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I am going to use a nMOSFET (NTMFS5C670NL) which will a varied power dissipation. The current going from drain to source is determined by a control signal. The maximum power dissipation for the MOSFET is 1.29 W.

The junction-to-case thermal resistance Rjc is 2.4 ℃/W, and the junction-to-ambient thermal resistance Rja is 41 ℃/W.

The actual junction temperature formula is Tja = (Rja*Pd)+Ta. If the ambient temperature is 25 ℃, then Tja will be equal to 77.89.

Case temperature Tc will be (-Rjc*Pd) + Tj = 75 ℃.

These values are only valid if the MOSFET is surface mounted using a 650 mm^2, 2 Oz Cu pad. This is the only scenario the datasheet for the MOSFET gives.

Is there any way to give an educated guess on what the case temperature will be with a bigger or smaller Cu pad? I have failed to find any white papers that can give a decent estimation.

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Yes, you can compute the \$R_{th}/cm^2\$ of that copper geometry in terms \$^\circ C/W ~per ~cm^2\$ in an open convection area then add Rth for the enclosure.

Then define the added subscript letter to abbreviate each interface or number for internal or external ambient layers.

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  • \$\begingroup\$ So Rja will be 266.5 (℃/ W * cm2) in an open convection area? In other words the top and bottom layer? So if the Cu area is 9 cm^2, then Rja will be 29 ℃/W? It was my understanding that you could not do anything about the junction-to-case thermal resistance. Or is that thermal resistance affected by the inner layers in the PCB? \$\endgroup\$
    – Tungstein
    Nov 29 '21 at 17:18
  • \$\begingroup\$ Yes it assumes you have the required amount of thermal vias. \$\endgroup\$ Nov 29 '21 at 18:49

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