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thanks,If I use a material with the same thermal conductivity as copper and fill the lampshade completely, will the diode be damaged, or the standard for judging is to see the heating condition of the diode. If the current is 3A and the voltage is 6v, will the nubm44 be damaged? I am sure It will not work for more than ten seconds.That is pushing the ability of those batteries. These diodes expend a great deal of heat in addition to the 6 watts of output power. I would get hardier batteries. That heat sink doesn't look large enough either.
I build a 7W laser some time ago for a friend in a pen host for experimental purposes it used to get so hot later the day he comes back and i switched it to a bigger host even the push button meltedThe nubm44 will not draw 6v if the driver is limited to 3A it will draw about 4.5V if the driver is limited to 3A that's 13.5W and the laser diode will put out about 4.5W of light @ 3A, so waste heat at the laser diode will be about 9W........if duty cycle is 15 seconds then you won't need much of a heat sink, but that's if duty cycle is only 15 seconds on and a couple minutes off.
That said your heat sink should make good thermal contact, the loose fitting junk from China is not going to work very well.
It's best to error on the side of too big with your heat resivor/heat sink.
thanks very helpful!!!The nubm44 will not draw 6v if the driver is limited to 3A it will draw about 4.5V if the driver is limited to 3A that's 13.5W and the laser diode will put out about 4.5W of light @ 3A, so waste heat at the laser diode will be about 9W........if duty cycle is 15 seconds then you won't need much of a heat sink, but that's if duty cycle is only 15 seconds on and a couple minutes off.
That said your heat sink should make good thermal contact, the loose fitting junk from China is not going to work very well.
It's best to error on the side of too big with your heat resivor/heat sink.
Seal with glass glue and inject heat transfer oil into itNo, that was not what I meant. Six volts at three amps can only supply 18 watts. The heat output plus the optical output is greater than that. That is what I meant. Copper is a better heat conductor than aluminum, so that would help. I'm not sure what you mean by "filling the lampshade completely".
OKThat is pushing the ability of those batteries. These diodes expend a great deal of heat in addition to the 6 watts of output power. I would get hardier batteries. That heat sink doesn't look large enough either.