Radiant Intensity from Radiant Power and Intensity Distribution

In summary, the infrared led TSAL7600 has the following properties:- Phi = 35mW- I_e = 25mW/sr- I_r(\theta) = cos^{4.818}(\theta)- I(\theta) = I_e \cdot I_r(\theta)- Integrating over the relative radiant intensity should give the radiant power- However, finding the actual radiant intensity is eluding me.
  • #1
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TL;DR Summary
Calculate Radiant Intensity from Intensity Distribution and Radiant Power. All 3 from datasheet, but integration over the intensity distribution does not yield the correct Radiant Power.
Consider the infrared led TSAL7600 which has the following properties:

$$ \Phi = 35 mW $$
$$ I_e = 25 mW/sr $$

The half angle is ## 30^\circ ## and:

$$ I_r(\theta) = cos^{4.818}(\theta) $$

is a good approximation for the relative radiant intensity.
tsal7600-cos-4.818-vs-datasheet.png


However, finding the actual radiant intensity is eluding me. I'm looking for something like:

$$ I(\theta) = I_e \cdot I_r(\theta)$$

where ## I_e ## is the peak intensity. If it's correct, then integrating over the surface of the relative radiant intensity should give the radiant power:

$$\Phi = \int_{0}^{2\pi} d\phi \int_{0}^{\pi/2}I_e cos^{4.818}(\theta) sin(\theta) d\theta $$

$$\Phi = 2\pi I_e \int_{0}^{\pi/2} cos^{4.818}(\theta) sin(\theta) d\theta $$

And thanks to integral-calculator.com

$$ \Phi = I_e \cdot 1.0799 $$
$$ \Phi = 26.997 \neq 35 $$

That's obviously nowhere close.

The above integral came out of a paper called Improvement of the Approximation Accuracy of LED Radiation Patterns, but I had already found ## cos^{4.818}(\theta) ## by playing around in octave so I didn't use a nice round number like 5 (with a nice solution). I presume the integral finds the surface area of the radiation pattern since I've seen similar elsewhere, but I don't claim to know what I'm doing. I'm an Engineer. but this is not my area of expertise.

In the paper, they use the radiant power to solve for ## I_o ##, the maximum intensity, which I believe is the same thing as ## I_e ##.

This integral:

$$\Phi = 2\pi I_e \int_{0}^{\pi/2} cos^{4.818}(\theta) sin(\theta) d\theta $$

evaluates to ## \approx 1 ## and I need ## \approx 1.4 ## to get the correct radiant power. That means one of four possibilities:

1. the integral $$\Phi = \int_{0}^{2\pi} d\phi \int_{0}^{\pi/2}I_e cos^{n}(\theta) sin(\theta) d\theta $$
is not correct

2. the datasheet distribution does not match the real distribution. If the actual half angle were more like ## 35^\circ ## for which ## cos^{3.487}(\theta) ## is a better fit, the integral evaluates to ## \approx 1.4 ##.

3. the actual distribution is not approximated well enough by ## cos^{4.818}(\theta) ##

4. the required 'fudge factor' is ## 1.4\approx \sqrt{2} ##. Is it possible that ## I_e ## needs conversion by ## \sqrt{2} ## to become ## I_o ##?

Can anyone offer up an explanation as to why the radiant power is not correct?

The goal here is to go from ## \Phi=35mW ## to ## I_e=25mW/sr ## using the estimated radiant distribution. I would like to eventually calculate the radiant power over a smaller angle.
 
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  • #2
Isn't there there a factor ## \cos \theta ## missing? I'd expect the flux to be given by $$ \Phi = 2 \pi \int_0^{\pi/2} d\theta \sin \theta \ I(\theta) \cos \theta \ , $$ because the intensity refers to the projected area.
 
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  • #3
InTheWorks said:
The goal here is to go from ## \Phi=35mW ## to ## I_e=25mW/sr ## using the estimated radiant distribution. I would like to eventually calculate the radiant power over a smaller angle.
I'd think that only one of the quantities is reasonably accurate. Have you checked with the company?
 
  • #4
WernerQH said:
Isn't there there a factor ## \cos \theta ## missing? I'd expect the flux to be given by $$ \Phi = 2 \pi \int_0^{\pi/2} d\theta \sin \theta \ I(\theta) \cos \theta \ , $$ because the intensity refers to the projected area.
I think ##I(\theta)=cos^n(\theta)## is the far field radiation pattern which assumes a point source. I don't know if that explains it. There's also the assumption that the radiator is lambertian.

In more basic terms, adding a cos term would only increase n. Increasing n gives a smaller number from the integral which is going the opposite way.

Going by the units, the radiant intensity is power/solid angle. That solid angle I think comes about via ##d\phi d\theta##. Perhaps. I haven't yet convinced myself.

Getting back to the lambertian radiator, if I understand it correctly the integral of ##I(\theta) = cos(\theta)## should give a result of pi:

$$ 2\pi \int_{0}^{\pi/2} cos(\theta) sin(\theta) d\theta $$

And it does.

So perhaps the explanation is that neither the point source nor lambertian radiator are valid assumptions for this LED?

WernerQH said:
I'd think that only one of the quantities is reasonably accurate. Have you checked with the company?
I have sent an email off to Vishay. I don't expect to get a reply though.
 
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  • #5
InTheWorks said:
Going by the units, the radiant intensity is power/solid angle. That solid angle I think comes about via ##d\phi d\theta##.
Correct. If you assume a Lambertian radiation pattern, i.e. ##I## independent of ##\theta##, you get a factor ##\pi##, as you have checked yourself. The peak radiance of ## \rm 25~mW~sr^{-1} ## would then imply a flux of ## \rm 78.5~mW ##.

You can make flux and radiance consistent, if you assume ## I(\theta) \propto \cos^n\theta ## with ## n \approx 2.5 ##. (The effective solid angle is ## 2 \pi \over n+2 ##.) I'm not sure if the diagram could be viewed as consistent with that.

InTheWorks said:
So perhaps the explanation is that neither the point source nor lambertian radiator are valid assumptions for this LED?
Photometry doesn't depend on such assumptions.
 
  • #6
WernerQH said:
Correct. If you assume a Lambertian radiation pattern, i.e. ##I## independent of ##\theta##, you get a factor ##\pi##, as you have checked yourself.

Thank you for confirming that.

WernerQH said:
The peak radiance of ## \rm 25~mW~sr^{-1} ## would then imply a flux of ## \rm 78.5~mW ##.

As I would expect for ##I(\theta)=cos(\theta)##. I haven't found a 5mm LED like that to even check though.

WernerQH said:
You can make flux and radiance consistent, if you assume ## I(\theta) \propto \cos^n\theta ## with ## n \approx 2.5 ##. (The effective solid angle is ## 2 \pi \over n+2 ##.) I'm not sure if the diagram could be viewed as consistent with that.

I was able to make the radiant power (flux) and radiant intensity (not radiance) agree with ##I(\theta) \propto \cos^3.487(\theta)##, but this significantly deviates from the published relative radiant intensity plot from the datasheet.

WernerQH said:
Photometry doesn't depend on such assumptions.

Perhaps not, but that specific integral does. I took a look at a 5mm IRED under the microscope at work (this is not the Vishay IRED).

img230307194740o.jpg


It doesn't look like a lambertian emitter. Maybe the central die does (it's 0.5mm x0.5mm), but that reflecting ring doesn't?
 
  • #7
I found a document by Vishay that describes how they measure leds: led_measure.pdf.

It says they measure the Radiant Intensity using a detector that is ##1cm^2## at a distance of ##100mm## which is a ##0.01sr## solid angle (##6.5^\circ##).

##100mm## isn't very far away (though probably far field) and the same as the Laser safety standard measurement distance (the minimum distance for eye accommodation). ##0.01sr## is close to the LED safety standard minimum subtended angle for long exposures (which is ##0.011sr##). Is that coincidence, I wonder?

They calculate the Radiant Power based on the detector current and surface area (and get mW?). Then they convert that to Radiant Intensity by dividing by the solid angle (##0.01sr##).

The Radiant Intensity and Radiant Power seem to come from the same measurement. So there should be no reason to doubt one over the other.

For the Radiation Angle, they use a similar fixture for Radiant Intensity. Apparently the Radiation Angle measurement is a manual measurement performed with a rotary scale. They measure the intensity at ##0^\circ## and call this the maximum. They rotate the LED left (and then right) until the Radiant Intensity drops to ##\frac{1}{2}## of the max.

They don't claim to map out the whole radiation pattern, just the ##\frac{1}{2}## angles which doesn't instill a lot of confidence in the datasheet plot.

They don't say what distance this is done at. But if they are using a detector of ##1cm^2##, I don't see how they will get a precise measurement of the ##\frac{1}{2}## angles at ##100mm##. So I don't know how similar the fixture is to the Radiant Intensity measurement.

This leads me to believe that the Radiant Intensity measurement isn't the same as the maximum intensity measured for the Radiation Angle. The Radiant Intensity is, for lack of a better word, the average over ##6.5^\circ##. The maximum intensity for the Radiation Angle could be the same average over a smaller angle.

If I knew what that angle was I should be able to calculate a conversion factor. Or do I already know the angle based on the discrepancy between the Radiant Power, Radiant Intensity, and the Relative Intensity Distribution?

##I_e = 25mW/sr##, but if the integral is correct and the Radiant Power is distributed over the integral's area, then ##I_{peak}## could be ##\approx 35mW/sr##.

We can get that with either a higher average intensity or the same average intensity over a smaller angle, like ##0.007sr## which is ##4.64^\circ##. With the same setup, they would just need a bigger distance than ##100mm## between the LED and the detector.

I really shouldn't be doing math this close to bed time. Does any of my above rambling make any sense?
 
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  • #8
InTheWorks said:
I found a document by Vishay that describes how they measure leds: led_measure.pdf.
Interesting. But I'm a theorist, and I had never heard about the "Ulbricht sphere". :-)
InTheWorks said:
If I knew what that angle was I should be able to calculate a conversion factor. Or do I already know the angle based on the discrepancy between the Radiant Power, Radiant Intensity, and the Relative Intensity Distribution?
Intensity proportional to a power of ## \cos \theta ## is not a physical law, but merely a convenient parametrization of the radiation pattern. A different pattern is quite possible. Unless you can measure the angular dependence yourself, with the data at hand you can only obtain ballpark estimates.
 
  • #9
WernerQH said:
Interesting. But I'm a theorist, and I had never heard about the "Ulbricht sphere". :-)
When you see a heading called "Vishay Telefunken" one might assume there was some translation error from German to English ;)

WernerQH said:
Intensity proportional to a power of ## \cos \theta ## is not a physical law, but merely a convenient parametrization of the radiation pattern. A different pattern is quite possible. Unless you can measure the angular dependence yourself, with the data at hand you can only obtain ballpark estimates.
I really appreciate this. I've been going a little crazy trying to understand the issue.

I shouldn't have to measure the angular dependence (and I'd need a calibrated detector). The manufacturer should provide that information. I normally treat all the information contained in a datasheet as law. It's been my experience, at least with the components I regularly use, that datasheets rarely have errors. The data at hand should be accurate.

The Radiant Intensity Distribution graph should have a disclaimer on it though. Something about it being a crude estimate.

From the picture I posted above, this chip clearly has a bond dot in the centre of the die. And Figure 9 shows the difference in radiation patterns between a chip with and without a bond dot.
Screenshot at 2023-03-09 05-45-18.png


So it's probably safe to assume that the intensity exactly on-axis is actually lower than expected. And I guess this could agree with the observation that the shape of the bond dot radiation has more surface area than ##cos^{4.818}(\theta)## with matching ##\frac{1}{2}## half angles.

The troubling part for eye safety is that the peak Radiant Intensity is likely higher than the average around the bond dot as measured on axis over ##0.01sr##. I should probably err on the side of caution and use a higher Radiant Intensity, perhaps 1.4 times higher? Maybe 2x higher?

The issue is that the LED safety standard sets a limit for Radiance at the Retina and gives no guidance on how to calculate it or measure it.

There is an appnote from OSRAM (AN090) that estimates the Radiance using:

$$ L_{ir} \approx \frac{I_e R(\lambda)}{Z^2}$$

where ##R(\lambda) = 0.33## is a weighting function from the standard and ##Z^2## is the emitter area rather than the virtual source area. They then go on to show that the estimated Radiance agrees with the Radiance "calculated from real spectral data". There is no source given for this approximation nor an explanation of how it works. The LED in their example is not a 5mm LED (with a bond dot) and I question if it's applicable to all LEDs.

Now I don't particularly care that they use the emitter size or the virtual source size, except that when two LEDs are placed side by side they can overlap on the retina creating a hazard when there is no hazard with each LED alone. And that's the issue I'm really grappling with. It's the apparent size of the whole LED that will form an image on the retina not the emitter size.

I can't blindly apply a formula like that and I'm trying to assess the eye safety of an infrared LED lamp made up of 36 5mm LEDs set tightly together. Worse is that these lamps are coming from China with no real LED specs. They give viewing angle of the LED and the total current. Based on that I can make a reasonable estimate of Radiant Power and Radiant Intensity. I figured I'd start with a well specified LED to make things easier...

The application is a modified camera and lamp for Aerodrums. One forum member posted instructions on how to make them and I don't feel comfortable sitting in front of an infrared lamp when I know nothing about it's safety. If it's dangerous I would like to let the others know.

I'll need to think about this more, but if you have any ideas or comments I would really appreciate hearing them.
 
  • #10
Here's what I think.

The standard model for the human eye has a focal distance to the retina of ##17mm## equivalent in air [1].

The ##5mm## spacing imposed by the LED body means that the smallest subtended angle between neighbouring apparent sizes of ##1.8mm## is:

1. ##0.032 rad## for a ##100 mm## distance
Screenshot at 2023-03-11 20-59-29.png

2. ##0.016 rad## for a ##200 mm## distance
Screenshot at 2023-03-11 20-58-58.png

At these distances there is no overlap on the retina. And I guess a decent safety margin. The closer you get the farther apart the images on the retina and the larger the image becomes. So the danger there is not from overlap. The farther you get from the LEDs, the closer the images on the retina become and the smaller the images become. So the danger there is not from overlap either. Maybe this is pretty obvious, but I needed a diagram.

In either case, Radiance is conserved. And that's the danger. So back to that approximation...

According to wikipedia, the formula for Radiance is:
$$ L_e = \frac{\partial^2\Phi_e}{ \partial \Omega \partial A cos(\theta)} $$
and
$$I_e = \frac{\partial \Phi_e}{\partial \Omega} $$
Which means:
$$ L_e = I_e\frac{\partial\Phi_e}{ \partial A cos(\theta)} $$
and I'm not clear on how:

$$ \frac{\partial\Phi_e}{ \partial A cos(\theta)} = \frac{1}{A} $$

The ##cos(\theta)## term vanishes because the angle of view is directly on axis. I think remaining partial derivative just conveys how evenly lit the emitter is. For flat chip LEDs, the die is approximately uniformly lit so I suppose we can replace ##\frac{\partial \Omega_e}{\partial A} ## with an unchanging area ##\frac{1}{A}## times an unchanging flux, but where did the flux go? Or rather why does ##\frac{\partial \Omega_e}{\partial A} ## become ##\frac{1}{A}##?

If the above assumptions are true, then the Radiance really does become:

$$ L_{ir} = \frac{I_e R(\lambda)}{Z^2}$$

where ## Z^2 ## should be the virtual size of the emitter area. At least the uniformly lit area. In the OSRAM app note they are using a very wide angle uniformly lit flat chip LED and the wider the angle, the less magnification the emitter area gets. So very likely the virtual area is equal to the emitter area. This must be an assumption built into the equation and why their "spectral measurement" agrees with their calculation? Their flat emiiter probably more closely approximates ##cos^n(\theta)## too.

This is a problem for the ##5mm## LED which has a reflecting bowl that the flat chip LEDs don't have. And there is a bond dot that occupies some of the die area. In other words, ##\frac{\partial \Omega_e}{\partial A} ## doesn't magically disappear. We can make it disappear if we assume that all of the flux is emitted by the virtual die area and we're happy with that over estimation. I think.

Using the whole diameter of the virtual size (i.e. ##1.8mm##) probably shouldn't be used. At least not without a proper ##\frac{\partial \Omega_e}{\partial A} ##.

If the above logic isn't blatantly wrong (please correct me if it is!) then the last piece of the puzzle is what to use for ##I_e##?

I'm not certain that the way Vishay measures ##I_e## at ##100mm## is a good representation of the actual peak ##I_e## over the small subtended angle (##0.011 rad##) that the standard requires. If it's a bigger number and truly an over estimation, then I would happy. Without knowing the actual Radiant Intensity Distribution, can I know?
 
  • #11
InTheWorks said:
In either case, Radiance is conserved. And that's the danger. So back to that approximation...
I find it a bit difficult to understand your reasoning. And predicting the radiant intensity at the retina is quite a different problem from estimating the flux density (## \rm mW/cm^2 ##) in the vicinity of the eye. And that's probably the quantity you need to consider when you are concerned about possible hazards to your eyes. (I suppose you have already found literature specifying safety limits? And do they give values that cannot be converted to ## \rm mW/cm^2 ## ? If I haven't miscalculated, for sunlight we have about ## \rm 130~mW/cm^2 ##, and you shouldn't stare at the sun.)

Radiant intensity ## \rm W/sr ## is used for a global characterization of a light source: how much it emits in all directions, and in principle it is irrelevant whether it is measured ## \rm 10~cm ## or ## \rm 5~m ## away from the source. (The source is idealized to be point-like, and the flux diminishes with the distance squared, which is compensated by integrating over a larger sphere. Right?)

It is also true that radiance ## \rm W/(m^2 sr) ## is constant along the line of sight. It has the same value where it enters the eye as it had where it left the source in a specific direction. And you can interpret the solid angle in two ways: either the apparent size (in steradians) of the receiver as seen from the source, or the apparent size of the source as seen from the receiver. Both ways of calculating the transferred energy give the same results. You can reduce the flux density at the receiver by moving it farther away (diminishing the solid angle subtended by the source), or by adding a diaphragm at the source (also effectively diminishing the solid angle).

For the purpose of estimating the potential hazards to your eyes I think it's reasonable to consider only averages over the radiating surface.
 
  • #12
WernerQH said:
I find it a bit difficult to understand your reasoning.

All of it or something in particular?

WernerQH said:
And predicting the radiant intensity at the retina is quite a different problem from estimating the flux density (## \rm mW/cm^2 ##) in the vicinity of the eye. And that's probably the quantity you need to consider when you are concerned about possible hazards to your eyes.

The Retinal hazard is part of the standard. All they use is the Radiance and the subtended angle. Which I think is a consequence of:

1. Radiance is conserved
2. Everything is relative to outside the eye

There is a limit for the Radiance at infrared wavelengths that the standard provides based on the subtended angle. That limit, for long exposure, is:

$$L_{ir} = \frac{6000}{\alpha} \frac{W}{m^2sr}$$

Note that it's slightly simplified. The actual formula in the standard is a sum over all wavelengths, but the IRED is 940nm with a 50nm spectral bandwidth.

There is also a limit for cornea which uses Irradiance:

$$ E_{ir} \leq 100 \frac{W}{m^2} $$

The calculation for Irradiance is straightforward:

$$ E_e = \frac{I_e}{d^2} $$

where ##d## is the source to eye distance. Although with my newfound knowledge of datasheet ##I_e##, I question the validity of that calculation too.

WernerQH said:
(I suppose you have already found literature specifying safety limits? And do they give values that cannot be converted to ## \rm mW/cm^2 ## ? If I haven't miscalculated, for sunlight we have about ## \rm 130~mW/cm^2 ##, and you shouldn't stare at the sun.)

Googling around I found the Radiance of the Sun at ##10 \frac{MW}{m^2sr} ## as seen from Earth, which should be ##10,000 \frac{mW}{mm^2sr}##, but the wikipedia article on the Sun claims that the power delivered to the retina is only 4mW (staring at the sun, with the iris doing it's job). I don't know how to make sense of those numbers.

The standard is based on long exposure (not more than 8 hours) and a dilated pupil (7mm). Using the standard, we can find the value of alpha that will satisfy this limit:

$$L_{ir} = \frac{6000}{\alpha} \frac{W}{m^2sr}$$

The standard (IEC 62471:2006 Photobiological safety of lamps and lamp systems) says that for subtended angles smaller than 0.011 rad, to use 0.011 rad. The subtended angle (according to AN090) is:

$$\alpha = \frac{Z}{d}$$

where ##Z## is the diameter of the die and ##d## is the minimum eye accommodation distance. The minimum eye accommodate can be 100mm or 200mm depending on where you look. The IEC 62471:2006 specifies 200mm for all lamp measurements, but AN090 uses 100mm which I believe comes from a laser safety standard. I believe that 100mm is a worse case than 200mm. For ##Z=0.5mm## and ## d=100mm##:

$$\alpha = \frac{0.5}{100} = 0.005 rad $$

This is less than the minimum ##\alpha## so the standard instructs us to use ##alpha = 0.011##. Plugging that into the limit:

$$L_{ir} = \frac{6000}{0.011} = 545 \frac{mW}{mm^2sr}$$

The AN090 application note gives an approximation that I've already explained is not applicable to 5mm LEDs. I can calculate a worser case Radiance by using the virtual die area (0.5mmx0.5mm) only and the total ##I_e##, but I don't have a lot of confidence that this is even right given the nature of how ##I_e## is calculated. Still, if we use the values for the TSAL7600:

$$L_{ir} = \frac{I_eR(\lambda)}{Z^2} = \frac{25*0.33}{0.5^2} = 33 \frac{mW}{mm^2sr} $$

Does that clear anything up? The standard sets the limits in a way which I presume could be easily measured with the right equipment. The application note AN090 tries to arrive at the limit using formulas that it has correlated to real measurements (for some LEDs).

WernerQH said:
Radiant intensity is used for a global characterization of a light source: how much it emits in all directions, and in principle it is irrelevant whether it is measured or away from the source. (The source is idealized to be point-like, and the flux diminishes with the distance squared, which is compensated by integrating over a larger sphere. Right?)

Right. But as I've shown, the LED beam pattern is not point-like. This is why I can't calculate the Radiant Power from the Radiant Intensity or vice versa using a ## cos^n(\theta)## beam pattern approximation.

What I'm questioning is not how to do the math, I'm questioning if the math is valid. In particular the math coming from that application note (AN090) that rely on assumptions that are not explicitly stated.
 
  • #13
InTheWorks said:
There is a limit for the Radiance at infrared wavelengths that the standard provides based on the subtended angle. That limit, for long exposure, is:

$$L_{ir} = \frac{6000}{\alpha} \frac{W}{m^2sr}$$

Note that it's slightly simplified. The actual formula in the standard is a sum over all wavelengths, but the IRED is 940nm with a 50nm spectral bandwidth.
Oh, obviously you have done a lot more research on this than I have. I thought that irradiance was the most important quantity, but it seems necessary to also consider the maxima of radiance to prevent "burning holes" into the retina. All this depends very much on the wavelengths and direction of gaze.

InTheWorks said:
Googling around I found the Radiance of the Sun at ##10 \frac{MW}{m^2sr} ## as seen from Earth, which should be ##10,000 \frac{mW}{mm^2sr}##, but the wikipedia article on the Sun claims that the power delivered to the retina is only 4mW (staring at the sun, with the iris doing it's job). I don't know how to make sense of those numbers.
The radiance is the same near the surface of the sun as it is here on Earth, only the solid angle is smaller. With these numbers I am more familiar: with a solid angle of ## \rm 68 \mu rad ## and a solar constant of ## \rm 1361~W/m^2 ## I get a radiance of about ## \rm 20 {MW \over m^2 sr} = 20 {W \over mm^2 sr} ##. "Google's numbers" are obviously rounded, but the number ## \rm 4~mW ## seems to be in the right ballpark (with a pupil size of a few square millimeters).

InTheWorks said:
What I'm questioning is not how to do the math, I'm questioning if the math is valid. In particular the math coming from that application note (AN090) that rely on assumptions that are not explicitly stated.
Sorry, from your original post I had the impression that it was mainly about math. But I can't really say anything about physiology and safety standards.
 
  • #14
WernerQH said:
Oh, obviously you have done a lot more research on this than I have. I thought that irradiance was the most important quantity, but it seems necessary to also consider the maxima of radiance to prevent "burning holes" into the retina. All this depends very much on the wavelengths and direction of gaze.

Right.

WernerQH said:
The radiance is the same near the surface of the sun as it is here on Earth, only the solid angle is smaller. With these numbers I am more familiar: with a solid angle of ## \rm 68 \mu rad ## and a solar constant of ## \rm 1361~W/m^2 ## I get a radiance of about ## \rm 20 {MW \over m^2 sr} = 20 {W \over mm^2 sr} ##. "Google's numbers" are obviously rounded, but the number ## \rm 4~mW ## seems to be in the right ballpark (with a pupil size of a few square millimeters).

Either way, the Radiant Intensity of the Sun is far higher that the standard allows for an "8 hour work day". As I understand it, these limits come from the data collected from eye injuries (ie. cataracts) sustained by glass blowers and iron works who are subjected to quite a bit of infrared radiation. At least the exposure limits for cornea which is ##100\frac{W}{m^2}##. The Sun is an order of magnitude higher than that.

WernerQH said:
Sorry, from your original post I had the impression that it was mainly about math. But I can't really say anything about physiology and safety standards.

The question still is about physics (radiometry) and math. Thanks to your help, I understand the problem better now. However, I still don't know what to do about it.

We know that the Radiant Intensity and Radiant Distribution don't add up. That was the source of my problem. I couldn't confirm that the AN090 math worked for ##5mm## LEDs. I know now that it's not appropriate. I've been picking away at this problem for months.

Vishay doesn't provide a means to do the calculations, but they provide a table of data saying that their LEDs are safe. They don't list the LED I've chosen to model with, but they do list others. The TSAL6400 has a maximum intensity of ##125\frac{mW}{sr}## and they list that same number in the eye safety document, but they don't compute the Radiance anywhere. The eye safety document also has this interesting note:

In case of IR Emitters the dominating limit is the cornea/lens risk in the wavelength range from ##780 nm## to ##3000 nm##. This limits the irradiance to## E_e = 100 \frac{W}{m^2}## which is expressed as intensity a value of ##I_e = 4 \frac{W}{sr}## with the measurement condition of that standard with ##0.2 m## distance in mind (##I_e = E_e r^2## ).

Evaluating the other limiting conditions as the thermal retinal risk and blue light hazard result in not limiting higher values for wavelengths ## \lambda \gt 850 nm## and therefore are not to be taken into account. Only for ##\lambda = 830 nm## a little reduction to ##I_e = 3.77 \frac{W}{sr}## is given by the thermal risk.

I'm not sure how they can make such a statement given the standard's requirements. How does ##I_e = 4 \frac{W}{sr}## map to Radiance? Is it ##\lt 0.54 \frac{W}{m^2/sr}## that the standard calculates?

At this point I think I need to estimate a worst case Radiance from the LED parameters that I do know. Maybe you can help fill in the gaps?

I know that the ##I_e## from the datasheet is measured using a ##0.01sr## angle. It's a ##1cm^2## detector placed ##100mm## away. This is not quite the ##0.011 sr## that the standard requires, but it's pretty close. The problem is that at 200mm away, the required detector for a ##0.01sr## angle would be smaller and so the sample of the beam pattern (that I don't know) would be different. Based on the presence of a bond dot, I would expect the measured Radiant Intensity at 200mm to be smaller, but could it (theoretically) be larger?

Without measuring (because I'm not equipped to) is there any way (theoretically) to put bounds on the difference? I need to account for the difference in a beam pattern that I don't know, but can't (due to physics) be worse than something with a viewing angle (half power) of ##60^\circ##. Is this possible and, if so, what is that something?
 
  • #15
InTheWorks said:
At this point I think I need to estimate a worst case Radiance from the LED parameters that I do know. Maybe you can help fill in the gaps?
Sorry, I don't feel confident to "fill in the gaps".
 
  • #16
I contacted Vishay support and they were quite helpful in understanding the problem.

The key pieces of information from my correspondence with Vishay:

* the Radiant Intensity is measured on all parts in final test
* the Radiant Intensity tolerances include positioning tolerance and off axis emission too
* the Radiant Power is measured on a single part in an integrating sphere
* the Directivity is usually also measured on only a few parts
* they do not have a tolerance for Radiant Power or Directivity
* there's no guarantee that the same parts are used in any measurement
* the datasheet values always reflect the values of the most typical parts
* the TSAL7600 datasheet is from 1999

It was also pointed out that since the total tolerance on the Radiant Intensity is 15 to 75 mW/sr, or +/-250%, the difference calculated in this thread for Radiant Intensity, 17%, shouldn't be alarming.

Using the same approach on VSMY2943SLX01 by manually picking points from the Directivity curve (it's an ugly shape) every 5 degrees gave an answer 5% away from the average. Had this been the case with TSAL7600, I would have never asked the question.

The lesson here is that my starting assumption that ##typical \equiv average## was not correct.

There's no way to separate out the tolerance on Radiant Power or Directivity. As such there's no way to determine the worst case due to beam pattern in an optimal way and the most sensible thing to do is use the typical Directivity with the max Radiant Intensity times 2. At least that's how I plan to approach it.
 
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  • #17
That's quite a spread: +/-250%. It's obviously not the precision you can hope for in other areas of science. But I'm glad that at last you have crude estimates for the necessary safety margins.
 
  • #18
InTheWorks said:
As such there's no way to determine the worst case due to beam pattern in an optimal way and the most sensible thing to do is use the typical Directivity with the max Radiant Intensity times 2
Yes I usually prodeed using the fact that the beam shape is similar for each "package" type and angular width rating. So any 3mm thruhole LED rated at 30deg width will be similar to any other and search up a polar graph (on fact sheets wherever I can). Best you can do IMHO unless you talk to the supplier/vendor. Of course they are happier to talk when your order is for 10^5 units.
 

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