PCB Copper Thickness Versus Rth - is this graph correct?

Hi

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

Currently I am using 0.5 Oz PCB thickness, but it is possible to increase t hat 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 incr ease of the copper thickness would be insignificant with respect to the Rth .

Found this graph, figure 3 on page 2:

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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 th e center of the board, the increased thickness would reduce the thermal res istance 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 transfe r 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

Reply to
Klaus Kragelund
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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.

Reply to
RobertMacy

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 lateral heat spreading.

1 oz copper has a sheet thermal resistance of about 70 K/watt. That's the theta from opposite of a square of copper foil of any size.

Example: a 1206 resistor with normal pads and traces.

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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.

--
John Larkin                  Highland Technology Inc 
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    
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Reply to
John Larkin

umber of dissipating components are spread out on the PCB to produce an uni form temperature across the PCB.

that to 1 Oz.

of PCB versus the copper thickness. My initial feeling would be that the in crease of the copper thickness would be insignificant with respect to the R th.

ce from 260K/W to 180K/W

the center of the board, the increased thickness would reduce the thermal r esistance from the device to the rest of the board, so the temperature woul d be close to uniform.

a number of devices spread evenly on the PCB and dissipating individually t he same amount of power, the heat would then also be uniform. But the trans fer of the heat to the surroundings are convection and conduction, and thes e should not be affected by the thickness of the copper layer.

m thicker copper thickness. Is this a valid assumption?

Hi Klaus, To my mind what's important is how the heat is being removed fro m the pcb. Is there some thermal connection to the outside world? (like br ass 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 co oling, and approximately uniform temperature across the pcb already, then t hicker copper won't do much.

George H.

Reply to
George Herold

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 spread out).

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

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Seven Transistor Labs 
Electrical Engineering Consultation 
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Reply to
Tim Williams

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.

--
John Larkin                  Highland Technology Inc 
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    
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Reply to
John Larkin

number of dissipating components are spread out on the PCB to produce an un iform temperature across the PCB.

e that to 1 Oz.

of PCB versus the copper thickness. My initial feeling would be that the i ncrease of the copper thickness would be insignificant with respect to the Rth.

nce from 260K/W to 180K/W

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 wou ld be close to uniform.

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 tran sfer of the heat to the surroundings are convection and conduction, and the se should not be affected by the thickness of the copper layer.

om thicker copper thickness. Is this a valid assumption?

hermal

an get

e

theta

PG

g

y well.

al vias

. Lots

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

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into 12x 1206.pdf

Cheers

Klaus

Reply to
Klaus Kragelund

number of dissipating components are spread out on the PCB to produce an u niform temperature across the PCB.

se that to 1 Oz.

a 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.

ance from 260K/W to 180K/W

n 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 wo uld be close to uniform.

e 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 tra nsfer of the heat to the surroundings are convection and conduction, and th ese should not be affected by the thickness of the copper layer.

rom thicker copper thickness. Is this a valid assumption?

rom the pcb. Is there some thermal connection to the outside world? (like brass standoffs.) Or is it just cooled by air conduction/convection? In t he former 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.

I has only limited contact to the enclosure, regretfully

Cheers

Klaus

Reply to
Klaus Kragelund

"John Larkin" wrote in message news: snipped-for-privacy@4ax.com...

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

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Seven Transistor Labs 
Electrical Engineering Consultation 
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Reply to
Tim Williams

That link doesn't work for me.

We've found that resistors from 0603 to 1206 can all dissipate a half watt or so if their end caps are soldered to big copper pours. The central hot-spot temperatures are the same.

--
John Larkin                  Highland Technology Inc 
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    
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Reply to
John Larkin

You can get serious PCB cooling with a Bergquist-type thermal pad between a PCB and a metal (or even plastic) enclosure.

Here's a couple of pages from a thermal study I did...

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We're about to order a batch of custom die-cut silicone-free pads from Bergquist, roughly $12 each.

The board has a lot of internal copper, and has "thermal antenna" pours on the bottom side to extract heat from critical parts, into the Bergquist pad.

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--
John Larkin                  Highland Technology Inc 
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    
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Reply to
John Larkin

Which translates to "You can't do the equations either."

There is a reason that people use Spice and thermal FEM software. And measurements.

--
John Larkin                  Highland Technology Inc 
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    
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Reply to
John Larkin

That URL has spaces. Try one or more of the following:

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Matt Roberds

Reply to
mroberds

A number of dissipating components are spread out on the PCB to produce an uniform temperature across the PCB.

ease that to 1 Oz.

rea of PCB versus the copper thickness. My initial feeling would be that th e increase of the copper thickness would be insignificant with respect to t he Rth.

stance from 260K/W to 180K/W

in the center of the board, the increased thickness would reduce the therm al resistance from the device to the rest of the board, so the temperature would be close to uniform.

ave a number of devices spread evenly on the PCB and dissipating individual ly the same amount of power, the heat would then also be uniform. But the t ransfer of the heat to the surroundings are convection and conduction, and these should not be affected by the thickness of the copper layer.

from thicker copper thickness. Is this a valid assumption?

g thermal

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the theta

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very well.

ermal vias

lem. Lots

s possible to provide lateral heat spreading:

This should work (no white spaces):

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t or so

I have 900mW for 12 resistors, but that's because I have a maximum 15 degre es delta hotspot temperature requirement

Cheers

Klaus

Reply to
Klaus Kragelund

This generalization gives an average temperature rise. Thicker copper gives a reduction in hotspots, or improvement in uniformity of temperatures across the surface.

For a uniform surface dissipation, your ballpark gives a 60% overestimate of rise, which from my experience is closer to (delta)t = 1degC per milliwat per cm^2 (+/-20%)

- another easily remembered relationship, without multipliers (if you ignore milli and centi...).

I suspect that another ballpark constant is available, giving a hotspot (delta tH/ delta tAv) estimate vs radius/thermal conductivity. RL

Reply to
legg

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