Circuit Board Heat Spreading

8/9/2019 Circuit Board Heat Spreading

1/48

RPS 2010 April Corporate Research & Development Packaging Technology

Circuit Board Heat Spreading

Roger Stout, P.E.

Research Scientist

Corporate R&D: Packaging Technology


2/48

Axisymmetric Board Model RPS April 2010 Corporate R&D : Packaging Technology2

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


3/48


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


4/48


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


5/48


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


6/48


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


7/48


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.)


8/48


Network properties

They can be concatenated

3

3

22

22

11

11

1

1

q

T

q

T

1q1T 3T3qZ X


9/48


Determinant of transmission matrix is unity

1

1det

101110

1R

R

Clearly true for the resistor example


10/48


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


11/48


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


12/48


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


13/48


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.


14/48


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


15/48


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


16/48


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.


17/48


Problem description

A circular fin allowing for heat flow at the tip


18/48


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 )(


19/48


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,


20/48


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.)


21/48


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


22/48


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


23/48


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.


24/48


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.


25/48


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?


26/48


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).


27/48


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


28/48


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 !!


29/48


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)


30/48


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


31/48


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


32/48


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


33/48


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).


34/48


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


35/48


Results of finite-element model


36/48


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


37/48


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


38/48





0 um gap in spreader at 0.021 m gap radius


0

10

20

30

0 0.01 0.02 0.03 0.04 0.05


theta-ba[

C/W]

0.00100.0020

0.0030

0.0040

0.0050

2D-axisymmetric spreader with 0 um break


39/48





200 um gap in spreader at 0.021 m gap radius


0

10

20

30

0 0.01 0.02 0.03 0.04 0.05


theta-ba[

C/W]

0.00100.0020

0.0030

0.0040

0.0050

2D-axisymmetric spreader with 200 um break


40/48


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


41/48


QFN 4x4 mm example

spreader on

back of board

76 mm

board

traces


42/48


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


43/48


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


44/48



45/48


Things not to do with spreaders

ground plane interruptedaround device footprint

spider vias


46/48


Buried vs. surface spreaders

kA

t

Rcond

hARconv 1

hARconv

1

hARconv

1

kA

tRcond

2

kA

tRcond

2

hARconv

1


47/48


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!


48/48

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

Documents

Circuit Board Heat Spreading