Junction-to-case thermal resistance top vs bottom: which do I use for calculating junction temperature?

Question

1. Question Summary

I have put questions under heading 3, and summarised them below:

1. I want to calculate the junction temperature for a MOSFET that has \$R_{\theta JC}\text{(Bottom)} = 1.2\;^{\circ}\text{C/W}\$, \$R_{\theta JC}\text{(Top)} = 31\;^{\circ}\text{C/W}\$ and \$R_{\theta JA} = 35\;^{\circ}\text{C/W}\$.
   Which thermal resistance metric do I use between junction and case?

2. The datasheet lists the thermal resistance from junction to ambient and junction to case - when calculating heat generated by devices bonded to copper pads on a PCB, what other thermal radiation paths do I need to consider, and how can I find their thermal resistances?

2. More information

The relevant MOSFET is a IRFH7440PbF.

Max \$R_{DS} = 2.4\;m\Omega\$. My maximum current is 20 A. The MOSFET will be on a PCB, soldered onto 1 sq inch of 2 ounce copper. I will be driving this via PWM however I will simply consider resistive losses for now to simplify this question. Maximum ambient temperature for my applications is 60 °C.

3. My working

1. Power dissipated by the MOSFET is \$\text{I}^2 \cdot r = 20^2 \cdot 2.4\;m\Omega = 960\;\text{mW}\$

2. The rise above ambient temperature caused by 20 A of current in the source-drain channel is \$960\text{mW} \cdot 1.2\;^{\circ}\text{C/W} = 1.152\;^{\circ}\text{C}\$
   Have I used the correct thermal resistance value?.

   I have used the lowest value (the thermal resistance between junction and the bottom of the case), but there is another thermal resistance value nominated in the datasheet that is significantly larger (between junction and top of case) at 31 °C/W.

3. Final maximum temperature is 60 °C + 1.152 °C = 61.152 °C
   What else am I missing here?

Alarm bells are going off in my head at this point. I do not have extensive circuit design experience, but I would expect 20 A through a MOSFET attached to a copper landing on a PCB to have a hotter junction than this at 60 °C ambient.

I feel like I have missed one or more thermal radiation paths. If the MOSFET were freestanding and not interfacing with any form of heatsink I would have simply used the junction to ambient thermal resistance and calculated the junction temperature as \$20^2 \cdot 2.4\;m\Omega \cdot 35\;^{\circ}\text{C/W} = 93.6\;^{\circ}\text{C}\$

Answer 1

I want to calculate the junction temperature for a MOSFET that has \$R_{\theta JC}\text{(Bottom)} = 1.2 C/W\$, \$R_{\theta > JC}\text{(Top)} = 31 C/W\$ and \$R_{\theta JA} = 35 C/W\$. Which thermal resistance metric do I use between junction and case?

You use the 1.2 °C/W junction-to-case thermal resistance if you are mounting on a heatsink. The top thermal resistance figure (31 °C/W) is in parallel for heat transfer but, it's a much higher figure and, can usually be ignored because it provides hardly any extra "comfort".

However, where you have gone wrong is not accounting for the fact that the device needs to use a heatsink to continue the "removal-of-heat" process. A typical heatsink interfaces between the case (bottom in your example) and true ambient. That heatsink might be (about) 6 °C/W (for a decent sized sink) and, up to 30 °C/W (for a piddling little heatsink).

If you chose a 6 °C/W heatsink then, that adds to the 1.2 °C/W just like resistors add when they are in series. It transports the "heat" on the following journey: -

Junction -----> (1.2 °C/W) -----> Case (bottom) -----> (6 °C/W) ---> Ambient

And, for the other two specified thermal resistances: -

Junction --> (31 °C/W) --> Case (top) --> (top heatsink if used) --> Ambient

Junction ----------------------> (35 °C/W) ------------------------> Ambient

So, in this example, you need to use a value of 7.2 °C/W when calculating how hot the junction gets when using a 6 °C/W heatsink in such-and-such an ambient temperature.

But, don't forget that inside an enclosed space, the heat generated will "warm" the local ambient temperature by several degrees and, this needs to be considered in many, many situations.

But, with your specific device, the heatsink is the amount of copper you have on the circuit board. This will be difficult to get as low as 6 °C/W and, I advise you to find an online calculator to tell you what a certain copper area could provide AND, you may choose to supplement the heatsinking with a nice slab of copper or aluminium on the top face too. Possibly with an array of fins if junction temperature is looking a bit tight.

Answer 2

To add to Andy's answer, if you read note (8) in the datasheet, it explains that thermal resistance to "ambient" is manufacturers estimate of just your situation (that is component mounted on the small copper pad on PCB and figure is total thermal path). There could be some application note detailing how exactly is this number measured, but generally 35 K/W would be first estimate for calculating junction increase unless you improve heatsinking in your design in some way. (Note that your particular situation can be even worse if you have not enough heat transfer away form board and/or if there is more heat dissipating components.)

The junction-to-case figures in the datasheet are intended as an input to calculating complete thermal resistance on your own. These are only "under the control" of component manufacturer, but you are expected to add bottom-to-ambient (or top-to-ambient) calculated on your own depending on PCB and device design.

Usually SMD components are cooled through PCB and as you have already noticed junction-to-bottom is so low that actual thermal resistance will be mostly determined by thermal paths in your PCB design. Junction-to-top could be useful if you would for example decide to glue a heatsink on top of mounted package. You can then count this path in parallel with bottom and PCB path.