Hello,
Is it possible to correct my thermal resistance calculation of a radiator with fan? It’s just an exercise and checking if my calculations are correct.
Thank you
I have a closed drawer of dimension 403x600x130.55 mm. Inside there are 16 modules (assembled printed circuits) each with 2 Mosfet transistors. The power dissipated is at the level of the transistors. (For a module the distribution of power in the 2 transistors is 48% and 52%).
There is a single heatsink capable of supporting a total power of 4800W. The thermal resistance of the heatsink is Rth=0.015°C/W and at the entrance to the drawer there are 3 fans with a flow rate of 310m³/h and an outlet to evacuate the hot air blown by the fans. The ambient temperature is 26°C in the room. Inside the drawer, the temperature is 45°C. The air leaving the drawer is around 60°C.
Characteristics of Mosfet transistors (datasheet):
Pd=700W
Tjmax=150°C
Rthjc= 0.18°C/W
Rthcs=0.05°C/W
The power distribution of the modules is done in rows of 4 modules:
→ 1st row: 4 modules of 375W each
→ 2nd row: 4 modules of 300W each
→ 3rd row: 4 modules of 275W each
→ 4th row: 4 modules of 250W each
For a total of 4800W
The goal is to check the value of the thermal resistance of the radiator by calculation.
I have a question concerning the ambient temperature to be taken into account in the calculations, is it the ambient temperature inside the drawer (45°C) or that outside (26°C)?
If it is the indoor ambient temperature that I must take into account in my calculations, is it the one with radiator and fan, ie 45°C or the one without radiator and fan (55°C-60°C)?
Results for ambient temperature Ta=45°C:
Knowing that for a module, the distribution of power in the 2 transistors are 48% and 52%. So we have for the 375W module, a distribution of 180W for transistor 1 and 195W for transistor 2.
Tj = Tambient + (Rthsa cooler * Total power in the cooler) + Rthcs contact paste + (Rthjc of the transistor from which the greatest power passes * Power dissipated by this transistor) <= 150
150=45+(Rthsa*4800)+(0.05+0.18)195
150=45+(Rthsa4800)+44.85
Rthsa=(150-44.85-45)/4800 = 0.0125°C/W
The value seems to be closer to that given at the start, which is 0.015°C/W.
Does my reasoning seem correct?