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Apr 21, 2017 at 23:32 vote accept Alian4life
Mar 21, 2017 at 21:58 answer added WhatRoughBeast timeline score: 1
Mar 21, 2017 at 20:37 answer added bobflux timeline score: 2
Mar 21, 2017 at 20:11 answer added StainlessSteelRat timeline score: 0
Mar 21, 2017 at 19:15 comment added Rohat Kılıç Yes, it's possible. You can use 2 or more LM338s connected in parallel. But you may need a huge heatsink for each.
Mar 21, 2017 at 18:52 comment added Alian4life Thanks for correction. So it seems like I would definitely need a converter. Can't I parallel 2*24V 1.5A regulators and some heatsink on them for that half amount of heat to take care of?
Mar 21, 2017 at 18:30 comment added Rohat Kılıç +1 for StainlessSteelRat 's answer. Just to correct your calculation: Junction-to-Ambient thermal resistance, \$R_{th(j-a)}\$, is more meaningful because it gives you directly the felt temperature of the heating element. Assuming the device has TO-220 package, total temperature rise will be \$\Delta T = (V_i-V_o) \cdot I_{LOAD} \cdot R_{th(j-a)} = 6V \cdot 3A \cdot 22.9°C/W = 412°C\$ and the felt temperature is \$T_f = \Delta T + T_{amb} = 436°C\$. Of course this is too much. Go for a DC/DC converter or, as SSR stated, decrease the input voltage.
Mar 21, 2017 at 18:15 comment added StainlessSteelRat 18W * 22.9 = 412.2 deg Celsius and maximum = 260. Junction to ambient. Add ambient temperature. Decrease voltage to 27V. 3V means 1/2 of load, so it will work. Load dissipation is not part of calculation. Personally, I'd go with a DC/DC convertor.
Mar 21, 2017 at 18:12 history edited Alian4life CC BY-SA 3.0
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Mar 21, 2017 at 18:09 comment added Alian4life It was TO220 but power dissipation was so high that it shuts down automatically
Mar 21, 2017 at 17:49 comment added StainlessSteelRat Was LM338 TO3 or TO220? Input-to-output voltage differential is a minimum of 3V. So you'd need 27V minimum for the regulator to work.
Mar 21, 2017 at 17:44 history asked Alian4life CC BY-SA 3.0