© 2010 Eric Pop, UIUCECE 598EP: Hot Chips Three Modes of Heat Transfer 1 “conduction” “radiation” “convection”

© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 1

Three Modes of Heat Transfer

“conduction”

“radiation”

“convection”

© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 2

Conduction

TH

dTQ kA

dxL

RkA

© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 3

Convection

= Cooling by mass motion (diffusion + advection) in a fluid

TH

Q

R

© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 4

Radiation

Note: Usually nothing is a perfect “black body” and parts of the emissive spectrum may be missing (ex: photonic band gap crystals).

4 41 1 2 1 1 2( )Q F A T T

20

50

2( )

exp( / ) 1bbB

hce

hc k T

Linearize (when and why?):

8 3 24(5.67 10 )(300) 6 W/m Kradh

For black body (Ɛ=1) at 300 K:

© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 5

• All problems have boundaries!

• Heat diffusion equation needs boundary conditions

• Dirichlet (fixed T):

• Neumann (fixed flux ~ dT/dx):

• When is it OK to “lump” a body as a single R or C?

• Biot number:

Boundaries and Lumped Elements

© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 6

Transient Cooling of Lumped Body

Source: Lienhard book, http://web.mit.edu/lienhard/www/ahtt.html (2008)

© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 7

What if Biot Number is Large

• Bi = hL/kb << 1 implies Tb(x) ~ Tsurf (lumped OK)

• Bi = hL/kb >> 1 implies significant internal Tb(x) gradient

© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 8

Lumped Body Examples (Steady State)Boundary conditions:

TL = 400 oC, TR = 100 oC

1) Assume NO internal heat generation

(how does the temperature slope dT/dx

scale qualitatively within each layer?)

2) Assume UNIFORM internal heat generation

0ln( / )

2i

cyl

r rR

lk

slab

LR

kA

© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 9

Contact Resistance

• RC = 1/hCA

• BUT, also remember the fundamental solid-solid contact resistance given by density of states, acoustic/diffuse phonon mismatch ~Cv/4! (prof. Cahill’s lecture)

© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 10

Notes on Finite-Element Heat Diffusion

RC

T0T0

TN

TiTi-1

Ti+1

Δx

L

0)(')( 0 TTgpTkA

1 1

2i i idu u u

dx x

21 1

22

2i i i id u u u u

dx x

T1

Boundary conditions:(heat flux conservation)

1 011

C

T TdTk A

dx R

1 02 11

C

T TT Tk A

x R

=

M T b

T1

TN

Matlab:T = M\b

b1

bN

… …

M11 M12 0 …

MNN

………M21 M22 M23 0 …

© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 11

More Comments on “Fin Equation”

• Same as Poisson equation with various BC’s

• BC’s can be given flux (dT/dx) or given temperature (T0)

• Very useful to know:– Thermal healing length LH (Poisson: screening length)

– General, qualitative shape or solution

2

02( ) 0

d T hpT T

dx kA

Si

toxSiO2

d

T0

L

W

x x+dx

generalsolution

/ /0 1 2

H Hx L x LT T C e C e sinh, cosh, tanh … etc.

© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 12

Fin Efficiency (how long is too long?)

• Fin efficiency η = actual heat loss by fin / heat loss if entire fin was at base temperature TB

• Actual heat loss:

• Here

-2 -1 0 1 2-4

-2

0

2

4

6

8

x/LH

sinh,

cosh

, ta

nh,

exp

sinh

cosh

tanhT=TBdT/dx ≈ 0

L

W d

tanh( / )

/H

H

L L

L L Not worth making cooling fins much >> LH !

exp

© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 13

Poisson Equation Analogy

• Thermal fin is ~ mathematically same problem as 1-D transistor electrostatics, e.g. nanowire or SOI transistor

• L < λ short fin, or “short channel” FET

• L >> λ long fin (too long?!), or “long channel” FET

with solution

nt nt

ox

d

C

and electrostatic screening lengthLiu (1993)Knoch (2006)