I'm using a 9V unregulated DC wall-wart to supply a 5V regulator. The 5V
line is going to provide 250mA approx.
I'm using a standard TO220 voltage regulator (1.7V drop IIRC)
What spec of heatsink do I need in order to keep the regulator at reasonable
temperature, say below 50degC/120degF? I guess it depends on the maximum
voltage output of the unregulated supply under the load - is there a way to
work out what this is?
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Yes. Either get a copy of the spec's for the wall-wart or load it down
with 250mA and measure its output. It would also be nice to know what
the output voltage ripple under load looks like, or at least the size of
the filter cap in the wall-wart.
Having said that, though, we can make an educated guess that the
wall-wart's output won't rise above 15V when it's unloaded on nominal
mains, and with mains high, it'll rise to 16.5V, tops. Assuming that
the wall-wart's current output capability is much greater than 250mA
allows us to say that its maximum output voltage will be 16.5V, no
matter what, and we can use that for the rest of our calculations.
With 16.5V on the input of the regulator and the output at 5V with 250mA
going into a load, the regualtor will be dissipating the difference
between the two voltages multiplied by the current into the load, or
P = (Vin-Vout)Iload = 11.5V*0.25A = 2.875W ~ 3 watts
The reason for using a heat sink in the first place is to keep the
junction of the regulator's pass transistor from rising to a temperature
which could damage it, but most regulators now (certainly the 7805?
you'll be using) have internal overtemp protection which will shut the
regulator down if it gets too hot, but this gets to be a PITA if you
don't want your circuit to stop operating when the regulator shuts down.
So, the job then becomes one of selecting a heat sink which will keep
the regulator from shutting down under worst case conditions. That means
under maximum load with high mains and a high ambient temperature.
The 7805 has a thermal resistance of 5°C/W from junction to case, so its
junction temp will be 15°C hotter than its case when it's dissipating 3
watts.
That also means that its case temperature needs to stay below 135°C when
the regulator is dissipating 3 watts, so in order to do that you need to
find a heat sink with a thermal resistance which will keep the case
below 135°C when the ambient temperature rises to its highest. Assuming
50°C ambient, that means the heat sink has to get rid of 3 watts worth
of heat when the ambient temp is 50°C without letting the regulator's
case get to 135°C. That's an 85°C difference, so the heat sink's
thermal resistance needs to be 85°C/3W = 28.33°C/W from the surface of
the heat sink to ambient.
You can check:
http://www.aavidthermalloy.com/bin/standarda.pl?Pnum=&PgNum=1&NumPerPage=v3&Device=10&SortBy=V4
For what you need. The higher the page number, the higher the thermal
resistance, and you'll find you won't need much of a heat sink at all.
Maybe none at all, depending on what your wall-wart's output looks like
with a 250mA load on it.
Also check:
http://www.aavidthermalloy.com/products/standard/pdf/Page4.pdf
for more information on how to choose a heat sink.