IF you are new to LED and want help choosing what to buy, POST HERE!

[JimmyIndica]

If you need little more? go with AZ says and try the PAR38! I wanna try sometime too!

 

[lmoore2680]

should be alright! your round 33 watt per. What type a setup you usin? Wherever your putting the 100 ,it might be little shy on that end. I would put the 100 in the middle!

Yea it's in middle it's in secret jardin Lodge120

Sent from my HTC6500LVW using Rollitup mobile app

 

[sunny747]

I know this has been discussed somewhere, but I can't remember where.. I won't be able to keep my hid's cool all summer. I inquired about an Area 51. The person there told me that they have a new light coming out in a few months or I can get on the list to buy an RW... Does anyone know what the difference in the new light will be?

 

[Timboy5]

Hey guys was looking for a 1000w replacement originally was looking at the Spyder 1200 but I havent heard much about it in this thread, a lot of people are liking the Area51 lights so I am looking towards those now.

What you guys think? Spyder 1200, 2 of the Area 51 RW150, Wait for the new Area 51 or DIY it with some Crees?

 

[alesh]

Hey guys was looking for a 1000w replacement originally was looking at the Spyder 1200 but I havent heard much about it in this thread, a lot of people are liking the Area51 lights so I am looking towards those now.

What you guys think? Spyder 1200, 2 of the Area 51 RW150, Wait for the new Area 51 or DIY it with some Crees?

If you can DIY then it's your best option.

 

[Doer]

If you can DIY then it's your best option.

Thanks, DIY is a good option for some. I like DIY but I can't decide which way to go. This thread is for that discussion, yes?

They have these new COBs that cool the junction better. Flip Chip Opto. And there is one that I am trying to calculate the efficiency for. I thought I had that down. But, it is all new to me. My degree is from 1975 for crying out loud. I thought I knew more than I did. But, I have forgotten what that was.

So, can someone please help me?

I have the LER of this COB calculated by a fine member here, who I really like. He said it is 325w LER for this one. How you get that is unknown by me. I think you have work the Spectral Distribution curves. New math, no doubt or, <gulp,> calculus?

Anyway, I am missing the last piece. How do you convert LER to efficiency?. I know it has been said before but, LER is a concept I have not compeletely grasped. So I need help. And I am asking my freindly stoner brothers; Peace.

 

[alesh]

Thanks, DIY is a good option for some. I like DIY but I can't decide which way to go. This thread is for that discussion, yes?

They have these new COBs that cool the junction better. Flip Chip Opto. And there is one that I am trying to calculate the efficiency for. I thought I had that down. But, it is all new to me. My degree is from 1975 for crying out loud. I thought I knew more than I did. But, I have forgotten what that was.

So, can someone please help me?

I have the LER of this COB calculated by a fine member here, who I really like. He said it is 325w LER for this one. How you get that is unknown by me. I think you have work the Spectral Distribution curves. New math, no doubt or, <gulp,> calculus?

Anyway, I am missing the last piece. How do you convert LER to efficiency?. I know it has been said before but, LER is a concept I have not compeletely grasped. So I need help. And I am asking my freindly stoner brothers; Peace.

Let's call the SPD function Jλ. Let's call the luminosity function [LF] yλ. Then LER = ∫yλ * Jλ dλ / ∫Jλ dλ .

In reality we cannot simply approximate the SPD function. So first you have to digitize the SPD. This should get you corresponding value for each wavelength in 380nm-780nm range. Same applies for LF with the exception that you can simply download it, ie from here. Since these are not exactly functions (in the mathematics sense), we gotta calculate integrals of both SPD function and SPD function * LF somehow. This can be easily done as the sum of all values from 380nm to 780nm of SPD (let's call that sum B) and SPD * LF (call it A) respectively. Now the LER is A divided by B. Make sure you luminous function is absolute (maximum is 683lm/W at 555nm) and not relative (max 1).

I hope this make sense, Mr. degree.
I'm not a good wordsmith. In English of course.

 

[Doer]

That is a very wonderful contribution to the common understanding. I think many of us have questions. And the questions are covered by layers and layers of other questions.

So, it is hard to ask the right questions. And getting them asked in the proper order is also tough. Thanks.

And if English is not your first language that is a total surprise to me.

 

[Doer]

BTW, just more help for the community. Here is a good write up on LER and why it is a better standard.

http://www.ledlighting-eetimes.com/en/alternative-calculation-method-will-improve-thermal-analysis-of-leds.html?cmp_id=71&news_id=222908534&page=2
Since recent advancements in the production process of phosphor-converted white LEDs have enabled accurate color point control for LEDs within the same product category, LER values are much more consistent for LEDs within the same product family. In fact, leading manufacturers, such as Philips Lumileds, offer excellent color control for their latest illumination-grade LEDs (within a 3-step MacAdam ellipse). This allows manufacturers to define a typical LER value that is representative for all LEDs within the same product family, targeting a specific CCT.

 

[churchhaze]

Units are your friend.

 

[churchhaze]

LER (luminous efficacy of radiation) is not a better standard... It's just a way to convert lumens to watts.. It would be great if lighting manufacturers gave us radiant watts and par watts and uMol/s instead of just giving us lumens. That would have been a nice standard so we don't have to digitize an SPD chart and do calculus against the luminosity curve.

You have lm/W and you want to convert it to W/W (which cancels out and becomes unitless). You need to know how many watts a lumen is for a given SPD. After that, it's just a conversion factor.

You don't have a fucking physics degree. Stfu.

 

[churchhaze]

I just so happen to be an expert at dividing 2 numbers together. Here's an example.

150lm/W
------------ = .4615
325lm/W

You see how the units cancel out? If you aren't ending up with the right units (efficiency has no units), you probably failed physics... and chemistry for that matter...

The calculation has been made easy since LER was already calculated... Hell, I'm not going to do the calculus! lol. That's for college students!

 

[Doer]

Let's call the SPD function Jλ. Let's call the luminosity function [LF] yλ. Then LER = ∫yλ * Jλ dλ / ∫Jλ dλ .

In reality we cannot simply approximate the SPD function. So first you have to digitize the SPD. This should get you corresponding value for each wavelength in 380nm-780nm range. Same applies for LF with the exception that you can simply download it, ie from here. Since these are not exactly functions (in the mathematics sense), we gotta calculate integrals of both SPD function and SPD function * LF somehow. This can be easily done as the sum of all values from 380nm to 780nm of SPD (let's call that sum B) and SPD * LF (call it A) respectively. Now the LER is A divided by B. Make sure you luminous function is absolute (maximum is 683lm/W at 555nm) and not relative (max 1).

I hope this make sense, Mr. degree.
I'm not a good wordsmith. In English of course.

OK, Mr. Wizard. I get a table like this for download. This is the LF table? Do I need to sum that?
Luminous efficiency
nm V
390 4.07678E-04
395 1.07817E-03
400 2.58977E-03
405 5.47421E-03
410 1.04130E-02
415 1.71297E-02
420 2.57613E-02
425 3.52955E-02
430 4.69823E-02

 

[Doer]

Here is some interesting stuff about this subject. It will take me a while to get up to speed.
http://www.digikey.com/en/articles/techzone/2011/feb/spectral-design-considerations-for-white-led-color-rendering

Symbol Description CCT (K) Duv Ra R9 R(9-12) LER(lm/W)
CW FL Cool white fluorescent lamp 4290 0.001 63 -89 13 341
DL FL Daylight fluorescent lamp 6480 0.005 77 -39 13 290
TRI-P Triphosphor fluorescent lamp 3380 0.001 82 17 47 347
MH Metal halide lamp 4280 0.007 64 -120 19 296
MER High-pressure mercury lamp 3750 0.000 43 -101 -29 341
HPS High-pressure sodium lamp 2070 0.001 20 -214 -43 380
3-LED 1 3-chip LED model 3300 0.000 80 -90 27 409

 

[Doer]

So, choosing what to buy. Each brand of 3000k seems to have a different SPD curve according to actual Kelvins and CRI.

 

[Doer]

So, my question is when a new LED technology happens how do we understand the SPD if they is no complete data. I looked on the datasheet again and it has no graphs.

All it says is 80 CRI and 3000K (nominal)

So, how do you get 325 LER from that?

 

[Doer]

Oh, btw,for people moving from HPS and get confusing information about heat, here is some info.

HPS puts off lots of IR radiation, but that is not the point. IR heats surfaces. It doesn't affect the air as much as the surface it strikes.

But, there is also Convected heat, the air is heated directly by the source, and the air moves around. It is the difference between an oven and a fireplace let's say.

What is the difference? When we put a current thru the junction of the LED, we can still keep it 25c, room temp. We take the watt of work, one direction and the light goes the other.

But, to just get the arc going in HPS it take over 240c or 464 degrees Fahrenheit.

Well, I see the problem right there.

The difference? LED does not need to maintain 464 degrees to even operate. And it sheds almost no IR.

 

[churchhaze]

But your friend sativied loves IR... I wonder if he realizes that if those photons were visible, they'd still heat the canopy, but they'd also produce photosynthesis. 2 birds with one stone.

Put your hand under one of these cobs and it will get hot very fast from the visible light.

 

[stardustsailor]

LER =Luminous Efficasy of Radiation . Every light source,no matter of what kind ,has a LER
(From a tea candlelight to a giant star ) .

LER denotes this simple thing : How many Lumens are perceived by human vision per Watt of emitted light .

Luminous Efficiency is another thing and denotes how many lumens are perceived by human vision per
electrical Watt .

If Luminous Efficiency is divided by LER ,the result will be a unitless value that denotes how many Watts of emitted light
a light source has per electrical Watt .
Also known as Radiometric Efficiency .

In order for LER to be calculated ,a spreadsheet is needed along with :

-The relative spectral power distribution of the light source
-The CIE 1924 Photopic Human vision sensitivity graph .
http://www.cvrl.org/cie.htm

The relative spectral power distribution ,has to be carefully digitized into numbers (y axis values per nm ) .
Say for example a 3000K 80 Ra LED light source is given .

If the highest relative y value ( =1 ) is at 600 nm ,then say at blue 450 nm you get a value of 0.6 ...
That means that light energy /power at 450 nm is 60% of the light energy /power at 600 nm .

So ,if you add up all the relative y values -from say 380-780 nm range - you will end up with a number over 1 .

Actually every nm of light has a power value .
In our example at 600 nm that power would be z*1 ,while at 450 nm would be 0.6*z ...

If we assume that the whole radiated power is of 1 Watt then :

1 =( y380*z) + ( y381*z)+( y382*z)+....+( y779*z)+( y780*z) =>
z= 1/ SUM y 380-780

The CIE 1924 Photopic curve (standard ) denotes how many lumens are perceived by human vision ,
for every Watt of radiated light per nm .
At 555 nm 1 Watt of radiated light is perceived as 683 lm.

Multiplying the relative spectral power distribution y values with the CIE 1924 Photopic y values ,
for every single nm and then summing up ,the result is how many lumens are perceived by human vision
from 1 Watt of radiated light ( not electrical Watt ) .Aka the LER of the light source .


All these is done to take out the lumens of the equation.
Or more precisely to calculate the radiometric power of a light source ,
when the photometric power is known .

And then turn that into umols/sec for every nm ,using this equation :

PPF (λ)= Rad. Watts (λ) * λ (nm ) / 119.708 (one hundred nineteen point seven hundred eight )

Simple ,enough ....

Cheers.

 

[Doer]

I really like that very clear post, Sir.

So, if you don't have an SPD for that COB, how can you get it?

And each light has slightly a different SPD, it seems.