circuit board heat spreading
TRANSCRIPT
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RPS 2010 April Corporate Research & Development Packaging Technology
Circuit Board Heat Spreading
Roger Stout, P.E.
Research Scientist
Corporate R&D: Packaging Technology
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Outline
Development of a 2D-axisymmetric board model
What is a thermal resistance?
Two port networks
A thermal test board as a two-port net
Application
Comparison with some real life data
Answering some common questions
Breaks in spreader planes
Simplified finite-element approach
Utilizing the 2D-axisymmetric model
Thermal vias
Buried vs. surface spreaders
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What is a thermal resistance?
An abstraction of part of a thermal system whichpermits one to relate the heat flow (through it) to thetemperatures at its ends
T1 T2
qin qout
Tamb
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We dont usually think of the resistors
behavior as depending on ambient, and the
heat in simply equals the heat out. Indeed, weusually just say
But certainly, we could also say it this way:
q
TTR 21
qTTTTR ambamb )()( 21
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Two port networks
Linear circuit elements can be treated in
a general way as two-port networks
Z1q1T 2q 2T
2
2
1
1
qT
qT
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All we have to do is
figure out whatgoes in the matrix
For example, the
resistor:
T1 T2
q1 q2
Tamb
the Tequation
the q equation 21
121
221
2
2
1
1
)()(
10
1
qq
qRTT
qRTTTT
q
TTR
q
TT
aa
aa
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Note the symmetry:
if
2
2
1
1
10
1
qTTR
qTT aa
1
1
1
1
1
2
2
10
1
10
1
q
TTR
q
TTR
q
TT
a
aathen
(In this case, the sign change onR is indicative ofthe direction of heat flow as defined in the model.)
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Network properties
They can be concatenated
3
3
22
22
11
11
1
1
q
T
q
T
1q1T 3T3qZ X
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Determinant of transmission matrix is unity
1
1det
101110
1R
R
Clearly true for the resistor example
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E.g. two resistors in series
T11
q1
2T3
q3
3
321
3
3
2
121
3
321
1
1
10
1
1100110
11011
10
1
10
1
q
TTRR
q
TT
R
RRR
q
TTRR
q
TT
a
a
aa
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Elements can be anything that fits thetwo-port schema
T1
T2
T3
T4
q1
q4
q3
q21
1
2
2
q
T
q
TX
3
3
4
4q
T
q
TY
1
1
4
4
q
T
q
TXY
you just have to know what goes
into the transmission matrices X and Y
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What is a circuit board?
Simply, a leaky resistance
Heat flows in one end, and out the other
Its characterized by the two end temperatures,
some internal properties, heat flow in and out, anda convection loss in between
But since the convection loss depends only on thelocal internal temperature profile and the externalambient, it doesnt depend on any more external
parameters than a simple resistor at least withrespect to a two-port net
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A thermal test board with a device beingtested (a heat source) can thus be cast
as a two port network as follows:
One end (the input) has a heat source, the
device, and some local temperature (crudelyspeaking, the lead temperature)
The other end (the output) is the distant
edge of the board; it has some temperatureprobably not ambient, and some heat flowout, probably nearly zero.
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Weve said nothing about the shape of the
board, or required anything specific aboutconstancy of properties over the board
It could be square, rectangular, circular, acapped cylinder; its thickness could vary, its
conductivity could vary, its convectioncoefficient could vary
All we need to do is come up with the four
matrix elements to define it as a two-portnetwork
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Circular fins
In many cases, a plausible approximationfor a thermal test board with a single heatsource at the center, is a circular plate
You can actually look up equations for acircular fin
What youll find relates input temperatureand heat flow to fin properties and
convection values Usually the fin tip is adiabatic
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What you probably wont find is a two-portnetwork model of a circular fin; that is, youcant chain together the tip end of one fin to
the inside edge of another, because the heat
flow out the tip is not a parameter. If you had such a model, clearly you could
approximate a much wider variety of circularcircuit boards, because you could treat small
regions as having constant properties, thenchange the properties for the next region, etc.
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Problem description
A circular fin allowing for heat flow at the tip
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Mathematical statement
Boundary conditions are temperatures and heatflows at inner and outer radii.
Temperatures are easy, but heat flows come fromthe Fourier heat conduction equation (in
cylindrical coordinates):
Governing equation in cylindrical coordinates:
021
)(2
2
TTkt
h
dr
dT
rdr
Td
dr
dTrkAq )(
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Closed-form solution (see Ap NoteAND8222/D for the gory details):
Where weve defined the board parameterm as
)()()()()()()()(
0101
0101
bebe
ee
b mrKmrImrImrK
mrKmrImrImrK
TT
TT
kt
hm
2
And those Ks andIs are the modifiedBessels functions of the first and second
kinds, respectively,
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So lets see it work with some
real thermal data
A real customer board (a hard drive card)not very circular, only sort-of uniform,packages not exactly centered on the
board.
(But hey, youve got to start somewhere.)
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U13p5
U13p7
U18p7
U18p5
TC loc 1
TC loc 2
TC loc 3-1
TC loc 4
0
13
16
20
24
16 5
8
0
2
38
TC loc 3-2
drive motor hole
5
5
locationsof 8-lead
SOICs
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Best fit of:
psi-board-ambient
2-sided convection
0
10
20
30
40
50
60
70
0 5 10 15 20 25 30
TC distance from heat source [mm]psi-XA[C/W]
U13p7 heatedU13p5 heatedU18p7 heated
U18p5 heatedbest f it board model
top ofcase TC's
back ofboard
TC's
top of board TC's
psi-board-ambient
1-sided convection
0
10
20
30
40
50
60
70
0 5 10 15 20 25 30
TC distance from heat source [mm]psi-XA[C/W]
U13p7 heatedU13p5 heatedU18p7 heated
U18p5 heatedbest fit board model
top ofcase TC's
back ofboard
TC's
top of board TC's
)()()(
)(
)()()(
)(
0
1
1
0
0
1
1
0
mrKmrK
mrImrI
mrKmrK
mrI
mrI
Q
TT
Q
TT
e
e
b
e
e
b
b
)()()(
)( 01
1
0 mrKmrK
mrImrIc
Q
TT
e
e
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outer boundaryconditions:
temperature, Te
heat, qe
inner boundary conditions:temperature, Tb
heat, qb
T1, q1 T2, q2 T3, q3
As previously suggested, the power of the two-
port network model is in being able to chaintogether one annular region after another.
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The two-port network transmissionmatrix result (again, see Ap Note
AND8222/D for details):1
11
00
11
00
)()()()(
)()()()(
jjjj
jj
iiii
iiij
zKGzIG
zKzI
zKGzIG
zKzIT
hktrG ii 22where
and the superscripts and subscripts i,j denotethe two radii of the region of interest.
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With the axisymmetric two-port board model,we can easily answer these questions
1) How much does package performance differbetween a min-pad test board and a 1 padtest board?
2) In general, how does package performancedepend on copper spreader area?
3) How does boards contribution to Theta-JA
depend on package size?4) What is the temperature profile (as a functionof radius) in the circular board?
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To answer the first question, a one region
model can be used to answer the min pad
part, and a two region model used to answerthe 1-inch pad part.
Or, a two region model can answer both
parts, simply by changing the properties ofthe inner region from plain circuit board, to
copper-covered circuit board.
Well see how to actually carry this out in
answering the second question (next).
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To answer the second and thirdquestions, we need a two region model.
We then vary either the inner radius orthe interface radius.
0
esebsb T
Q
T
TT
0
TT
Q
TT eb
)(
)(
TTQ
TTTT
e
eb
Q
TTb
sebs
TT
1)(Q
TTe
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Solving for temperature profile
Suppose youd like to know the temperature
at the interface
0
esebsb T
Q
TTT
0
TT
Q
TT e
bebe
bebeb
0
TT
q
TT e
sese
sese
s
s
entire solution
interface solution
be
ses
Q
TTeliminating Te note Q qs !!
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Sample coding in Excel
gamma2 =MMULT(MMULT(A21:B22,MINVERSE(C21:D22)),C25:D26) =MMULT(MMULT(A21:B22,MINVERSE(C21:D22)),C25:D26) nu2
beta2 =MMULT(MMULT(A21:B22,MINVERSE(C21:D22)),C25:D26) =MMULT(MMULT(A21:B22,MINVERSE(C21:D22)),C25:D26) delta2
matrix for computing values at radius r1 given values at radius r3
matrix at r1, zone1 matrix at r2, zone1 matrix at r
=BESSELI(mval1*rad1,0) =BESSELK(mval1*rad1,0) =BESSELI(mval1*rad2,0) =BESSELK(mval1*rad2,0) =BESSELI(=-cval1*rad1*BESSELI(mval1*rad1,1) =cval1*rad1*BESSELK(mval1*rad1,1) =-cval1*rad2*BESSELI(mval1*rad2,1) =cval1*rad2*BESSELK(mval1*rad2,1) =-cval2*ra
gamma1 =MMULT(E21:F22,MINVERSE(G21:H22)) =MMULT(E21:F22,MINVERSE(G21:H22)) nu1
beta1 =MMULT(E21:F22,MINVERSE(G21:H22)) =MMULT(E21:F22,MINVERSE(G21:H22)) delta1
matrix for computing values at radius r2 given values at radius r3
matrix at r1, zone1
=BESSELI(mval1*rad1,0) =BESSELK(mval1*rad1,0)
=-cval1*rad1*BESSELI(mval1*rad1,1) =cval1*rad1*BESSELK(mval1*rad1,1)
=MMULT(E21:F22,MINVERSE(G21:H22))
=MMULT(MMULT(A21:B22,MINVERSE(C21:D22)),C25:D26)
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board thermal resistance
for different 3"x3" FR4 boards
0
50
100
150
200
250
0.000 0.005 0.010 0.015
package "radius" (m)
(Tb-Tamb)/Q
[C/
minpad
1"sq 1oz-Cu
1"sq 2oz-Cu
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3"x3" board thermal resistance
for 2mm package "radius"
as function of amount of copper
0
20
40
60
80
100
120
140
160
0.002 0.007 0.012 0.017
copper "radius" (m)
(Tb-Tamb)/Q
[C/
1oz Cu
2oz Cu
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axisymmetric board temperature profiles
heat input radius is 0.002 m
0
20
40
60
80
100
120
140
160
180
0 0.006 0.012 0.018 0.024 0.03 0.036 0.042
distance from center of board [m]
(Tb
-Tamb)/Q
[C
/W]
min pad
1-in 1-oz
1-in 2-oz
board has area of 3"x3"
"1-in" heat spreaderhas area of 1"x1"
R ll h l i i
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Recall thermal reciprocity
heat input here
sameresponse
here
responsehere
The temperature response of a remote point (r)
in a system to heat input at a some given source(s), will be identical to the temperature response
at (s) given an identical heat input at (r).
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Gaps in spreaders
k=1 W/m/C
thickness 1.5E-3 m
conductance = 0.0015 W/C
FR4
2 oz Cu
20x as much heat flows in this
thickness of copper than in this
thickness of FR4
k-equiv = ?conductance = ?
FR4
2 oz Cu
essentially all heat is forced to flow
around the break, through FR4 only
2 oz Cu
k=380 W/m/C
thickness 70E-6 mconductance = 0.027 W/C
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Results of finite-element model
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2-port gap analysis
E
EAgapBA
Q
TR
Q
T
22
22
11
11
10
1
gapelement
spreaderbetween device
and gap
spreader /boardbeyond
gap
board-ambienttemperaturerise
device
power
temperaturerise at edge
edge heatloss -
assumedzero
0
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2D-axisymmetric spreader with no break
interaction strength vs. distance
0.042 m radius board
0.028 m radius 2 oz copper spreader
no break in spreader
curve parameter is heat source radius [m]
0
10
20
30
0 0.01 0.02 0.03 0.04 0.05
distance from center of heat source [m]
theta-ba[C/W]
0.00100.0020
0.0030
0.0040
0.0050
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interaction strength vs. distance
0.042 m radius board
0.028 m radius 2 oz copper spreader
0 um gap in spreader at 0.021 m gap radius
curve parameter is heat source radius [m]
0
10
20
30
0 0.01 0.02 0.03 0.04 0.05
distance from center of heat source [m]
theta-ba[
C/W]
0.00100.0020
0.0030
0.0040
0.0050
2D-axisymmetric spreader with 0 um break
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interaction strength vs. distance
0.042 m radius board
0.028 m radius 2 oz copper spreader
200 um gap in spreader at 0.021 m gap radius
curve parameter is heat source radius [m]
0
10
20
30
0 0.01 0.02 0.03 0.04 0.05
distance from center of heat source [m]
theta-ba[
C/W]
0.00100.0020
0.0030
0.0040
0.0050
2D-axisymmetric spreader with 200 um break
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Thermal vias
Probably not the answer to many thermalproblems
they effectively increase the conductivity of the PCB
but only through its thickness
they only work if theres a good heatsink at the other
end of the via
which could mean a buried spreader plane
Ref ON Semiconductor Ap Note AND8432/D
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QFN 4x4 mm example
spreader on
back of board
76 mm
board
traces
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Vias under an exposed pad
OD
ID
pad size
boardmaterial
via lengthvia wall
material
via fillingmaterial
spreader plane
via
modelbob
iow
ifperpeq
tt
srnk
sr
srnk
srnkk 2
2
2
2
2
2
2
2
1
bplaneineq kk
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Random possibilities
Possible 9-via patternfor 2.75 mm pad
1.0 mm
2.75 mm
0.3 dia
0.8 mm
0.57 mm
Possible 18-via patternfor 2.75 mm pad
Cm
W32perpeqk copper filled
Cm
W24perpeqk solder filled
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Things not to do with spreaders
ground plane interruptedaround device footprint
spider vias
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Buried vs. surface spreaders
kA
t
Rcond
hARconv 1
hARconv
1
hARconv
1
kA
tRcond
2
kA
tRcond
2
hARconv
1
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Compute centered / surface ratio
k
htk
hARcentered
4
21
htk
htk
hARsurface
2
1
)()(
)(
141
14
21
2
4
2
1
22
Bi
Bi
Bi
Bi
htk
htk
k
htk
R
R
surface
centered
where
k
htBi , the Biot number.
If Bi = 0.1 (typical board in free convection), difference is only 0.2%
- and radiation emissivity may actually swing the result in favor of buried!
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Summary
The 2D axisymmetric board model provides a simple,
physics-based approximation for exploring interactionsbetween major variables in simple thermal test boardsystems
Along with the two-port network approach, many basic
and commonly asked questions about packageperformance as a function of board environment maybe answered.
Thermal vias arent all theyre cracked up to be
Take full advantage of buried planes Avoid unnecessary gaps