So, I decided I wanted to make my very first foray into some basic electronics. I decided I was going to attempt to make a laser mag-lite, as per this hack here. The steps seem pretty straight forward, so I figured it would make a good first project.

I wound up getting (not thinking about possible differences) the laser diode mount for the DVD reader in the XBox 360. I'm now in post-mortem mode. I extracted the diode with no issue, and through some trial and error I figured out the correct 2 pins (out of four on this diode). Once I got those extended and plugged into the mag-lite, it glowed red for a few seconds, then I started to smell some ozoney-smelling "burning" and then it went out, and now it is apparently dead, and will not light again.

So I'm looking for confirmation/input why it failed, as I have my theory. My working theory is that since this was a "reader" diode as opposed to a writer, it has a much lower energy usage... and so the 3V of the mag-lite batteries were way too much for the poor little diode, burning it out. Is this a likely theory?

If so, would the speed of the burner (16x, etc.) being used make a difference? I don't know if the laser is any different based on the speed or if that just indicates differences in the speed of the spindle and the lens actuators in the laser assembly.

As one related question, in project directions for this kind of project, most of them talk about "finishing" the end project with a laser control circuit. What exactly does the LCC do?

P.S. If someone with the rep could maybe create and tag this Q with the following tags: dvd, mag-lite, post-mortem I'd appreciate it.

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    \$\begingroup\$ It's probably for the better you destroyed it, because those lasers are strong enough to seriously damage your eyes. \$\endgroup\$ – starblue Apr 7 '11 at 8:06
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    \$\begingroup\$ You'll... shoot your eye out? \$\endgroup\$ – Mark Apr 7 '11 at 20:20
  • \$\begingroup\$ @Mark; LOL! Good Ol' "Christmas Story"! :) . Yeah, the LiveLeak link makes that pretty clear at top, for anyone who didn't realize. Great reference though! B) \$\endgroup\$ – eidylon Apr 7 '11 at 20:31
  • \$\begingroup\$ Sorry to have to tell you, the 'LCC' is what prevents the laser diode from burning out, by managing the voltage on the diode so that it draws a constant current. \$\endgroup\$ – JustJeff Apr 8 '11 at 0:03
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    \$\begingroup\$ With DVD lasers, it's not just about personal eye safety, either. Of course you wouldn't intentionally point it at anyone else, but you have to worry very seriously about stray reflections finding their way to people, too. \$\endgroup\$ – JustJeff Apr 8 '11 at 11:15

Laser diodes are like LEDs in that the more current you put through them, the hotter they get, and the hotter they get, the more current will go through them, which makes them get hotter... and then fzzt.

So yes, as BarsMonster says, without anything to regulate the current (in the simplest case, a current limiting resistor) the laser diode will die immediately.


In the simplest (acceptable) case you need stabilised current source - for example built from LM317/1117. For a diode from DVD-RW, you can start from 100ma. If it's from reader - not more than 3-5ma. Yes, reader diodes are 100x weaker.

Without any regulation any diode would die immidiately.


From what I've read, most lasers have a light sensor built into them and require it for reliable operation. There is a very fine line between putting in too little current and getting no output, or putting in too much and destroying the laser. This amount of current required will vary depending upon temperature, phase of the moon, etc. and so the only way to consistently find it is to vary the current so the laser is right at the threshold.

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    \$\begingroup\$ That is probably what the other terminals are connected to... a photodiode to provide feedback to the controller. \$\endgroup\$ – RBerteig Apr 8 '11 at 1:21

This may be way off the mark because a laser diode I-V characteristic may be very different from the ordinary p-n junction diode, but if a regular semiconductor diode is forward biased with nothing to limit the current (such as a resistor) it's generally going to act like a short circuit. The current is going to become very large, and therefore the power dissipation is going to be very high, and the diode will probably burn out.

If you have connected 3.3V 3V across the diode's terminals with no series resistance, then the scenario described above may be what has happened...


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