heatlinear-regulatorpower-dissipationtemperaturethermal
If I use a linear voltage regulator as LM317:
From the datasheet, I read that maximum operating temperature = 125°C, junction = 150°C
How I can calculate if this component is thermally suitable, instead of a switching regulator?
This is certainly a technique I have used a few times to overcome the limited power dissipation abilities of the diminutive 78L05. I've known the range of currents that the load is taking and placed a dropper resistor in series with the power feed to the device.
Why didn't I use a switching regulator?
I couldn't - I was sending power and data down a 50 m cable (phantom power) and the extreme complication of filtering out the switching regulator's current surges meant it just wasn't feasible.
Just for reference, I'm going to use °C/W because it's what I'm used to.
In the datasheet for your regulator under the thermal characteristics there are typically two values.
\$R_{\theta JA}\$ - Junction to ambient
\$R_{\theta JC}\$ - Junction to case
If not using a heatsink then the \$R_{\theta JA}\$ is the value you use to calculate how warm the regulator is going to be.
When using a heatsink you take your \$R_{\theta JC}\$ and add this to your heatsink rating (20°C/W in this case) also remembering to add in any other thermal resistances. The thermal pad between the device and the heatsink is one such example.
Now this 22°C/W is in a perfect world where it has a super perfect thermal connection so take your result as an estimate rather than a "It will always be 60°C"
When you calculate the temperature, you are correct in thinking this will be the temperature of both the component and heatsink. However, remember that there's a lot of difference between a TO-220 package and a block of aluminium and it will take time for both items to reach the same temperature.
Edit: Because I remembered I had it to hand, here's a thermal image of a PCB I'm working on. This is about 5 minutes after switch on, notice the component itself is 43°C and yellow on the thermal image but the surrounding large copper heatsink pad for the device is a purple and closer to 20°C.
P.S - Please ignore the white hot supernova approaching from the right, this is a work in progress board
Best Answer
What you need to know is the Junction-to-Ambient thermal resistance. There is a table for that on page 4 of the DATASHEET.
It has the values for each of the packages.
So, you know your voltage drop (19V), you know the required current (480-700mA). With this information, you can now find out your power dissipated (P=IV) and use this value to see how much your IC will heat up.
For an example, lets assume you have a TO-263 package. You calculate your power to be 3.68W. You see that the thermal resistance of this package is 38°C/W, thus the temperature will rise by 139.8°C. Now, while you may think that is fine, because it is under 150°C, you also need to add on the ambient temperature of the environment. Assuming this is 25°C, this will give you a total of 164.8°C. This now exceeds the maximum.
There are other factors involved, such as the current drawn by the device itself, not just your load, some environmental factors etc, but this is the easiest way to calculate what your temperature could be. You can use this method for any IC, not just the LM317, and you should find all the information here for you to calculate this yourself.
Further reading