A question about reading and calculating thermal resistance of a voltage regulator from its datasheet

Question

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 10 kΩ:

As you see I want to set the excitation voltage for the vane pot to around 9 or 10 V where the op-amp will be supplied directly from a 12 V switch-mode power supply. For this purpose I will use an LM317 to obtain 9 or 10 V for the vane excitation, so I will set the LM317 for 9 or 10 VDC.

I will mount the vane in a box around 50 m high which will stay there for 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 can't figure out which model I should use:
   
   
   
   Mine is T0-220, but in the datasheet there are two types namely 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 = V² / R = 10² / 10000 = 0.01 W
   
   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. Did I calculate correctly and does the duration (three months) matter when it comes to using a heatsink?

Answer 1

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.