Heatsinking calculation

Discussion in 'Homework Help' started by Steve1992, Jun 9, 2012.

  1. Steve1992

    Thread Starter Senior Member

    Apr 7, 2006
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    0
    Hello,

    Calculated Power dissipated from a LM317 voltage regulator (9V output, 0.5A load current) = 4.5W
    Ambient temperature, 30 degrees
    Max. junction temp. (from data sheet) 125 degrees

    125 - 30 = 95
    Junction-to-ambient thermal resistance = 95/4.5W = 21 degrees/Watt

    With a Heatsink added, specified thermal resistance of 3.3 degrees/Watt:

    3.3 x 4.5W = 15 degrees.
    30 + 15 = 45 degrees = heatsink temperature = correct?
     
  2. KJ6EAD

    Senior Member

    Apr 30, 2011
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  3. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,800
    The power dissipated by the regulator is the voltage across it times the current through it. So you want (Vin-Vout)*Iout.

    Assuming that that were to still come out to 4.5W so that we can go forward.

    Is the 3.3 K/W (kelvins/watt, which is the proper unit since degrees celsius is not an absolute scale, but they are the same thing and degC/W is more commonaly used, so don't worry about it) the specified thermal resistance from the junction to ambiant with the heatsink in place, for just the thermal resistance of the heat sink?

    If so, then the calculations are correct except that the 45C at the end is not the heatsink temperature, it is the junction temperature.

    Also, your calculation of 21C/W should be noted that it is not the junction-to-ambient thermal resistance, but rather the maximum permissible value of it.

    I would recommend carrying your units throughout your calculations.
     
    Steve1992 likes this.
  4. Steve1992

    Thread Starter Senior Member

    Apr 7, 2006
    100
    0
    Thanks.

    My next question was going to be how can I calculate the case to junction temp. now with the heatsink used? It looks like I inadvertently answered it: 45 C.
     
  5. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,800
    No, 45C is the junction temperature (assuming 4.5W dissipated in the device, which is not going to be the case here unless the input voltage is 18V). The case-to-junction temperature is the difference between the temperature of the junction and the temperature of the case. To find that, you need either the junction-to-case thermal resistance (which is generally assumed to be the same as the junction-to-ambient thermal resistance without a heatsink for many packages) or the case-to-ambient thermal resistance for the heat sink.

    Some IC's (including many TO-220 packaged parts) have a very low junction-to-pad thermal resistance where the pad is an exposed thermal pad specifically designed for heatsinking. If you're using a heatsink that mates well to that pad, then you can really keep junction temperatures in check with a lot less heatsinking effort.
     
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