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
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 firstname.lastname@example.org if you need further assistance.