Re: revelation

On Sat, 10 Jun 2006 08:14:13 +0100, Tony Williams
<tonyw@xxxxxxxxxxxxxxxxxxx> wrote:

In article <1149876123.420238.159010@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Bill Sloman <bill.sloman@xxxxxxxx> wrote

[snip]
such resistors have an area of 1.6mm x 0.8mm or 1.28mm^2.
Eighteen of them get up to 23.04mm^2. The copper area is 50.8mm
by 25.4mm or 1290mm^2, some 56 times larger.

More resistor/copper ratio? 30 resistors or more in
simple // is possible with a re-jig of the layout.

|<--------------2"---------------->|
+----------------------------------+
| A Cu Area-1 |
| ______________________________|
| | | | | | | | | | | | | | | | |
| | R R R R R R R R R R R R R R R
| | |_|_|_|_|_|_|_|_|_|_|_|_|_|_|
| | | |
IN----o | | Cu Area-2 o----OUT
| | |____________________________|
|<A>| | | | | | | | | | | | | | | |
| | R R R R R R R R R R R R R R R
| |_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_
| |
| A Cu Area-1 |
+----------------------------------+

Area-1 = 1xA + 2x(2-A)xA. Area-2 = (1-2A)x(2-A).

If dimension A is about 0.22" then the two copper
areas are equal.


Yes, other geometries are interesting. The limit on the number of
resistors is of course cost; these thinfilms cost about 30 cents each
in quantity, and copper is free. Paving the entire 2 square inches of
available board area with thinfilms would be too expensive, and a
sparse array doesn't heatsink the resistors much, since FR4 is such a
rotten heat conductor. So we go for a large ratio of copper area to
resistor area, and the copper does basically all the cooling.

The immediate problem is to use 2 sq in of board area to optimally
create a high-precision, high-power resistor at reasonable cost. There
are no real limits on "precision" or "power" (the product just gets
better) but 150 ppm self-heating error at 1 watt is pretty good, and
survival without longterm effects should be at least 3 watts.
Wirewounds and TO-220 resistors would both be a pain, so we're trying
to stick to available surface-mount parts.

One square of 1 oz copper has a thermal *** resistance of about 70
K/W. Coincidentally, the *** resistance of the typical surfmount
alumina resistor substrate is about the same. We can't afford AlN
thinfilms!

If you thermally ground both end-caps of a resistor, the self-heat
hot-spot is obviously in the middle. The temperature rise factor, in
K/W at the hot spot, depends on resistor aspect ratio, not size, and
is about same for 2010, 1206, and 0603, maybe even better for the
smaller sizes because of relative end-cap geometry. This alone pushes
one to a larger number of small resistors, since each resistor then
gets a smaller share of the available watts, so hot-spot temperature
rise goes as 1/N where N is the number of resistors.

The resistor's thermal profile from self-heating (again, end-cap
cooled) is roughly gaussian, peaking at the center.

For a given N, if the copper is thick, namely isothermal, the copper
topology doesn't matter. All that matters is that a certain copper
area is available, and that each end of each resistor gets to use the
same effective copper area. But the thermal resistance of the copper
is non-trivial, so something like Tony's geometry is probably better
than the two 1-in square blocks, even with the same N.


There are lots of secondary effects, including ...

The solder and copper on/under the end-caps reduces effective aspect
ratio

The solder on the pads helps spread heat and reduce thermal crowding.

There is some thermal conduction from the bottomside of the resistor
to the board, some of that to copper. Unless you glue the resistors
down, there's typically about an 8 mil air gap, which doesn't conduct
much heat.

Heat can be conducted through FR4 to power/ground planes, and vias can
conduct heat to pours on other layers, which can then also help dump
heat into planes. On a six-layer board, theta between layers is
something like 10 K/w per square inch, not trivial.


You'd need about $100K of software and another $100K of time to model
this accurately. Experiments using a thermal imager would be a big
help here, so we're considering one.

I just thought this was an interesting problem.


John


.

Quantcast