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Verification of approach to calculating power dissipation, temperature and use of heatsink

Andrew J

Hi,

 

This is a generic question about determining need for and selection of a heatsink for a part - am I on the right lines?

 

From the data sheet for a part:

Tj - Junction Temperature: 125c

Rja - Junction-to-ambient: 40c/w

Rjc - Junction-to-case: 3 c/w

 

For the part, if I measure:

Vin: 2.93v

Vout: 1.04v

Iout: 1.53A

That would give me power dissipated of (2.93 - 1.04) * 1.53 = 2.8917W

With ambient temperature set at 25c, then:

  • No heatsink, junction temperature will be: Pdiss * Rja + ambient = 2.8917 * 40 + 25 = 140.668c [heatsink definitely needed]
  • With heatsink, sink-to-ambient resistance Rha 26 c/w, junction temperature will be: Pdiss * (Rjc + Rha) + ambient = 2.8917 * (3 + 26) + 25 = 108.8593c  [way too close for comfort]
  • With heatsink, sink-to-ambient resistance Rha 16 c/w, junction temperature will be: Pdiss * (Rjc + Rha) + ambient = 2.8917 * (3 + 16) + 25 = 79.9423c [much better - could improve with a bigger heatsink]

 

I haven't taken into account thermal paste between the part and the heatsink.  I'm assuming that the thermal resistance of the paste is taken off the thermal resistance of the heatsink to improve it's efficiency?

  • I'm confused about how to do this as the ratings I've seen are  xW/mK (e.g. 2.4W/mK) or areaC/W (e.g. 0.061cm²C/W).  However something like: Pdiss * (Rjc + Rha - Rpaste) + ambient.

 

I've tried to distill this from reading a variety of sources so I'm just checking that I'm on the right track.  

 

Thanks,

 

Andrew

Parents

  • I think I may have gone off track when I said above that the MOSFET may be the part to give me some grief - I think I was doing the thermal calculation wholly incorrectly (looking at the voltage drop Drain to Source * current).  I think I should have been doing it as follows, could someone confirm?

     

    IRF3205 - datasheet

    Rds(on) = 8mOhm MAX

    Tj = 175c

    Pdiss = 200W @ 25c  / linear derating 1.3 w/c

    Rja = 62 c/w

    Rjc = 0.75 c/w

    Rcs = 0.5 c/w (case to heatsink!)

     

    So given a current of 3A, then P = 3x3x0.008 = 0.072w

    With no heatsink, junction temperature = 0.072 * 62 = 4.464c + ambient.

     

    Ambient would have to rise to around 88c to cause a problem 175 -((88c-25c) * 1.3) = 93.1c max junction temperature vs 88+4.64 = 92.64 actual junction temperature.

     

    Either I'm still off track, or I don't need a heatsink.

  • in reply to Andrew J

    That sounds about right - when fully on the FET won't need to dissipate much heat.

    However the biggest heat source can be when the FET is turning on or off - the slew rate and the frequency can cause the power during switching to be very significant because the resistance in the FET is going from a very high value to a low value or vice versa.

    So the heat will usually depend on how often the FET is switched on or off and how quickly the switching takes place. The gate capacitance can slow down switching so there is often an emphasis on driving the gate with a low impedance driver.

  • in reply to dougw

    Thanks Doug.

     

    Calculating this seems to be fairly complicated but from the (limited) research I've done, then...

     

    Switching frequency, F, of 200kHz

    Current: 3A

    Vin: 24V

    Rise time, Tr: 101ns

    Fall time, Tf: 65ns

     

    Power = (Tr*I*V + Tf*I*V) * F * 0.5 = (0.000000101*3*24 + 0.000000065*3*24) * 200000 * 0.5 = 1.2W  [this formula I'm not sure on TBH, but the answer seems not unreasonable]

     

    Temp with no heatsink would be 1.2*62 = 74.4 + ambient.

    Ambient would have to rise to around 56c to cause a problem 175 -((56c-25c) * 1.3) = 134.7c max junction temperature vs 56+74.4 = 130.4c actual junction temperature.

     

    56c might be conceivable depending upon the temperature rise of other components, ventilation in a case etc... so a small heatsink would be useful.

Reply
  • in reply to dougw

    Thanks Doug.

     

    Calculating this seems to be fairly complicated but from the (limited) research I've done, then...

     

    Switching frequency, F, of 200kHz

    Current: 3A

    Vin: 24V

    Rise time, Tr: 101ns

    Fall time, Tf: 65ns

     

    Power = (Tr*I*V + Tf*I*V) * F * 0.5 = (0.000000101*3*24 + 0.000000065*3*24) * 200000 * 0.5 = 1.2W  [this formula I'm not sure on TBH, but the answer seems not unreasonable]

     

    Temp with no heatsink would be 1.2*62 = 74.4 + ambient.

    Ambient would have to rise to around 56c to cause a problem 175 -((56c-25c) * 1.3) = 134.7c max junction temperature vs 56+74.4 = 130.4c actual junction temperature.

     

    56c might be conceivable depending upon the temperature rise of other components, ventilation in a case etc... so a small heatsink would be useful.

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