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For a range of LEDs I'd like to model their performance and predict the flux density across an area using manufacturers datasheets. Eventually I'd like to take various power measurements and calculate efficiency when it either isn't provided or isn't fully trusted.

Looking at a specific LED such as the Luminus SST-20-B and the angular pattern of radiant power is provided. I've read values from this graph into a pair of arrays (using more samples than provided below), fitted it to a poly curve and then represented it as a polynomial.

import numpy as np
import matplotlib.pyplot as plt

a = [0, 10, 20, 30, 40, 50, 60, 70, 80, 90]
i = [1., 0.985, 0.949, 0.878, 0.788, 0.661, 0.507, 0.349, 0.19, 0.043]
curve = np.polyfit(a, i, 4)
poly = np.poly1d(curve)

I plotted this polynomial against the datasheet values when deciding to use a 4th degree fit. I've seen some other methods in a paper using sums of cosines and guassians however, I'm unsure how to fit them, they seemed to require more coefficients and maybe a polynomial will be fine for my purposes. Angular pattern of Radiant Power

I used this polynomial to calculate the relative power at each point across a plane by calculating the angle and distance from the LED for each point. I used the polynomial function to find the relative power at the relevant angle and then divided that by the radius squared to plot an image of relative flux density which looks pretty much like I would expect.

img = np.zeros((101,101))
height = 10.0
led_location = [50,50]
for x in range(img.shape[0]):
    for y in range(img.shape[1]):
        distance = np.linalg.norm(np.subtract(led_location, [x, y]))
        angle = np.arctan(distance/height)*180/np.pi
        img[x][y] = poly(angle)/np.hypot(height,distance)**2

plt.imshow(img)

Relative Flux density

What I'd like to be able to do now is attached some real units to this flux density pattern as the datasheet for the SST-20-B provides a radiometric power of 710mW or photon flux of 2.68 μmol/s.

It's a long time since I've done maths at this level so I'm happy to believe what I attempt next is just plain wrong.
If I integrate to find the area between 0° & 90° can I multiply that by 2π in order to find the volume of all radiated flux and then scale the relative flux by this number?

integ = poly.integ()
volume = 2*np.pi*(integ(90)-integ(0))
fig, ax = plt.subplots(figsize=(13, 10))
im = ax.imshow(img*2.68/volume)
fig.suptitle('Flux Distribution from single LED')
cbar = fig.colorbar(im, ax=ax)
cbar.set_label('μmol/s')

Scaling the distance and height so the LED is 10cm away and 1px = 1cm gives values which seem logical but I'm quite concerned this might just be a fluke. I should have labelled the scale µmol/(m2s) in the linked image

Photon Flux Density

Have I made any fundamental errors and if so what alternative methods should I be researching and looking at to correct them?

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    $\begingroup$ In terms of things to research: very likely the data you are getting from the datasheets is a plot of Intensity, or flux per unit solid angle, not "radiant power" which usually means flux. (Luminous Intensity vs Radiant Intensity just means whether the spectrum was integrated with the Photopic curve or not.) And the data you are trying to generate is Irradiance, or flux per unit area. Hopefully these terms help you research things. $\endgroup$ – Sean Skelly Oct 27 at 23:57
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    $\begingroup$ Typical units for Radiant Intensity is W/sr (watts per steradian), and for Irradiance is W/m^2 or W/cm^2. (Units are different in the Photometric world - Luminous Intensity becomes 'Candela', which is Lumens/sr, and Illuminance becomes Lumens/m^2.) $\endgroup$ – Sean Skelly Oct 27 at 23:59
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    $\begingroup$ Your general approach is OK. Specifically: irradiance = intensity * cos(theta) / r^2, where theta is the angle away from normal/perpendicular (theta is 0 for your center pixel, for example). $\endgroup$ – Sean Skelly Oct 28 at 0:22
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    $\begingroup$ Now hopefully you can see how to set your units. I agree that you should scale your intensity values by integrating them. When you integrate all your solid angles, you get 2*pi if you integrate a hemisphere (4*pi if you integrate the full sphere). Just remember that your polynomial is a single cross-section of a hemisphere. It's assumed that the LED behavior is rotationally symmetric. (These are left as comments instead of an answer, as this is more of a guide for your research.) $\endgroup$ – Sean Skelly Oct 28 at 0:35
  • $\begingroup$ Thanks for the insight. I've been working away on this and have definitely made what feels like progress. I performed a manual shell integration of the radiant distribution in the datasheet and then normalized that with the volume of a cylinder (as a hypothetical LED with perfectly even distribution would have had a relative intensity of 1 across all angles). The exact numbers are at home but I think the peak relative intensity came out about 2.3 times higher than if the light had been evenly distributed. $\endgroup$ – Jake Hawkins Oct 29 at 18:51

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