DiY LEDs - How to Power Them

NoFucks2Give

Well-Known Member
if you or someone else can tell us the conversion factor for each kind of COB that grower use here
I have some. I have a StellarNet Blue Wave spectrometer that can measure in Lumens, Watts, and PAR with fairly decent accuracy.
I only purchase those that have a red peak greater than 620nm or if they have no SPD graph and they look like they may have an abundance of deep red. Like 3000K or less with a CRI 90 or greater.
 

wietefras

Well-Known Member
No it is NOT! Could you not follow the math? Again?
You are wrong. It IS the QER as T-Time already indicated. Do you even know what QER means?
First off there is no CoB with 61% radiometric efficacy.
Wrong again. Triple wrong in fact.
1) It's about efficiency and not efficacy.
2) Many COBs and leds can be run at 61% efficiency. Just run it very soft. It's going to be expensive, but it can be done.
3) Cree released a royal blue led which can easily hit even 81% efficiency. Citizen also.
C-C-C-C-COMBO BREAKER!!!!!
Additionally PAR is Quantum and counts photon particles. No such thing as PAR watts.
Wrong again. You really don't understand what PAR actually means. PAR does not stand for PPFD as you keep confusing those two terms. PAR simply means that they measured "all wavelengths between 400nm and 700nm". You can measure all kinds of things only in that range. Among others, the radiant power which is then called PAR watt.

What @T-Time was attempting to do was converting from wall watts to PPFD. That can be done if you have the conversion factor for that specific CoB which takes in to consideration the radiance at each wavelength.
He was trying to go from PAR watts to PPF and and that conversion uses the QER. Which is exactly what he was using. Although I understand something else went wrong in his calculations. The numbers he subsequently got from "the calculator" are fine.

I really feel that people should use lumen to PPF conversion factors though. Much easier and less error prone. Simply divide lumen/lux values by this factor and you get equivalent PAR numbers.
See here: https://www.rollitup.org/t/par-multiplier-thread.928907/page-2

Anyway, you can keep pretending everybody has got it wrong , but you simply just don't know what you are talking about. Just read some of Alesh's threads on how to calculate efficiency and all these terms will be explained.

Or at the very least start here: https://en.wikipedia.org/wiki/Photosynthetically_active_radiation

What is really baffling is why you insist on NOT learning the basics first. What's the point of buying an expensive bit of tech when you don't even want to learn how to use it properly?
 

NoFucks2Give

Well-Known Member
What is really baffling is why you insist on NOT learning the basics first. What's the point of buying an expensive bit of tech when you don't even want to learn how to use it properly?
That's your opinion and you know where I stand on that. This is what I do for a living. I have been doing this shit too long, specifically Horticulture LED lighting, to put up with with the bullshit from the blind leading the blind. Not a fuck, not a rats ass. If you cannot not cite a peer review study to back your bullshit, then it is just that, BULLSHIT! I do not want to hear your anecdotal ramblings. Wikipedia is NOT a peer reviewed source.

Why are you such a fuckin jerk off? Talking through your ass and being rude.

I don't understand PAR??? If you measure PAR what units are used with the measured value? The same as PPFD. From your link:
Micromole:
per second and square meter (μmol m-2 s-1). This term is based on the number of photons in a certain waveband incident per unit time (s) on a unit area (m2) divided by the Avogadro constant (6.022 x 1023 mol-1). It is used commonly to describe PAR in the 400-700 nm waveband.​

If not, then do me a favor, contact StellarNet and tell them they got it wrong.

ScreenshotPAR.jpg


While you are at it come on down to the University of Florida Horticulture Department and tell all the professors and researchers us we are wrong too.

Energy based PAR is not used in Horticulture. Have you ever seen a horticulture study that measured PAR in W/m²? LED manufacturers use watts.

No such thing as PAR Watts in horticulture. When converting from radiometric watts the flux energy (watts) is removed to get the number of photons. Plants do not care about the wavelength energy. The radiometric watts measures the energy of the photons. Where PAR is a quantum measurement of the number of photons independent of wavelength.

The observation that the overall quantum yield of photosynthesis is nearly independent of wavelength (see Figure 7.12)- Plant Physiology 6th Edition, chapter 7.
Fig 7-12Untitled.jpg


.

If you had a "PAR Watt" what could you do with it? I showed you where PAR was used to express PPFD, now you show me where the term "PAR Watt" is used.

Your use of the term lumens is incorrect as well as PPF.

It is not lumen to PPF it's Illuminance to PPF and Luminance to PPFD.
Lumen is a measure of energy, like watt except it's photometric flux.

Show me the Math. Not tables of numbers, show me the math.
In your link the numbers look good that Malocan took with his spectrometer.

Just because the Cree XP3G was announced as having 81% wall watt efficacy says nothing for CoB efficacy. And they say, the XP3G Royal Blue has a Photosynthetic Photon Flux (PPF) efficiency of up to 3.2 μmol/J at steady-state. Not 4 or 5 like these CoBs here.

So you take a LED cover it with phosphor converting the wavelengths and you think there's not going to be considerable loss?

Keeping in m ind that PPF is not a real number that can be used. PPF is Photon Irradiance a useless number. As I said previously there is no way to convert PPF Irradiance to PPFD radiance. Cree also says:

"Cree royal blue XLamp LEDs are over 50% efficient and white XLamp LEDs are over 40% efficient. That is, under normal operating conditions, approximately 50% to 60% of the input power is output as heat, while the rest of the input power is converted to light." --Thermal Management of Cree XLamp LEDs​

And that is mono LEDs not phosphor coated CoBs with shit loads of LED dies crammed into a tiny space raising the temperature way beyond the rated 25° C.

So are you going to stick with the CoB at issue is 61% efficient? And the µMole/J is 4.64? Is that irradiance or radiance? What is that 4.64? Show me the math!

And the digitized SPD numbers, what unit of measure are they?

Here's my math if I were to use the digitized SPD numbers:

PPFwl = wavelength x 0.00836 x (the SPD Number)

PPF = the sum of PPFwl where wl = 400-700

But that is just an approximation because you cannot use the SPD or PPF to get PPFD.



I know how to calculate the numbers, I'm not sure Alesh does, but I am trying to get him to explain some things.

So 5 things,
1. Show me the math
2. What is the unit of measure of the digitized SPD numbers..
3. Is the CoB at issue 61% efficient (or any CoB for that matter)
4. Show me and example of where PAR Watts is used in horticulture.
5. What is the end result number and its unit of measure, e.g. the 4.64

Optionally call or email Dr. Thomas Colquhoun who heads all the LED research at the University of Florida and tell him he is wrong and that PAR is not measured in µMoles.
Phone: (352)273-4584
Email:ucntcme1@ufl.edu

Google "QER LED conversion" and see how common it is used other than on this site.






 
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wietefras

Well-Known Member
For the record, I have been trying to be polite trying to help you understand the correct math, terms, units etc etc, but you started acting like an utter ass from the get go. So yeah, you reap what you sow.

That's your opinion and you know where I stand on that. This is what I do for a living. I have been doing this shit too long,
You have not been doing THIS for long. Seeing how pretty much every statement you make is incorrect, it's clear that the concepts which you are posting about are clearly beyond your scope of understanding.

I won't go through the whole post again trying to correct and help you understand everything which you get wrong. It really is just to much work to correct all the mistakes you make.

I'll try to help you on three basic items:
1) Nothing is measured in "µMoles" as you keep saying. It's either µmol/s for PPF or µmol/s/m2 for PPFD.
2) 81% efficiency in normal conditions impossible? To prove this you quote a marketing line from a product release in 2015!
Lets zip forward to current times and we get:
Cree, Inc. (Nasdaq: CREE) announces the new XLamp® XP-G3 Royal Blue LED, the industry’s highest performing Royal Blue LED. The new XP-G3 LED doubles the maximum light output of similar size competing LEDs and delivers breakthrough wall-plug efficiency of up to 81 percent.
That product IS a COB. Just without the extra phosphor coating. It's the fact that it's a COB which makes it a COB not the coating.
If you want 61% on a "regular" COB, then just run that COB really soft. That's around 200lm/W for an 80 CRI COB. That's expensive perhaps, but it's perfectly attainable.
3) I gave you a Wiki link where the concept of PAR is explained. Including "PAR watt". No one said people use this in "horticulture". So no need to keep whining about that. It's used in these calculations. Also, as I said before, look for the thread Alesh created on how to calculate and use QER and LER.

Bonus tip, since you have a spectrometer, just let it measure both lux and PPFD and divide the two to get the easy correction factor. No need to understand what QER, LER an "PAR watt" are.
 

mahiluana

Well-Known Member
50% to 60% of the input power is output as heat, while the rest of the input power is converted to light.
I measured a minimum of 70-80% heat of the input power

"Cree royal blue XLamp LEDs are over 50% efficient and white XLamp LEDs are over 40% efficient.
0.500 A ..................... i measured the heat of XT-E @ 500mA with 3,08 V for ten minutes
888.5 Led lm............. and i received 1,3W of heat power. That means around 80% heat power
574.7 lm/W................ from input.
3.09 Vf
1.546 Led W



So the reference, when you say up to 50% efficiency can`t be the total input of electrical power.
Maybe your 100% is part of the electric power, that is ! not ! converted to heat.
 

Attachments

ManCaver

Member
Hi guys quick question, since all those quantum boards are out of stock I plan to build my own.
came up with these things, which are the same diodes , figured the 3500k are the best for flowering + veg.
https://www.digikey.com/product-detail/en/samsung-semiconductor-inc/SI-B8U05128HUS/1510-1585-ND/6149750

Plan to use 14 of these , so each strip at about 75% would use around 450mA if I've read the spec sheet right , so I need total of 6.3 amps on around 24.2 volts to run this setup am I correct ?
that means the constant-current Mean Well HLG-150H-24 would fit perfectly for supplying exactly 6.3 amps. providing around 150 watts for this setup, which would run these 14 strips at around 75% each.
http://www.mouser.co.il/ProductDetail/Mean-Well/HLG-150H-24/?qs=sGAEpiMZZMt5PRBMPTWcaWBIn1H1DpLz0npuskIDd%2bM=
Can n1 confirm my calculations are correct ? :bigjoint:

They also fit perfectly on those cheap ebay 300x140mm heatsinks, each houses 7 strips perfectly.
You basically get 336 diodes compared to the 288 diode version or even the 304 diode version Horticulture Lightning group uses in their boards. for around the same dollar bill. :peace:
 

NoFucks2Give

Well-Known Member
No need to understand what QER, LER an "PAR watt" are.
That I already know it is not necessary for me to understand what is unnecessary.

If you were trying to help you'd answer me regarding your claims. But no, you just hit and run.

From previous post:

So 5 things,
1. Show me the math
2. What is the unit of measure of the digitized SPD numbers..
3. Is the CoB at issue 61% efficient (or any CoB for that matter)
4. Show me and example of where PAR Watts is used in horticulture.
5. What is the end result number and its unit of measure, e.g. the 4.64
 

wietefras

Well-Known Member
Holy crap, from all the tons of bullshit that you post that's the thing you are going on and on about whether it's technically a COB or not? It has the same die as the COB, but then without the phosphor coating like the "white" one next to it. You seriously want to contest that that is something completely different?

How about the royal blue Citizen COB then? Does that make it all better?
 

wietefras

Well-Known Member
That I already know it is not necessary for me to understand what is unnecessary.

If you were trying to help you'd answer me regarding your claims. But no, you just hit and run.

From previous post:

So 5 things,
1. Show me the math
2. What is the unit of measure of the digitized SPD numbers..
3. Is the CoB at issue 61% efficient (or any CoB for that matter)
4. Show me and example of where PAR Watts is used in horticulture.
5. What is the end result number and its unit of measure, e.g. the 4.64
Come on man. I didn't hit and run, I explained what you did wrong. Instead of actually reading and comprehending, why do you need pile on another heap of other irrelevant nonsense. Why would I be obliged to answer all that as well? It's truly impossible to try and correct all the mistakes you make.

Why can't you simply try to understand what you wrote wrong instead of trying to deflect all the time? It's really incredibly tiring.

For instance, you really should have been able to read this in the posts above:
1) You actually replied to a post containing "the math". I also directed you to Alesh's thread explaining "the math". There really isn't any more explanation I can give you.
2) If it's a proper chart then the units are mentioned on the sides of the chart. How is that relevant anyway?
3) Again, how is that relevant? I replied to your incorrect claim which said:
First off there is no CoB with 61% radiometric efficacy.
And yes, pretty much any current Cree, Citizen, Bridgelux etc COB can run at 61% efficiency. Try to understand that efficiency is not something that's set in stone. It depends largely on the number of watts going through the COB.
4) Again, how is that relevant. Terms like QER and LER are also not commonly used. So what? It's used in "the math". That was the point since you said:
No such thing as PAR watts. Watts is radiometric and is a measurement of Radiant Flux. PPFD, Photon Flux, is quantum measured in moles. Watts is neither PAR or a measurement of intensity, irradiance or radiance.
Which is simply utterly wrong. As explained it's the term for the radiant flux in the PAR range (400-700nm) and therefore it actually is used in agriculture too, but that's completely besides the point. Just use Google if you want to see where and how.
5) The 4.64 has been explained many times to you now. It's the QER. Unit is umol/J. As was also explained several times, it's not the "end result" in itself, but instead it's what you use to go from PAR watts to PPF.
 

sforza

Well-Known Member
:peace::eyesmoke:Imagine the way of the heat - outside of your chip.
Your heat meats your thermal grease - and even if it is a very good one - the heat conductivity compared to eg. copper is very low.

OK. - you learned to keep it very thin and you put it only on very flat and plane surfaces.
But all your heat meet on materials with heat resistance.
Thickness of these materials: glue, grease, aluminium- or copper-heatsinks influence
this heat resistance in a linear function -
until heattransfer happens ( aluminium, copper --> to the air - or water)

I only break down this wall with a drilled hole and - started to dig also on the other side.
The chip-platine itself is a thermal resistance and the area directly under the light emitting surface is the center of heat production.
Scatching the surface over there can double or trippple this area and bring the water closer to the fire :fire:

half a year ago I meassured a 50W cob @1500mA mounted on my coolmac without hole.
At 25°C watertemp. my powermeter showed 50W.

Same chip(scratched) and driver mounted on a hole with direct contact to water of 25°C
pushed my powermeter close to 55W.

As you know low Tj has great influence on light efficacy and longlife - but the greatest
benefit is to use the heat in the shower or elsewhere.
***********************************************************************************************


I doubt a little bit, that there are no oil or gas waterheaters in whole Florida.
But US-citizens are known to consume 4 times more energy than eg. europeans.
That means in summer you switch on your aircondition and some take out the electric-heater because they catch cold feet.
Together with your orange-blond, stupid superleader you are in a turbo tripple lose-lose-lose situation. America - first in hell...:dunce::dunce::dunce:

If you need a watercooler working without any additional energy -
have an eye on my cooling-tower. You need to vaporize ~1L / 600W of heat



here eg. you can find a heatexchanger:

http://www.ebay.de/itm/80-100-120-150-Kombi-Elektro-Boiler-Warmwasserspeicher-mit-Warmetauscher-Solar-/381284978091?var=&hash=item0
No need to bring politics into it bro. Since we didn't go all green like you Germans, our electricity is less expensive, so we can use more of it, plus our taxes are lower, so we have more money left over to spend on electricity.

Besides, when you live in a nice big house on a large lot like a free independent human being, as most of us do living here in the USA, you are going to use more electricity than someone living in a little apartment, like a rabbit in a warren, jammed up with strangers living beside, above, and below you.

I realize that you Germans are all butt hurt about Uncle Sugar not wanting to carry the financial burden of defending your ass from the Russians for decades while you spend your money giving handouts to Muslims. And you are really mad about our pulling our ass out of the Paris non-treaty and thus no longer being on the hook to pay billions of dollars to the third world so that they can go Green e.g. putting Greenbacks into their leaders' Swiss bank accounts, but you will get over it eventually.

What choice do you have? Are you going to stop shipping cars to us? There is a big ocean between us and the Russians, while they are not far where you and your commie East German three term going on four term Chancellor live. You need us, to buy your shit and to protect your ass. We don't need you for anything.

You Europeans must take your destiny into your own hands and stop sucking the tit of the USA. After 70 years, it is about time, don't you agree? You and the French have fun together, after all, you can depend on the French to do the right thing, can't you?
 

NoFucks2Give

Well-Known Member
Your heat meats your thermal grease - and even if it is a very good one - the heat conductivity compared to eg. copper is very low.
I do not know. Convection is so complex there is no way to calculate the results it must be done with empirically through experimentation. That's what the US Dept. of Energy says.

You mentioned you are using a very low watt pump (4 watt??) so I assume your flow is less than 250 liters per hour and the tube looks to have a fairly large cross sectional area. So the forced convective cooling may be minimal with such slow flow. Where natural convection may have more to do with the cooling than forced. I also question the heat transfer between aluminum and water. Water has such a high thermal resistance and high specific heat that I am having a hard time wrapping my head around the dynamics of the water's contact with the aluminum pad.

I really like your design. But I cannot stop my mind from questioning if it is better than some more conventional methods.

I do not use thermal grease. I use a soft annealed copper foil 0.0007" thick. I then torque down the PCB to the heatsink squeezing the copper foil into the voids. I have a lot to do on my plate so I have not yet tested the foil against something like a silver based thermal compound.

I think TIM is a non-issue as it is so thin and if done correctly with the correct material, it is negligible when compared to the other thermal resistance in the system

Same chip(scratched) and driver mounted on a hole with direct contact to water of 25°C
pushed my powermeter close to 55W.
I do not know what that means. All I want to know is the temperature of the case.
 

mahiluana

Well-Known Member
our electricity is less expensive, so we can use more of it, plus our taxes are lower, so we have more money left over to spend on electricity.
Specially when someone lacks awareness,education, responsability and overview - politics could be helpfull to compense.

You !!! - USA-citizens - don`t have money, because your sovereign debt will pass this year
20.000.000.000.000,- US$. One of the reasons is - not to put enough taxes on electricity and fosile energy - and with your brandnew-orange-blond-super-leader - even less.

defending your ass from the Russians for decades while you spend your money giving handouts to Muslims.
Seems that USA is not even able to defend the ass and bowels of the US-president inside the whitehouse from the russians "intruders". That happens when stupid people(you?) elect stupid leaders. Putin, russian`s secret service (and me) is laughing about you and your "defenders". ---



NATO is obsolete as long as the captain of the USA is called Trump.

Or do you really think that Germany will go with you to North-Korea to have a look whether Kim Jong U. is more stupid than Donald T.

You need us, to buy your shit and to protect your ass. We don't need you for anything.
WHAT ??? :clap::clap::clap::clap::clap::clap: :cuss::finger::wall: YOU PROTECT MY ASS ???
DISGUSTING :spew:STOP THAT !!!
 

NoFucks2Give

Well-Known Member
Specially when someone lacks awareness,education, responsability and overview
Our US history books are a bit distorted. They say the US won WWII, discount the eastern front, and dropping the A-Bomb was to save lives. Nothing about Truman being an stupid asshole.

Here in the US is also a very common chest thumping saying in the US that "anyone can become president" as part of the plan to brainwash kids into believing the US is the greatest country in the world. Well Trump is a product of taking that "anyone can become president" a bit too far. But it is was inevitable that when the Cubs won the World Series, that Biff had to become President. (Back to the Future II).
 
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