Help in selecting a Heatsink.

Discussion in 'General Electronics Chat' started by FrancescoC, Jun 20, 2016.

  1. FrancescoC

    Thread Starter Member

    Nov 22, 2014
    30
    5
    Hi,
    i have a simple LM317H regulator with a pass transistor TIP32C.
    I need to select a heat-sink for the TIP32C.
    Normally i would look in the datasheet for the thermal resistance (something like 4C/Watt), but this value is missing.

    I had a look at some other charts in the datasheet but I have not been able to deduce how to select the heat-sink I need.

    I have calculate the power dissipation, of the TIP32C, by multiplying the voltage across it, times the current.
    In my case is 45V x 0.4A=18W.

    Can you please give me some help?

    regards

    Francesco C
     
    Last edited: Jun 20, 2016
  2. ian field

    Distinguished Member

    Oct 27, 2012
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    782
    45V into a 7805 is probably exceeding its rating - or at the very least, cutting it fine.................

    18W probably means a bigger heatsink than you had in mind - you could always cheat and use a leftover CPU cooler + fan.
     
  3. FrancescoC

    Thread Starter Member

    Nov 22, 2014
    30
    5
    Hi Ian,
    there was an Edit to my post. The regulator is an LM317H. I have 50Vdcin and 5Vout.
    At that difference in voltage the LM317H can only supply 100mA. Therefore the TIP32C is used for the current, up to 0.4A.

    Regards

    Francesco C
     
  4. crutschow

    Expert

    Mar 14, 2008
    12,991
    3,227
    This data sheet states that a TIP32C can dissipate 40W@25°C case temperature.
    Assuming this is at the 150°C maximum junction temperature, than the junction-to-case thermal resistance is (150-25) / 40 = 3.125 °C/W.
    So for 18W dissipation, and assuming a maximum ambient of 50°C, the maximum junction-ambient thermal resistance you can have is 100°C/18W = 5.556 °C/W.
    Subtracting the 3.125 °C/W of the transistor case thermal resistance gives a maximum allowed heat sink thermal resistance of 5.556 - 3.125 = 2.431 °C/W, the lower the better.
    That's assuming you have a good thermal grease layer between the case and heat sink so it doesn't add significantly to the thermal resistance.
    If you need an electrical insulating washer between the case and heatsink than you need to subtract its thermal resistance from the heatsink requirement.
     
    Last edited: Jun 21, 2016
    FrancescoC likes this.
  5. FrancescoC

    Thread Starter Member

    Nov 22, 2014
    30
    5
    Hi crutschow,
    thank you very much for your help.
    Your answer is excellent. It gives a very good explanation on how to get the final result.

    Regards

    Francesco C
     
  6. bertus

    Administrator

    Apr 5, 2008
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  7. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
    4,769
    969
    I don't know why anyone would use a linear regulator these days.. even more so when dropping so much voltage..
     
  8. ian field

    Distinguished Member

    Oct 27, 2012
    4,413
    782
    A hysteretic buck regulator would be the simplest option. For the off the shelf route - it might be possible to get away with a 48V telecom gear switcher. A string of diodes could always go in series with the input pin to drop a couple of volts.
     
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