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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.

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.

$$ \Phi = 35 mW $$

$$ I_e = 25 mW/sr $$

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

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

is a good approximation for the relative radiant intensity.

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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