MOSFET thermal resistance

I am trying to calculate the total thermal resistance of my FET to see what heat sinks are required, are the following calculations correct?

$$Power = \frac{T_{jmax} - T_a}{\theta_{JC} + \theta_{CA}}$$

$$\frac{175 - 25}{0.29 + 40} = 3.7W$$

The MOSFET I am using is IRFP4468PBF.

• The total thermal resistance is strongly influenced by layout; it is not possible to help with this without that information. – Peter Smith Dec 28 '18 at 15:23
• I have added the thermal distance parameters – Daniel Bashy Dec 28 '18 at 15:50
• The parameters have notes and in particular both Ja and Jc refer to note 8: " When mounted on 1" square PCB (FR-4 or G-10 Material). For recommended footprint and soldering techniques refer to application note #AN-994" App note link infineon.com/dgdl/… – Peter Smith Dec 28 '18 at 16:01
• You have calculated the maximum power dissipation for the device without a heat sink when mounted on a 1" square pad; what we really need to know is the application circuit so the actual power dissipation can be evaluated against the possible need for a heatsink. – Peter Smith Dec 28 '18 at 16:11
• the total power dissipation I have calculated to be 235W, so i need a heat sink which can handle 231W right ? – Daniel Bashy Dec 28 '18 at 16:13

You have calculated that your application circuit will dissipate 235W. To keep the junction below a reasonable temperature (I do not like to get closer to the maximum than perhaps 50C) lets set a maximum junction of 125C.

For your device, that means for an ambient condition of 25C and a maximum junction of 125C (100C temperature rise), you will need a sink that has a total thermal resistance of 0.4255C/W.

Unfortunately, the junction to case and case to sink parameters, when added together exceed this by a significant margin (0.53C/W).

The best in this situation would be to use multiple devices in parallel if possible.

The contribution of the above parameters alone will cause a temperature rise of 124.55C (for a junction temperature of 149.55C for 25C ambient), so the heat sink would need to have a thermal resistance of no more than 0.1C/W and even then the device will be running at the maximum permitted junction temperature which is highly not recommended.

Operating at absolute maximum ratings will definitely shorten the device life quite apart from being hot enough to be dangerous.

Although such heat sinks exist, they would have forced air and be very bulky, as you can see from this list and as you can see, they are quite expensive.

Note that this analysis does not take into account a heat sink that has imperfect fitting, so no margin is available.

So my suggestion is to go back to the design and find a method of achieving your requirements without such a large amount of heat in one place.

Maximum power handling

This is an area that is often misunderstood; the part you are using has a maximum power capability limited by the internals of the part, but the actual maximum power dissipation allowed is determined by the thermal resistance.

From this app note from Infineon there is a graph of maximum power handling at ambient with no heat sinks:

As you can see, the D2PAK maxes out at just over 5W for 25C ambient.

• thank you for the reply, If i were to put to MOSFETs in parallel does that mean the 235W would split equally between them ? – Daniel Bashy Dec 30 '18 at 10:17
• so would each MOSFET dissipate 117W and would require a heatsink with a thermal resistance of 0.7C/W mounted on each device ?? – Daniel Bashy Dec 30 '18 at 10:36
• Good answer. I added an answer in response to your "Although such heat sinks exist, they would have forced air and be very bulky, as you can see from this list and as you can see, they are quite expensive." – Misunderstood Jan 24 '19 at 5:14

The thermal resistance of standard copper foil (1 ounce/foot^2) is 70 degree Centigrade / watt per square of foil, the heat flowing from edge to opposite edge.

For any size square.

Your calculation for the termal resistances for MOSFETS can be found in the datasheet. The following table displays some typical ones.

These for the IRFP4468PbF are indeed Rjc = 0.29 and Rja=40.

If you want to calculate the sink you require, you should calculate Rcs + Rsa ( thermal resistance case to sink + thermal resistance sink to ambient). These are both defined by the sink you are going to use. You can see that on the following picture:

You also need to define or have defined what your power dissipation will be. Otherwise you have 2 unknowns.

If you know the power and you have calculated Rcs + Rsa, you can arbitrarily choose a heatsink that fits the limitations. (look at their datasheets)

A sidenote: Like you see on the picture, you always go from junction to ambient. This means that if you use Rja that you will not use Rjc anymore, because Rja = Rjc + Rca. So although Rca is rather small, compared to Rjc, this is a mistake in your calculation.

• my power dissipation is 235W, for the heatsink i just have a big block of aluminium – Daniel Bashy Dec 28 '18 at 15:52
• what are the dimensions of your block aluminium? How are you going to attach your sink to the block? – J. Joly Dec 28 '18 at 15:57
• the formula you have just stated is the same one as mine, I have re arranged it for pd ? – Daniel Bashy Dec 28 '18 at 16:02
• I know that you rearranged yours, but it is not the same, look at the sidenote i added. You made a mistake, reading the values from the datasheet. – J. Joly Dec 28 '18 at 16:06
• Im not sure what you mean, how have I calculated it wrong is from the data sheet – Daniel Bashy Dec 28 '18 at 16:14

Although such heat sinks exist, they would have forced air and be very bulky, as you can see from this list and as you can see, they are quite expensive.
- @Peter Smith

This may not be true. Depends on space and enclosure requirements.

I design LED strips. Thermal management is a huge issue.

What I do here is place the component at the the edge of the PCB with the hole towards the edge.

I would put a strip of copper, as thick, long, and wide as possible on the component side.

Add mounting holes to this copper strip.

I would then attach a bar of aluminum on to this copper strip.
I then attach a heatsink to the aluminum bar.
The two 12" heatsinks cost $5 eack at heatsinkusa.com (pay by the inch). The 560 mm" aluminum bar cost under$2.

Here I used a copper bar attached to a copper water pipe with ice water running through it. Materials cost about \$3 per foot.
This strip has 3W x 16 LEDs every 0.7"
Temperature differential between pipe and LED case was a couple of degrees.

Cree XP

Luxeon Rebel

This is what happens when the screws are not tight

My thinking here is mounting the heatsink to the component side rather than using thermal vias keeps the thermal resistance very low. There is a copper path from the LED's thermal pad to the copper pipe. Screwing the PCB to the copper bar reduces thermal resistance significantly. I did not need thermal paste. If I would have needed better thermal performance I would have use a soft thin copper film rather than a traditional TIM.

• Interesting solution. It's always a compromise between thermal relief to comply with reflow soldering and heatsink connection. But it seems you don't solder the LED to the heatsink pad. Do you? Another question: Why don't you extend the copper pad further under the copper bar? – Fredled Jan 24 '19 at 11:53