On Thu, 09 Jan 2014 07:08:19 -0700, RobertMacy
<robert.a.macy@gmail.com> wrote:
>On Thu, 09 Jan 2014 03:55:55 -0700, Klaus Kragelund
><klauskvik@hotmail.com> wrote:
>
>>> ...snip to keep Aioe happy
>> 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.
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 by Klaus Kragelund●January 10, 20142014-01-10
On Friday, January 10, 2014 5:47:45 PM UTC+1, John Larkin wrote:
> On Fri, 10 Jan 2014 00:27:59 -0800 (PST), Klaus Kragelund
>=20
> <klauskvik@hotmail.com> wrote:
>=20
>=20
>=20
> >On Thursday, January 9, 2014 4:24:51 PM UTC+1, John Larkin wrote:
>=20
> >> On Thu, 9 Jan 2014 02:55:55 -0800 (PST), Klaus Kragelund <klauskvik@ho=
tmail.com>
>=20
> >>=20
>=20
> >> wrote:
>=20
> >>=20
>=20
> >>=20
>=20
> >>=20
>=20
> >> >Hi
>=20
> >>=20
>=20
> >> >
>=20
> >>=20
>=20
> >> >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.
>=20
> >>=20
>=20
> >> >
>=20
> >>=20
>=20
> >> >Currently I am using 0.5 Oz PCB thickness, but it is possible to incr=
ease that to 1 Oz.
>=20
> >>=20
>=20
> >> >
>=20
> >>=20
>=20
> >> >So I was looking for a graph of the thermal resistance on a certain a=
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.
>=20
> >>=20
>=20
> >> >
>=20
> >>=20
>=20
> >> >Increasing the copper from 0.5oz to 1oz would reduce the thermal resi=
stance from 260K/W to 180K/W
>=20
> >>=20
>=20
> >> >
>=20
> >>=20
>=20
> >> >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 therm=
al resistance from the device to the rest of the board, so the temperature =
would be close to uniform.
>=20
> >>=20
>=20
> >> >
>=20
> >>=20
>=20
> >> >If on the other hand, with a PCB with decreased copper thickness, I h=
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.
>=20
> >>=20
>=20
> >> >
>=20
> >>=20
>=20
> >> >So, for the actual design, evenly spaced components would not benefit=
from thicker copper thickness. Is this a valid assumption?
>=20
> >>=20
>=20
> >> >
>=20
> >>=20
>=20
> >> >Cheers
>=20
> >>=20
>=20
> >> >
>=20
> >>=20
>=20
> >> >Klaus
>=20
> >>=20
>=20
> >>=20
>=20
> >>=20
>=20
> >> It's complex.
>=20
> >>=20
>=20
> >>=20
>=20
> >>=20
>=20
> >> The surface area of the board is the convective path to the air.=20
>=20
> >>=20
>=20
> >>=20
>=20
> >>=20
>=20
> >> Copper pours and planes spread the heat out from a component. Spreadin=
g thermal
>=20
> >>=20
>=20
> >> resistance is usually important. A surface-mount resistor or transisto=
r can get
>=20
> >>=20
>=20
> >> very hot if it can't spread the heat laterally into the board surface.
>=20
> >>=20
>=20
> >>=20
>=20
> >>=20
>=20
> >> The thicker the copper, and the more un-interrupted planes, the better=
the
>=20
> >>=20
>=20
> >> lateral heat spreading.=20
>=20
> >>=20
>=20
> >>=20
>=20
> >>=20
>=20
> >> 1 oz copper has a sheet thermal resistance of about 70 K/watt. That's =
the theta
>=20
> >>=20
>=20
> >> from opposite of a square of copper foil of any size.
>=20
> >>=20
>=20
> >>=20
>=20
> >>=20
>=20
> >> Example: a 1206 resistor with normal pads and traces.
>=20
> >>=20
>=20
> >>=20
>=20
> >>=20
>=20
> >> https://dl.dropboxusercontent.com/u/53724080/Thermal/V220_1206/1206.tx=
>=20
> >>=20
>=20
> >>=20
>=20
> >>=20
>=20
> >> The resistor is a hot spot, because the heat doesn't spread laterally =
very well.
>=20
> >>=20
>=20
> >> Theta would be much lower if the pads were bigger, or if there were th=
ermal vias
>=20
> >>=20
>=20
> >> to other-layer copper pours or planes.
>=20
> >>=20
>=20
> >>=20
>=20
> >>=20
>=20
> >> So just physically spreading out parts doesn't solve the hot-spot prob=
lem. Lots
>=20
> >>=20
>=20
> >> of copper is the best lateral heat spreading mechanism on a PC board.
>=20
> >>=20
>=20
> >
>=20
> >I have the resistors spread out and all components have as much copper a=
>=20
>=20
> We've found that resistors from 0603 to 1206 can all dissipate a half wat=
t or so
>=20
> if their end caps are soldered to big copper pours. The central hot-spot
>=20
> temperatures are the same.
>=20
I have 900mW for 12 resistors, but that's because I have a maximum 15 degre=
es delta hotspot temperature requirement=20
Cheers
Klaus
Reply by ●January 10, 20142014-01-10
John Larkin <jjlarkin@highnotlandthistechnologypart.com> wrote:
> On Fri, 10 Jan 2014 00:27:59 -0800 (PST), Klaus Kragelund
> <klauskvik@hotmail.com> wrote:
>
>> https://www.dropbox.com/s/y55jw3urqgr5329/900mW into 12x 1206.pdf
>
> That link doesn't work for me.
On Fri, 10 Jan 2014 04:47:06 -0600, "Tim Williams" <tmoranwms@charter.net>
wrote:
>"John Larkin" <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote in
>message news:afsuc9p4usebvl10l8hjpil2v26738ck6e@4ax.com...
>> 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
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
Precision electronic instrumentation
Reply by John Larkin●January 10, 20142014-01-10
On Fri, 10 Jan 2014 00:29:10 -0800 (PST), Klaus Kragelund
<klauskvik@hotmail.com> wrote:
>On Thursday, January 9, 2014 4:46:46 PM UTC+1, George Herold wrote:
>> On Thursday, January 9, 2014 5:55:55 AM UTC-5, Klaus Kragelund wrote:
>>
>> > 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/uploads/2013/10/Using-PCB-copper-area-to-dissipate-the-heat-produced-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 the 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
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...
https://dl.dropboxusercontent.com/u/53724080/Thermal/ESM_HX_6-7.pdf
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.
https://dl.dropboxusercontent.com/u/53724080/PCBs/ESM_rev_B.jpg
--
John Larkin Highland Technology Inc
www.highlandtechnology.com jlarkin at highlandtechnology dot com
Precision electronic instrumentation
Reply by John Larkin●January 10, 20142014-01-10
On Fri, 10 Jan 2014 00:27:59 -0800 (PST), Klaus Kragelund
<klauskvik@hotmail.com> wrote:
>On Thursday, January 9, 2014 4:24:51 PM UTC+1, John Larkin wrote:
>> On Thu, 9 Jan 2014 02:55:55 -0800 (PST), Klaus Kragelund <klauskvik@hotmail.com>
>>
>> wrote:
>>
>>
>>
>> >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/uploads/2013/10/Using-PCB-copper-area-to-dissipate-the-heat-produced-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
>>
>>
>>
>> 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.
>>
>>
>>
>> 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
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
Precision electronic instrumentation
Reply by Tim Williams●January 10, 20142014-01-10
"John Larkin" <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote in
message news:afsuc9p4usebvl10l8hjpil2v26738ck6e@4ax.com...
> 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
--
Seven Transistor Labs
Electrical Engineering Consultation
Website: http://seventransistorlabs.com
Reply by Klaus Kragelund●January 10, 20142014-01-10
On Thursday, January 9, 2014 4:46:46 PM UTC+1, George Herold wrote:
> On Thursday, January 9, 2014 5:55:55 AM UTC-5, Klaus Kragelund wrote:
>=20
> > Hi
>=20
> >=20
>=20
> >=20
>=20
> >=20
>=20
> > 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 u=
niform temperature across the PCB.
>=20
> >=20
>=20
> >=20
>=20
> >=20
>=20
> > Currently I am using 0.5 Oz PCB thickness, but it is possible to increa=
se that to 1 Oz.
>=20
> >=20
>=20
> >=20
>=20
> >=20
>=20
> > So I was looking for a graph of the thermal resistance on a certain are=
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.
>=20
> >=20
>=20
> >=20
>=20
> >=20
>=20
> > Increasing the copper from 0.5oz to 1oz would reduce the thermal resist=
ance from 260K/W to 180K/W
>=20
> >=20
>=20
> >=20
>=20
> >=20
>=20
> > But, is this valid. If we take the example of a single hotspot device i=
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.
>=20
> >=20
>=20
> >=20
>=20
> >=20
>=20
> > If on the other hand, with a PCB with decreased copper thickness, I hav=
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.
>=20
> >=20
>=20
> >=20
>=20
> >=20
>=20
> > So, for the actual design, evenly spaced components would not benefit f=
rom thicker copper thickness. Is this a valid assumption?
>=20
> >=20
>=20
> Hi Klaus, To my mind what's important is how the heat is being removed f=
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. =20
>=20
I has only limited contact to the enclosure, regretfully
Cheers
Klaus
Reply by Klaus Kragelund●January 10, 20142014-01-10
On Thursday, January 9, 2014 4:24:51 PM UTC+1, John Larkin wrote:
> On Thu, 9 Jan 2014 02:55:55 -0800 (PST), Klaus Kragelund <klauskvik@hotma=
il.com>
>=20
> wrote:
>=20
>=20
>=20
> >Hi
>=20
> >
>=20
> >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 un=
iform temperature across the PCB.
>=20
> >
>=20
> >Currently I am using 0.5 Oz PCB thickness, but it is possible to increas=
e that to 1 Oz.
>=20
> >
>=20
> >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 i=
ncrease of the copper thickness would be insignificant with respect to the =
Rth.
>=20
> >
>=20
> >Increasing the copper from 0.5oz to 1oz would reduce the thermal resista=
nce from 260K/W to 180K/W
>=20
> >
>=20
> >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 wou=
ld be close to uniform.
>=20
> >
>=20
> >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 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.
>=20
> >
>=20
> >So, for the actual design, evenly spaced components would not benefit fr=
om thicker copper thickness. Is this a valid assumption?
>=20
> >
>=20
> >Cheers
>=20
> >
>=20
> >Klaus
>=20
>=20
>=20
> It's complex.
>=20
>=20
>=20
> The surface area of the board is the convective path to the air.=20
>=20
>=20
>=20
> Copper pours and planes spread the heat out from a component. Spreading t=
hermal
>=20
> resistance is usually important. A surface-mount resistor or transistor c=
an get
>=20
> very hot if it can't spread the heat laterally into the board surface.
>=20
>=20
>=20
> The thicker the copper, and the more un-interrupted planes, the better th=
e
>=20
> lateral heat spreading.=20
>=20
>=20
>=20
> 1 oz copper has a sheet thermal resistance of about 70 K/watt. That's the=
>=20
>=20
>=20
> The resistor is a hot spot, because the heat doesn't spread laterally ver=
y well.
>=20
> Theta would be much lower if the pads were bigger, or if there were therm=
al vias
>=20
> to other-layer copper pours or planes.
>=20
>=20
>=20
> So just physically spreading out parts doesn't solve the hot-spot problem=
. Lots
>=20
> of copper is the best lateral heat spreading mechanism on a PC board.
>=20
I have the resistors spread out and all components have as much copper as p=
ossible to provide lateral heat spreading:
https://www.dropbox.com/s/y55jw3urqgr5329/900mW into 12x 1206.pdf
Cheers
Klaus
Reply by John Larkin●January 10, 20142014-01-10
On Thu, 9 Jan 2014 17:23:31 -0600, "Tim Williams" <tmoranwms@charter.net> wrote:
>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.
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
Precision electronic instrumentation