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Do I need a heatsink?

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wuchy143

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Hi All,

Currently I have a board which has a 15.5 volt to 5V regulator(LM317M / To-220). I'm going to change it to a 15.5V to 3.3V output to change to a cheaper micro. I know that I can just populate the parts and see if the heatsink gets too hot for the part but I am not familiar with the calculations and would like a more theoretical approach.

I looked online to try and find ways of doing this but most examples require the current and currently I don't know that and am not sure what it'll be.

I thought about measuring the current(lifting a leg of the to-220 LM317) to see what the power dissipation currently is and maybe that would help me...

Anyone have a trick up their sleeve on this one? Or what would you do in this situation?
 
You need to know the current.

The power dissipation is the voltage drop times the current. In your case the voltage drop is 15.5 - 3.3 so it's 12.2

The LM317M has a thermal resistance of 50 dec C / W with no heatsink, according to the STM data sheet, and a maximum temperature of 125 deg C. If you are running in air at 40 degrees C, then you can let it heat up by 85 deg C, so it can dissipate 85 / 50 = 1.7 W, so it can take 1.7 / 12.2 = 139 mA with no heat sink.

If you are taking more than that, the thermal resistance of the junction to the case is 3 deg C / W, so you should add that to the rating of any heatsink you use.

If you had a heatsink rated at 10 deg C / W, the total would be 13 deg C / W, and you could go to 85 / 13 = 6.5 W, so about half an amp.

You should always go for a larger heatsink if in doubt.
 
The regulator now drops 15.5-5 = 10.5V. With a 3.3V micro the reg will drop 15.5-3.3 = 12.2V. So if it now dissipates X watts, it will dissipate X times 12.2/10.5 watts (an increase of ~ 20%), assuming the new micro takes the same current as the old. If the reg doesn't get hot now it will probably handle the extra 20% ok.
 
To calculate it you need to know the current, but if the current is the same as you have with the setup you have now the power will only be about 20% higher. So if you can hold your finger on it now (50C) it should be ok.
 
thanks a lot guys. I appreciate the help on this sort of an odd problem. You all confirmed that ultimately I NEED to know the current to get a "real" value. But as you all have said. This part will be fine @ 3.3V under it's current load. Thank you all.
 
You do need to know the current, but you do not need to know it accurately. If it is less than 100 mA, that is all you need to know, it will be fine without a heatsink. If the current design doesn't burn your finger, the new design will be OK, and there is a good chance that it will take less current at 3.3 V than the old one took at 5 V.
 
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