How does one calculate the spectral irradiance from an array of IR LEDs?

Question

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:

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

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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

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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

Answer 1

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.

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.