Welcome to Laser Pointer Forums - discuss green laser pointers, blue laser pointers, and all types of lasers

Buy Site Supporter Role (remove some ads) | LPF Donations

Links below open in new window

445nm power output question...

Benjiman46

Benjiman46

0
Joined
Jan 21, 2014
Messages
20
Points
3
Hi everyone, as you might have seen, I've just bought and built my first laser, now I want to build another one for my friend who was infatuated by my laser, the bug is spreading rapidly haha

Being a very competitive friend, he wants one that is bigger, better and more powerful than mine in every way... SO my question is;

If I take a 1.8a driver, copper driver pill, copper extended heatsink and C6 from Survival Laser and team it with one of DTRs 445nm M140 diodes with G2 lens... What power output can I expect, and is there a formula to work out an average?

My laser is approx 1w +/- 200mw, and is built with the M140 mentioned above, copper heatsink, C6 and a 1.25a driver with two 3v CR123a Energizer non-rechargeable batteries...

Also, having 2 batteries in the laser like this, does it mean my power output is 6 volts?

AND if I changed to 3.7v rechargeable batteries, would it make any difference to power output?

Okay so it was more like 6 questions, whatever lol

Any help will be appreciated :)

Ben
 





The thing with laser diodes ... is that their output can not be guaranteed.

At 1.8A the m140 with G2 lens can do well over 2W but it can also do much less than that. Sometimes you get the inefficient diodes ... and it sucks but ... That how it is :)

Try to get better 3.7V batteries. AW Batteries are recommended.

:beer:
 
Last edited:
Those 3v batteries aren't doing you much good now. I would definatley get some IMR 3.7 v. Right now your input power is 6v, not output. I have a pen at 1.25a and a g2 that's putting out 1.6w. I also have another laser at 1.8a with a g2 that's putting out about the same. The pens diode is efficient, and the other lasers not so much. So it's impossible to determine output without an lpm, by going off the current used.
 
Tmack said:
Those 3v batteries aren't doing you much good now. I would definatley get some IMR 3.7 v. Right now your input power is 6v, not output. I have a pen at 1.25a and a g2 that's putting out 1.6w. I also have another laser at 1.8a with a g2 that's putting out about the same. The pens diode is efficient, and the other lasers not so much. So it's impossible to determine output without an lpm, by going off the current used.
Click to expand...

Okay cool, but if I get one 18650 IMR, will that only input 3.7v?

What is the smaller version of the 18650 that you can fit two of in a C6 will this input 7.4v? Or am I being daft... lol I don't really understand series and parallel...
 
You need two 18350 or 16340. The 16340 will be slimmer and may have some play. The 18350 will be thicker and have a more snug fit and also have a larger capacity. They are half the length of the 18650 so either will fit fine.
 
Tmack said:
You need two 18350 or 16340. The 16340 will be slimmer and may have some play. The 18350 will be thicker and have a more snug fit and also have a larger capacity. They are half the length of the 18650 so either will fit fine.
Click to expand...

Okay, so just to be sure... ONE 18650 is the same as TWO 16340s? Same power output/input into the driver and everything? Meaning One 18650 at 3.7v will have the same power as two 16340s?
 
Last edited:
One 18650 is 3.7-4.2v

Two 18350 is 7.4-8.4v

Your doubling the voltage, which is what you need for your driver.
One cell would not even run your laser. And the two cell your using now probably is under powering it. So once you get two proper cells you'll really see what she can do.
 
Last edited:
Tmack said:
One 18650 is 3.7-4.2v

Two 18350 is 7.4-8.4v

Your doubling the voltage, which is what you need for your driver.
One cell would not even run your laser. And the two cell your using now probably is under powering it. So once you get two proper cells you'll really see what she can do.
Click to expand...

Clear as crystal! :D I'll get four Efest IMR 18350 3.7v and a nitecore intellicharge i4 v2... Would you recommend the charger? (Getting four batteries so I can charge two and use two)
 
Last edited:
You just picked the best cells and the best charge :D

That's exactly what most of us have including myself

Just make sure the efest aren't too big. I know some are a bit too fat.

I think the efest should be fine though .

I have a good friend with tons of c6. Ill shoot him a pm.
 
Last edited:
Tmack said:
You just picked the best cells and the best charge :D

That's exactly what most of us have including myself

Just make sure the efest aren't too big. I know some are a bit too fat.

I think the efest should be fine though .

I have a good friend with tons of c6. Ill shoot him a pm.
Click to expand...

Awesome :) all ordered :D
 
Ive been checking threads for you and aw 18350 fit . And as far I know aw and efest are the same size, so your good to go. I just got loads of efest cells and they've been some f the best I have .
 
Last edited:
I use Efest 18350 cells for all my C6 builds. It fits just perfectly. It sticks out of the battery tube by about 1mm but the tailcap will fully close without much effort.
 
To answer the original question, you can expect anywhere from 1.8 - 2.3 watts... (on average, this figure does not include rare cases of efficiency/inefficiency)

:beer:
-Matt
 
OR you could use a 9mm diode at 2.3A and get close to 3W :thinking: If this is a pissing match, why not shoot for the moon?
 
You must log in or register to reply here.


Back
Top