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two IGBT thermal resistances for Rjc

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jeffed1969

Mechanical
Aug 7, 2001
3
Introduction:
I am working on thermal management of electronic equipment. As you may know, these folks deal in terms of thermal resistance rather than conductivity as it more closely resembles an electronic circuit with this approach. I have been using resistance = delta temp over heat power or R = AT/Q (for lack of symbol font). For the thermal circuit from the junction to the case to the sink to the ambient air, Rjc + Rcs + Rsa = (Tj - Ta)/Q.

Problem:
The data sheet for one particular IGBT lists TWO thermal resistances for the junction to the case, one for the transistor (.16 C/W) and one for the diode (.30 C/W). Other data sheets list only one value for this property.

Question:
How does one use this information in the general equation above?
 
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Well, you have to go back to the electric guy who designed the circuit and ask him for the power dissipated in the transistor and and the power dissipated in the diode.

Then use the following equations:

Tjtrans= Ptrans*Rthjctrans+(Ptrans+Pdiode)*(Rthcs+Rthsa)+Ta
Tjdiode= Pdiode*Rthjcdiode+(Ptrans+Pdiode)*(Rthcs+Rthsa)+Ta

Then choose Rthsa in a way that neither the transistor nor the diode gets to hot. Usually both power dissipations vary with some electrical parameters. So be sure to catch the worst case for both components.
 
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