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7230 Views 1 Reply Latest reply: Jan 30, 2013 10:28 AM by larrygrover RSS
Currently Being Moderated

Jan 29, 2013 7:15 PM

Heat Sink Size

How do i figure out the specs of the heat sink size i need.

I am using voltage regulator LM7805 to regulate  filtered 24 VDC Average( peak will be around 35 volts) down to 5 volts.

my maximum current draw will be 800 mA.

i like to cool semiconductor devices down so that they're only fairly warm (around 40 degrees Celcius) even though they're rated at 100 degrees celcius.

parts that run so hot look like they'll have a short lifespan. but i could be wrong.

what is a good operating temperature?

 

 

attached are the voltage regulator specs:

 

i would appreciate any help

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Attachments:
  • larrygrover Novice 1 posts since
    Jan 30, 2013
    Currently Being Moderated
    Jan 30, 2013 10:28 AM (in response to davidghofer)
    Heat Sink Size

    The maximum power dissipation will be 800 mA x (35-5)V = 24 Watts

     

    You need to know the maximum ambient temperature where the heat sink is.

    The maximum device junction temp is given as 150C and  thermal resistance junction to case as 3C/W

    Using 130 for the junction temp. and assuming the ambient is 25C, then  24 W x Tr = (130-25) = 2.52C/W

    OPS! this is less then the thermal resistance of the device alone!

    This is telling us that you need a regulator with much lower thermal resistance junction to case.

     

    If you can limit the upper voltage input range, you can reduce the the maximum power dissipated by the device

    and reduce the size of the heatsink required.

     

    You may contact me at linton627@yahoo.com if you need further assistance.

    Lawrence Grover

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