much of the input energy is lost in phosphor conversion.
As I already said, you can approximate it by comparing radiometric output of the phosphor converted LED to a royal blue LED of the same line BUT there's one big assumption - both are actually using the exact same chip (as in the exact same bin). Which I think rarely is the case.
And no you can't assume that a blue pump of a phosphor converted LED that produces 3 µmol/J at given current also produces 3 µmol/J at that current and call the delta of energy phosphor losses. If that's what you mean.

Ya, you'd want to know what they are. You'd want to know how much energy you're wasting leaving on the phosphor (at least I do). I don't think you can calculate that without a base pump EE.

Yeah why not. Always better to know more than less.
However in the end overall electrical efficiency is what matters and that's what you're given by a manufacturer. You don't make the decision because of high phosphor losses. You make the decision because of overall efficiency. Which may or may not be greatly affected by phosphor losses.

When I meant linear, I meant it scales at a constant rate. 119.6.

Trying to say that WV energies are proportional to their nm And can be found by multiplying by a constant. That's not the same as linear and I was wrong.

(650nm)/(450nm) = 1.44
^^^ same as vvv
(650/119.6) = 5.43μmol/J; X
(450/119.6) = 3.76μmol/J; Y
X/Y = 1.44

True true. Energy per photon is inversely proportional, photon per energy - which is what we need there - is of course directly proportional - linear. A dumb moment of mine.

I stand by my first calculation. I didn't specify effeciency as "base pump EE", but that is what I'm referring to, and that is what I'm solving for. Knowing base pump EE, will allow you to determine phosphor losses. Also, was pointing out what the difference in base pump peak WV means initially for chip performance.

No. For a white phosphor converted LED you need to use the whole SPD. Not just the peak wvl.

 

[PrometheanLeaf]

I'm going to have to read this thread again, when I'm more stoned or more sober. Right now it is feeling like 12th-grade science class.
Fuck I even married someone who has probably never heard or read the name, Avogadro.

Sometimes you are my favorite lol.

 

[ChiefRunningPhist]

Well if you want to convert output of an LED specified as photon flux (in µmol/J) to radiant power (in mW) you need to calculate how many photons there are in each joule of the output.

That's great but thats not what I'm after, or what I was explaining what I was calculating, and also, the output of the chip (including the phosphor) in μmol/J will be the exact same as base pump μmol/J. With CE =1, it means there's no photon qty loss only energy loss.

As I already said, you can approximate it by comparing radiometric output of the phosphor converted LED to a royal blue LED of the same line BUT there's one big assumption - both are actually using the exact same chip (as in the exact same bin). Which I think rarely is the case.

That's not my method. Why are you dismissing my method? I'm finding the peak WV of the base chip via the measured SPD, not by assuming what the base pump WV is based off of some other chips SPD.

And no you can't assume that a blue pump of a phosphor converted LED that produces 3 µmol/J at given current also produces 3 µmol/J at that current and call the delta of energy phosphor losses. If that's what you mean.

That's exactly what I mean. CE = np × nq. If CE = 1, then there is no photon qty loss. Only enegy left on phosphor from downgrading photons to the desired WVs.

You don't make the decision because of high phosphor losses. You make the decision because of overall efficiency. Which may or may not be greatly affected by phosphor losses.

Well I disagree. PM me.

No. For a white phosphor converted LED you need to use the whole SPD. Not just the peak wvl.

For total chip effeciency I agree, but not for base pump EE, or, [total chip - (phosphor losses)].

 

[Renfro]

Lots of pretty high level info in this thread. I am a little slow, can you tell me how I use this to grow better weed? lol Just asking for the simple version that I can apply in my endeavors.

 

[alesh]

That's great but thats not what I'm after, or what I was explaining what I was calculating, and also, the output of the chip (including the phosphor) in μmol/J will be the exact same as base pump μmol/J. With CE =1, it means there's no photon qty loss only energy loss.

That's exactly what I mean. CE = np × nq. If CE = 1, then there is no photon qty loss. Only enegy left on phosphor from downgrading photons to the desired WVs.

I see what you mean now. I didn't realize that you assume than every single blue photon either passes through or is converted to a lower energy level - CE = 1.
Is that a real world value though? I always assumed that CE is an unknown number <1.

 

[ChiefRunningPhist]

I see what you mean now. I didn't realize that you assume than every single blue photon either passes through or is converted to a lower energy level - CE = 1.
Is that a real world value though? I always assumed that CE is an unknown number <1.

I guess so, I've been looking for the data for awhile and that's pretty much what I've come across.

Here's a quick link from 2008...
https://www.laserfocusworld.com/lasers-sources/article/16554852/leds-phosphorconverted-white-leds-have-almost-no-conversion-loss

 

[alesh]

I guess so, I've been looking for the data for awhile and that's pretty much what I've come across.

Here's a quick link from 2008...
https://www.laserfocusworld.com/lasers-sources/article/16554852/leds-phosphorconverted-white-leds-have-almost-no-conversion-loss

Well if that's true, then your method is reasonably good and I'd have to apologize. I find it hard to believe but I might be wrong as your link suggests.

Although I can't still figure out in which scenario does it make more sense to compare efficiency of the base pump instead of the whole LED.

 

[PrometheanLeaf]

Well if that's true, then your method is reasonably good and I'd have to apologize. I find it hard to believe but I might be wrong as your link suggests.

Although I can't still figure out in which scenario does it make more sense to compare efficiency of the base pump instead of the whole LED.

Likely more effective with densely grouped diodes like cobs that are ran harder, mimicing single source dynamics, more than strips I'd think.

At a certain point, the data with have diminishing returns in regards to what performance efficacy returns you can recieve. What is effective strain to strain and type of training methods will probably effect the end product more than trying to build a 'perfect' light.

A bx2 Blue Dream sativa dominate cut will require much different levels than say a Ghost Og indica dominate cut.

I never see anyone building these lights building for strain specific requirements in mind, I've always wondered why.