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Can batteries burn out a laser?

OpTic Emu

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I got a new laser recently and it was working really well. The battery it came with died(a 18650) and I charged it but now the laser won't work at all. It hasn't been dropped or anything the laser itself should be fine. Can the battery burn out the laser diode or internal wiring? I'm not really sure how this stuff works yet.
 

Alaskan

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Or you forgot which direction it goes into the laser, some laser pointers allow the battery to go in backwards without damage, others fry right away.
 

Alaskan

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I don't think a laser pointer should be made which can fry if the batteries are put in backwards, a simple diode could protect against reverse polarity, but of course then you loose some of the battery voltage and power due to the added loss. Some designs, if the driver isn't a boost driver, might not run at the reduced voltage if only one battery is used. So, there are reasons some don't use that protection but then I blame it on the choice of constant current driver being unable to protect against reverse voltage, a reverse voltage protection diode could be incorporated into the driver itself.
 

ElectricPlasma

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I don't think a laser pointer should be made which can fry if the batteries are put in backwards, a simple diode could protect against reverse polarity, but of course then you loose some of the battery voltage and power due to the added loss.
I'm glad I'm not the only one that's thought about this. :eek: I've always thought of just putting a diode on one of my drivers to protect against reverse polarity, I don't think it eats up much power (<10mA ?) but it will drop the voltage a bit. I think of that as an advantage for unprotected batteries though, because then the voltage needs to be higher for the laser to run at full power so you won't let the battery discharge so far as to damage the battery.
 

Alaskan

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The diode will pass what ever current the driver is pulling from it, and the voltage drop/loss through the diode at that current (different diodes have different voltage drop curves) will consume power. The power consumed is the voltage drop multiplied by the current, so if you use a common silicon diode with a .7 volts of drop, multiplied by 1 amp, you are consuming an extra 700mw of power, so it is not insignificant. At 2 amps you are loosing 1.4 watts from the voltage drop through that diode which is converted into heat. You can choose lower voltage drop diodes such as a Schottky diode and reduce the loss to about half of what a silicone diode produces. If I were to solder a diode in series with the power lead to the driver I would use a diode rated for at least double the current you will be pulling through it and use a Schottky diode of some kind too. If you could clamp the diode to some kind of heat sink, that would help to keep the diode cool at higher current levels.

Yea, not having that diode in there is an advantage, just got to be sure you watch which way you put the battery in or get a driver which is designed so it doesn't care if the voltage is reversed.
 

ElectricPlasma

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I had no idea that diodes have that kinda multiplier, looks like I got some homework to do :p Thanks for the info!
 

Alaskan

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I worked with a top tier technician once who thought the voltage drop of a diode did not consume power too, sure surprised me he thought that and he was a full time technician with 25+ years of experience. I guess some tech's just don't get into component level work enough.
 




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