0
\$\begingroup\$

I would appreciate any help. The question may seem similar to the question asked at this link: Help me calculate spectral irradiance of an LED from a LED spec sheet , but the answer was never given or posted.

So here is my questions:

I am building a device that illuminates the human eye from a distance of approximately 120mm, with an array of 40 IR LEDs for around 7 seconds. The purpose of this is beyond the scope of the question.

In order to have theoretical evidence that the IR LED's will not cause any hazardous harm to the eye, the total Infrared Radiation exposure should satisfy the limits of the IEC62471 standard. I have posted an image/snippet from the standard of the equation that I need to use below:

enter image description here

My problem is that I do not know how to calculate the spectral irradiance of the IR LED from its datasheet.

Here is the datasheet: https://4donline.ihs.com/images/VipMasterIC/IC/OSOS/OSOS-S-A0002483803/OSOS-S-A0002483803-1.pdf?hkey=52A5661711E402568146F3353EA87419


The parameters of my setup are the following (Some values retrieved from the datasheet):

IR LED wavelength: 950 nm

IR Centroid wavelength: 940 nm

Number of LEDs: 40

Shape of array: A square of 30 x 30 mm that is positioned in between both eyes (at a distance of 120 mm)

Typical Radiant Intensity per LED: 11 mW/sr

Spectral bandwidth at 50% (∆λ): 42 nm (I am unsure if this is what is needed in the equation)

Distance from eye: approximately 120 mm

Duration of Illumination 7 seconds

Half angle: 60°

Total radiant flux: 35 mW


The units for spectral irradiance is W⋅m^-2⋅nm^-1. Therefore, if I am correct, I can get the W/m² by saying:

  • 11 mW/sr*40(Num of IR LEDS) = 440 mW/sr
  • W/sr/m² = 0.44/sr/(0.12²) = 30.55 W/m² = 30.55 W.m^-2

However, that's where I get stuck. Therefore, I would like to know the following:

  1. Am I on the right path?
  2. If so, how do I convert W/m² to W/m²/nm (W.m^-2.nm^-1) ?

Thank you for the help

\$\endgroup\$
2
  • \$\begingroup\$ 1) Consider the average radiated power flux for 120 deg arc surface vs 1 steradian. 2) Don't need the nm ∆λ just mW/m2 at distance with beam angle. Reducing Lambertian Beam angle of 160 deg by 2 roughly doubles the Luminous intensity except for lens loss in plastic so a 10 deg LED could be just as bright at 1/16th the current. \$\endgroup\$ – Tony Stewart EE75 Jul 8 '20 at 23:38
  • \$\begingroup\$ Hi Tony Thank you for your response. I do apologize, however, I am struggling to understand what you are saying in this comment. - Do you mean my calculation of 30.55 W/m² must just be ≤ 18000(7^-0.75)? - If so then my LED's produce 30.55 W/m² ≤ 4182.62 W/m² which should be safe for the human eye at that distance? The rest of your comment has me kind of lost. I will need to research this to understand what you are saying. \$\endgroup\$ – Scott Bruton Jul 9 '20 at 7:01
1
\$\begingroup\$

Let me give you some perhaps useless or useful background info, but you are on the right track. I presume you chose this design using 40 wide-angle LEDs to increase the intensity over a very wide area much bigger than the head when at this range you could have used fewer brighter 5mm LEDs to achieve the same intensity at low current or much more intensity and still be under the safety limit.

enter image description here

The steradian is a useful term for computing the flux at a distance to normalize the radiant power into a standard lens cone angle regardless of the lens used or not. When no lens is used, ie. a flat emitter, the beam pattern is said to be a Lambertian response of 160 deg which is defined by the -3dB point or half power.

The plane aperture angle of 2θ ≈ 1.144 rad or 65.54° for 1 steradian which is smaller than your IR LED. Thus the 35 mW of IR power is reduced to 11 mW/sr.

The total beam width for optical LEDs are given as 2θ while IR LED's often narrow beam in a 5mm Lens are much brighter but if it was 6 deg, it could be off centre by as much as 1 deg so they centre the peak power and specify the half-angle, θ instead.

\$\endgroup\$
3
  • \$\begingroup\$ Thank you for your knowledge. I really appreciate it. I would supply you with an answer to why the designer picked 40 IR LED's, however, I have taken over the project, previous engineer has left, therefore, I cannot answer this : ) Regarding my question. So, according to the standard my array of IR LEDs are safe for illuminating the eye? I have seen there is another standard "IEC 60825-1 SAFETY OF LASER PRODUCTS - PART 1: EQUIPMENT CLASSIFICATION AND REQUIREMENTS". In your opinion, would my IR LED array be considered a laser? \$\endgroup\$ – Scott Bruton Jul 10 '20 at 10:22
  • \$\begingroup\$ does not apply.. But you would learn more by describing purpose . ...iris scanner? \$\endgroup\$ – Tony Stewart EE75 Jul 10 '20 at 10:24
  • \$\begingroup\$ Similar to that. I am using the device to measure the diameter change of the pupil during a papillary light reflex. So the IR LEDs illuminate the eye, a white LED shines for 0.25 seconds, and the pupil constricts and re-dilation. I use a IR camera to record and analyse the pupil change. So, would you agree with the following: According to the standard: My exposure limit is: 18000.7^-0.75 = 4182.63 W/m² And the 40 IR LEDs that I have produce: 40 * 0.011W/sr/0.120² = 30.55 W/m² (at a distance of 120mm) Therefore, this is well below the limit and therefore safe? \$\endgroup\$ – Scott Bruton Jul 10 '20 at 13:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.