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Q&A

Case temperature of MOSFET

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Homework problem

A MOSFET has a power dissipation of $P_{d} = 10\text{W}$. The MOSFET is mounted on a heatsink. There is an isolation pad between the MOSFET case and the heatsink. We are given the thermal resistances: -

Junction to case: $R_{th,jc}= 1.7 : \frac{°\text{C}}{\text{W}}$

Isolation pad: $R_{th,cs}= 1.3 : \frac{°\text{C}}{\text{W}}$

Heatsink to ambient air: $R_{th,sa}= 2.0 : \frac{°\text{C}}{\text{W}}$

The ambient temperature around the MOSFET is 25°C. What is the temperature of the MOSFET case?

My attempt

The scenario must look something like this: -

Image alt text

Here I have represented the thermal resistances as resistors, the dissipated power as a current source, and the temperatures as potentials. From this I calculate the following: -

$$P_d = \frac{T_c-T_a}{R_{th,cs}+R_{th,sa}} \Leftrightarrow$$ $$T_c = P_d(R_{th,cs}+R_{th,sa})+T_a = (10\cdot 3.3)°C + 25°C = 58°C$$

So the case must be 58°C.

Is my thought process/calculations correct or have I misunderstood something? Moreover, are calculations like these useful/used when designing circuits or are they only good for "back of the envelope" calculations?

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1 comment thread

Is this homework? Can you provide a link to the datasheet for the MOSFET? (2 comments)

1 answer

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Yes, you seem to understand the concept.

From the case to the ambient air, there will be 1.3°C per watt across the pad, and another 2.0°C per watt across the heatsink. In total, there will be 3.3°C per watt from case to ambient air. Since the part is dissipating 10 W, the case will be 33°C higher than ambient, in steady state. Since ambient is 25°C, the case will be at 25°C + 33°C = 58°C.

To take this a step further, we can also compute the junction temperature. It will be another (10 W)(1.7°C/W) = 17°C higher, so 75°C.

Yes, it really is this simple.

are calculations like these useful/used when designing circuits or are they only good for "back of the envelope" calculations?

They are definitely calculations you do during circuit design whenever you think heat might be an issue. Most of the time, a device is specified for maximum internal operating temperature. 150°C is a common upper limit for silicon semiconductors, for example. It is common to find the junction to ambient thermal resistance, then use that to find the maximum power dissipation assuming an ambient temperature.

Or, you work it backwards to find the maximum ambient temperature your circuit will work in without frying the device. If that is too low, then you have to do something different, like use a bigger heatsink, spread the dissipation across multiple devices, or find a way to dissipate less power in the first place.

Of course you always want some margin. Stuff happens, and some things are not under your control. The heat sink spec is probably what it does when freshly out of the box. A layer of dust after a year of operation can significantly increase the thermal resistance. And the manufacturer spec will be with some assumptions of how air is moving. Those may not be valid inside your box, for example.

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1 comment thread

Works for me (1 comment)

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