Well, if you assume that all heat generated by the triac or SSR is going
to be transferred to whatever material you attach to it, then a simple
application of Newton's Law of Cooling would probably provide an
adequate upper bound for the area needed.  According to Fundamentals of
Engineering Thermodynamics (Shapiro and Moran, 5th ed.) the law is as
follows:

Q = h x A x (Tb - Tf)

Where,
Q - Heat Dissipated
A - Area of body
Tb - Temp. of body
Tf - Temp of surrounding fluid (air in this case)
h - Convection Heat Transfer Coefficient

In this case Q is the heat generated by your device (in Watts), Tb would
be its operation temperature (units don't really matter here since the
equation just uses the difference of two temps), and Tf would be the
ambient temperature of the air surrounding it.  The magical number in
this case is h, which is not a thermodynamic property, but an empirical
parameter.  For free convection (no fan being used) h is listed as being
between 10 - 100 W/m^2 K (Watts per square meter Kelvin).  To get an
upper bound I would choose h to be 10, just to be safe.  Solving for A
will give you the necessary area for adequate cooling.  Of course you
can increase the surface area greatly by adding fins like PC heat sinks
have.

Hope this helps and wasn't too poorly worded.
-----------------------------------------------------------------
Ryan Etwell


-----Original Message-----
From: pic microcontroller discussion list
[mailto:PICLIST@MITVMA.MIT.EDU] On Behalf Of Dan Oelke
Sent: December 23, 2003 12:26 AM
To: PICLIST@MITVMA.MIT.EDU
Subject: Re: [ee]: Are solid state relays TTL devices?

Along this line - can anyone tell me, a electronics hobbyist, how to do
calculations as far as how big a heatsink is needed for a SSR or triac?

For instance I found the power dissipation curves for a triac.  From
that
and my system knowledge I know how many watts of power need to be
dissipated.  What I don't understand is what the C/W numbers are used
for.

In one application I am kicking around in my head I am thinking of just
bolting an SSR to a big hunk of steel.  What I didn't know was how to
figure out if that would be sufficient or if I should use the piece of
scrap aluminum I have here instead.  I'm sure I could post it here and
have
half the people yell at me for being stupid and saying it will burn up
and
the other half saying no problem - but I'd like to be able to do the
math
myself and have some level of confidence that it will work.

Please be gentle as I'm a newbie in this area.  Even a couple of good
references to books or web sites would be great.

Thanks,
Dan


>Yes, it will burn up and fail "on" unless you put a good-sized heatsink
>on it. The 40A is sort of a theoretical maximum, but it should be okay
>at 10A with an adequate HS. You need to keep the metal base plate of
>the unit as cool as practical, pick the number of degrees C/W from the
>maximum temperature spec and the maximum ambient, and try to do better.

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