Another Heat Sink Theory Question ???

J

jalbers@bsu.edu

Guest
What is wrong with my reasoning?

Consider a square piece of aluminum 4cm x 4cm x 2 mm (thick) . The
thermal resistance of this piece of aluminum is .005 deg C/W according
to http://www.novelconceptsinc.com/calculators-slab-thermal-resistance.cgi
using .04, .04, .002, and 250 (thermal conductivity).

Look at a commercial designed heat sink for example:
http://parts.digikey.com/1/parts/988299-heat-sink-3-1-25-compact-501603b00000g.html
This heat sink has a thermal resistance of 7.8 deg C/W (no forced
air).

It looks like the simple square piece of aluminum .005 deg C/W would
be better than the 7.8 deg C/W commercial heat sink. I know that this
can't be true, but why?
 
"jalbers@bsu.edu" <jalbers@bsu.edu> writes:
It looks like the simple square piece of aluminum .005 deg C/W would
be better than the 7.8 deg C/W commercial heat sink. I know that this
can't be true, but why?
Click to expand...
hmmm... wouldn't you want a heatsink to dissipate *more* heat per
watt, not *less* heat per watt?
 
DJ Delorie <dj@delorie.com> writes:
hmmm... wouldn't you want a heatsink to dissipate *more* heat per
watt, not *less* heat per watt?
Click to expand...
Of course, the key is "dissipate". If you were looking at just
*conduction*, yeah, bigger is better. But you need surface area to
get the built-up heat out of the heatsink, too.
 
On Thu, 16 Jul 2009 12:02:38 -0700 (PDT), "jalbers@bsu.edu"
<jalbers@bsu.edu> wrote:

What is wrong with my reasoning?

Consider a square piece of aluminum 4cm x 4cm x 2 mm (thick) . The
thermal resistance of this piece of aluminum is .005 deg C/W according
to http://www.novelconceptsinc.com/calculators-slab-thermal-resistance.cgi
using .04, .04, .002, and 250 (thermal conductivity).

Look at a commercial designed heat sink for example:
http://parts.digikey.com/1/parts/988299-heat-sink-3-1-25-compact-501603b00000g.html
This heat sink has a thermal resistance of 7.8 deg C/W (no forced
air).

It looks like the simple square piece of aluminum .005 deg C/W would
be better than the 7.8 deg C/W commercial heat sink. I know that this
can't be true, but why?
Click to expand...
The calculation appears to be the thermal resistance from one side of
the aluminum plate to the other.

The heatsink theta is the thermal resistance from the heatsink to
still air.

The theta of the 4 cm aluminum chunk to still air would be huge,
vaguely in the 150 K/W sort of ballpark.

John
 
jalbers@bsu.edu wrote:
What is wrong with my reasoning?

Consider a square piece of aluminum 4cm x 4cm x 2 mm (thick) . The
thermal resistance of this piece of aluminum is .005 deg C/W according
to
http://www.novelconceptsinc.com/calculators-slab-thermal-resistance.cgi
using .04, .04, .002, and 250 (thermal conductivity).

Look at a commercial designed heat sink for example:
http://parts.digikey.com/1/parts/988299-heat-sink-3-1-25-compact-501603b00000g.html
This heat sink has a thermal resistance of 7.8 deg C/W (no forced
air).

It looks like the simple square piece of aluminum .005 deg C/W would
be better than the 7.8 deg C/W commercial heat sink. I know that this
can't be true, but why?
Click to expand...
Good question.
It's because they are not actually the same figure, even though they have
the same units, it's like comparing apples to oranges.
Proper heatsinks are rated as the dissipation into free air, as that's what
matters for most electronics designs. Straight thermal conductivity of a
material is not the same thing, it's not taking into account how the heat
gets out of the heatsink.
It's all about surface area, hence the heatsinks have those "fins" that
increase the surface area.
If your design uses another techniques to extract the heat, like liquid
cooling for example, then that's a different ballgame, it would be more the
thermal conductivity that matters in this case.

Dave.
--
---------------------------------------------
Check out my Electronics Engineering Video Blog & Podcast:
http://www.alternatezone.com/eevblog/
 
On 16 Jul., 22:01, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Thu, 16 Jul 2009 12:02:38 -0700 (PDT), "jalb...@bsu.edu"



jalb...@bsu.edu> wrote:
What is wrong with my reasoning?

Consider a square piece of aluminum 4cm x 4cm x 2 mm (thick) .  The
thermal resistance of this piece of aluminum is .005 deg C/W according
tohttp://www.novelconceptsinc.com/calculators-slab-thermal-resistance.cgi
using .04, .04, .002, and 250 (thermal conductivity).

Look at a commercial designed heat sink for example:
http://parts.digikey.com/1/parts/988299-heat-sink-3-1-25-compact-5016...
This heat sink has a thermal resistance of 7.8 deg C/W (no forced
air).

It looks like the simple square piece of aluminum .005 deg C/W would
be better than the 7.8 deg C/W commercial heat sink.  I know that this
can't be true, but why?

The calculation appears to be the thermal resistance from one side of
the aluminum plate to the other.

The heatsink theta is the thermal resistance from the heatsink to
still air.

The theta of the 4 cm aluminum chunk to still air would be huge,
vaguely in the 150 K/W sort of ballpark.

John
Click to expand...
would it really be that bad? I was looking at a datasheet for LM117
and they claim 1in^2 copper on a pcb will get a to252 from ~100K/W to
~50K/W

-Lasse
 
On Thu, 16 Jul 2009 14:11:24 -0700 (PDT), "langwadt@fonz.dk"
<langwadt@fonz.dk> wrote:

On 16 Jul., 22:01, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Thu, 16 Jul 2009 12:02:38 -0700 (PDT), "jalb...@bsu.edu"



jalb...@bsu.edu> wrote:
What is wrong with my reasoning?

Consider a square piece of aluminum 4cm x 4cm x 2 mm (thick) .  The
thermal resistance of this piece of aluminum is .005 deg C/W according
tohttp://www.novelconceptsinc.com/calculators-slab-thermal-resistance.cgi
using .04, .04, .002, and 250 (thermal conductivity).

Look at a commercial designed heat sink for example:
http://parts.digikey.com/1/parts/988299-heat-sink-3-1-25-compact-5016...
This heat sink has a thermal resistance of 7.8 deg C/W (no forced
air).

It looks like the simple square piece of aluminum .005 deg C/W would
be better than the 7.8 deg C/W commercial heat sink.  I know that this
can't be true, but why?

The calculation appears to be the thermal resistance from one side of
the aluminum plate to the other.

The heatsink theta is the thermal resistance from the heatsink to
still air.

The theta of the 4 cm aluminum chunk to still air would be huge,
vaguely in the 150 K/W sort of ballpark.

John

would it really be that bad? I was looking at a datasheet for LM117
and they claim 1in^2 copper on a pcb will get a to252 from ~100K/W to
~50K/W

-Lasse
Click to expand...
I did say "vaguely" and "ballpark." The value you cite corresponds to
about 100 K/W. It all depends on power density, orientation,
clearances for convection, all that stuff. This thermal stuff is very,
very messy.

John
 
On Jul 16, 3:02 pm, "jalb...@bsu.edu" <jalb...@bsu.edu> wrote:
What is wrong with my reasoning?

Consider a square piece of aluminum 4cm x 4cm x 2 mm (thick) .  The
thermal resistance of this piece of aluminum is .005 deg C/W
Click to expand...
As others have said, the problem is that 'thermal resistance'
assumes a source and sink of heat which (in this case) are
flat, 4x4 cm in size, and uniform in temperature.

It isn't similar at all to an electronic package heatsink, unless you
imagine some kind of chip that's REALLY big, and are using something
other than airflow to remove the heat.
 
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