King Cob Led grow light

GrowLightResearch said:

Typically at 200% the flux is usually about 175% of the test flux. At 50% of test current, almost always, the flux is also 50%.

Click to expand...

sorry you are wrong, im not going to count pixels but your gross assumptions from those graphs are not accurate. everybody knows that efficiacy increases steadily with reduced currents. look at something more accurate like any of the lumen calculators provided by the mfr. heres 4 examples for you

cxb3590, nominal current, 2400 mA vs 1400 mA



CXM22, nominal current 1100 mA vs ~200 mA


vero 29B, nominal current 1800 mA vs 900 mA





Cree XPE far red



i could go on and on

what you are suggesting is that no chip is any more efficient below nominal current. that efficacy graph would look like this (exaggerated):


as opposed to the common trend seen by EVERY led you can test or find photometric data for. they always start to go slightly asymptotic at low currents which is the exact opposite of what you are describing.

CobKits said:

i could go on and on

Click to expand...

This is what I said. Let's take your Vero 29.
@ 900mA lm = 8,140 the test flux
@1800mA lm = 14,554 at 200% the flux
14,554 ÷ 8,140 = 1.78 ≈ 175% is usually about 175% of the test flux

No charts graphs needed. Your pretty charts and graphs did make you look like you knew what you were talking about.

The math? So so. The red arrow on the CBX3590, 2,400 is not 200% of 1,400.

CobKits said:

those graphs are not accurate.

Click to expand...

CobKits said:

more accurate like any of the lumen calculators

Click to expand...

I have no practical use for those calculators. I have a datasheet, I look at twice the test current and take note of how close the line is to 200%.
The numbers you are using have no value to me. First off luminous numbers (e.g. lm/W) are worthless to me. I need to know the number of photons, mW/W is what I need to easily convert to photon quanta.

The efficacy below test is of no value either.

The decision I need to make when looking at the 175% is whether or not to push the LED beyond its test current or to buy two LEDs and run them both at test current or below.

That decision is based on the performance between test and twice test. Your numbers are from zero to max. That does not help me at all in this specific decision.

When I compare, what I typically see is by pushing the current I am getting 25% less photons. The temperature will increase significantly, the higher temperature will require additional thermal management, and the forward voltage is going to rise costing me more in electricity while getting less photons. Thermal management is expensive. So I would buy two LEDs. More LEDs gives better uniformity allowing the fixture to be closer to the canopy. As I lower the height of the fixture I can reduce the current increasing the efficiency of the system.

Deep red is a different story, it typically runs at better than 190% at 200% test. It also has a low Vf and a very small spread in Vf. So I run deep red over test.

BTW Cree far red is spec'd at 25° not 85°.

The graph in the datasheet is easier. It's a simple glance at the 200% test current point and see how close the curve to 200% flux.

Do they give out the raw data that these calculators use? If I could get the datasheet curves in a tabular data format, that would be great. I could completely automate the decision process.

GrowLightResearch said:

This is what I said. Let's take your Vero 29.
@ 900mA lm = 8,140 the test flux
@1800mA lm = 14,554 at 200% the flux
14,554 ÷ 8,140 = 1.78 ≈ 175% is usually about 175% of the test flux

No charts graphs needed. Your pretty charts and graphs did make you look like you knew what you were talking about.

The math? So so. The red arrow on the CBX3590, 2,400 is not 200% of 1,400.





I have no practical use for those calculators. I have a datasheet, I look at twice the test current and take note of how close the line is to 200%.
The numbers you are using have no value to me. First off luminous numbers (e.g. lm/W) are worthless to me. I need to know the number of photons, mW/W is what I need to easily convert to photon quanta.

The efficacy below test is of no value either.

The decision I need to make when looking at the 175% is whether or not to push the LED beyond its test current or to buy two LEDs and run them both at test current or below.

That decision is based on the performance between test and twice test. Your numbers are from zero to max. That does not help me at all in this specific decision.

When I compare, what I typically see is by pushing the current I am getting 25% less photons. The temperature will increase significantly, the higher temperature will require additional thermal management, and the forward voltage is going to rise costing me more in electricity while getting less photons. Thermal management is expensive. So I would buy two LEDs. More LEDs gives better uniformity allowing the fixture to be closer to the canopy. As I lower the height of the fixture I can reduce the current increasing the efficiency of the system.

Deep red is a different story, it typically runs at better than 190% at 200% test. It also has a low Vf and a very small spread in Vf. So I run deep red over test.

BTW Cree far red is spec'd at 25° not 85°.

The graph in the datasheet is easier. It's a simple glance at the 200% test current point and see how close the curve to 200% flux.

Do they give out the raw data that these calculators use? If I could get the datasheet curves in a tabular data format, that would be great. I could completely automate the decision process.

Click to expand...

Don't forget to account for higher vf at higher drive currents when looking at data sheets. To get an honest picture you must figure the VF at the drive current you are estimating. Then adjust the VF for temp droop. Then calculate flux. Take that number and adjust for temp droop. That will get you pretty close. From what I can tell you'll be within =/- 10% of actuall results.

Stephenj37826 said:

Don't forget to account for higher vf at higher drive currents when looking at data sheets. To get an honest picture you must figure the VF at the drive current you are estimating. Then adjust the VF for temp droop. Then calculate flux. Take that number and adjust for temp droop. That will get you pretty close. From what I can tell you'll be within =/- 10% of actuall results.

Click to expand...

yeah any relative flux graph on a datasheet is isothermal and does not account for temp droop

Stephenj37826 said:

Don't forget to account for higher vf at higher drive currents when looking at data sheets.

Click to expand...

That is not what the post was about. I made a comment to someone about whether to run a strip at test or at max current. In my comment I said :

GrowLightResearch said:

Manufacturers always spec their products at the most efficient current in the IV curve.

Click to expand...

Then @CobKits came with this out of nowhere

CobKits said:

false

regardless of arbitrary test current, every single LED in existence is more efficient than that at lower currents. as seen by relative flux vs current curve on every single datasheet

Click to expand...

Things spiraled downhill @CobKits and went to shit

RE: Don't forget to account for higher vf
This is what I said on that

.

When I compare, what I typically see is by pushing the current I am getting 25% less photons. The temperature will increase significantly, the higher temperature will require additional thermal management, and the forward voltage is going to rise costing me more in electricity while getting less photons. Thermal management is expensive. So I would buy two LEDs. More LEDs gives better uniformity allowing the fixture to be closer to the canopy. As I lower the height of the fixture I can reduce the current increasing the efficiency of the system.

Stephenj37826 said:

Then adjust the VF for temp droop

Click to expand...

The temp droop would be wiped out by the drop in flux due to higher temperature.
I certainly put the temperature coefficient into the formula when comparing LEDs
And I'm careful to check that the LEDs are all compared at the same temperature.

Before final selection is made I measure the flux or the top candidates with my spectrometer.

CobKits said:

yeah any relative flux graph on a datasheet is isothermal and does not account for temp droop

Click to expand...

Every datasheet I've seen addresses thermal with a Vf vs temp curve, flux vs. temp, temperature coefficient, and most have thermal resistance. Some will put multiple current v flux curves where each curve represents a different temperature.

look im not going to go back and forth with you but literally every single post is nonsensical. its like you talk just to talk

GrowLightResearch said:

The temp droop would be wiped out by the drop in flux due to higher temperature.

Click to expand...

this makes zero sense. the drop in flux due to higher temperature IS temperature droop. how does it cancel itself out?

GrowLightResearch said:

Every datasheet I've seen addresses thermal with a Vf vs temp curve, flux vs. temp, temperature coefficient, and most have thermal resistance. Some will put multiple current v flux curves where each curve represents a different temperature.

Click to expand...

yes, none of which you referenced until you had to backpedal on being called out on your BS

CobKits said:

look im not going to go back and forth with you but literally every single post is nonsensical. its like you talk just to talk

Click to expand...

Exactly. It's just bizarre how he can be so wrong pretty much all the time. At some times I think he is just trolling, but he puts in an awful lot of effort and he does get emotional when he gets corrected. It's just utterly bizarre.

CobKits said:

the drop in flux due to higher temperature IS temperature droop.

Click to expand...

Higher temperature decreases forward voltage which decreases watts increasing efficacy lm/W . This advantage decrease in Vf / increase lm/W is wiped out by the decrease in flux.

wietefras said:

It's just utterly bizarre.

Click to expand...

wietefras said:

It's just bizarre how he can be so wrong pretty much all the time.

Click to expand...

What I find bizarre is how you still cannot understand the concept of the inverse square law.
Beyond bizarre is you thinking reflection wipes out ISL. And overlap??

GrowLightResearch said:

Higher temperature decreases forward voltage which decreases watts increasing efficacy lm/W . This advantage decrease in Vf / increase lm/W is wiped out by the decrease in flux.

Click to expand...

Nope, it worked this way:
Lower current means lower voltage, lower wattage and therefor lower temperature and better lm/w effiency.
Higher current means higher voltage and higher temperature but lower lm/w effiency.
The lowest possible current leads to the best possible effiency.

If vf goes down because of higher Tj. temperature there is no increase in lm/w performance. Output wounld also decrease only more heat will be produced. For this reason, you need better cooling when you run the LED's with more power.

Compare H- and F-Series.
Both use the same diodes. H-Series use 80mA per diode which leads to 180lm/w(3000°k) and 50° Tc*. F-Series use 124mA per diode which leads to 168lm/w but at 65° Tc*.(*datasheet)
Lower current leads to lower temps and better effiency. A voltage droop caused by higher temperatures will also lower effiency. Drivers also work more efficiently when cooled properly. This applies to many areas ..

Randomblame said:

Higher current means higher voltage and higher temperature

Click to expand...

The current vs. flux is spec'd at a constant temperature. With good thermal management, and or low ambient, the temperature may not rise with increased current. Therefore you cannot say higher current always means higher temperature even though it is often true it does not always apply.

Keep it simple, take current out of the equation.
If temperature rises due to ambient two things will happen,

1. Vf will decrease lowering wattage, raising efficacy
   
2. Flux will decrease lowering efficacy

Which has more impact? This was my point in the original post. One wipes out the other and then some. Work the numbers on an Amber LED for the most effect.

Lowering current below test current is not of any significant consequence.

In a production environment I am going for the most photons at the least cost where running below test current is not likely. I will set the max current to max efficiency which is a balance between the fixture's total flux vs the wall watts. That max efficiency is typically just above the test current current point. The current draw on the power supply and its efficiency at that current also factors in. That was also a major consideration in my original post on whether to use a 40 or 60 watt driver.

In a research environment the fixture will almost always be run below test current, and only a single fixture, so the efficiency there is moot.

It may be true in some cases, but in most cases lower currents improve efficiency significantly. The constant 25°C can hardly be achieved in reality, except with water cooling. As a rule, the Tj sinks. with the current and the efficiency improves significantly. LM561c datasheet says eg. 65mA/195lm/w, but H-Series runs at 80mA/180lm/w and F-Series at 124mA/168lm/w. All these numbers are from the datasheets.

Well, the difference is that you use monochromatic diodes and I only white diodes. Datasheet from LM561c calls a test current of 65mA. I have calculated my lamp so that they get a maximum of 63mA. I'll take ⅔ CRI80 and ⅓ CRI90 for a bit more red/far-red and mix the strips, but basically it stays at 3000°K.

You have to admit that the stripes are unlikely to behave like your yellow LED's.
Incidentally, 1900-2200°k could be better here than yellow.
I use Samsung's online calculator to calculate my lamp and I think I can trust the numbers. My goal is to keep the distance as low as possible and only use as much current as necessary. When I'm done I have up to 400w of well spreaded strips, which I can use on an area of 10ft² and as I know me, it will work just the way I planned it, LOL!

Randomblame said:

Well, the difference is that you use monochromatic diodes and I only white diodes.

Click to expand...

In respect to flux and temperature white LEDs are the same as monochromatic deep blue.

I mentioned Amber because it's flux is the most affected by temperature and same for its Vf (least affected). Still the same applies to deep blue (and white). NOTE: Far red Vf will likely be the least affected

Randomblame said:

The constant 25°C can hardly be achieved in reality, except with water cooling.

Click to expand...

Some datasheets use 85° and the results remain the same. For the sake of this discussion we can just say the temperate has stabilized at a steady state.

I believe at any reasonable operating temperature the the decrease in flux will have a greater influence then the decrease in Vf.

GrowLightResearch said:

What I find bizarre is how you still cannot understand the concept of the inverse square law.
Beyond bizarre is you thinking reflection wipes out ISL. And overlap??

Click to expand...

It is indeed bizarre how you ignore all proof that you are wrong and still keep insisting you are right. Or how you cannot even understand that simple principles like reflection will prevent light from spreading.

I will set the max current to max efficiency which is a balance between the fixture's total flux vs the wall watts. That max efficiency is typically just above the test current current point.

Click to expand...

Stuff like this is just complete bullshit.

If you actually were a researcher like you pretend to be, then at least you would have the ability to test these harebrained ideas. Just run the fixture with twice the number of leds and at the same total power. Do you then get more light per watt from the wall or not? Everybody here knows you do, but only you keep insisting that you don't.

Besides, just look at the "efficiency vs watts" charts that Cobkits posted. What more proof do you need?

The only reason not to keep adding leds is for economical reasons. At some point increasing efficiency starts to cost more than the savings you will get on electricity. It also depends on the price of electricity, so the most economical wattage per led won't even be the same for everybody.

just muddy the waters already and go for multiple point source interference...........

wietefras said:

Besides, just look at the "efficiency vs watts" charts that Cobkits posted. What more proof do you need?

Click to expand...

I don't even need that as proof. I have no problem that needs to be fixed.

I was only explain how I make one tiny decision. No one has to agree with me. That how I do it. I believe it to be the right thing to do. You can mount an ad hominem with slanderous attacks but no one cares. Move on. Subject over long time ago.

At this point I am not sure if your level of delusion is a symptom of a personality disorder or whether you are genuinely psychotic.

GrowLightResearch said:

At this point I am not sure if your level of delusion is a symptom of a personality disorder or whether you are genuinely psychotic.

Click to expand...

I think you need to take a long look in the mirror.
BTW you are totally wrong. I'm no expert but you are really out there. Take a deep breath and read any led datasheet.

What are you guys even talking about at this point?

nc208 said:

What are you guys even talking about at this point?

Click to expand...

Same thing.