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I am working on a project that involes an inverter made with IGBTs. The device I selected for builing the H bridge is this: STGW60H65DFB. One of my roles in this project is also to select a suitable heat-sink for the transistors.

I have done all the calculation needed for the power losses and I found that a single device (STGW60H65DFB) dissipates about 60 W, in particular about 50 W are dissipated on the switch and 10 W on the antiparallel diode. About the thermal data, looking at the datasheet, it says:

RthJC Thermal resistance junction-case IGBT 0.4 °C/W

RthJC Thermal resistance junction-case diode 1.14 °C/W

Now my questions are: should I do the thermal calculations separately for the switch and the diode? Or can I assume an "average" thermal resistance of both?

If I should do separately, should I consider separated powers for diode and switch or use in both the cases the same power (the sum, so 60 Watts)?

Thank you.

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  • \$\begingroup\$ At 60 watts, you will be using air cooling? What voltages are you converting? High voltages need more respect, to avoid arc-over as dust builds up over the years. This requires more spacing between traces. \$\endgroup\$
    – analogsystemsrf
    Jul 14, 2017 at 13:41
  • \$\begingroup\$ I am converting 425 Vdc to 240Vrms to inject a power of 10kW to the power line with an inductor filter of 2mH and a current through it of 60A peak to peak. My question is about how to choose the right heat sink. Yes, I will use air cooling with a fan. \$\endgroup\$
    – Simus994
    Jul 14, 2017 at 13:50

1 Answer 1

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You have to do it separately because IGBT and diode will have different junction temperatures. In a first approximation you can assume that the heat sink temperature for both devices is the same.

dT,ibgt,jc = 50W * 0.4K/W = 20K

dT,diode,jc = 10 * 1.124W = 11.2K

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