# Unregulated DC to a voltage regulator - how much heatsink?

Discussion in 'Electronic Basics' started by Jim, Feb 12, 2004.

1. ### JimGuest

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?

Many thanks,

Jim

2. ### JimGuest

Thanks for the reply, Jamie.

Do I need to cope with the fact that the *unregulated* 9V supply could be
supplying more than 9V, e.g. 12V or whatever?

Jim

3. ### bgGuest

A heatsink is rated in degrees per watt. That means that the temperature
will rise so many degrees above ambient due to the watts of power flowing
thru it. Your regulator will need to dissapate
its wattage thru the heatsink.
watts = volts times amps
12 volts - 5 volts = 7 volts across the regulator
amps = .25 amps thru the regulator
watts = 7 times .25 = almost 2 watts dissapated by the regulator
example ---
If the heatsink is rated at 10 degrees per watt, the heatsink will rise 20
degrees above ambient, which would be about 90 degrees at room temerature.
Can you get to where you want to be with this info?
bg

Jim wrote in message ...

4. ### JamieGuest

not much at 250 ma's
a little u-channel peice of alliminimum will do it.
WDf = 5*0.250 is good enough for selecting the heat sink.

5. ### JimGuest

Hi BG,

Thanks very much for a very clear and helpful answer.

The 12V max from the supply was a guess though - I still don't know what the
unregulated supply could put out in terms of voltage with the 250mA load. I
think I read that an unregulated supply with no load could output up to 1.7
* nominal = 15.3V in this case. But in that case there will be no current
anyway, so no heat.

Is there a way of estimating the voltage from the unregulated supply when
the 250mA load is in place, or do I need to cope with the possiblity of
15.3V at 250mA? BTW, I want it to work with pretty much any decent 9V
unregulated supply, so I can't just measure the one I have here.

Thanks,

Jim

6. ### bgGuest

Jim wrote in message ...
All supplies will have an output resistance. It is the output resistance
that drops the supply voltage as current is pulled from the supply. There is
no standard value for a 9 volt wart or any other supply for that matter.
Many of the three terminal regulators have protection built in to shut them
down if an over current or over power condition exists. If you heatsink the
regulator so it can handle it's full rated power dissapation, you will
probably be safe with just about any typical 9 volt wall wart.
bg

8. ### John FieldsGuest

---
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.

9. ### John FieldsGuest

---
It'll need to be a little better than that, since the space between the
regulator and the heat sink will impose its own penalty.

Check
http://www.aavidthermalloy.com/products/standard/pdf/Page4.pdf

for more details.

10. ### JimGuest

Many thanks John for taking the time to write that comprehensive reply. I
think I understand fully now. The links are very useful too.

Jim

11. ### JamieGuest

i assume your using a 7805 type reg.
to be more clear most heat sinks that you get
made have a Tempiature per watt rating..
if you are using this type of reg then most of the
heating you should be concerned about is in the IC it
self and like i said, if you want to contruct one you can
get a peice of small U-channel from your local hardware store
and that will work just fine..
to get proper wattage calculations one would need to subtract
the out from the in.
for example
W= 9V-5V=4*0.250= 1;
now lets assume a 16 volts in.
W= 16V-5V=11*0.250= 2.75
the above are just theory answers and not exact because a
7805 type reg like many other circuits do have a minimum of input
to concider.
so using these rough calculations a piece of Small Gauge U-Channel
still fits the bill!

12. ### Dr Engelbert BuxbaumGuest

Jim wrote:

If in doubt, don't estimate - measure instead. You can use a power
transistor (2N3055 or so) or a power MOSFET as variable load. Measure
the voltage across the output of the supply unit as a function of
current flowing.