I have to dissipate \$2W\$ from a voltage regulator. It's a 7805 in a TO-220 package. The datasheet is here.
It's the first time I have to pick those so I would like to have a review of the following decision because I'm afraid of missing something as this sound really complicated for me. So I will put you here my entire reasoning.
\$R_{thJC}\$ is 5 C°/W
for the TO-220 package and \$R_{thJA}\$ is 50 C°/W
(table 3, page 7). As I will need to dissipate 2W
I will have without heatsink 100°
of heat on the chip. The room is around 21°
. \$T_{op}\$ is 0° to 125°
so to be safe I definitely need a heatsink. In that case it will just go around 31°
based on that formula \$Max_{ambiant}+ R_{thJC} \times W_{dissipated}\$ or 21 + 50 * 2
But now I'm blocked. For the example I will take this heatsink. He's rated as 40 K/W
. I assume K is for kelvin degrees. In that case does it means he's rated as 233°C/W
? I've found that formula :
\$Max_{JunctionTemp} >= Max_{AmbientTemp} + (W_{Dissipated} \times (R_{thJC_{C°/W}} + R_{thHeatSink_{C°/W}}))\$
Which give me :
\$ 21 + (2.5 \times (5 + 233) = 595°C\$
So, there's something wrong as this would mean that the junction between chip and heatsink will be 600° hot ... What have I missed ?