# candles vs. watts



## HankM (Mar 11, 2004)

I know what candles/ft-candles are in the visible spectrum but how can I relate this to "watts" output in the near-IR (LED) spectrum? i.e. If I have 2000 millicandle LED in the visible region, what is the equivalent brightness, expressed in watts output for an IR-LED in the 800 to 1000 micron region for the same apparent brightness with NVGs? Assume the beams are the same for each.


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## SilverFox (Mar 11, 2004)

Hello HamkM,

Welcome to CPF.

I do not believe there is a good conversion, but I am standing by ready to learn...

The SureFire M1 is rated at 10 mW and is supposed to be good to 100 meters. This rating must be with Gen II or better, because my night vision camera will only find it useful to about 10 meters. I think SureFire is stretching it a bit, but I have not tested the light with the better NV equipment.

Tom


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## Doug S (Mar 11, 2004)

How many foot-pounds in 100 degrees C? Similar question. These are simply not comparable units.


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## idleprocess (Mar 11, 2004)

Watts is just a measure of energy dissipation. I think most people relate watts to light because numerous residential & commercial light bulbs are sold on the basis of wattage. Even compact flourescent bulbs are marketed as "comparable to [x] watts."

A 100 watt florescent bulb, for example, would put out substantially more visible light than a 100-watt incadescent bulb.


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## LEDMANIA (Mar 12, 2004)

If I'm going to recall my "rusty" knowhow in Physics,Watts is a unit of power.It can be converted into BTU,Calories,Horse Power,Jouls/hr,ft-lbs/hr etc.etc..but there is no direct relation to luminousity..
The light manufacturer only use the word watts on their light products as power consumption and not for brightness.
confuse? Lets say a 20watt bulb is brighter than 10watt bulb.They are both compared in power but not in brightness..
That said....

-----------------------------------

Peter Gransee Said; "Just Inovate not Litigate"


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## LEDependent (Mar 12, 2004)

I can see where that idea came from, as the Lumileds datasheet lists the light output in mW instead of Lumens for Royal Blue.


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## HankM (Mar 12, 2004)

Thanx to all who responded. The answer is still not there. I agree that most if not all visible LED mfrs state the power consumed in watts or mw and they give a millicandle output plus a beam divergence (shown as 50% intensity points). IR-LED mfrs state the "Radiant Intensity" usually expressed as mw/SR (milliwatts per steradian). Visible-LED specs are are also sometimes in mw/sr. What I'm trying to get a handle on is some equivalent of millicandles to those radiant milliwatts/sr, not the power dissipated watts, as most of you implied. I guess if I could determine what a radiant SR envelope means relative to a specific beam angle at its 50% intensity beam divergence points, then I would have the approximate conversion I need.
(PS -- This has been my first experience with this forum and I'm surprised at the amount of feedback in such a short time. Thanx again.)


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## gwbaltzell (Mar 12, 2004)

I'm not sure that the conversion would do what you intend. NVG actually amplify light but how much depends a lot on the wavelength of the light and the particular NVG itself. I don't see what can be achived. Output of a visible LED could be expressed in watts. Watts out divided by watts in would give % efficiency. But I think a glance at Don Klipstein's page may be in order. Note: He says "One watt of light at any *single wavelength (or in a very narrow band)* is, in lumens, 681 times the official photopic function of that wavelength." (Photopic - daytime vision, Scotopic - naked eye *very* low light vision). This in and of itself would negate any IR comparsion since the photopic function of IR would be zero. Zero times anytime is still zero.

edit: The SI definition: "The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540 x 10^12 hertz and that has a radiant intensity in that direction of 1/683 watt per steradian."
So a 7 lumen 555 nm LED is putting out 10 mW sr (if HTML was availble it is preferably written 10 mW&middot;sr). If the forward voltage drop is 3.2 volts and the current is 20 mA its drawing 64 mW so it is 15.6% efficient.
Ok, I originally posted 7000 mcd even when I knew better. I don't trust my math but I think 7 lumens in a 30 degree angle would be about 1,500 mcd? (My fourth answer. Or is it my fifth?) Based on a perfect lambertian radiation pattern. And you wonder why no one tries to do this? And Google™ has this one completely wrong in their calculator. And their "once in a blue moon" is questionable but amusing.

The lumen is cd sr (candela per steradian).


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## HankM (Mar 13, 2004)

Thanx. After digging into it a little more, I've essentially reached the conclusion that there is no simple comparison. I was hoping there was a simple solution. The mfrs I've contacted say about the same thing, so the issue is now officially dead. /ubbthreads/images/graemlins/mecry.gif


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## idleprocess (Mar 14, 2004)

My understanding of mcd is that you need to know the beam angle to really get a handle on how bright the light source is. A cheap laser pointer would have an insane mcd reading, but lasers have a tiny beam angle. Spread that tiny angle out a bit and you find that the absolute amount of light produced is miniscule compared to even a dim indicator LED. A high-flux "spider" LED might look like it's dim at ~200 mcd, but that's typically over 120-140 degrees - enough light for a decent flashlight if it's concentrated.

I once found a formula that would convert mcd*angle to lumens. I think it involved determining the number of steridians a given beam angle covered, then factored the given mcd figure to get lumens. It worked allright against the few datasheets I've found that quoted lumens and mcd/cd.


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## jtr1962 (Mar 14, 2004)

[ QUOTE ]
*idleprocess said:*
I once found a formula that would convert mcd*angle to lumens. I think it involved determining the number of steridians a given beam angle covered, then factored the given mcd figure to get lumens. It worked allright against the few datasheets I've found that quoted lumens and mcd/cd. 

[/ QUOTE ]

Here's what I use:

lumens= 0.00096*(candles)*(beam half angle in degrees)²

For example, a 10,000 mcd LED with a 10° half angle would put out 0.00096*10*(10)² = 0.96 lumen. Note that mcd must first be converted to candles (1000 mcd = 1 cd). Also note that this equation doesn't take into account the light emitted by the LED past the half angle. In the case of narrow beam LEDs this can be significant, and the equation may underestimate the light output by up to 50% or more. A perfect example of this is one manufacturer's 15° and 30° LEDs using the same chip (and hence the same light output). The 15° LED is rated at 10,000 mcd while the 30° one is rated at 6,000 mcd. Going by the equation the 15° LED puts out 0.54 lm while the 30° one puts out 1.296 lm. It is obvious that the equation grossly underestimates the light output of the narrower beam LED.

Once the lumens is known, finding the efficiency is relatively straightforward. For example, assuming a 3.3 volt drop at 20 ma, the efficiency in the previous example would be 1.296 lm/(0.02)(3.3)W = 19.6 lm/W. If the LED is colored, converting this to mW is fairly straightforward using the lm/W of the peak emitting wavelength. If it is white then the exact spectrum must be known, although 250 lm/W is a fairly good approximation. If we assume this, then the LED is emitting 1.296 lm/ 250 lm/W, or 0.00518 W = 5.18 mW, so the wall-plug efficiency is 5.18/66 or 7.85%.


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## idleprocess (Mar 16, 2004)

Here's the website where I dug up that equation I've used:
*(1-abs(cos[beam angle]))*2*pi*[cd]*

On a spreadsheet:
*=((1-ABS(((COS([beam angle]/2*(PI()/180))))))*2*PI())*([mcd]/1000)*

Here are some of the results I generated (likely complete with a current consumption fat-finger on that "1.3W" Nichia):

<font class="small">Code:</font><hr /><pre>
Manuf.	Model Format	Vf	mA	mW	color	mcd	angle	area	mcd*area	Lumens

Lumex	SSL-LX5093XUWC	T1-3/4	3.60	30.00	108.00	white	2300	15	0.0545 125.2379	0.1236
Lumex	SSL-LX5093XUWC	T1-3/4	3.60	30.00	108.00	white	2300	15	0.0545 125.2379	0.1236
Lumex	SSL-LX3054UWC/A	T1	3.60	30.00	108.00	white	1100	20	0.0977 107.4435	0.1050
Nichia	NBCW011 SMD	3.60	80.00	288.00	white	2200	120	9.4248 20734.5115	6.9115
Nichia	NACW019 SMD	3.60	40.00	144.00	white	1100	120	9.4248 10367.2558	3.4558
Nichia	NCCW022 SMD	3.80	350.00	1330.00	white	1850	80	2.2120 4092.1228	2.7195
Nichia	NSCW017 SMD	3.60	20.00	72.00	white	600	120	9.4248 5654.8668	1.8850
</pre><hr />

The "area" and "mcd*area" fields were attempts to formulate a metric before I found the mcd-lumens equation I'm using.

I don't even begin to assume that this calculation is anywhere near dead-nuts - I used it to get a rough comparison between LEDs and other light sources for work once, and have used it to get a feel for general LED brightnesses.


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## idleprocess (Mar 28, 2004)

(bump)

Can anyone tell me if my formula works? Looking at the spreadsheet formula, it looks a bit... redundant in some aspects.


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## gwbaltzell (Mar 28, 2004)

I think you'll find the formula pi * (angle/2)^2 * cd comes close also. Particularly in the narrower angles.
on a spread sheet in mcd it looks like: =PI()*(RADIANS(angle/2))^2*mcd/1000 
in Google™ in mcd: pi * (("angle"/2) degrees)^2 * "mcd"/1000 =
a Google™ result: (pi * (((80 / 2) * degrees)^2) * 1 850) / 1 000 = 2.83267219


The rightmost column below is the expect output based on watts in and lumens per watt.
<font class="small">Code:</font><hr /><pre>Manuf.Model Format Vf mA mW color mcd angle Lumen lm/w Lumen
Lumex SSL-LX5093XUWC T1-3/4 3.6 30 108 white 2300 15 0.124 1.5 0.162
Lumex SSL-LX5093XUWC T1-3/4 3.6 30 108 white 2300 15 0.124 1.5 0.162
Lumex SSL-LX3054UWC/A T1 3.6 30 108 white 1100 20 0.105 1.5 0.162
NichiaNBCW011 SMD 3.6 80 288 white 2200 120 7.579 20.0 5.76
NichiaNACW019 SMD 3.6 40 144 white 1100 120 3.790 20.0 2.88
NichiaNCCW022 SMD 3.8 35 133 white 1850 80 2.833 20.0 2.66
NichiaNSCW017 SMD 3.6 20 72 white 600 120 2.067 20.0 1.44</pre><hr />
My formula is for a point source, yours is for a lambertian (flat sheet) which of course comes closer to the larger or more wide angle LED.


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