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How to calculate dissipated heat?

  1. Oct 11, 2009 #1
    I'm trying to figure out how big of heatsink I'm going to need for my switching regulator, or if it needs one at all. The IC is LM2678-5 which is rated at ~87% efficiency for the maximum load I might have to drive with it. It's stepping down from up to 20V down to 5V (battery is rated at ~18) and the maximum current that I might have draw from it is 3A though that's highly unlikely. In the case that this happens though, how do I figure out the specs of any heatsink I might have to purchase?

    Is the power dissipated just 13% * voltage drop * current? That would be a whopping 5+ watts for the most extreme case. The datasheet indicates the following in terms of thermal resistance:

    T Package, Junction to Ambient 65 C/W(Note 5)
    T Package, Junction to Ambient 45 C/W (Note 6)
    T Package, Junction to Case 2 C/W

    Note 5: Junction to ambient thermal resistance (no external heat sink) for the 7 lead TO-220 package mounted vertically, with ½ inch leads in a socket, or on a PC board with minimum copper area.

    Note 6: Junction to ambient thermal resistance (no external heat sink) for the 7 lead TO-220 package mounted vertically, with ½ inch leads soldered to a PC board containing approximately 4 square inches of (1 oz.) copper area surrounding the leads.

    Would appreciate any help making sense of this. Thanks!
     
    Last edited: Oct 11, 2009
  2. jcsd
  3. Oct 13, 2009 #2
    5 watts is not a large power loss for a TO-220 package. Do you have room for an external N-channel transistor?
    Bob S
     
  4. Oct 13, 2009 #3
    Yes, space is not really an issue at this point.
     
  5. Oct 13, 2009 #4

    berkeman

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    Staff: Mentor

    What is the airflow like? What is your rated ambient temperature range (Commercial to 70C or Industrial to 85C)? How much heat is generated in your enclosure by the current you are supplying at 5V? Is it all inside your enclosure, or do you supply power to some external load?

    I'd be inclined to use a moderate heat sink on it, and make sure that you get reasonable ambient airflow past it (like via convection, with any other hot stuff above the regulator heat sink). Have you looked at heat sink selector guides for their thermal resistance (both with and without grease)?
     
  6. Oct 13, 2009 #5

    vk6kro

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    Science Advisor

    It's stepping down from up to 20V down to 5V (battery is rated at ~18) and the maximum current that I might have draw from it is 3A though that's highly unlikely. In the case that this happens though, how do I figure out the specs of any heatsink I might have to purchase?

    Is the power dissipated just 13% * voltage drop * current? That would be a whopping 5+ watts for the most extreme case. The datasheet indicates the following in terms of thermal resistance:


    The device is 87% efficient, so the power input for 15 watts out must be 15 * 100/87 or 17.24 watts. So the loss would be 2.24 watts.
    Even 5 watts wasn't too bad, but this is even better.

    However, note 5 indicates that the chip would get a very hot junction without a heatsink even with 2.24 watts dissipation. Note 6 suggests 4 sq inches of PCB copper would be OK.
    A heatsink of aluminum with small fins and about 2 sq inches or mounting it on the side of a diecast box would be suitable.
     
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