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

K

#### Klaus Kragelund

Jan 1, 1970
0
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

R

#### RobertMacy

Jan 1, 1970
0
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

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.

G

#### George Herold

Jan 1, 1970
0
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 increasethat 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?
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.

T

#### Tim Williams

Jan 1, 1970
0
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

K

#### Klaus Kragelund

Jan 1, 1970
0
It's complex.

The surface area of the board is the convective path to the air.

Copper pours and planes spread the heat out from a component. Spreading thermal

resistance is usually important. A surface-mount resistor or transistor can get

very hot if it can't spread the heat laterally into the board surface.

The thicker the copper, and the more un-interrupted planes, the better the

1 oz copper has a sheet thermal resistance of about 70 K/watt. That's thetheta

from opposite of a square of copper foil of any size.

Example: a 1206 resistor with normal pads and traces.

https://dl.dropboxusercontent.com/u/53724080/Thermal/V220_1206/1206.txt

https://dl.dropboxusercontent.com/u/53724080/Thermal/V220_1206/DSC06287.JPG

https://dl.dropboxusercontent.com/u/53724080/Thermal/V220_1206/IR_0056.jpg

The resistor is a hot spot, because the heat doesn't spread laterally very well.

Theta would be much lower if the pads were bigger, or if there were thermal vias

to other-layer copper pours or planes.

So just physically spreading out parts doesn't solve the hot-spot problem.. Lots

of copper is the best lateral heat spreading mechanism on a PC board.

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

K

#### Klaus Kragelund

Jan 1, 1970
0
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 the former the thickness of the copper would help... where if it's just air cooling, and approximately uniform temperature across the pcb already, thenthicker copper won't do much.

I has only limited contact to the enclosure, regretfully

Cheers

Klaus

T

#### Tim Williams

Jan 1, 1970
0
John Larkin said:
Theta depends on the part size, too. If you dump heat into, say, a
circular
patch on an infinite metal sheet, theta depends on the patch area. Theta
goes to
infinity as the contact area goes to zero. Getting the heat out locally,
close
to the part, is often the bottleneck.

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

V
Replies
3
Views
12K
Jasen Betts
J
S
Replies
12
Views
4K
josephkk
J
P
Replies
8
Views
2K
Rich Grise
R
Replies
5
Views
94
Replies
7
Views
176