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New member
Hello, I have been recently trying to build a burning laser, so I bought a laser module and a 200mW red laser diode from aixiz (it did not mention the voltage that it should be run on). I used the module mainly for the lens, and then hooked the diode to two rechargable eneloop AAA batteries (don't worry, this has nothing to do with asking help for the kipkay flashlight hack project).

You say (I've read it somewhere in the introductory posts) that a little more voltage that what the diode is made for could kill it. I've also read somewhere that heat can also be just as lethal. I just have these simple questions to ask to help me understand a few things about how things operate!

1) How can I know what voltage the diode is made for (the optimal voltage that will output a lot of power without the risk of killing it)? I have a multimeter if that helps.

2) If I apply more voltage than the specified, wouldn't that give me more power output (to some extent, not for ever of course)? What's a typical graph of voltage-power for a common laser diode?

3) Is the damage from excess voltage to the diode immediately fatal (ie, the diode gets killed right away, so there's no chance of it shining again), or could it still be producing bright light even if it can't reach its maximum capacity? I'm asking this because I'm getting the feeling (from what I read) that any mistake will kill the diode right away (as described above), but I've hooked my 200mW diode to 2 AAA batteries, for a total voltage of less than 3Volts, then to 2 AA alkalines for a total voltage of about 3Volts, and then to 3 AAA batteries for a total voltage of about 3.5Volts, and it's still shining pretty bright (able to light matches). So how is this possible?

4) I've read somewhere on these forums that diodes are current dependent and that common light bulbs are voltage dependent. I know that the voltage-current graph for a laser diode is not linear, but what do you mean when saying that it's current dependent (*anything* is current dependent, if you think about it)? You either need to specify the current OR the voltage, you shouldn't say something like "This diode operates at 3 Volts with 250mA of current" because either of the two parameters is already fixed when the other is provided. If 250mA of current flow through a diode when applying 3 Volts to its terminals, you can never get any other value for the current when the same amount of voltage (3Volts) is applied.

5)What's the difference between a 200mW laser diode and, say, a 5mW laser diode? Couldn't I force the 5mW one (with high voltages, and extreme cooling) to output just as much power as the other one?

Thanks for your help!

rog8811

New member
Too many questions at once :

I will work on the following ones

1) How can I know what voltage the diode is made for (the optimal voltage that will output a lot of power without the risk of killing it)? I have a multimeter if that helps.

2) If I apply more voltage than the specified, wouldn't that give me more power output (to some extent, not for ever of course)? What's a typical graph of voltage-power for a common laser diode?
Do not think "voltage" with LD's, think "current"

Build a LM317 driver with a pot in the circuit,
http://www.laserpointerforums.com/forums/YaBB.pl?num=1185701612
supply it with 7.2v minimum and after testing it to see if it works ok
http://www.laserpointerforums.com/forums/YaBB.pl?num=1197651171
put your LD and  a 1ohm resistor into the circuit  measure the mv across the resistor...the reading you get is = to the ma being supplied to the LD....the voltage to the LD is what it is....

Find out what your LD is (make model etc) use the search on the forum making sure you change the time from 1 week to 1 year....all the info you need is there....

You will then know how many ma you can feed it without it going poof

Regards rog8811

FireMyLaser

Well-known member

It's the heat that kills the diode. The heat will allow more (to much!) current to flow though it.

The rule is that a LD/LED's brightness is controlled by the current and a light bulb is controlled by the voltage.

Hope that helps.

New right? Welcome to the forums.

Zom-B

New member
5)

No, because the microscopic output facet mirrors of the 5mW are not made for a massive amount of photons like 200mW. The sheer bombardment will cause something they've made a term for: Catastrophic Optical Damage (COD). Think about it, the facet is a few tens of nanometers, and the light emitting from it can burn things in the scale of millimeters!

Things

New member
3) When you hooked your diode to the batteries, you are lucky that they have some internal resistance, or you wouldnt have that diode anymore, it would be a LED. Typically the power output when a diode dies, drops DRAMATICALLY, and the dot usually goes a weird shape. If your lasers dot has "lines" in it, than thats a sign it has Catastrophic Optical Damage (COD). Once a diode is dying, there is no way you wan get it back to normal, it is dead. (Some Blu-rays may still work though).

4) Common lightbulbs ARE NOT voltage dependent! If you sent 22kv through a regular lightbulb, it would not light up one bit. If you sent say 5v at 3A through the light bulb, then chances are it will light up. Diode's are basically direct shorts, so if you let it have as much current as it wants, it will take all that current, but kill itself in the proccess. LED's are different in the sense that yo can give them as much current as you want, a long as you stay within the voltage limit they will be fine, even if you fed them 200A! Yes making lasers would be one heck of alot easier if you could just hook em to batteries and let it run.