# How to calculate the effective junction temperature of the FPGA?

I am using the Xilinx based FPGA. Xilinx datasheet suggest the theta ja value is 7.3 °C/W and power dissipated is nearly 50W and ambient temperature at which it is operated is 85

Junction temperature= ambient temperature + thermal resistance (theta ja)power dissipation

                = 85+ 7.3*50

= 450°C


Is this junction temperature is expected or some mistake in the consideration. Anybody who worked on thermal analysis can help me out. Xilinx XPE shows the effective theta ja is 1 °C/W. I am not understanding why they have consider 1 in xpe and 7.3 in datasheet. How Theta ja will vary?

Also i want to understand what is the relation between the theta ja, jc, jb

Also as per my understanding,

Materials that are good conductors of heat (metal) have a low thermal resistance

Materials that are poor conductors of heat (plastics) have a high thermal resistance

But i want to understand why we are adding this instead of subtracting to reduce the thermal resistance

Theta ja = Theta Jc + Theta ca

Any insight would be greatly appreciated.

• Are you absolutely sure the ambient temperature is 85C? That's quite hot for this kind of device. Nov 4, 2020 at 11:29
• Yes.. I am working on avionics based application. It is operated at 85 degree Celsius ambient. Nov 4, 2020 at 11:36

## 1 Answer

Is this junction temperature is expected or some mistake in the consideration.

It's not a mistake unless you want to run the device without considering the junction to case thermal resistance. $$\\theta_{JC}\$$ is the main route for heat to travel (to a heatsink/fan) whereas $$\\theta_{JA}\$$ is a parallel route that is not anything like as good for removing heat.

Any insight would be greatly appreciated

Noted!

• Thanks for your quick reply.. My card will be conduction cooled one. Hence it will be having casing and heat spreader around them. May I know how effectively we can consider this Theta JC along with Theta JA & Theta JB. I request you to explain with example considering the above parameters.. Nov 4, 2020 at 11:35
• I've made my explanation and answered your question. Just look at the numbers and it should be obvious. If the "case" is at 40 °C and the power is flowing exclusively from junction to case at 40 watts, then if $\theta_{JC}$ is 0.02, the junction will be 1 °C warmer than the case (based on the numbers in your question). Nov 4, 2020 at 11:49