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:

enter image description here

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:

    enter image description here

    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?

  • \$\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, 2017 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, 2017 at 21:26
  • \$\begingroup\$ Is my using thermal resistance using only "Junction to ambient" (θJA) is enough ? \$\endgroup\$
    – user1245
    Sep 9, 2017 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, 2017 at 21:31
  • 1
    \$\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, 2017 at 21:32

1 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.

  • \$\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, 2017 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\$ Sep 9, 2017 at 21:50
  • \$\begingroup\$ The circuit has the voltage-setting resistors missing, also the important input and output capacitors are missing, all are shown in the datasheet. \$\endgroup\$
    – Audioguru
    May 30, 2021 at 23:48

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