Thermal resistance for DPAK/IPAK MOSFET

Discussion in 'General Electronics Chat' started by eigenvictor, Dec 9, 2014.

  1. eigenvictor

    Thread Starter New Member

    Jul 16, 2014
    Hello, I am new on this forum, and I have a very specific question about mounting DPAK/IPAK mosfets. I am mounting them on a perf board so I am considering soldering separate copper pads 15mm x 15mm in size which is smaller than the recommended area of 6cm^2 to achieve 50K/W. However, they are much thicker (0,5mm). Since calculations in the datasheet are given for a standard PCB copper thickness (2oz?) which is probably like 10 times thinner than what I have, my question is this: for a fixed copper pad area, is there a linear dependence between thickness and the resulting thermal resistance? Basically, instead of 6cm^2 at 0.07mm I have about 2cm^2 at 0.5mm. What will be the RthJA in this case? Thanks for any insight.
  2. Dr.killjoy

    Well-Known Member

    Apr 28, 2013
    Ok I will give this a go here... You are trying to cool a mosfet with thicker material with smaller surface area vs a thinner piece with larger area.. For me think of using a blow torch instead of the mosfet... The thinner piece will heat up faster cause of the lack thickness at the heat source but will be cooler futher away from direct heat cause the edges will dissipate some of the heat.. With the thicker material it will take longer to heat up cause of the thermal mass .. But it will stay hotter cause the heat has a smaller surface and no where for to dissipate.. Also it take longer for it to cool cause of the thermal mass ..

    I am sorry if this make no sense or helps out cause I am trying and just really tired .. So please delete if felt necessary..

    Jay Sr
  3. crutschow


    Mar 14, 2008
    It's not linear. Generally as long as the copper is thick enough to distribute the heat, then any thickness beyond that has only a small effect on the thermal resistance. It's the radiating area that's the primary determination of the thermal resistance.
  4. ronv

    AAC Fanatic!

    Nov 12, 2008
    Very little difference due to thickness. It's all about area until the area gets large. You can come close using 50/sq. root of the area in cm.
  5. Papabravo


    Feb 24, 2006
    When it comes to heat there are only three ways to get rid of it:
    1. Radiation
    2. Conduction
    3. Convection
    Design accordingly.
  6. eigenvictor

    Thread Starter New Member

    Jul 16, 2014
    So when you guys are saying it is all about the surface area, you are assuming Rcond<<Rconv? Calculations in the application notes are given for a typical FR4 PCB with a really thin copper layer. I would guess that for a typical 2oz copper, Rcond (L/kA) would be higher than or at least comparable to Rconv for the "recommended" surface area of the copper pad. But if you make the copper pad several times thicker to increase the cross-sectional area (assuming conductive flux is lateral), that will make Rcond negligible compared to Rconv and we just have Rconv which is given as R = 1/hA. I don't know how to calculate h, but you guys said Rconv could be approximated as 50/sq. root of the area in cm. So it follows then that even for the area as small as 1 cm^2, it would still be within the required 50K/W. Make it 1.5cm^1.5 (what I have) and you are down to 30! Sweet. That's what I want. Of course if I wanted to go much lower than that, then at some point it is all about the surface area. Does it make sense? Anything important I forgot to consider? I suspect the Rconv could turn out even lower than I calculated, since convective dissipation occurs on both sides of the copper plate (however one side is attached to the board so it is not that simple). And just to clarify, I am only considering natural convection.