I have this voltage regulator from this supplier and the exact model name is given as: "LM317T/NOPB".

I plan to use this circuit to power a windwane's potentiometer which is around 10k:

enter image description here

As you see I want to set the excitation voltage for the vane pot to around 9 or 10V where the opamp will be supplied directly from a 12V switch mode power supply. For this purpose I will use an LM317 to obtain 9 or 10V for the vane excitation. So I will set LM317 for 9 or 10V DC. I will mount the vane in a box around 50m height which will stay there at least 3 months.

Since I will use a linear voltage regulator I decided to find out whether I need a heatsink.

I have encountered two challenges which I put here as questions:

1-) In the datasheet when I check the thermal resistances I cannot figure out which model I should use: enter image description here

Mine is T0-220, but in the datasheet there is two types namely as KCT and KCS. But my model is "LM317T/NOPB". And on it it is written "LM317T +P". So should I use the thermal resistance for KCT or KCS?

2-) Here is how I calculate the ΔT:

ΔT = θJA * P

In my case:

P = V^2 / R = 10^2 / 10000 = 0.01W

But for junction to ambient temperature I used to use the following:

Junction to ambient = Junction to case + Case to Ambient

But in the datasheet there is no Case to Ambient but there is Junction to ambient. Does "junction to ambient" already include "case to ambient"? Should I in my case only use Junction to ambient? If so in my case ΔT becomes like 4°C, and I conclude I don't need heatsink. Do I calculate correct and does it matter the duration like three months when it comes to using heatsink?

  • \$\begingroup\$ With a 10k pot, the LM317 only needs to source 1 mA. So its power consumption is only 2 mW. You're way over-thinking this. \$\endgroup\$ – The Photon Sep 9 '17 at 21:18
  • \$\begingroup\$ If the part you bought actually has the National Semi logo on it, that means its been sitting around in RS's stockroom for at least 5 years. If you were using enough power to actually care about the answer to this question, it would be better to buy new parts so you can get a datasheet that actually corresponds to your part. \$\endgroup\$ – The Photon Sep 9 '17 at 21:26
  • \$\begingroup\$ Is my using thermal resistance using only "Junction to ambient" (θJA) is enough ? \$\endgroup\$ – user1245 Sep 9 '17 at 21:29
  • \$\begingroup\$ And where in the datasheet I can find out the minimum input voltage needed for this regulator to output 10V? Thanks \$\endgroup\$ – user1245 Sep 9 '17 at 21:31
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    \$\begingroup\$ If you have multiple questions, ask them with multiple posts. But search first to see if the question has already been asked. (Hint: the term to search for is "drop-out voltage") \$\endgroup\$ – The Photon Sep 9 '17 at 21:32

You have calculated the power dissipated in the wind gauge - what you need to know is the power dissipated in the LM317. The wind vane will draw 1 mA, and you have 3 volts across the LM317, for a power dissipation of 3 mW due to the load. However, the LM317 requires a minimum load current of 5 mA for proper operation - the voltage-setting resistors are usually selected to draw that current.

With 1 mA for the wind gauge, and 5 mA for the LM317 operation, the LM317 will dissipate about 6 mW - no need for any heat sink at such a low power. You could use a TO-92 packaged LM317 without concern.

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  • \$\begingroup\$ Oh thanks I understand now. But couple of things 1-) Where did you find 5mA in the datasheet? In datasheet at 6.3 in Recommended Operating Conditions min output current is 10mA 2-) So as thermal resistance using only "Junction to ambient" (θJA) is enough right? thanks in advance \$\endgroup\$ – user1245 Sep 9 '17 at 21:13
  • \$\begingroup\$ The TI datasheet shows "Minimum load current to maintain regulation" with a typical value of 3.5 mA, and maximum of 10 mA. I got the 5 mA figure out of my head - but it is reasonable. My answer doesn't really answer the calculation question, but does show the power dissipation is so low that there's no need to worry about heatsinking. \$\endgroup\$ – Peter Bennett Sep 9 '17 at 21:50

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