martedì 17 giugno 2014

IR LED lamp design

LEDs have many advantages over traditional lighting, in multi-spectral imaging the main advantage is that they have a narrow well-know output spectrum. Using a small source allows a better control over it, reducing Stray light.
Design a LED lamps is not easy as it seems. In radiometric experiments power consumption is not a problem but homogeneous output and stability (stable irradiance and spectral band) are fundamental parameters to deal with. Spectral output depends on temperature in my case I had $\lambda_{p} $ of 0.2 nm/K so if the temperature rise to 20 K you have a 4 nm shift. To get some more infos I asked a question on electronics.stackexchange.

Parallels or series?

The first thing you need to know is that not all the LEDs are equal, some of them are brighter some dimmer, so you should search for binned LED to start with the optimal condition from the beginning. The second problem is to find the best arrangement for the LEDs. If you google a little bit you will find a lot of video showing how to build LEDs matrix or lamps. The most common advice is to put one resistance for every LEDs, this reduce problems related to the difference between the LEDs. However this arrangement also required a large amount of power and so produce a large amount of heat so I prefer to find another solution for a prototype. You can place LEDs in series (second arrangement in the figure) this lead to a more efficient arrangement but it has some disadvantages too. We will see them in the next part.
The two antipodes full-parallel and full-series arrangements of six LEDs.

Calculating resistance value

If you have choose a LEDs and a source the only thing that you have to calculate are the resistances shown in figure. It is very easy, you should think that you are forced to distribute ALL your energy (V) to your components. There are components like LEDs that can't take too much energy so you have to add another component that take this surplus: resistor. So if you have to distribute 12 V and the LEDs can take only 9 V you have to find a resistor that take 3 V, keeping the current fixed to a value that doesn't destroy the LEDs , indicated as $I_{f} $ . $$ R= \frac{Supply~Voltage-(LED_{V} \times n_{LEDs})}{Current}$$.
Whit a DC supply of 12 V if the LED voltage drop (indicated as $ V_{f} $) is 1.35 V and the $ I_{f} $ is equal to 100 mA you have for the first arrangement :$$\frac{12-(1.35 \times1)}{0.1}=106 \Omega$$ for the second: $$\frac{12-(1.35 \times 6)}{0.1}=39 \Omega$$. It seems that the second arrangement is better but if we think that the voltage can fluctuate we can see that using a higher resistance can lead to a more stable output. This is is easy to see from the next graph that is a simple plotting of the Ohm law: $$i=\frac{V}{R}$$.
Response to voltage fluctuation of the two arrangements.
You can see that when you have a small resistor (e.g. 39 Ohm) a small voltage fluctuation cause a larger current flow fluctuation hence a larger variation in LED irradiance. Of course maybe is better to stabilize the voltage before and use the second arrangement where all the LEDs share the same current flow.

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