IGBT Heat sink

Discussion in 'Homework Help' started by g20313225, May 17, 2012.

  1. g20313225

    Thread Starter New Member

    May 17, 2012
    1
    0
    Hey can anyone please help me. I've atached my problem. I thought i_{c} was the colector curent?? Can someone please explain me how this works?
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    I guess the (rather ambiguous) question is what maximum heat sink thermal resistance is allowed for the stated conditions.

    One bit of information missing from the attachment is the maximum permissible junction temperature - Tjmax

    Suppose Tjmax=150° C

    At Ic= 50 amps the piecewise model tells you

    Vce=0.1+0.05*Ic= 0.1+0.05*50=2.6V

    The collector dissipation at 50 A will then be Vce*Ic=2.6*50=130W.

    With Tjmax=150° C and Tamb=45° C the permissible temperature differential from junction-to-ambient is Tja=150-45 or 105° C.

    With 130W collector dissipation the maximum thermal resistance from junction-to-ambient is therefore Rth(j-a)=105°/130W=0.808 °C/W

    The junction-to-(heat)sink thermal resistance Rth(j-h) is the summation of the thermal resistances from junction-to-case and case-to-(heat)sink

    Rth(j-h) = 0.2 °C/W + 0.3°C/W = 0.5 °C/W.

    So the maximum allowable heat sink thermal resistance is the total allowable junction-to-ambient thermal resistance less the IGBT's junction-to-(heat)sink thermal resistance

    Rth(heatsink)=0.808 °C/W - 0.5 °C/W= 0.308 °C/W.
     
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