I know it sounds dumb but I think im missing something. If P=IV then a 3.5V battery running 600mA would put out 1785mW assuming 85% driver efficiency. But this is not the case right? So is there a diode inefficiency as well or am i missing something?
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FrozenGate by Avery
Noob question about calculating output power?
- Thread starter vivioo7
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aryntha
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Of course there's diode inefficiency, and that varies by diode. Those diodes get VERY hot, so a lot of that energy is being lost in heat. Which diode are you talking about? 445? 405 (PHR,6x,8x,12x?) 655?
Im going to be making a 405nm but I recall seeing this difference in a lot of builds for example this thread
http://laserpointerforums.com/f65/post-your-ma-mw-51432.html
I assume a lot of these are using the 3.7V 18650 which would give a diode efficiency of about 30% i think?
assuming all but the driver and diode are 100% efficient I know big stretch
http://laserpointerforums.com/f65/post-your-ma-mw-51432.html
I assume a lot of these are using the 3.7V 18650 which would give a diode efficiency of about 30% i think?
assuming all but the driver and diode are 100% efficient I know big stretch
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