Help on 808nm 300mW

[Cyparagon]

There are too many things wrong with that schematic to list. I'd start over if I were you if your laser diode isn't already dead.

 

[chipdouglas]

i also see your 9v batter as the problem assuming it is one of the rectangle ones. if it is, then you aren't getting enough current to reach 300ma. and if you are on a fresh batt. it wil drain out of regulation quickly.

 

[MA257]

i also see your 9v batter as the problem assuming it is one of the rectangle ones. if it is, then you aren't getting enough current to reach 300ma. and if you are on a fresh batt. it wil drain out of regulation quickly.

So then what so i use as a power source?

Oh, and thanks for the find i got it on watch

 

[chipdouglas]

use a battery/batteries that have at least a 600mah rating.

 

[MA257]

use a battery/batteries that have at least a 600mah rating.

isn't a 9v inbetween 490-620mAh?

 

[rhd]

If alkaline maybe, inside they're essentially just 6x AAAA in series. I think AAAAs could hit 600mAh if Alkaline. Probably more like 400mAh if NiMH.

 

[chipdouglas]

i believe the rectangualr 9v's are like less than 100mah. either way it is proven they are junk to laser hobbyists. now if you are fixed on that batt because you have the connector soldered in, then go to radio shack and buy a cartridge that that takes 2 14500's that has the 9v connector on it.

michael.

 

[rhd]

If you don't have 14500s, you could even grab a 3x AA cartridge and use 3x alkaline AAs. Even with the huge voltage drop of the LM317, you're still powering a diode with like 2.2 Vf - you'd be ok.

 

[JLSE]

either way it is proven they are junk to laser hobbyists.

Not entirely... Our LaserBee LPM's make good use of them

 

[MA257]

Thanks i'm starting to get it now but thank you.
Once i get back i'll work on it and post get back on the results.

 

[MA257]

OK, since my reading around it seems live current is the amperage of the electric output going to the laser. Therfore the current laws state that, I=(V/R).
(V=voltage), (I=current in watts), (R=resistance)
And if i measure out my battery at 8.56v and around min. 40ohms that means the max output is very low running at 200mW max but is not guaranteed.

So, Which will give me a higher output of current in watts, a higher amp battery or a higher volt battery?

 

[rhd]

There's a lot wrong with what you just said (so much, that I'm not sure I can get at your meaning, and reply properly).

"which one will get the laser at a higher mAh output."
- Lasers don't output mAhs. They output light that we generally measure in Watts, and this level depends as much on the driver as the batteries powering it.

"the max output is very low running at 200mAh"
- Output, of anything (laser, battery, driver, whatever) can't be measured in 200mAh because mAh is a CAPACITY measure, not a RATE measure. So whatever formula your were trying to use, it's not right.

"40ohms that means..."
- Where did 40 ohms come from? What's 40 ohms?

 

[MA257]

"which one will get the laser at a higher mAh output."
- Lasers don't output mAhs. They output light that we generally measure in Watts

i know i was mistaken i mean that the measure of watts going to the laser diode runs at a min of 200mW

40ohms that means...

ohms is a measure of resistance from the battery to the end result. The lowest resistance reading in ohms (Ω) i can get my driver running is 40ohms.

And lastly, amperage is a rate because it is technically distance/time (amp=cycles/seconds), while voltage is the "pressure" of the current, while watts is just the equivalent.