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Homework Help: Optimum mounting angle for three LEDs in a 240V LED light bulb

  1. Nov 6, 2014 #1
    1. The problem statement, all variables and given/known data
    The asker needs to know the angles to mount three LEDs inside a traditional mains powered light bulb shaped enclosure "to get even radiation from the LEDs", and also the position of the covering plastic sphere in relation to the LEDs. (I was thinking that the position of the cover would have little effect on the light distribution.)

    It seems obvious that the LEDs should be evenly spaced around the inside of the globe, as viewed from the top of the globe. The question therefore is reduced to what angle away from the axis of the bulb should the three LEDs be facing? I am imagining that if the LEDs were placed at 90 degrees from the axis (i.e. so they are all pointing out sidweays), there will be a darker spot in the middle. At some point, as the angles of the LEDS are shifted towards the top of the bulb, the darker spot will disappear, and this point is the optimum angle. Is that correct to assume?

    How would this point best be calculated? And is that point actually the optimum solution or would there be any reason to make the central point even brighter , and the edges dimmer, by moving the LEDs angles to aim closer together towards the axis of the bulb?

    2. Relevant equations
    A graph was given of the light intensity against angle for the LEDs. It looks almost exactly like a cos function, so it can be assumed to be modelled as one (except that it goes to zero beyond +/- 90 degrees, rather than into the negative.

    3. The attempt at a solution
    I havent done any 3D trig or calculus for a long time, esp in spherical coordinates, though I'm familiar with high-school level one dimensional calculus. I plotted two of the LEDs along one line (i.e. a 2D simplfication) using microsoft mathematics, with the LEDs at different angluar distances from each other, and then plotted the sum of the two graphs. This was the first thing I could think of doing. The only problem is that the cos or sin functions go into the negative, and this affects the graph of the functions summed. I'm looking at whether microsoft mathematics is capable of handling a custom function like this...

    Next I considered the problem in 3D and I had to look up some stuff on how to do spherical co-ordinates. I could try to plot the functions using Matlab, which I used to know, so it should come back pretty quickly, hopefully..

    So my two ideas of how to solve , so far, are

    1. use Matlab to 3D spherical plot the light distribution and look at the graph to see at what angle does the central dark spot disapper, and

    2. use 3D calculus to work it out analytically. from stationary points or something in 3D
  2. jcsd
  3. Nov 6, 2014 #2


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    You haven't mentioned the beam pattern of the LEDs themselves. If they are round or square there is no way to achieve perfectly uniform illumination - you can't stick squares or circles of paper onto a ball without them overlapping and/or leaving gaps.
  4. Nov 6, 2014 #3
    Thanks, sorry, I didn't describe the beam pattern very well. The beam of each LED falls off from the centreline in a pattern that looks very close to the graph of cos x, between -90 and +90 degrees, if that makes sense.

    The pattern is given as a graph drawn onto a product data sheet, however I think it's probably close enough to cos x to adequately model it as that. For e.g., at + and - 30 degrees, its about 0.86 (and cos 30 degrees is 0.866). At + and - 60 degrees, its about 0.47 (and cos 60 is 0.5). At 15 degrees its 0.97 and cos 15 is 0.97, at 75 degrees its 0.2 and cos 75 is 0.26, at 45 degrees its 0.7 and cos 45 is 0.707.

    The major difference between the LED beam pattern and an actual cos graph being that, the beam is 0 outside + and - 90 degrees, and the graph of cos x goes into the minus, so when adding the intensity from more than one LED it's necessary to modify the cos function, such as by using a piecewise definition so that its cos x for -90 < x < 90 and 0 everywhere else.

    Is that what you meant?
  5. Nov 6, 2014 #4


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    My bad. I didn't mean the graph of intensity vs angle, I meant the physical shape of the LED and the beam it projects (eg circular, square etc).

    A circular LED located at the centre of a sphere will project a circular spot of light on the surface of the sphere. The edges of the spot will be fuzzy or soft because the beam does not have sharp edges, instead the brightness tapers off at the edges as per the "cos" curve that you describe.

    You can arrange for two such spots to overlap so a point midway between the two has the right brightness but you can't cover a sphere in circular spots without them either overlapping (brighter areas) or leaving gaps (dark areas) somewhere.

    Actually now that I think about it you CAN do it if you only use two LEDs pointing in opposite directions, one for each hemisphere. They would need to have a beam angle of at least 180 degrees which could be a problem. You can't do it with three or more LEDs, however you might approximate uniform illumination using a large number of small circular LEDs and a diffuser?
  6. Nov 6, 2014 #5
    I made a graph of a 1D simplification, with two LEDs, using Matlab. (Microsoft Mathematics was unable to cope with a piecewise function for the intensity vs angle). This was the program in Matlab

    function Led02
    x = -200:0.1:250;
    separation = 120
    x5 = ledbeam(x,0) + ledbeam(x, separation)
    x6 = ledbeam(x,0)
    x7 = ledbeam(x, separation)
    axis([-200 250 -1.2 2])
    grid on

    function y = ledbeam(t,offset)
    y = cosd (t - offset) .* ((t < (90 + offset)) & (t > (-90 + offset)));

    This gave what I considered an optimum distribution with an angle of 120 degrees between the LEDs. (I dont know how to attach an image to this post without already having it on the web). However the piecewise function did not smoothly blend the "joins" in the graph where the two LEDs beams overlap. This led me to conclude the problem would be too hard to do analytically, and I'm focusing on trying to graph it in 3D in Matlab now.

    I've also found this rule for calculating the angle between any two points on a sphere, in spherical coordiantes, which is something like what I'll need to plot it in 3D , to calculate the intensity at various positions on the sphere based on their angular distance from any one LED loccated at a particular point on the sphere (and summing this, for all three)

    cos(psi) = sin(th1)*sin(th2)*cos(ph1-ph2) + cos(th1)*cos(th2).
    where th = theta, ph = phi in standard spherical coordinates.
  7. Nov 6, 2014 #6


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    What you need are LEDs with a beam roughly this sort of triangular shape. Four would illuminate a hemisphere, eight for a sphere.

  8. Nov 6, 2014 #7
    Thanks, CWatters, regarding the overlap, I was thinking of how when you add the graphs of two trig functions (considering only the peaks), if the separation between them is close enough, it appears to have one maximum in the middle, but , if the peaks are further away, there is a clear dip in the middle. For one pair of LEDs with the beam pattern as given (that falls off like cos x, to zero at 90 degrees away from the centreline) , it looks like the optimum separation is 120 degrees.

    If I think about this for long enough I should be able to calcuate/ plot it in Matlab, though its pretty complicated (for me)... like maybe 1-2 full days at the rate I'm going? (its really hard to tell how long these kinds of things will take...)
  9. Nov 6, 2014 #8


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    I think you are getting ahead of yourself. Clearly it might be possible (with the right beam pattern) to get uniform brightness along the line between the centre of two circular spots (the 1D case) but you can't extend that using circular LEDs to the 2D or 3D case.
  10. Nov 6, 2014 #9
    Thanks again, I was assuming that I had to use the LEDs as given in the data sheet, however this may (perhaps) not be the case. The ones in the data sheet have a wider beam than that, falling to 0 over a 180 degree field (+ and - 90 from the centre)
  11. Nov 6, 2014 #10
    I wasnt thinking completely uniform brightness, just the "optimum", Ill try and upload my 1D graph to the web so i can post it (is there a way to post images without having to do that?)
  12. Nov 6, 2014 #11
    This is the 1D pattern for two LEDs 120 degrees apart, in the middle this looked flatter than any other angle (just by trial and error in the matlab program above).. Im hoping to be able to do this in 3D with three LEDs next...

  13. Nov 6, 2014 #12
    Though you have a point about how this may not be so optimum in other locations, away from the midline between the two LEDS. I hadnt thought of that.

    However, doing it in full 3D would eliminate that problem (by showing the final brightness at all angles in the 3D graph)
  14. Nov 6, 2014 #13


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    With your 3 LEDs in space, you'll first need a definition of "optimal".

    The center will always be a maximum of the light intensity if my calculation is correct.
  15. Nov 6, 2014 #14
    Thanks mfb, I was thinking that what consitutes "optimal" may become more clear as I progressed... Which is kind of vague I know. eg in the 1D graph above that seemed optimal, since the central maximum would be either higher or lower than the other two peaks for any other separation. I thought something like that may happen with the 3D case which would then be considered the optimal. Or if that doesnt happen, there would be a range of realistically possible answers with a trade-off of some sort, like making the centre region more evenly lit vs having a greater range of illumination all round.

    How did you do your calculation? Is that true even for , say, the three LEDs spaced evenly 120 degrees apart on the "equator" , pointing out at right angles to the axis of the whole bulb unit, which means both the N and S poles would be equal, would they both be maxima?
  16. Nov 6, 2014 #15
    I've pretty much done it, in Matlab, I'll post some pics later. I'll tune it a bit more first...
  17. Nov 7, 2014 #16
    Coming later...
    Last edited: Nov 7, 2014
  18. Nov 7, 2014 #17
  19. Nov 7, 2014 #18


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    That appears to be a result of the "1/2 brightness beam angle" being around 120 degrees eg..

    Cos(theta) = 0.5
    theta = 60 degrees

    For the 3 LED case perhaps try the "1/3rd brightness beam angle" around 140 degrees...

    Cos(theta) = 0.3333333
    theta = 70

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