Question about LED ratings

[mike_va 0]

Osram "Eye Safety of IREDs used in Lamp Applications" has this bit (attached) that I have a question about.

 

 

 

If the LED's are fitted with optics (reflector) to focus the beam does this increase the mW/sr that is used in the application. i.e. these are +/-60 deg without optics, if one used a reflector of +/-10 does Ie typ go up?

[Soundy 1]

I don't know most of those units, but... yes, generally, the tighter the beam, the more powerful - and potentially more damaging - it will be. This applies to any sort of electromagnetic radiation, as well as things like sound waves (I DID take a bunch of theory in shaping sound coverage and calculating throw distances and whatnot, many many years ago... inverse-square law applies to pretty much any sort of wave propagation).

[Stanislav 0]

In theory sharpening angle of radiation from 60 deg to 10 deg can increase axial radiant intensity in 32 times (if the power of radiation will remain the same). In practice the differense will be less because increasing optical loss.

[mike_va 0]

In theory sharpening angle of radiation from 60 deg to 10 deg can increase axial radiant intensity in 32 times (if the power of radiation will remain the same). In practice the differense will be less because increasing optical loss.

Thanks, is the radiant intensity the same as the Ie term?

 

edit: ok I answered that myself (yes) on Wikipedia

 

Did you mean 36, not 32? Not trying to give you a hard time just don't have a lot of experience with these calcs and especially steradians...

 

http://www.light-measurement.com/calculation-of-radiometric-quantities/

[Stanislav 0]

In theory sharpening angle of radiation from 60 deg to 10 deg can increase axial radiant intensity in 32 times (if the power of radiation will remain the same). In practice the differense will be less because increasing optical loss.

Thanks, is the radiant intensity the same as the Ie term?

 

edit: ok I answered that myself (yes) on Wikipedia

 

Did you mean 36, not 32? Not trying to give you a hard time just don't have a lot of experience with these calcs and especially steradians...

 

http://www.light-measurement.com/calculation-of-radiometric-quantities/

 

I got value 32.36 - ration between Axial Light intensities (!not equal Average Light intensities). The ratio depends also on the form of Light intensity distribution curve. I assumed the ratio of the light intensity on border of the cone by the light intensity on axis=0.5.

 

I used the Illuminator calculation tool in VideoCAD. All necessary formulas with steradians were programmed in this calculator.

 

But in practice the difference will be less because the less the angle of radiation is the more optical loss.

You can measure this difference in practice using the CCTVCAD Lab Toolkit. See here detailed description of this procedure.

[Stanislav 0]

There is a simplified way for manual calculation of the ratio between the Average Light intensities in dependance of the cone angle (without taking into account the form of Light intensity distribution curve and the difference in optical loss):

 

1. Solid angle As=2*PI*(1-Cos(An/2))

 

where An - the cone angle in degrees

 

For 10 degree As10=2*3.14*(1-Cos(10/2))=0.023897 steradian

For 60 degree As60=2*3.14*(1-Cos(60/2))=0.84136 steradian

 

2. Average Light intensity Li=F/Ab

 

where F - light flux which is the same for 10 degree and for 60 degree angles

 

For 10 degree Li10=F/As10=F/0.023897

For 60 degree Li60=Li=F/As60=F/0.84136

 

3. The ratio between the Average Light intensities of 60 deg illuminator and 10 deg illuminator with the same Light flux, without taking into account the form of Light intensity distribution curve and the difference in optical loss.

 

R=Li60/Li10=(F/0.023897)/(F/0.84136) =0.84136/0.023897=35.2

[mike_va 0]

There is a simplified way for manual calculation of the ratio between the Average Light intensities in dependance of the cone angle (without taking into account the form of Light intensity distribution curve and the difference in optical loss):

I am trying for my own amusement to calculate eye safety levels. Here is one I did on an Raytec. I get a very different ratio than you when adjusting for the angle though.

[Stanislav 0]

Please see my corrections of your calculations:

 

[Stanislav 0]

Also P radiant=47.1W is unreal for Raymax100. Its consumption is 70W, thus its expected P radiant can be no more than 20W minus optical loss.

 

It seems new parameters are much better. Just seen spec of SFH 4232 Platinum Dragon Package, half angle ±60°, 850nm, typ. 530mW at 1A dc. Its total radiant flux=530mW with 1.8W power consumption.

 

 

See an interesting IR illuminators design guide by OSRAM High Power Emitters for Illumination Applications

 

See Page 5. There the difference of 60 deg LED SFH 4232 and the same LED with a 10 deg Lens is compared.

Pay attention it is HALF ANGLE no FULL angle.

 

"A suitable lens for the Platinum Dragon

SFH 4232 to meet the ±10° FOV

requirement is for example the Lisa-SS lens

from LEDIL which will be used in this

example.

Test measurements show that the radiant

intensity of SFH 4232 is increased by a

factor of 7.6 to 1400 mW/sr at 1 A. Without

lens we only get 180 mW/sr.

"

Thus the practical result is 7.6 times. (for 20 deg and 120 deg full angle)

The difference from the theory arises because of optical loss.

[rory 0]

Osram "Eye Safety of IREDs used in Lamp Applications" has this bit (attached) that I have a question about.

Just never look at IR ... problem solved.

I cant look at it now, too much eye damage from years of working with it in the pitch dark in the field, especially all the Extreme gear - UF500 would literally burn your skin at up to 3 feet, nevermind the eyes. I dont care about what safe distances all the manufacturers claim, even they dont really know.

[rory 0]

I am trying for my own amusement to calculate eye safety levels. Here is one I did on an Raytec. I get a very different ratio than you when adjusting for the angle though.

Dont forget its not just about the distance but how LONG you look at it too.

Either way you dont want to stick your eyes 10" in front of any IR, unless you dont value your eyes.

[Fiona 0]

Hi Stanislav, sorry to butt-in at the end of another one of your tour-de-forces.

 

I wonder if you might be so kind as to comment on the affect that fog or smoke may have on reflecting or absorbing IR illumination. Any comments would be warmly welcomed.

 

Question: Does fog, smoke or heavy rain reduce the affect of IR illumination?

[Stanislav 0]

Hi Stanislav, sorry to butt-in at the end of another one of your tour-de-forces.

 

I wonder if you might be so kind as to comment on the affect that fog or smoke may have on reflecting or absorbing IR illumination. Any comments would be warmly welcomed.

 

Question: Does fog, smoke or heavy rain reduce the affect of IR illumination?

 

I am sure that fog, smoke and heavy rain must reduce the affect of IR illumination, as they reduce conventional light efficiency. IR is the same electromagnetic waves as the visible light and its behavior is very similar.

May be very small particles less affect to IR because of longer wavelength, but I don't know exact data.