Instructables

Step 17: Secondary light sources - Assembling light pipes

Picture of Secondary light sources - Assembling light pipes
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I did not want the LEDs to poke straight through to the front. For one thing, 5mm is too wide! Also, the substrate was too thick for an LED to reach all the way through. Instead, obtain a few meters of clear acrylic rod. It conducts light like an optical fiber, only it's 2-3mm in diameter.

Put together a pile of as many LEDs as you'll need, a pile of 2cm-lengths of heat-shrink large enough to go over an LED, and a pile of 2.5cm-lengths of acrylic rod. (You can quite easily cut the rod with a good pair of side-cutters)

Assemble the LED, heat-shrink and acrylic rod together, then using a soldering iron or a heat gun shrink them together. Try to avoid getting the LED too hot, as it can kill them. Repeat until you have a hundred or so of these assemblies. (This is, believe it or not, another step that takes a long time.)

Note: I initially tried melting acrylic rod directly onto the LED. The LED is a different type of plastic so the rod does not 'stick' properly, the acrylic starts to darken when it is heated, and lots of noxious gases are produced. Not recommended.
 
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newpeople2 years ago
My e-mail: exchange123@yeah.net
newpeople2 years ago
Hello, I am in China, your work well, and I want one, you give me a production of documents?
hopper423 years ago
Would the smaller fibers couple well enough into the acrylic rod that you could just use a small bundle of fibers taped to the end of an acrylic rod for the "big" stars? That would remove the need for involved LEDs at all.
MrTrick (author)  hopper423 years ago
You could, or you could just shove as many fibres into the larger hole as will fit.

I didn't take that option, because I didn't know how many fibres would be needed, and I was worried that it would use up too many.
Fake_Name3 years ago
You can *significantly* improve the regulation of your LED current source by putting a small-value resistor in series with the emitter of the transistor, between it and ground. You want it to drop about ~1V at the target voltage.

This way, f you feed the base 1.6V, it will regulate the current.

Basically, if the current increases, the vdrop across the emitter resistor increases, which decreases the base-emitter voltage, and the transistor turns off more.

Symmetrically, if the current decreases, the vdrop across the emitter resistor decreases, increasing the base-emitter voltage, and the transistor turns on more.

This is called Current Feedback, and is the basis of a simple current source, which is the ideal thing to drive LEDs with.

In your example schematic, simply move the 100R resistor.

The advantage of this is you get much tighter current regulation, any thermal effects of the transistor changing bets as a function of temperature is removed, and you can predict the amount of current flow from the base voltage without having to involve the transistor's beta.

Also, you can basically stick any transistor in the circuit without having to change anything, and the brightness should stay close to the same.
MrTrick (author)  Fake_Name3 years ago
That does sound like a better regulator design in general. In this star map though, I can't just move the 100R resistor, because that one transistor is regulating 12 parallel sets of LEDs - there are 12 x 100R resistors.

Using a 12 resistors R1-12, and single resistor R13 might be more appropriate, where the value of the two resistor types adds up to the original value. (100R)
R1-12 carry only a single LED chain's worth of current, but R13 will need to be more robust, probably at least 1W-rated. I think that will work well, and as you said improve upon the original design.
Yep. In this situation, the resistors in each string are the "current-sharing" resistors.

The emitter resistor does not have to be large, because of the way the circuit works. It'll work fine with only 0.1V across the emitter resistor, provided the transistor has a decent amount of gain (beta > 150-200). Alternatively, you could use a darlington, but you'd have to change the base voltage to compensate for the larger vdrop.

That way, with 1A of current, you only get 100mW of dissipation.

With a smaller emitter resistor, you should probably change the values of the pot about a bit, though. Since the relevant range for the base voltage is ~.5v - ~1V, you want to add scaling resistors so that the pot can only reach those voltages, which will give you finer control of the brightness. A 1k resistor to ground, a 1K pot, and 8K between the pot and 5V+ would work.

Otherwise, very small adjustments in the pot result in very large changes in brightness.
Hmmm. With a darlington, you base current is very low.

Therefore, you could get clever, and put a CdS photoresistor in series with the pull-up on the pot, and make the brightness vary with ambient light.

Put a pot across the photoresistor (value probably 1-4x the photoresistor, tweak to taste) to tune the sensitivity.

That way, it's visible in the daylight, but doesn't blow your eyes out at night.
tannerb9243 years ago
Instead of wiring a hundred LEDs it seems like you could use a string of white Christmas lights. They might be too bright but it's worth looking into, especially for the circuit building-impaired.