Hello to all makers and to the bustling community of Instructable.
This time Merenel Research will bring to you a pure research problem and a way to solve it with math.
I had this problem myself while I was calculating the LED fluxes of a RGB LED lamp I built (and which I will teach how to build). After extensively looking online I didn't find an answer, so here I post the solution.
Very often in physics we have to deal with curves that have the shape of the Gaussian distribution. Yes! It is the bell shaped curve used to calculate probability and was brought to us from the great mathematician Gauss.
Gauss curve is extensively used in real life physic applications, especially when we have to deal with radiation propagated from a source or received from a receiver, for example:
- the emission of the power of a radio signal (e.g the Wi-Fi);
- the luminous flux emitted from a LED;
- the reading of a photodiode.
In the manufacturer datasheet we are often given the actual value of the area of the Gaussian, which would be the total radiant power or luminous flux in a certain portion of spectrum (e.g. of a LED), but it becomes difficult to calculate the actual radiation emitted at the peak of the curve or even more difficult to know the overlapping radiation of two close sources, for example if we are illuminating with more than a LED (e.g. Blue and Green).
In this Instructable paper I will explain you how to approximate the Gaussian with a curve way more easy to grasp: a parabola. I will answer to the question: how many Gaussian curves are in a Parabola?
SPOILER → THE ANSWER IS:
The Gaussian area is always 1 unit.
The area of the corresponding parabola with the same base and height is 2.13 times bigger than the relative Gaussian area (see the picture for the graphical demonstration).
So a Gaussian is 46.94% of its parabola and this relationship is always true.
This two numbers are related in this way 0.46948=1/2.13, this is the strict mathematical relation between a Gaussian curve and its parabola and vice versa.
In this guide I will lead you to discover this step by step.
The only instrument we will need is Geogebra.org, a great online mathematical tool for drawing charts.
The Geogebra chart I made to compare a parabola to a Gaussian can be found at this link.
This instructable is long because is about a demonstration, but if you have to quickly sort out the same problem I had with LED luminous fluxes, or other phenomenon with overlapping Gaussians curves, please just jump at the spreadsheet that you will find attached at the step 5 of this guide, which will make your life easier and automatically make all the calculations for you.
I hope you like applied math because this instructable is about it.
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Add a Teacher Note to share how you incorporated it into your lesson.
Step 1: Understanding Light Emitted From a Monochromatic LED
In this analysis I will consider a series of colored LED, as you clearly see from their spectrum chart (first picture) their spectral power distribution really looks like a Gaussian which converges into the x axis at -33 and +33nm of the mean (manufacturers usually gives this spec). However, consider that the representation of this chart normalizes all the spectra on a single power unit, but LEDs have different power according to how efficiently are manufactured and how much electrical current (mA) you feed into them.
As you can see sometimes the luminous flux of two LED overlaps on the spectrum. Let's say that I easily want to calculate the overlapping area of those curves, because in that area there will be the double amount of power and I want to know how much power in tems of lumen (lm) we have there, well that is not an easy task we'll try to answer in this guide. The problem arose because when I was building the experimental lamp I really wanted to know how much the Blue and Green spectrum were overlapping.
We will focus only on monochromatic LEDs which are those that emits at a narrow portion of the spectrum. In the chart: ROYAL BLUE, BLUE, GREEN, ORANGE-RED, RED. (The actual lamp I build is RGB)
Let's rewind a bit and do a little bit of physics explanation at first.
Every LED has a color, or more scientifically we would say that has a wavelength (λ) that determines it and which is measured in nanometers (nm) and λ=1/f, where f is the frequency of oscillation of the photon.
So what we call RED is basically a (great) bunch of photons that oscillate at 630nm, those photons hit the matter and bounce in our eyes, which act as receptors, and then your brain process the color of the object as RED; or the photons could go directly into your eyes and you would see the LED that emits them glowing in RED color.
It was discovered that what we call light is actually just a small portion of the Electromagnetic Spectrum, between 380nm and 740nm; so light is an electromagnetic wave. What is curious about that portion of the spectrum is that it is precisely the chunk of the spectrum that more easily passes through water. Guess what? Our ancient ancestors from the Primordial Soup where actually in water, and it is in water where the first, more complex, live beings began to develop eyes. I suggest you to watch the video by Kurzgesagt I've attached to better understand what is light.
To sum up a LED emits light, which is a certain quantity of radiometric power (mW) at a certain wavelength (nm).
Usually, when we are dealing with visible light we don't speak about radiometric power (mW) but of luminous flux (lm), which is an unit of measure weighed at the response to visible light of humans' eyes, it derives form the candela unit of measure, and it is measured in lumen (lm). In this presentation we will consider the lumens emitted form LEDs but everything will apply to mW exactly to the same extent.
In any LED datasheet the manufacturer will give you these bits of information:
For example from this datasheet attached you see that if you power both led with 100mA you have that:
BLUE is at 480nm and has 11lm of luminous flux;
GREEN is at 530nm and has 35lm of luminous flux.
This means that the Gaussian Curve of Blue will be taller, it will spike up more, without modifying in its width and it will oscillate around the portion delimited by the blue line. In this paper I will explain how to calculate the height of the Gaussian that expresses the full peak power emitted by the LED, not only the power emitted in that portion of spectrum, unfortunately that value will be lower. Furthermore, I will try to approximate the overlapping part of the two LEDs to understand how much luminous flux is overlapped when we are dealing with LEDs that are "neighbors" in the spectrum.
Measuring the flux of LEDs is a very complex matter, if you are eager to know more I have uploaded a detailed paper by Osram that explains how things are done.
Step 2: Introduction to the Parabola
I will not go into much details about what is a parabola as it is extensively studied at school.
An equation of a parabola can be written in the following form:
ARCHIMEDES HELPS US
What I would like to underline is an important geometrical theorem by Archimedes. What the theorem says is that the area of a parabola limited in a rectangle is equal to the 2/3 of the rectangle area. In the first picture with the parabola you can see that the blue area is 2/3 and the pink areas are 1/3 of the area of the rectangle.
We can calculate the parabola and its equation knowing three points of the parabola. In our case we will calculate the vertex and we know the intersections with the x axis.
BLUE LED Vertex(480, ?) the Y of the vertex is equal to the luminous power emitted at the peak wavelength. To calculate it we will use the relation that exists between the area of a Gaussian (actual flux emitted by the LED) and the one of a parabola and we will use Archimedes theorem to know the height of the rectangle that contains that parabola.
Looking at picture that I have uploaded you can see a complex model to represent with parabolas several different LED luminous fluxes, but we know that their representation is not exactly like that as it resemble more of a Gaussian.
However, with parabolas, using math formulas we can find all the intersections point of several parabolas and calculate the intersecting areas.
In step 5 I have attached a spreadsheet in which I have put all the formulas to calculate all the parabolas and their intersecting areas of the monochromatic LEDs.
Usually, the base of the Gaussian of a LED is large 66nm, so if we know the dominant wavelength and we approximate the LED radiation with a parabola we know that the relative parabola will intersect the x axis in λ+33 and λ-33.
This is a model that approximate an LED total emitted light with parabola. But we know that if we want to be precise it is not exactly right, we would need to use a Gauss curves, which brings us to the next step.
Step 3: Introduction to the Gaussian Curve
A Gaussian it is a curve that will sound more complex than a parabola. It was invented by Gauss to interpret errors. In fact, this curve it is very useful to see the probabilistic distribution of a phenomenon. As far as we move towards left or right from the mean we have a certain phenomenon less frequent and as you can see from the last picture this curve is a very good approximation of real life occurrences.
The Gaussian formula is the scary one that you see as second picture.
The Gaussian properties are:
- it is symmetrical respect to the mean;
- x = μ not only coincide with the arithmetic mean but also with the median and mode;
- it is asymptotic at the x axis on every side;
- it decreases for x<μ and increases for x>μ;
- it has two inflection points in x = μ-σ;
- the area under the curve is 1 unit (being the probability that any x would verify)
σ is the standard deviation, the bigger the number the broader the Gaussian base is (first picture). If a value is in the 3σ portion we would know that it really moves away from the mean and there is less probability for it to happen.
In our case, with LEDs, we know the area of the Gaussian which is the luminous flux given in the manufacturer datasheet at a given wavelength peak (which is the mean).
Step 4: Demonstration With Geogebra
Firstly you have to create a couple of variable, clicking on the slider command:
The standard deviation σ=0.1 (the standard deviation defines how broad the Gauss curve is, I put a small value because I wanted to make it narrow to simulate an LED spectral power distribution)
The mean is 0 so the Gaussian is built on the y axis, where it is easier to work.
Click on the small wave function to activate the function section; there by clicking on fx you can insert the Gaussian formula and you will see popping up on the screen a nice tall Gaussian Curve.
Graphically you will see where the curve converges on the x axis, in my case in X1(-0.4;0) and X2(+0.4;0) and where the vertex is in V(0;4).
With this three point you have enough info to find the equation of the parabola. If you don't want to make calculation by hand feel free to use this website or the spreadsheet in the next step.
Use the function command (fx) to fill in the parabola function you have just found:
Now we have to understand how many Gaussians are in a parabola.
You will have to use the function command and to insert the command Integral (or Integrale in my case, as I was using the Italian version). The definite integral is the mathematical operation that allows us to calculate the area of a function defined between to x values. If you don't remember what a definite integral is, read here.
a=Integral(f, -0.4, +0.4)
This Geogebra formula will solve the defined integral between -0.4 and +0.4 of the function f, the Gaussian. As we are dealing with a Gaussian its area is 1.
Do the same for the parabola and you will discover the magic number 2.13. Which is the key number to do all the luminous flux conversions with LEDs.
Step 5: Real Life Example With LEDs: Calculating the Flux Peak and the Overlapping Fluxes
LUMINOUS FLUX AT THE PEAK
To calculate the actual height of the stirred Gaussian curves of the LED flux distribution, now that we have discovered the conversion factor 2.19, is very easy.
BLUE LED has 11lm of luminous flux
- we convert this flux from Gaussian to parabolic 11 x 2.19 = 24.09
- we use Archimedes Theorem to calculate the relative rectangle area that contains the parabola 24.09 x 3/2 = 36.14
- we find the height of that rectangle dividing for the base of the Gaussian for the BLUE LED, given in the datasheet or seen on the datasheet chart, usually around 66nm, and that is our power at the peak of 480nm: 36.14 / 66= 0.55
OVERLAPPING LUMINOUS FLUX AREAS
To calculate two overlapping radiation I will explain with an example with the following two LEDs:
BLUE is at 480nm and has 11lm of luminous flux
GREEN is at 530nm and has 35lm of luminous flux
We know and we see from the chart that both Gaussian curves converge in -33nm and +33nm, consequently we know that:
- BLUE intersects the x axis in 447nm and 531nm
- GREEN intersects the x axis in 497nm and 563nm
We clearly see that the two curves intersects as one end of the first one is after the beginning of the other (531nm>497nm) so the light of these two LEDs overlap in some points.
We firstly have to calculate the parabola equation for both. The attached spreadsheet is there to help you with calculations, and has embedded the formulas to solve the system of equations to determine the two parabolas knowing x axis intersecting points and the vertex:
BLUE parabola: y = -0.0004889636025x^2 + 0.4694050584x -112.1247327
GREEN parabola: y = -0.001555793281x^2 + 1.680256743x - 451.9750618
in both cases a>0 and, so the parabola is correctly pointing upside-down.
To prove that this parabolas are right just fill in a, b, c in the vertex calculator at this parabola calculator website.
On the spreadsheet all the calculus are already made to find the intersection points between the parabolas and to calculate the definite integral to obtain the intersecting areas of those parabolas.
I our case the intersecting areas of blue and green LED spectra is 0.4247.
Once we have the intersecting parabolas we can multiply this newly founded intersecting area for the Gaussian multiplier 0.4694 and find a very close approximation of how much power the LEDs are emitting together in total in that section of the spectrum. To find the single LED flux emitted in that section just divide by 2.
Step 6: The Study of the Monochromatic LEDs of the Experimental Lamp Is Now Complete!
Well, thank you very much for reading this research. I hope it will be useful for you to deeply understand how light is emitted from a lamp.
I was studying the fluxes of the LEDs of a special lamp made with three types of monochromatic LEDs.
The "ingredients" to make this lamp are:
- 3 LED BLU
- 4 LED GREEN
- 3 LED RED
- 3 resistors to limit the current in the LED circuit branches
- Embossed Acrylic Cover
- OSRAM OT BLE DIM control (Bluetooth LED control unit)
- Aluminum heatsink
- M5 bolds and nuts and L brackets
Control everything with the Casambi APP from your smartphone, you can switch on and dim each LED channel separately.
To build the lamp is very simple:
- attach the LED to the heatsink with double-sided tape;
- solder all the BLU LED in series with a resistor, and do the same with the other color for each branch of the circuit. According to the LEDs you will choose (I used Lumileds LED) you will have to choose the resistor size in relation to how much current you will feed into the LED and to the total voltage given by the power supply of 12V. If you don't know how to do this, I suggest you to read this great instructable about how to determine the size of a resistor to limit the current of a series of LEDs.
- connect the wires to each channel of the Osram OT BLE: the all the main positive of the branches of the LEDs goes to the common (+) and the three negatives of the branches goes respectively to -B (blue) -G (green) -R (red).
- Wire the power supply to the input of the Osram OT BLE.
Now what is cool about the Osram OT BLE is that you can create scenarios and program the LED channels, as you can see in the first part of the video I am dimming the three channels and in the second part of the video I am using some pre-made light scenarios.
I have extensively used math to deeply understand how the fluxes of this lamps would propagate.
I really hope that you have learnt something useful today and I will make my best to bring to instructable more cases of profound applied research like this one.
Research is the key!
Participated in the
Made with Math Contest