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18650 + m140 + X -drive 1.8 = No?

L

Lazer god

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Okay, I understand that the setup in the title will not work or be weak, and I have read a alot about it not working at all.

Can anyone tell me what I will experience, I am making an extender and will be done soon.
 





Pi R Squared

Pi R Squared

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It would work fine except that you need 2 batteries, two 16340 will fit in the space of one 18650.

Alan
 
dubious speculation

dubious speculation

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With a fully charged 18650 you will have reduced output, you need 2 li-ions as Pi said.

Take a look at DTR's test thread. According to the results an m140 at 1W has an operating potential difference of 4.2v.
However, in a not so perfect and inefficient reality you will most likely see something closer to 400mW with a single Li-ion.
 
jander6442

jander6442

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The X-drive is a buck driver you would need a higher voltage than what the diode will use so the driver could buck the voltage. You want to get a X-boost driver for that build unless you do as PI stated, however you'll have reduced runtimes... could get quality cells to help out I guess.
 
D

Davidx

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Hi Lazer God ( love the name). I'm completely new at building. Still at the solder a kit together stage.
But one of my kits used a M-140, driver set to 1.8. I use two 16340 batteries, and it really kicks ass. Over 2 watts using G2 lens on my LPM. Sometimes I switch to a 3 element lens just to change the divergence to something cleaner. Get a needle like beam too.

Congratulations. You're going to love it
 
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L

Lazer god

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Okay great, I picked up some 123As from the store, I will use those until I make my extender or until they go dead.
 
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