# Choosing a Heatsink for LM317 [duplicate]

I am using a LM317 to create 12V into 7V which is driving a load that draws 1A. So I have to dissipate 5W from my voltage regulator. Datasheet is here.

According to the datasheet Rja is 65 C°/W (Case 221A) so the heat on the chip is 65*5=325C° which is way too much obviously.

So I need to choose a heat sink but not sure how. One confusion I have is, each heat sink has a thermal resistance associated with it but I'm not quite sure what to do with that number or what to calculate next.

If I pick a heat sink with a thermal resistance of lets say 4 C°/W then does that mean the heat on the chip will simply be 4*5=20C°? So I've solved the temperature problem?

• At that power level, you're well into SMPS territory, me thinks. Commented May 2, 2016 at 23:11
• You'd actualy have to add the thermal resistance of the package, thermal compound AND heatsink, then add the delta temp to ambient temperature, thats the resulting temp of the package Commented May 2, 2016 at 23:11
• @WesleyLee Is thermal desistance of the package the 65 C°/W? Commented May 2, 2016 at 23:14
• junction-to-ambient= no heatsink Commented May 2, 2016 at 23:17
• Gah I am just so confused, all this thermal stuff is new to me and if I could just see the proper equation with numbers filled in it'd help so much in understanding. Commented May 2, 2016 at 23:18

Shortly after finishing school, I came across this book - I don't remember what its called, but it showed me to model thermal values as circuit elements.

simulate this circuit – Schematic created using CircuitLab

Calculate what your total power dissipation from your regulator is and sub in that value for Pd.

Rjc is the junction to case thermal resistance. This is given in the datasheet. Ta is the max ambient temperature the device may operate at.

Rsa is the surface to ambient thermal resistance. This is given in the datasheet as well, but this is the only parameter you can change. If the default values are not sufficient, you can add a heat sinks thermal resistance here.

The node labelled junction temperature, is maximum desired temperature. Basically what you are doing, is solving this circuit as you would if it was currents and voltages.

Voltages = temperature. Currents = Power

With this, you can calculate what minimum heat sink you would need.

Edit

$\theta_{JA}$ = thermal resistance from junction to air.

However $\theta_{JA} = \theta_{JC} + \theta_{CA}$.

In the schematic above Rjc is is only from the junction to the case. Rsa is the from the case to the ambient.

In the datasheet, they have given you RJC and Rsa ( $\theta_{JA}$ ) as well as they have given you $\theta_{JC}$

Edit again

Let's work through some numbers.

Maximum junction temperature of your device is 150C. So lets derate and lets say 100C.

That is our desired junction temperature.

Your power dessipiation is 5W. Rjc = 5W/C Max ambient is 35C.

Solve for Rsa.

$$Tj = R_{JC} * Pd + R_{sa}* Pd + Ta$$ $$\frac{Tj - Ta}{Pd} - R_{JC} = R_{SA}$$

Substituting values in

$$R_{SA} = 8W/C$$

Do we need a heat sink ? The datasheet says that the $\theta_{JA} = 65$ so that means that $\theta_{CA} = 60$.

The minimum value we need is 8W/C, so clearly we need a heatsink.

IF the calculated value of Rsa was 65 or more, then then no heat sink would be needed because the thermal resistance of the device itself would be sufficient enough to dissipate the heat.

• Okay so Pd would be my 5Watts dissipated in the regulator, Check. Junction Temperature Node is whatever I want the desired temperature to be? How about Max of 60C. Rjc is 5 C°/W from datasheet. Ta is 25C for my house temperature. And so my confusion is what to do with Rsa? Do I put the heatsink resistance here? Commented May 2, 2016 at 23:23
• @SuhDude Solve for Rsa. You have all the values you need. Commented May 2, 2016 at 23:24
• so how come I never take into account the Rja of 65 C°/W when calculating heat sink resistance? Commented May 2, 2016 at 23:26
• roughly, Rja = R(junction-to-case) + R(case-to-ambient) in case you want to use it in free air. If you use a heatsink then most of the heat goes through R(case-to-heatsink) ~roughly~ in parallel with R(case-to-ambient) [which becomes negligible after you add a heatsink] Commented May 2, 2016 at 23:30
• @SuhDude see edit. Does that make more sense ? Commented May 2, 2016 at 23:33