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I have this BUK9M12-60E MOSFET.

I have calculated the maximum power dissipation of the MOSFET to be 14W (Conduction loss, Switching Loss and Gate Loss). I believe that I don't require any heatsink.

I'd like to calculate the the Junction temperature of the device when it is working and when my ambient temperature is 85degC.

But in order to find, I don't have the Thermal Resistance (Junction to Ambient value). So, how to proceed with the calculation to find the junction temperature.

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  • \$\begingroup\$ There is simply no way that a package that small can dissipate 14W without a heatsink. \$\endgroup\$
    – Dave Tweed
    Apr 5, 2021 at 15:14
  • \$\begingroup\$ But the maximum power dissipation is capped at 79W at 25degC. Is it with heatsink? \$\endgroup\$
    – Newbie
    Apr 5, 2021 at 15:21
  • \$\begingroup\$ Yes, it means that the mounting base is held at 25°C. There's simply no way to do that without a heatsink of some kind. At 85°C, Fig. 1 shows that you'll need to derate the power to less than 60%, or about 50W. And again, that assumes that the mounting base is held at 85°C, which is very hard to do if your ambient is also 85°C. \$\endgroup\$
    – Dave Tweed
    Apr 5, 2021 at 15:31
  • \$\begingroup\$ Could you help to calculate the junction temperature? \$\endgroup\$
    – Newbie
    Apr 5, 2021 at 15:50

2 Answers 2

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You will find detailed data on the LFPAK33 package in this document LFPAK MOSFET thermal design guide.

As a rule of thumb, in an automotive environment with 80°C max Ta, the maximum power dissipation on FR4 laminate is 1W. At 85°C it will obviously be less.

So you can forget about your 14W without a very, very substantial heat sink or equivalent structure to remove the heat.

The 79W is not only with a heatsink, it's with an extremely unrealistic heatsink that holds the mounting base temperature at 25°C despite 79W flowing into it. You might be able to accomplish that by fast-flowing chilled water through a copper heatsink. In short, it's not a realistically useful number except as kind of a theoretical extrapolation.

As an aside, I fear we are losing the intuition of power dissipation with the phasing out of incandescent bulbs. Anyone who has unscrewed a 60W or 100W incandescent bulb just after it has been illuminated for some time will know what I mean.

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  • \$\begingroup\$ Can I increase the copper pad area in the FR-4 Laminate for the power dissipation instead of using a heatsink? If its possible, how big should the ground-copper pad for the MOSFET should be? \$\endgroup\$
    – Newbie
    Apr 5, 2021 at 15:58
  • \$\begingroup\$ @Newbie for 14W I don't think it's even close to possible with FR4 and 1oz or 2oz copper. Maybe with an aluminum or copper core PCB. But there is enough information in that document for you to do the detailed design work, I can only point you to it. \$\endgroup\$ Apr 5, 2021 at 16:02
  • \$\begingroup\$ what if I add some 20-30 vias along the pads of MOSFET \$\endgroup\$
    – Newbie
    Apr 6, 2021 at 6:52
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This is not a value that you calculate...it is given in the datasheet. For this particular device the manufacturer assumes that you will always mount it to a "mounting base", so they only specify the thermal resistance from the junction to that base. See the bottom of page 4.

On page 9 of the datasheet you will find a link to an NXP application note that will probably help a great deal.

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  • \$\begingroup\$ Thank you for the answer. But I think, for my application, I do not require to use a heatsink. So, I require the Thermal Resistance (Junction to ambient) value to calculate the actual junction temperature \$\endgroup\$
    – Newbie
    Apr 5, 2021 at 15:15
  • \$\begingroup\$ Then you need to contact NXP and ask for this value. \$\endgroup\$ Apr 5, 2021 at 15:23
  • \$\begingroup\$ The junction-to-ambient thermal resistance is the junction-to-base resistance, which they give you (max 1.89 C/W), plus the base-to-ambient resistance, which they have no control over. @Spehro Pefhany linked what looks like a good guide to designing with the package you have. A quick glance shows that you're an order of magnitude over what can be expected to be dissipated on copper pours alone. \$\endgroup\$
    – vir
    Apr 5, 2021 at 16:35

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