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FrozenGate by Avery

What kills lasers

As the Voltage increases across a component so does the current .

So say I have a resistor at 30Ohms and I put in 2V

Using V=IR

I=V/R
I=2/30
I=0.066666666666666666666666666666667

So if I increase the voltage to 9V

I get a current of 0.3A.

So if you want the current to increase, you either need to change the resistance or the input voltage . So no matter if you change voltage or current, one of the other variables must change .

-Adam
 





iewed said:
Heat would cause the crystals to fail.

But to fully answer your question, energy kills lasers.
Click to expand...

Ok, this is the correct answer. Well at least partially.

Energy kills lasers, the lack of energy will make a laser not operate (a "dead" battery) Kill-dead similar? :-/
Energy needs to be constantly regulated in both volts, and amps, to little or too much of either will effectively "kill" a laser, of coarse too much will literally kill it to where even when the correct volts & amps are provided will no longer work, when too little just won't let it lase until the correct energy is provided. Or energy provided in the wrong conditions will kill the laser, meaning if you turn it on in a static field, or under water.

Of coarse, you also have efficiency to consider, because you can tune a laser to be efficient for power conservation, or performance as well. I guess you could say, "quality of life", or how some lasers operate better then others.


“There are a million ways to die. Choose one.”
 
Still that doesn't proove current is proportional to voltage.A resistor's resistance cannot change(unless it's damaged) , but a diode is not a resistor, so in our case (that is, powering laser diodes) you can keep the voltage stable and feed it more current(?) I think, I'm right, but then again I don't know exactly how the different drivers we use work and how they behave. :-/


Turning it on under water would probably cause a short circuit.This would be the equivalent of "energy applied in the wrong place".Like reversing polarity or bypassing a component on the driver and delivering more energy to it.
 
ok i think i have gleaned the answer i was looking for..

when a energy is used to "pump" a crystal the energy is converted to photons. if the crystals mass or characteristics are not sufficient to convert all the energy, the energy becomes heat (like the IR in a dpss or the electricity in a diode).

as this heat is sinked away the system continues to lase, but if the system is unable to remove the heat fast enough, POP goes the laser. Also if too much energy is used, the crystal will not have a chance to sink the heat away even if it could (instantaneous catastrophic failure).

so to add power you must remove heat, but that will only help you so much.

Once a crystal reaches the point where it generates enough heat to destroy itself in less time than it can thermally transfer it (no matter how well you cool it), it has reached its own "critical mass" and ceases to be.

thnx..
 
IN DPSS its photons pumping crystals that get converted to photons of a different wavelength . Not just "Energy".

-Adam
 
Once a crystal reaches the point where it generates enough heat to destroy itself in less time than it can thermally transfer it (no matter how well you cool it), it has reached its own "critical mass" and ceases to be.
Click to expand...

So many wrong expresions used in that sentence. :-/ The crystal doesn't generate heat. It also doesn't just cease to be. And "to thermally transfer heat" sounds redundant.

Anyway, I guess you pretty much got it. :P
 
keeperx said:
ok i think i have gleaned the answer i was looking for..

when a energy is used to "pump" a crystal the energy is converted to photons. if the crystals mass or characteristics are not sufficient to convert all the energy, the energy becomes heat (like the IR in a dpss or the electricity in a diode).

as this heat is sinked away the system continues to lase, but if the system is unable to remove the heat fast enough, POP goes the laser. Also if too much energy is used, the crystal will not have a chance to sink the heat away even if it could (instantaneous  catastrophic failure).

so to add power you must remove heat, but that will only help you so much.

Once a crystal reaches the point where it generates enough heat to destroy itself in less time than it can thermally transfer it (no matter how well you cool it), it has reached its own "critical mass" and ceases to be.

thnx..
Click to expand...

Overheating your laser is 1 of a million ways to kill it. My RPL has an automatic shut-off protection when the unit gets too hot, but 1/2 the heat is created from the battery discharging so rapidly, so my laser shouldn’t die because of this. How you said "Pop goes the laser", is not actually a really effective way of killing your laser. Overheating your laser in attempts to destroy the diode or crystals in a laser with a good heatsink will be challenging, mostly will probably mode shift or drain the battery or crack the lens, unless the laser is a piece of crap and the diode might blow, providing that you run no duty cycle. It will be a little easier without a heatsink, but you will most likely dampen the performance rather then kill the laser, unless you really try. You will most likely shorten the overall life span of the laser then kill it. There are much more effective ways of killing your laser.
 
Overheating your laser is 1 of a million ways to kill it. My RPL has an automatic shut-off protection when the unit gets too hot, but 1/2 the heat is created from the battery discharging so rapidly, so my laser shouldn[ch8217]t die because of this. How you said "Pop goes the laser", is not actually a really effective way of killing your laser. Overheating your laser in attempts to destroy the diode or crystals in a laser with a good heatsink will be challenging, mostly will probably mode shift or drain the battery or crack the lens, unless the laser is a piece of crap and the diode might blow, providing that you run no duty cycle. It will be a little easier without a heatsink, but you will most likely dampen the performance rather then kill the laser, unless you really try. You will most likely shorten the overall life span of the laser then kill it. There are much more effective ways of killing your laser.
Click to expand...

Yea but I mean, a current overdose is still the result of heat, I mean it just heats up so fast that the diode can doesn't even have time to heat up before current excites particles in the dye so much it causes imperfections. :-/

I mean, if you shove something like 20 times the working current into your RPL diode the thermal sensor won't even pick it up before it's too late.So in this case it's still heat that kills it, but it doesn't matter about how big the heatsink is or how many TECs there are.It's just too fast and too much to dissipate that fast.
 
great.. thanks for the input, but im not interested in killing a laser.. im interested in the concepts involved so I can be sure to build efficient and stable systems and to simply have the knowledge there so it may be converted to wisdom someday.
 
Are you talking about DPSS lasers or just laser diodes? Very important distinction there.
 
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