IR LED Safety: Subtend Angle and Retinal Exposure Limit

Question

The relevant formulas from OSRAM AN090 "Eye safety of IREDs used in lamp applications" with slight modifications :

$$ Z = (l + w)/2 $$ $$ \alpha = Z/d $$ $$ L_{ir} \approx \frac{I_e R(\lambda)}{Z^2} $$ $$ L_{ir} < 6000/\alpha \hspace{0.3cm} W/m^2/sr $$ $$ or $$ $$ L_{ir} < 6/\alpha \hspace{0.3cm} mW/mm^2/sr $$

And a table for the subtense limits

Time range	\$\alpha_{min,eff},\gamma_{FOV,min}\$	\$\alpha_{max,eff},\gamma_{FOV,max}\$
t \$\le\$ 0.25 s	0.0017 rad	0.1 rad
0.25 s \$\lt\$ t \$\lt\$ 10 s	\$0.0017\sqrt{\frac{t}{0.25}} \$ rad	0.1 rad
t \$\ge\$ 10 s	0.011 rad	0.1 rad

Consider three ficticious LEDs at a distance of 20cm. All have the same \$I_e\$, but with different die sizes: Z=0.3m, Z=3mm, Z=30mm.

Subtend angles:

$$ \alpha_1 = 0.3/200 = 0.0015 $$ $$ \alpha_2 = 3/200 = 0.015 $$ $$ \alpha_3 = 30/200 = 0.15 $$

As I understand it, applying the table for t > 10s:

$$ \alpha_{1,eff} = 0.011 $$ $$ \alpha_{2,eff} = 0.015 $$ $$ \alpha_{3,eff} = 0.1 $$

If \$I_e\$ and \$R(\lambda)\$ are constant then the radiance:

$$ L_{ir} \propto \frac{1}{Z^2} $$

And that makes sense to me because the same "power" is spread out over a larger area.

If we then compute the \$L_{ir}\$ limit for the different subtend angles:

$$ L_{ir,1} < \frac{6}{\alpha_{1,eff}} = \frac{6}{0.011} = 545 ~ mW/mm^2/sr$$ $$ L_{ir,2} < \frac{6}{\alpha_{2,eff}} = \frac{6}{0.011} = 400 ~ mW/mm^2/sr$$ $$ L_{ir,3} < \frac{6}{\alpha_{3,eff}} = \frac{6}{0.1} = 60 ~ mW/mm^2/sr$$

As the subtend angle gets bigger, the image becomes larger on the retina. However, the limit allows more power (flux?) with smaller images.

This seems counter intuitive since \$L_{ir}\$ feels like a measure of "density". What's really going on?

Specifically, since \$I_e\$ is the same in all cases, the larger die will project a "less dense" image on the retina so I would think the allowed limit would be higher to account for distributing the heating over more area on the retina. But that's clearly not how it works.

There is another related question concerning arrays of IR LEDs. In this answer, the minimum subtend angle, \$\alpha_{min,eff}\$, is taken to be the minimum spacing between array elements. Is this correct? I haven't seen this documented anywhere.

The total heating effect of the array should be similar to the "less dense" image from a larger die. I would think that large die vs array of small dies should have similar limits when properly accounted for. However, an array of small dies will have a limit of \$ 545 ~ mW/mm^2/sr \$ for each element and really large die will have a limit of only \$ 60 ~ mW/mm^2/sr \$. It obviously doesn't work this way, but if both of those numbers were multiplied by the same amount of steradians, the resulting flux will be higher for the array LED. What am I missing?

I'm trying to understand if common CCTV IR LED rings, using 5mm LEDs, are safe for long exposures at shorter distances. Or if I need to build my own array with large die IR LEDs appropriately spaced. The application is motion capture facing at the IR LEDs.

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Update

I created this drawing to hopefully show my confusion a little better. First some assumptions.

1. the subtend angle per IEC 62471 is with respect to the outside of the eye. The angle would be different inside the eye.

2. area is not important in and of itself.

The drawing is not to scale and depicts a single emitting surface of \$ 0.1 ~ sr \$ in red. I gave it a radiance of \$ 60 ~ mW/mm^2/sr \$. There is a green section meant to respresent a subsampling of \$ 0.01 ~ sr \$ of the same emitter surface.

The irradiance is probably not important, but I'd like to know if it works the way I've indicated.

One could also interpret the drawing to show a 'large' array where the subsample is a smaller die within an array.

This is the crux of the problem for me. A small subtend angle, say from a single die, has a higher radiance limit. Building a 'large' array like this would result in a radiance far higher than allowed for the larger subtend angle.

This obviously can't be so what have I got wrong?

Answer 1

I'll give this a shot.

As the subtend angle gets bigger, the image becomes larger on the retina. However, the limit allows more power (flux?) with smaller images. This seems counter intuitive since Lir feels like a measure of "density". What's really going on?

Radiance is power per angle per area, so it is normalized for the angular extent of the source. Thus a \$ 60 ~ mW/mm^2/sr \$ source emits 10x more energy over 0.01 srad as it does over 0.001 srad. So if everything scaled directly with total energy emitted, your \$ 60 ~ mW/mm^2/sr \$ at \$\alpha_{eff}\$ of 0.1 would become \$ 5000 ~ mW/mm^2/sr \$ at \$\alpha_{eff}\$ of 0.011, but its only ~ 1/10 th that since you do make a much bigger spot on the retina.

There is another related question concerning arrays of IR LEDs. In this answer, the minimum subtend angle, αmin,eff, is taken to be the minimum spacing between array elements. Is this correct? I haven't seen this documented anywhere.

I read that as calculating the angular spacing between diodes to see if they were going to overlap on the retina or not.

The total heating effect of the array should be similar to the "less dense" image from a larger die. I would think that large die vs array of small dies should have similar limits when properly accounted for. However, an array of small dies will have a limit of 545 mW/mm2/sr for each element and really large die will have a limit of only 60 mW/mm2/sr. It obviously doesn't work this way, but if both of those numbers were multiplied by the same amount of steradians, the resulting flux will be higher for the array LED. What am I missing?

If the array is far enough away that you cannot see the spaces between the diodes, then it looks like a single very large source with averaged radiance over the diodes and empty space (so less radiance). If its close enough to see the spaces, then each diode makes a non-overlapping image on the retina, and so all heating at each point is from a single LED only (or at least the rest are negligible). So the small diodes in a sparse array are potentially more dangerous than larger diodes in a less sparse array, at least in terms of power per area per solid angle.

I'm trying to understand if common CCTV IR LED rings, using 5mm LEDs, are safe for long exposures at shorter distances. Or if I need to build my own array with large die IR LEDs appropriately spaced. The application is motion capture facing at the IR LEDs.

I didn't look up any similar LEDs or the calculations for an array, but if you mean through hole LEDs, those are putting out maybe 10mW of optical due to thermal constraint, possibly much less. At NIR wavelengths you can put ~ 1mW of collimated laser light onto the retina for hours at a time, so I expect you will calculate that almost any conceivable ring illuminator is far below safe limits if made with through hole LEDs. That might not be the case if done with higher power SMD diodes though, since those have heatsinking and can be driven much harder.

Answer 2

In case this helps anyone else, the problem stems from the approximation for Radiance in the app note:

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

Here they are using \$ I_e \$ from the apparent source and \$ Z^2 \$ from the emitter size. This approximation is at least true for their example.

Using the emitter size rather than the apparent size does give a more worse case result (smaller alpha) for a single IRED, but this is why there is a discrepancy between an array of small IREDs vs one large die.

If the approximation holds for any IRED and not just planar IREDs, then it's probably okay to use the apparent size to calculate \$ \alpha \$ and the emitter size to calculate \$ I_e \$. I don't know this as fact.