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I have a irradiance reading from an LED which does not seem to make sense. I was hoping someone could shed alittle light on the issue.

The LED is question has a relative radiant intensity which follows a cosine law. I therefore assume the LED can be considered a Lambertian emitter.

As such, for a Lambertian emitter radiance is constant $ L = \phi/(A\cdot\pi)$

The total radiant flux is $\phi$ = 142mW.

The radius of the LED is 3.2mm. So the radiance (L) from the LED can be calculated as: $ L = 0.142/(0.0016^2 \cdot \pi^2) = 5541 W/m^2 \cdot sr$

A light detector at 200mm with a 5.64mm radius detecting area should give an irradiance (E) of: $(5541 \cdot 0.00564^2 \cdot \pi)/0.2^2 = 13.8 W/m^2$

My new light meter shows the irradiance as $0.61\,W/m^2$.

This is 22 times lower than expected.

For clarity the detector is normal to the LED although moving the meter off axis does not change the reading. The 142mW is radiant flux given in the LED data sheet.

Also please note all ambient light has been removed.

Any thoughts?

LED data sheet Note the LED is being driven at 83mA hence the lower radiant flux.

M

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  • $\begingroup$ In addition to my answer, to make it more clear that your calculation cannot possibly be right, consider multiplying your answer of 13.8W/m^2 by just one steradian. You get .2^2*13.8=0.522W in that one steradian - about four times what was supposed to be the total power of the LED. $\endgroup$
    – AXensen
    Mar 30, 2023 at 16:28
  • $\begingroup$ In your second equation, I think you should multiply $L$ wirh the solid angle under which the source is seen by the detector. You multiplied with the solid angle of the detector. Put there the radius of the led, not the detector. The detector's area is only important to continue then from irradiane (W per m2) to received power (W). $\endgroup$ Mar 30, 2023 at 18:29
  • $\begingroup$ Thankyou for the comments. I obviously am missing something fundament here. Another way to calculate the irradiance at the meter is from the radiant intensity, which the data sheet shows as a cos theta curve. Peak radiant intensity is radiant flux/(4 x pi). Radiant intensity is W/sr. So Q/(4 x pi) / d^2 = 0.142/(4 x pi x 0.2^2) = 0.282W/m^2 $\endgroup$ Mar 31, 2023 at 9:41
  • $\begingroup$ Since you are mentioning “LED radius,” matching the 1/4" common packaging, I'm suspecting this is a dome-shaped TH package with axial leads, is it? Like this guy? Then you have a convex lens which focuses the flux along the package axis. If that's the case, ×22 sounds reasonable. Measure it on the circle in the sagittal plane with some reasonably small step, and integrate, accounting for spherical co-ordinate transform, i.e. assuming axial symmetry. If you need a better precision, you're up to a full sphere calibration. $\endgroup$ Mar 31, 2023 at 11:34
  • $\begingroup$ You've just added a link to the datasheet. Look at the Fig. 8! $\endgroup$ Mar 31, 2023 at 11:41

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I'm not entirely following the logic of your equations. I don't think the size of the LED should come into play at all. The irradiance equation you provided is probably valid just above the surface of the LED, but the LED should be treated as a Lambertian point source as long as the distance (200mm) is big compared to the size of the LED (3.2mm). For example, for a spherically symmetric emitter, the irradiance at a distance $r$ would be $142\text{ mW}/(4\pi r^2)$. For a Lambertian emitter, I'll use the same concept but ensure we also have a $\cos\theta$ distribution:

$$ 142\text{ mW}=L_0(r)r^2\int_0^{\pi/2}2\pi\cos(\theta)\sin(\theta)d\theta=L_0(r)r^2\pi $$ $$ L_0(r)=\frac{142\text{ mW}}{\pi r^2} $$ So I would say your irradiance is $$ L(r,\theta)=\frac{142\text{ mW}}{\pi r^2}\cos\theta $$ Assuming you meant your detector was straight above the LED ($\theta=0$), we get a power at 200mm of $1.13\text{ W/m}^2$. Which is only a factor of two off. And from there I'd question what you get this 142mW from. Is it the power $P=IV$ dissipated in the LED? Because not all of that goes to light. Is your detector at $\theta=0$ and is it perpendicular to the light source?

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  • $\begingroup$ I think what you are referring to as radiance is actually irradiance. But the result should be correct. $\endgroup$
    – Puk
    Mar 30, 2023 at 16:39
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    $\begingroup$ I'm sure you're right, since I only learned the name of this quantity just now from reading the question. Thanks for that, and for the critical edit you made. $\endgroup$
    – AXensen
    Mar 30, 2023 at 16:44
  • $\begingroup$ Yes the meter is normal to the LED, $\theta$ is zero. Note the meter value does not change if I move it off axis but maintain 200mm distance. $\endgroup$ Mar 31, 2023 at 9:50
  • $\begingroup$ Does it have a piece of curved plastic/glass above it to spread out the light so it's more isotropic and less Lambertian? That would reduce the peak power at least. $\endgroup$
    – AXensen
    Mar 31, 2023 at 10:06
  • $\begingroup$ I have added the data sheet for the LED in question. Yes it does have a dome but the relative intensity shows a cos theta curve. $\endgroup$ Mar 31, 2023 at 11:14
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I wonder if the LED is acting as a point source as suggested.

The irradiance reading on the light meter from the LED would be the radiant flux divided by the area of the hemisphere. $\phi = 0.142mW, r=200mm$

$$\frac{\phi}{2\cdot\pi\cdot r^2} = 0.565\frac{W}{m^2}$$

This would also explain why the meter reading does not change when moved away from the normal.

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Mar 31, 2023 at 14:46

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