Hello,

I've designed a power supply for two LM317 voltage regulators and specified a heat sink using very conservative values to avoid any issues -
The regulators share the heat sink so I've summed their power dissipations together

PD 1 = (26V-17V) * 0.4A = 3.6W
PD 2 = (17V-12V) * 0.1A = 0.5W
PD total = 4.1W
Rjc = 5°C
Rcs = 1°C
Max operating temp = 100°C
Ambient temp = 50°C (room temp is actually 21°C )

Required thermal resistance = 6.195°C

I chose the following heatsink rated at 2.5°C/W
https://uk.rs-online.com/web/p/heatsinks/7226928

I thought this should be more than enough, but the surface temperature of the heatsink is 51°C after being on for 30 minutes. This seems hot to me, given the size of the heat sink.

My questions are -

Am I right to just sum the power dissipation of the two regulators together and then calculate the heat sink required for them to share?

Is there a way to calculate the ball park expected surface temperature of the heatsink?

Likewise, knowing the surface temperature of the heatsink, can I calculate the actual junction temperature?

Thanks in advance

 

Hello,

Perhaps the following pages will interest you:
https://sound-au.com/heatsinks.htm
https://sound-au.com/articles/diy-heatsink.htm

Bertus

 

the surface temperature of the heatsink is 51°C after being on for 30 minutes.

The given heatsink thermal resistance is likely for it being mounted in open air with the fins vertical.
Is that its orientation?

Am I right to just sum the power dissipation of the two regulators together and then calculate the heat sink required for them to share?

Yes.

Is there a way to calculate the ball park expected surface temperature of the heatsink?

Yes, just use the terminal resistance of the heatsink times the total power being dissipated, and add that to the ambient air temperature near the heatsink.

Likewise, knowing the surface temperature of the heatsink, can I calculate the actual junction temperature?

Yes, you just add in the device junction-to-case, and the case-to-heatsink thermal resistances (are they mounted using thermal grease to reduce the case-to-heatsink resistance?) and multiply that by the individual device power being dissipated.

 

The heat sink temperature looks about normal for something with no forced air flow to remove heat faster from it. The 'resistance' from the heat sink surface to air needs to in decreased (by being in contact with something cooler) if you want the the thermal chain from device to air to be better than heated air rising flow from the surface of the sink.

 

Hello,

Perhaps the following pages will interest you:
https://sound-au.com/heatsinks.htm
https://sound-au.com/articles/diy-heatsink.htm

Bertus

These look great, thank you very much

The given heatsink thermal resistance is likely for it being mounted in open air with the fins vertical.
Is that its orientation?

Yes, it's mounted with the fins vertical

Yes, just use the terminal resistance of the heatsink times the total power being dissipated, and add that to the ambient air temperature near the heatsink.

Excellent, thank you. So in theory the heat sink should be between 30°C/40°C depending on the actual air temp around the heatsink

Yes, you just add in the device junction-to-case, and the case-to-heatsink thermal resistances (are they mounted using thermal grease to reduce the case-to-heatsink resistance?) and multiply that by the individual device power being dissipated.

Sorry, I may be misunderstanding, but would this not give me the temperature of the junction if the heatsink surface temp was that of the one calculated, not measured?

Thanks for your help

 

would this not give me the temperature of the junction if the heatsink surface temp was that of the one calculated, not measured?

Perhaps I wasn't clear.
You add the temperature increase due to the case-to-junction, and case-to-heatsink thermal resistances to the measured heatsink temperature for each device.

 

Perhaps I wasn't clear.
You add the temperature increase due to the case-to-junction, and case-to-heatsink thermal resistances to the measured heatsink temperature for each device.

Ah I see, so for regulator 1 -

((5°C + 1°C) * 3.6W) + Measured heatsink 51°C = 72.6°C ?

 

Ah I see, so for regulator 1 -

((5°C + 1°C) * 3.6W) + Measured heatsink 51°C = 72.6°C ?

Correct.

 

Amazing, just what I needed. Thank you!