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how much current ( mA) could drain my 300mw green handeld?

realista

realista

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hi.. i have a rayfoss fmk handheld at 300mw... i would know how could be the possible mA that this laser requires.

i know it has 2watt diode.... .. so it drains 2A ?
rayfoss specs.. says. 700ma but i think it is wrong.
 





0.7A * 4V = 2.8W of electrical power input. Diodes might operate at 50% efficiency with luck, but then the drivers and crystals will also waste power. According to that estimate, the pump power is about 1.4W (if you forget driver inefficiency). The diode might be rated up to 2A but it wouldn't be running at that if the current from the battery is only 700mA. It might be possible to get 300mW from 1.4W pump power with some tuning, so in theory 700mA is possible.

Whilst it is possible in theory, the only way to know for sure is to measure the current yourself. (Also, note my estimate of 4V - as the voltage drops the current would need to increase to provide the same power)
 
thanks... in fact i think 700 ma is an error.

what i asked is... GENERALLY....what's the mA required by a 300mw green laser? 1A? 1.3? 1.5?
 
It is a very difficult question to answer in general.

The pump diode will have an efficiency in the order of 1 mW/mA, but after that there is the upconversion process to green output.

Very, very, roughly it'd say you should multiply the green output by a factor of 4 to 5 to get the desired current. So if a module is rated for 100 mW green output power, its electrical input would be in the 400 to 500 mA range.

Since efficiencies vary so wildly among sets available, this is just a general pointer - and it might be off a factor of 2 in either direction depending on quality.
 
Why don't you measure it?

Conect you DMM in series with the batteries, this will be: +batt > +dmm cable > dmm > -dmm cable > laser case.

You set the DMM in current measure mode and you're done.
 
many multimeters have a 250mA current limit. Mine has a 10A current measure setting, the fuse for that is a piece of copper wire about 2 or 3mm in diameter!
 
BShanahan14rulz said:
many multimeters have a 250mA current limit. Mine has a 10A current measure setting, the fuse for that is a piece of copper wire about 2 or 3mm in diameter!
Click to expand...
Yep, I forgot to mention, you have to set the DMM to 10A or so, if you set it to the other fused setting, fuse will just blow.
And btw, that copper wire does not act like a fuse, lol
 
Well, the copper wire is a shunt, so voltage drop over that is measured.

It WILL act as a fuse at some point, provided the test leads, circuit board and plugs dont evaporate first :D
 
It's not a fuse it's the shunt ..... like when you measure the current on the resistor of a dummy load, the resistor is the "shunt" of that load ..... same principle here, that is a piece of constantane wire, with a known resistance (milliohms), and the DMM read the voltage at its sides, that is proportional to the current that pass through it.

EDIT: Benm beatened me in post (so i learn to not keep the tags opened too much time :p :D)
 
Wow! That's interesting! So the wire is a certain length to create a known resistance for it to measure voltage across. I doubt that the leads could hold up to as much current as that shunt can handle. After further research, it seems that these are usually made of constantan or manganin because they have extremely low temperature resistance coefficients, meaning big change in temp. results in almost no change in resistance. Well, neat! Thanks, LPFers.

Realista, if you didn't get to order those because they are backordered, AW sells Panasonic NCR18650's, they are true 2900mAh without having to charge to 4.3V or discharge to 2.4V. They're protected, and flat tops I think. Just search for aw 2900mah.
 
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