# PCB Copper Thickness Versus Rth - is this graph correct?

Discussion in 'Electronic Design' started by Klaus Kragelund, Jan 9, 2014.

1. ### Klaus KragelundGuest

Hi

I have a PCB, 50mm x 50mm, and I need to optimize cooling of the PCB. A number of dissipating components are spread out on the PCB to produce an uniform temperature across the PCB.

Currently I am using 0.5 Oz PCB thickness, but it is possible to increase that to 1 Oz.

So I was looking for a graph of the thermal resistance on a certain area ofPCB versus the copper thickness. My initial feeling would be that the increase of the copper thickness would be insignificant with respect to the Rth..

Found this graph, figure 3 on page 2:

http://www.iaasr.com/wp-content/upl...roduced-by-surface-mount-components.-Rev1.pdf

Increasing the copper from 0.5oz to 1oz would reduce the thermal resistancefrom 260K/W to 180K/W

But, is this valid. If we take the example of a single hotspot device in the center of the board, the increased thickness would reduce the thermal resistance from the device to the rest of the board, so the temperature would be close to uniform.

If on the other hand, with a PCB with decreased copper thickness, I have a number of devices spread evenly on the PCB and dissipating individually thesame amount of power, the heat would then also be uniform. But the transfer of the heat to the surroundings are convection and conduction, and these should not be affected by the thickness of the copper layer.

So, for the actual design, evenly spaced components would not benefit from thicker copper thickness. Is this a valid assumption?

Cheers

Klaus

2. ### RobertMacyGuest

Rule of thumb for heat dissipation of free standing surface, no fan is
1 C rise per watt over 100 sq in area.

That kind of implies that thicker copper, which is in series with your
copper/PCB to air transfer, doesn't make a lot of difference.

But gut feel is that thicker copper also gives you some thermal mass,
which might save a marginal part during a 'spike' of dissipation.

3. ### George HeroldGuest

Hi Klaus, To my mind what's important is how the heat is being removed from the pcb. Is there some thermal connection to the outside world? (like brass standoffs.) Or is it just cooled by air conduction/convection? In theformer the thickness of the copper would help... where if it's just air cooling, and approximately uniform temperature across the pcb already, then thicker copper won't do much.

George H.

4. ### Tim WilliamsGuest

The defining quantity is the spacing of said components relative to the
lateral diffusivity (i.e., how far sideways along the board the heat will

I believe it's around 3cm for 2oz copper (ah, such wonderful juxtaposition
of units ), so putting equal-dissipating components on a grid of around
6cm center-to-center (note a triangular mesh allows maximal packing) will
be about optimal between copper/board thickness and utilization. Such
spacing will allow about 2W per component.

Use proportionally smaller spacings for thinner material.

Not necessarily smaller for thinner foil only, but let's see. FR-4 is
0.81 W m^-1 K^-1 while copper is 400; the average board is 1600um thick.
0.5oz copper is 17um, or say 34um total (double sided). The conductivity
per square of copper is 0.0136 W K^-1, and of FR-4, 0.0013 W K^-1. So
even for thin plating, it's still true that copper dominates the lateral
conductivity.

Tim

--
Seven Transistor Labs
Electrical Engineering Consultation
Website: http://seventransistorlabs.com

Hi

I have a PCB, 50mm x 50mm, and I need to optimize cooling of the PCB. A
number of dissipating components are spread out on the PCB to produce an
uniform temperature across the PCB.

Currently I am using 0.5 Oz PCB thickness, but it is possible to increase
that to 1 Oz.

So I was looking for a graph of the thermal resistance on a certain area
of PCB versus the copper thickness. My initial feeling would be that the
increase of the copper thickness would be insignificant with respect to
the Rth.

Found this graph, figure 3 on page 2:

http://www.iaasr.com/wp-content/upl...roduced-by-surface-mount-components.-Rev1.pdf

Increasing the copper from 0.5oz to 1oz would reduce the thermal
resistance from 260K/W to 180K/W

But, is this valid. If we take the example of a single hotspot device in
the center of the board, the increased thickness would reduce the thermal
resistance from the device to the rest of the board, so the temperature
would be close to uniform.

If on the other hand, with a PCB with decreased copper thickness, I have a
number of devices spread evenly on the PCB and dissipating individually
the same amount of power, the heat would then also be uniform. But the
transfer of the heat to the surroundings are convection and conduction,
and these should not be affected by the thickness of the copper layer.

So, for the actual design, evenly spaced components would not benefit from
thicker copper thickness. Is this a valid assumption?

Cheers

Klaus

5. ### Klaus KragelundGuest

I have the resistors spread out and all components have as much copper as possible to provide lateral heat spreading:

https://www.dropbox.com/s/y55jw3urqgr5329/900mW into 12x 1206.pdf

Cheers

Klaus

6. ### Klaus KragelundGuest

I has only limited contact to the enclosure, regretfully

Cheers

Klaus

7. ### Tim WilliamsGuest

Yes. Or, since you "can't do equations", ;-)

For surfaces with no surface heat dissipation (lateral heat spreading
only), the thermal resistance between concentric cylindrical surfaces is:
Rth = ln(r2 / r1) / (2 pi sigma_th)

Which of course diverges for r1 --> 0.

When the surfaces dissipate heat linearly with temp difference (true of
solid conductors, but a poor approximation of actual convection or
radiation), solutions take the form of the complex Bessel function (i.e.,
T(r) = c1 * J_0(i*c2*r)). A closed form solution (albeit in terms of the
Bessel function) is left as an exercise for the student. ;-)

Tim