1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Thermal Resistance of Heat Sink

  1. Apr 3, 2016 #1
    1. The problem statement, all variables and given/known data
    Calculate the thermal resistance of the heat sink required for the regulator of the question above given the information below [from the datasheet] and the data in the table.

    Max Load Current (Ilm)= 1A
    Max Ambient Temp (Ta)= 75°C
    Thermal Resistance Case to Heat (θchs)= 0.2°C/W
    Control Section
    Thermal Resistance Junction to Case (θjc)= 0.6°C/W
    Max Junction Temp (Tjmax)= 125°C
    Power Section
    Thermal Resistance Junction to Case (θjc)= 1.6°C/W
    Max Junction Temp (Tjmax)= 150°C

    Data sheet:

    Details form previous question:
    Vout=1.25-15V (As calculated in the previous question. https://www.physicsforums.com/threa...tor-output-voltage-range.864877/#post-5428815)

    2. Relevant equations

    3. The attempt at a solution
    Okay so here is what I have so far....

    This value is well above the value of 1.6°C/W stated on the datasheet so it will require a heat sink.

    I have done the same for the power section, but I feel I'm not on the right track due the thermal resistance values for the heat sinks being so drastically high. Especially if I'm right in thinking they then have to be added together?
    Last edited: Apr 3, 2016
  2. jcsd
  3. Apr 3, 2016 #2
    Okay so after looking harder into this I've found this equation in the datasheet.



    [itex]θhs=\frac {Tj}{Ta+Pd}-θchs-θjc[/itex]

    [itex]\frac {125}{75+3}-0.2-0.6=0.8°C/W[/itex]

    This seems a better answer, but I feel there's still a gap in my knowledge I'm not quite grasping somewhere...
  4. Apr 3, 2016 #3


    User Avatar

    Staff: Mentor

    The worst-case scenario should occur when there is a maximum difference between the input and output voltages. You took the best case scenario where the output is maximum at 15 V.

    Check your "Transposed" equation's algebra. You're adding temperature and power units in the denominator of the first term.... that just can't be right!
  5. Apr 3, 2016 #4
    This is an equation supplied in the datasheets. Should I disregard it and stick with my original workings?
  6. Apr 3, 2016 #5


    User Avatar

    Staff: Mentor

    The datasheet version is fine. It's the algebra you performed on it that I have my doubts about.
  7. Apr 3, 2016 #6
    A FOOL I AM! the same damn equation if I'd bothered to transpose it correctly.

    [itex]Pd=1(18-1.25)=16.75W[/itex] (Allowing for the maximum difference in voltage.)



    Therefore the thermal resistance of the heat sink required is [itex]2.18+2.69=4.87°C/W[/itex]
    Last edited: Apr 3, 2016
  8. Apr 3, 2016 #7


    User Avatar

    Staff: Mentor

    That looks better (for the equation) :smile:

    Strangely, the final paragraph of the Thermal Considerations section of the application sheet doesn't consider adding the two power contributions when they determined that their heat sink was sufficient; they just noted that it's thermal resistance was adequate for each case individually. I don't know why this is o_O

    To combine the two, if its valid to do so and their model doesn't already take it into account, I think I would combine them in parallel, not series (i.e. add thermal conductances to carry away the combined heat).
  9. Apr 3, 2016 #8
    I take it they should be combined in parallel as the control resistors R1 and R2 are in parallel with the LT1083?
  10. Apr 3, 2016 #9


    User Avatar

    Staff: Mentor

    The thermal "circuit" is entirely separate from the electronic one (except that the heat generated is due to the electronic side of things). It's just that you have two heat sources (control and power) dumping heat into the chip substrate which both need to be conducted away via the case-to-heatsink-to-ambient path. Think of it like adding heat currents. The total "current" is the sum of the two heat powers.

    I don't understand their model if their example is correct: They seem to assume that the same total power (24 W in their case) is generated by both sections individually since they place that power across both junction thermal resistances. I'd think that the thermal circuit would look something like this:


    Perhaps they're modelling some worst case scenario where it is assumed that all the heat is being generated by one section or the other. In that case you'd just pick the higher of the two heatsink resistances for your answer.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted