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.

enter image description here

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}\$


2 Answers 2


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.

  • \$\begingroup\$ Excellent answer, very clear! As a follow up, you described the top thermal resistance as being in parallel - the mental image I have is a parallel resistor that is much larger than its counterpart (the bottom path), and from electronics 101 we know that the equivalent series resistance of R1 in parallel with a much larger R2 is roughly equal to R1 - is this an equivalent analogy that reflects what you are saying? \$\endgroup\$ Jan 28, 2023 at 12:25
  • 1
    \$\begingroup\$ That is correct @daviegravee \$\endgroup\$
    – Andy aka
    Jan 28, 2023 at 12:42
  • 1
    \$\begingroup\$ --> in fact, when it comes to more complex thermal modelling, we can use an electrical simulator like micro-cap or LTspice. Ambient is "ground" and, we inject a current into the resistance (where the current represents power in watts). Whatever voltage we see at the non-ambient terminal of the resistor is the temperature above ambient. You can probably see how this can grow with different thermal resistances in parallel and series. However, it really comes into its own when modelling both thermal resistances and thermal lag (a capacitor is used for lag). \$\endgroup\$
    – Andy aka
    Jan 28, 2023 at 12:48
  • \$\begingroup\$ I briefly encountered capacitors as a way of capturing the concept of thermal inertia just a few hours ago and haven't looked too much further into it. Do inductors play a role in thermal modelling too? \$\endgroup\$ Jan 28, 2023 at 12:54
  • 1
    \$\begingroup\$ @daviegravee not that I'm aware of. \$\endgroup\$
    – Andy aka
    Jan 28, 2023 at 13:04

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.

  • \$\begingroup\$ Excellent, i'm researching surface mount device heatsinking strategies - with regards to heatsinking the top of the device, does properly mounting a heatsink to the top of the MOSFET case effectively reduce \$R_{JC}(Top)\$ to some smaller value? \$\endgroup\$ Jan 28, 2023 at 12:32
  • 1
    \$\begingroup\$ No, it adds to it. Even with ideal heatsink you will never get lower thermal resistance than published Rjc. (So, except for very specific/corner-case situations, top-side heatsink is usually not the most useful or practical way to go.) Think as serial/parallel combination of resistors: R[jundtion-ambient] = 1 / ( 1 / (R[junction-bottom] + R[pcb-ambient]) + 1 / (R[junction-top] + R[top-ambient])) \$\endgroup\$
    – Martin
    Jan 28, 2023 at 12:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.