# Calculate junction temperature of MOSFET during pulse

I want to find out how hot the junction gets during a 300 us pulse of 400 A peak current.

Datasheet of MOSFET, Datasheet

From the datasheet I came across this graph,

For a pulse of 300 us or 3*10^-4 seconds, I found Zth = 0.04 K/W

So, if I = 400 A and Rdson = 0.07 ohms (From datasheet), then

P = I^2*R = 400^2 * 0.07 = 11200 W

Then, Zth = 0.04*11200 = 448 K = 175 degrees Celsius

Therefore Junction Temperature = 175 + Ambient(Assuming 30) = 175+30 = 205 C

Is that correct ?

• It looks reasonable. However, you are way out of the safe operating area curve above that graph. Commented Oct 20, 2017 at 17:38
• One other thing you'll have to worry about is the thermal resistance to your ambient. Is the MOSFET mounted to a heatsink? To a controlled baseplate? With what sort of attach (thermal paste, tape, etc.). Commented Oct 20, 2017 at 17:50
• Also, the 448 K is the rise above ambient Commented Oct 20, 2017 at 18:35
• The heatsink won't help for the reasons that Analogsystemsrf laid out. Basically, when heat is generated over a short time period, it does not have time to travel to a heatsink. All the heat has to be temporarily stored in the area immediately surrounding the silicon junction. Heatsinks help with steady-state heat dissipation. Commented Oct 21, 2017 at 4:56
• The easiest way to see the problem is to carry out all calculations in K until the end. 0.04 * 11,200 = a 448K temp rise. Ambient is 303K. Junction temp is therefore 303 + 448 = 751K. 751 - 273 = 478 C. Commented Oct 21, 2017 at 5:21

Your math looks reasonable.. however that device will not withstand anywhere near 400A. Especially not for 300uS.

Looking at the safe operating curve... The maximum you can do for 300uS is about 100A if your Vds is less than 20V.

So basically the temperature thing is mute. You will let out the smoke.

• Thanks for that answer, How would a heatsink fair ?, since its limited by Tjmax as I read in the datasheet Commented Oct 20, 2017 at 18:50
• Is this regardless of heat sink being used?
– GNZ
Commented Oct 20, 2017 at 18:50
• I think its without the heatsink, as its from the datasheet Commented Oct 20, 2017 at 18:51
• On the graph it says: "Tc = 25 C". It means to make the graph valid, you have to provide whatever heat sink is needed to keep the case at 25 C.
– GNZ
Commented Oct 20, 2017 at 18:53
• @Deadshot Heat-sink or no you will fry this device. Commented Oct 20, 2017 at 19:42

Lets put some genuine theory into this topic.

1 micron cube of silicon has Thermal Timeconstant of 11.4 nanoseconds.

10 micron cube of silicon has Timeconstant 100X slower, or 1140 nanoSeconds (1.14uS)

100 micron cube of silicon has Timeconstant another 100X slower, or 114 microSeconds.

With wafers often processed as 300 micron thick silicon discs, the Timeconstant is 3*3 slower, or 1000 uS or 1 millisecond.

Now let's put this 0.3mm thick slab of silicon atop a 3,000 micron (3mm) thick slab of copper (the TO-220 mounting tab?) which with 10X more thickness makes the Thermal Timeconstant yet another 10X10 slower, to 100 milliSeconds. We can do this because the Thermal Taus of silicon and of copper are nearly the same.

What does all this mean? Unless the pulse duration is > 0.1 seconds, almost all the heat HAS TO REMAIN inside the silicon/copper. The thermal capacity of that bi-metallic structure will be storing the heat during that 0.1 second (or shorter) pulse.

You can set up an Finite Element Model. Or use these following rules-of-thumb.

In 11.4 nanoseconds, most of the heat propagates less than 1 micron.

In 1140 nanoseconds, most of the heat propagates less than 10 microns.

In 114000 nanoseconds (114 microSeconds) most of the heat propagates less than 100 microns.

etc etc