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ArcticMyst Security by Avery

Help with a M140 diode

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Apr 26, 2013
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ok, Let´s go, i recently buyed a M140 diode from DTR and i buyed a Buck Type 1.8A driver from Survival Laser, my question is, the M140 diode can work with only one (1) 18650 battery?:thinking: because i have a C6 cree Host
Thanks!

PD: My english is regular because im Venezuelan :yh:
 





Joined
Jan 29, 2012
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Yes, but not with a buck. With a buck youll need 2 x 16340. If you want 1 x 18650 youll need a boost driver. :beer:

Edit: eg Xboost V7 or microboost.

Edit 2: Im guessing you just stuck one battery in there and it didnt work eh? Well if you have 2 x 18650 and a piece of wire you could just put the 2 batteries in series ie one after the other in the host and you can use the piece of wire to bridge the contact (ie instead of the tail cap) and check to see if its working. You can also get extension tubes for the C6 if you dont want to change the driver, they allow the use of 2 x 18650.
 
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sinner

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It is due to the Survival Laser buck driver needs atleast 6V-8.4V+ to regulate current and give out a smooth voltage/current to the diode..

Your (1)18650 is 4.2V at max charge.

This means you are not providing enough voltage to the driver so in turn, the diode doesnt lase as expected.

The Solution: Is very simple, Cheap, and $5 to buy (2) small batteries of same voltage (4.2 each)

Most of our C6 Buck builds use 16340 or 18350 batteries which are Half sized than a 18650..
 
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It is due to the Survival Laser buck driver needs atleast 6V-8.4V+ to regulate current and give out a smooth voltage/current to the diode..

Your (1)18650 is 4.2V at max charge.

This means you are not providing enough voltage to the driver so in turn, the diode doesnt lase as expected.

The Solution: Is very simple, Cheap, and $5 to buy (2) small batteries of same voltage (4.2 each)

Most of our C6 Buck builds use 16340 or 18350 batteries which are Half sized than a 18650..

ok, and its supposed if i put the battery and turn it ON the diode no work really? this do no can still damage the diode? and if i try putting the other battery (second battery) and try to turn it ON without the extend battery tube from survival laser, the diode don´t work really?
 

Blord

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One 18650 battery wouldn't hurt the M140 diode if you are using the buck driver. The output will be very low or nothing at all.
 
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One 18650 battery wouldn't hurt the M140 diode if you are using the buck driver. The output will be very low or nothing at all.

And if i try to put (x2) 18650 without the extended battery tube?
 

Blord

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The answer has been given post #2. The buck driver need two 3.7V cell to regulate.
 

norbyx

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Since you were not getting any power output, did you by any chance test the laser with the batteries in the other way arround? With the polarity iverted? If so you might have burnt the driver.
If you have a adjustable power supply you could test the diode alone and see if it works or not.
 
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Hi!

I can`t really imagine right now, how you will be able to do that without the extension tube? :can:


Cheers brunes01

As I mentioned above, with a bit of wire...;) Obviously it would only serve to test the laser and see if it works...:beer:
 
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In series means one after the other ie end on end + bat - + bat -. This increases the voltage (ie they add) so 2 x bats gives 8.4 V.
If its still not clear type "batteries in series" into google or wiki and there will be lots of images.

Just replied to your PM too.:beer:
 
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Please read these before doing anything. Your reply shows you need to learn a little first, or you will kill your diode. :beer:

If you're new to lasers and want some info on how to build a laser, please read this thread: "I want to build a laser"

Likewise, if you need info on which goggles to buy or where to get them, please read this thread: "Get some safety goggles now!"
 
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Just replied to your PM. Forget about what diode and host. I still need to know whether you have an anodised heat sink and which buck driver you bought.:beer:
 




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