A Sealed Enclosure That Cannot Get Rid of 20 W
An enclosed 400 W power supply contains a switching MOSFET dissipating 11 W; the rest of the unit adds 9 W. The MOSFET is a TO-220 (RθJC = 1.4 °C/W) bolted through a 0.25 mm pad (RθCS = 0.6 °C/W) to an extruded heatsink rated 4.5 °C/W, a figure the vendor measured in still air at a 75 K rise. The assembly is sealed to IP54; the outside air is 40 °C and a thermocouple inside the box reads 62 °C. The MOSFET's Tj(max) is 150 °C, but the reliability plan requires Tj ≤ 110 °C at end of life. Three options are on the table: a much larger extrusion inside the box, a filtered fan cut into the cover, or a lower-loss device that reduces the MOSFET to 6 W.
1. Check the present design against the 110 °C requirement and explain why the 4.5 °C/W datasheet number is not the number to use here.
Read the worked answer
Total junction-to-ambient resistance = 4.5 + 0.6 + 1.4 = 6.5 °C/W, with the sink's ambient taken as the measured 62 °C internal air:
Tj = 62 + 11 × 6.5 = 133.5 °C
The reliability target is 110 °C, so the design misses by about 23 K - it is not marginal, it is out of spec. And 133.5 °C is the optimistic estimate, for three reasons:
- 4.5 °C/W was measured at a 75 K rise; here the fin-to-air rise would be only ~48 K, where buoyancy-driven flow is weaker and the true RθSA is higher, perhaps 5–5.5 °C/W.
- The measured 62 °C is an average of box air. Air entering the sink sits in the plume of the other 9 W and is hotter; the sink's effective ambient is above 62 °C.
- Dust loading and end-of-life fan/interface degradation are excluded from the datasheet figure but required by the reliability plan.
So the honest conclusion is that the thermal chain must be re-designed, not re-measured.
2. Is a larger heatsink inside the box a sound fix? Quantify what it would have to achieve.
Read the worked answer
Required total resistance at 110 °C with 62 °C internal air: (110 − 62)/11 = 4.36 °C/W. Allowing the interface and junction terms (2.0 °C/W), the extrusion itself would need 2.36 °C/W while dissipating 11 W in a closed box.
That is a very large natural-convection extrusion - several tenths of a metre of a deep profile - and it fails the system test as well as the arithmetic: in a sealed, fan-less box the sink's own buoyant plume recirculates, so the air it draws is progressively hotter than the 62 °C average. Halving the sink's datasheet resistance inside a sealed box typically buys far less than half the rise in practice.
The real bottleneck is the enclosure, not the extrusion. With 20 W total inside and a 22 K internal-to-external rise, the box behaves like ≈1.1 °C/W; no sink inside it can beat that. Reject this as the sole fix.
3. Compare the filtered-fan option and the 6 W device option on numbers, then recommend one.
Read the worked answer
Filtered fan. Bringing 40 °C air in and moving ~1 m/s over the sink puts the sink's local ambient near 45 °C and drops RθSA to about 2.0 °C/W. Then Tj = 45 + 11 × (2.0 + 0.6 + 1.4) = 45 + 44 = 89 °C - a 21 K margin.
6 W device. With the existing 6.5 °C/W chain and 62 °C air: Tj = 62 + 6 × 6.5 = 101 °C, a 9 K margin. Allow for the sink being slightly less efficient at the smaller rise (say 7.0 °C/W total): 62 + 42 = 104 °C - still inside 110 °C.
Recommendation. Take the 6 W device if it is genuinely available at acceptable cost. It keeps IP54 intact, adds no moving part, no filter maintenance interval, and no dependency on fan health - and 9 K of margin is real. Choose the filtered fan only if the loss cannot be reduced: it gives the larger margin (21 K) but converts a passive problem into a maintenance schedule, and a standard fan destroys the IP54 rating unless an IP54-rated filter-fan is used, whose filter must then be inspected on a defined interval. Do not accept the 'bigger sink' option, and do not accept a design whose only barrier to 150 °C Tj(max) is the reliability derating - the 110 °C target is the specification.