LM338 12W heat dissipation

Discussion in 'The Projects Forum' started by x1222, Aug 24, 2012.

  1. x1222

    Thread Starter Member

    Oct 22, 2011
    31
    0
    I'm thinking of using the 12V rail on a computer power supply and regulating it down to 6V and split that into 3x LM338, each receiving 2 A.

    Would it be tough to dissipate 12W? Right now I have these heatsinks, not sure how big i'll need them. My local store has some a bit bigger heatsinks.
    http://dipmicro.com/store/HSINK220C

    They're a few mm wider than the LM338 and about the same thickness. No data sheet though. I could get some computer case fans blowing on them too. Switching regulators or battery packs will probably be my last resort.
     
  2. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    I can pretty much guarantee you that those are too small.
    How to Size Heat Sinks for Semiconductors is a nice tutorial. Reputable mfrs have thermal resistance specs on all their heat sinks. Unfortunately, those are often not passed on to the consumer when they are sold by hobbyist parts vendors.
     
    x1222 likes this.
  3. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    Based on similar sinks I have used, I would guess ballpark of about 10C/W for those. Add the 4C/W for attaching a TO-220 to a heatsink, and your 12W will get your junction temp up to about 170C rise above ambient and they will cook (or go into thermal shutdown). You need a heatsink with a theta of less than or equal to about 5C/W to dissipate 12W in TO-220.

    The AAVID 5297 in the attached reference in the post above mine MIGHT work but the 5298 would be better. Adding airflow would also be better.
     
    Last edited: Aug 24, 2012
    x1222 likes this.
  4. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    I think the number is closer to 2.5C/W, especially if you don't want to run the junction at 125C. It also depends on what your max ambient temp is.
     
  5. bertus

    Administrator

    Apr 5, 2008
    15,641
    2,344
    x1222 likes this.
  6. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    Yeah, I let them go to 140 worst case, about 110C above ambient.
     
  7. THE_RB

    AAC Fanatic!

    Feb 11, 2008
    5,435
    1,305
    Unless I misunderstood you, that is 6v*2A *3 which is 36 watts total.

    That is more than just a heatsink issue, you now have a total device /enclosure etc needing to dissipate 36W!

    To me, 36W continuous is a LARGE heatsink AND a fan, and an enclosure that is correctly ventilated for air flow AND located somewhere where it can source cool air and pump out all that nasty hot air.

    Maybe if you say what you needed 6v and 6A output for, and what type of 12v 6A supply you have, we can suggest better options. :)
     
    x1222 likes this.
  8. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    I think he is buying a separate heatsink for each 338 device so it's 12W per regulator/sink. The calculation gets more complicated using a single heatsink. He said he is going to use a fan in the case.
     
  9. x1222

    Thread Starter Member

    Oct 22, 2011
    31
    0
    Thanks for the replies.

    I'm building a robotic arm with 6 servos, a couple of them are bigger sizes. 2A probably won't be the typical current, but I want to be on the safe side. Like bountyhunter said I plan to separate the 3 LM338 with 3 separate heatsinks, so I was thinking they wouldn't affect each other. They'll be in open space so airflow should be no problem, but an enclosure might look nice.

    I calculated 36 C/W lol, so I'm way off. I'm guessing I don't have the right power value.

    Pmax dissipated wasn't found on the datasheet so I calculated:
    With assumed max ambient of 35 C
    Pmax = (Tj - Ta) / (junction to ambient resistance from datasheet LM338T)
    = (125 - 35) / 50
    = 1.8

    From Ron's link
    qsa (Thermal resistance of heatsink) = qja - qjc - qcs

    qja = (Tja - Ta) / P = 125 - 35 / 1.8 = 50
    qjc = Tja - Tc / P = 125 - 100 (assumed) / 1.8 = 14
    qsa = 50 - 14 - 0 (cpu thermal paste resistance negligible) = 36

    To find the size, still kind of confused on this one, I can use Harry's quick and dirty formula from bertus's link

    Thermal Resistance = 50 / √A
    where A is in cm^2, not sure if 50 is a constant or it is the power dissipated, I'd guess latter.

    A = (50/ 36) ^2 = 2 cm^2 so my heat sinks work!

    Am I at least on the right track?
     
  10. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    I think you misinterpreted the app note. The Pmax you calculated is for no heat sink. It has nothing to do with the required heat sink size.

    The P in the equation is the maximum power you expect the regulator to have to handle. In your case, this is 12W. I personally would not run the chip ("junction") at 125°C. I would set the limit at 100, or maybe 110, but that's just me, although I'm sure I'm not alone.
    I think I see, in the app note, where you thought it necessary to calculate Pmax (although you calculated the wrong Pmax). In your case, qjc is given in the datasheet as 4°C/W. You don't have to calculate it, as was done in the app note.
    You can calculate qja using the equation in the app note. Remember, P=12W.:) You can assume qcs=0.1 to 0.2.
     
    x1222 likes this.
  11. x1222

    Thread Starter Member

    Oct 22, 2011
    31
    0
    Ah that looks better.

    So qja = 100 - 35 / 12 = 5.42

    qsa = 5.42 - 4 - 0.1 = 1.32 C/W

    So I have to look for a heatsink with thermal resistance of 1.32 or less.

    Then from harry's dirty quick formula, assuming heat sink will have thermal resistance = 1.32

    1.32 = 12 W / √A

    A = 9 cm by 9 cm is roughly the surface area of heatsink I could look for, ignoring thickness.

    And it ain't being too easy to find one with ~1.32 C/W, unless I add some pretty good air flow.
     
    Last edited: Aug 26, 2012
  12. bertus

    Administrator

    Apr 5, 2008
    15,641
    2,344
    Hello,

    If the LM would be a TO3 part, the looking for a heatsink would be much more easy.

    You can also use a LM in combination with a TO3 power transistor.
    Look in the data sheet of the LM317 how this is done.

    Bertus
     
    Last edited: Aug 26, 2012
  13. x1222

    Thread Starter Member

    Oct 22, 2011
    31
    0
    Sorry, could you please elaborate more on what you mean by this?

    I took a look at a LM1084 5A regulator, which is easily available at my electronics store. Qjc for the control section is only 0.65 C/W,

    so I get a qsa of 4.67 C/W , which is way better and seems rather doable. Something like the avid 5298 like bountyhunter said would then seem to suffice.
     
  14. bertus

    Administrator

    Apr 5, 2008
    15,641
    2,344
    Hello,

    You can combine the regulator with a TO3 PNP transistor.

    [​IMG]

    (the schematic comes from the LM340 datasheet).

    Bertus
     
    x1222 likes this.
Loading...