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I need to choose a heatsink and a fan for cooling a three phase rectifier. Let's say that the total power dissipation is 1.8kW and that the rectifier is made of three half bridge IGBT/diode modules where each switch dissipates 300W.

With Rth,jc (junction to case) for the IGBT and diode and Rth,cs (case to sink) I have calculated Rth,sa (sink to ambient) thermal resistance of a heatsink that should be used.

However, since there are six IGBT/diode modules I was wondering if I need to divide the final Rth,sa by 6. If I do this I get unreasonably (in-my-not-a-big-expert opinion) low Rth,sa of 0.015K/W and would require a heatsink of 240x120x300mm size with forced air cooling in which case the air velocity should be 20m/s.

This particular heatsink weighs 9kg and this air velocity on a Beaufort wind scale corresponds to number 9 for which the land conditions are: Twigs break off trees; generally impedes progress according to Wikipedia. So I have strong doubts about this...

Is this value of Rth unreasonable or I should stick to 0.9K/W ?

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    \$\begingroup\$ 20m/s is 72km/h that's pretty windy. \$\endgroup\$ – Jasen Jan 24 '18 at 8:51
  • \$\begingroup\$ If the average dissipation per diode is 300 watts then you need 6x the thermal conductance. \$\endgroup\$ – Andy aka Jan 24 '18 at 9:26
  • \$\begingroup\$ @MarkoP: Are you building a SMPS for an electric locomotive? ...no. That wouldn't be 3 phase; but what else? \$\endgroup\$ – Curd Jan 24 '18 at 11:29
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300W and 0.9K/W already gives 270K temperature difference.

Given the most power semiconductors are specified for 125°C to 150°C junction temperature and ambient temperature may reach 30°C at most locations, your headroom is at most 120K, not 270K.

Recalculate your heatsink.

Oh, and if you had to dissipate 1.8kW, think of a hair dryer and the storm it creates.

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A typical "budget" for the temperature drop sink-to-ambient is e.g. dT_sa = 40K (depends on chosen T_junction, T_ambient, Rth_jc, and Rth_cs). For 1.8kW semiconductor losses you need

Rth_sa = 40/1800 = 0.022 K/W for your heat sink.

With forced convection (strong fan as e.g. used in CPU cooling) and commercial heat sinks (aluminium extruded) you can estimate the heat sink size (including fan) as

Vol[dm3] = 1/(CSPI*Rth_sa) with CSPI = 2..5 for forced convection

Vol = 1/(5*0.022) = 9dm3

Your heat sink has 8.6dm3 and looks well chosen, but recalculate with your real "temperature budget" dT_sa.

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