Comparing mW to MCD equivalent

fazgood

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I am helping my son set up a science project and I am trying to make a best guess at LED intensity. His project requires several LED light sources of similar intensity. Problem is, when you get to UV they are rated in mW since they can't be seen by human eyes. I can not find a way to convert them (not possible?). Is there a way to try to approximate the same intensity in UV as in blue and violet?

Blue - clear -5000 mcd - 15° - 470 nm
Violet - clear – 1000mcd - 15° - 420 nm
UV - clear --- 12 mW --15° - 400 nm

Using these LEDs my thought would be to use one BLUE and five VIOLET to achieve similar intensity. My question is, can I approximate the number of UV required to have the similar intensity?

Thanks

Faz
 
mW is the amount of output and MCD is the amount of throw or projection and they are usually not interchangeable unless you know the area and distance of where all the 12mW of light goes.

Would it be better for you to buy a high power UV LED and use a reflector/optic to direct the light?
 
That could work but would increase the cost of the project substantially and limit the number of differing wavelengths that are available. I have access to a LUX meter and a PAR meter and was hoping to be able to make some sort of "best guess"

Great source by the way. I have not been to that site before. It looks like it will cost me $$

Thanks

Faz
 
mW is a radiometric measurement, essentially absolute power. candela is a photometric measurement and hence takes into account the sensitivity of the eye.

As one person mentioned, you need to take into account beam angle and pattern to make a rough conversion, but you also need to take into account the eye sensitivity at that wavelength. Since you are talking UV, the eye sensitivity is near 0 so the conversion is really not possible hence why the UV led is in mW and not millicandela.

That said, what are you trying to do? Is this related to the eyes perceived brightness or something else where mW would make more sense?

Semiman
 
His project involves the effects of growing algae using different colored light. We plan to make several identical chambers that have a light source comprised of specific colored LEDs. In order to be as objective as possible all of the light sources (LEDs) need to emit the same intensity. I am not an engineer so I am learning as we go along but from the info provided in this thread it appears we should use the radiometric measurement in mW to measure the power. The LEDs I ordered all have a 30 degree beam angle so that should help with consistency.

On the spec sheet for the blue LED is lists Power Dissipation of 120mW. It also has a luminous Intensity of 4600mcd at 20mA but no mention of light output. On the spec sheet for the UV LED it lists Power Dissipation of 80mW. However, the light output is rated in Absolute Irradiance of 12mW at 20mA. Spec Sheet links below.

So, with the limited equipment that I have access to I am finding it hard to try to put together an objective way to measure to get a fairly close amount of light to each subject. The Lux meter seems to be able to read the UV light but it might be reading just the visible blue part. I need to see if I can find the specs on it. I have access to a PAR meter from a friend but I have not had time to research if that would do what I need it to do.

Any further thoughts would be appreciated.

Thanks

Faz

http://www.superbrightleds.com/specs/uv1230_specs.htm

http://www.superbrightleds.com/specs/b4630_specs.htm
 
mW is the amount of output and MCD is the amount of throw or projection and they are usually not interchangeable unless you know the area and distance of where all the 12mW of light goes.
...

I think you probably mean, "mW is the amount of INPUT POWER".

Toshi
 
TMorita, no from his previous post, that 12mW is output power.

You will have to do a two step process:

1) Convert the millicandela and angle into an absolute lumen measurement. There are a few calculators on the web. It is an approximation:

http://led.linear1.org/lumen.wiz
http://www.gizmology.net/LEDs.htm (Also a good article)

One thing to watch here is whether they are using half angle or full angle, i.e. one 15 degree led may be the same as another 30 degree. I noticed you have mentioned 30 degree and then you list 15 degree. The data sheet links you gave are 30 degree full angle. The second calculator above assumes full angle so I will use 30 degrees.

2) Once you have the lumens, you will need to convert that to mW. Most LEDs lumen measurements are based on the 1931CIE curve but this was updated in 1988. As you are working with blue LEDS, you need some good resolution on a conversion chart. I found one for 1988 with 1nm resolution, but not for 1931. This one may still be ok.

Look up the wavelength of the LED on the chart:
http://members.misty.com/don/photopic.html

It is in 5nm on the left, then 1nm along the top. This will get you a photopic conversion efficiency. You will notice it is 1 at 555nm where the eye is most efficient. At an efficiency of 1, you get 683 lumens/watt by definition essentially.

Hence, if you had a 1 lumen green led at 555nm:
(1 lumen) / (683 * 1.00efficiency) = 1.46mw optical power.

The blue, 5000mcd, 30 degree LED you describe puts out about 1.08 lumens.

From the chart, 470nm is 0.091 on the efficiency curve. Hence the output is:

1.08 / (683 * 0.091) = 17.2mw

The violet one is about 0.216 lumens and about 18mw

Those numbers seem a touch high, but are within the realm of believeability. I find the millicandela to lumen calculation high in most cases due to the shape of the LED beams.

Unless the lumen meter you have is really expensive (> $2,000) it will be wildly inaccurate in the blue. You really need a spectrophotometer to be accurate.

My calculations are likely accurate enough for a kids science experiment. There are such huges variances in the LEDs that you buy that without an expensive meter you are not going to get close.

One thing you could do is find a sensor that has a sensitivity chart versus wavelength. Since you know the wavelength of your source, this would allow you to measure for comparative purposes much more accurately then the light meter.

Good luck.

Semiman
 
Semiman,

Thanks for this info. I somehow missed your reply but this is excellent. Now I just need to digest it to be able to help my son understand it.

I have aquired all of the LEDs and plan to put together the circuits this weekend. I will post back our results.

Faz
 
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