Referring to the MP8795HGLE-Z data sheet, Page 8 Efficiency with Vo=2.5V, Io=20A is 90%.
This means MP8795HGLE-Z will dissipate 5.56W.
On same page Thermal results shows 85C case temperature which 60C temperate rise.
Using these values to compute thermal resistance results in 10.8 C/W.
However thermal resistance junction to ambient given in datasheet is 29 C/W.
What explains this discrepancy? Is the external inductor included in the efficiency graph?
If so how do I find out what the power dissipation of MP8795HGLE-Z only?
Hello Joel,
The power loss 5.56 W calculated describes the power loss of the entire circuit which includes conduction and switching losses as well. The 29 C/W temperature is the measured temperature of the part on eval board. More information is attached below.
In order to calculate the power dissipation of the part only, first find calculate the input power and output power of the part(probe switch node voltage and current from the part) then the difference of the two will represent the power dissipation of the part.
Hope this helps.
Best,
Saquib
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