Using the formula: Power = E
2 / R (Voltage squared divided by Resistance equals Watts).
So a 510 atomizer from what people are saying the resistance should be will consume 3.9 watts = (3.7 volts)
2 / 3 ohms
However I just tested my atty at 1.7 ohms resistance, so my atty is consuming 8 watts of power. No wonder my battery life stinks! But the vapor is awesome!
I just tested a disposable "smoking everywhere" I just acquired at 2.45 ohms, so it's consuming 5.5 watts assuming the voltage is the same. I have no way of testing the voltage as the battery is an automatic.
I think the total power output of the laser should be 3 to 5 watts to be effective. Perhaps as little as 1 watt would work as a test...
Right now I'm not taking into consideration the the amps needed as I'm not certain as to the exact diode. The calculation for that is I = P / E (Amps = Watts divided by Voltage). This will be required for finding the correct way to supply power to the laser. An approximation is 3-5A at 1.8V.
Time for work, more later?