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Problem 1.2-28

Figure P1.2-28 illustrates a power conversion electronic chip.

lead ribbon

Wc = 1 cm

Wc = 1 cmWs = 1.5 cm

Ws = 1.5 cm

metal vias & underfill

thv = 1.2 mm, Dv = 0.5 mm, vd= 100 vias/cm2

kv

= 25.3 W/m-K, kuf

= 2.1 W/m-K

dielectric

thd = 0.35 mm, kd = 0.85 W/m-K

chip

thc = 3.2 mm, kc = 121 W/m-K

5 Wg =

spreader

ths = 7.5 mm, ks = 85 W/m-K

5 28 2x10 m -K/W

cR . =

2458 W/m -K

20 Cw

w

h

T=

=

= 0.9

237 2 W/m -K

30 Ca

a

h .

T=

=

Figure P1.2-28: Power conversion chip.

The chip is Wc = 1 cm x Wc = 1 cm in area and thc = 3.2 mm thick with conductivity kc = 121W/m-K. Inefficiencies in the power conversion process result in the generation of thermal

energy at a rate g = 5 W. This generation is distributed throughout the chip but can be modeled

as occurring at the upper surface. The lower surface of the chip is attached to a dielectric

material with thickness thd = 0.35 mm and conductivity kd = 0.85 W/m-K. The dielectric

material is attached to a spreader block; this attachment is characterized by contact resistance Rc"= 8.2x10-5 K-m2 /W. The spreader is Ws = 1.5 cm x Ws = 1.5 cm in area and ths = 7.5 mm with

conductivity ks = 85 W/m-K. The bottom surface of the spreader is cooled by water. The water

temperature is Tw = 20C and the heat transfer coefficient is wh = 458 W/m2-K. Above the chip

there are many metal vias used for signals that lead to traces on a ribbon lead. Each metal via is

a cylinder with diameter Dv = 0.5 mm and thickness thv = 1.2 mm. There are vd= 100 vias/cm2

and the conductivity of the via material is kv = 25.3 W/m-K. Between the vias is an underfillmaterial with conductivity kuf = 2.1 W/m-K. You may neglect the thermal resistance of the lead

ribbon. The top surface is cooled by convection with air,a

h = 37.2 W/m2-K and Ta = 30C. The

top surface also radiates to the air temperature. The emissivity of the surface is = 0.9.

This problem is clearly three-dimensional. However, it is possible to develop simple resistance

network models of the problem that can be used to obtain an upper and lower bound on the chip

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temperature (i.e., the temperature at the interface between the vias and the chip where thethermal energy generation occurs).

a.) Obtain a lower bound for the chip temperature using a resistance model. Your solution

should include a resistance network that is clearly labeled with the numerical value of eachresistance for the conditions shown in Figure P1.2-28.

The inputs are entered in EES:

$UnitSystem SI Mass Radian J K Pa

W_c=1 [cm]*convert(cm,m) "chip width"th_c=3.2 [mm]*convert(mm,m) "chip thickness"g_dot=5 [W] "rate of thermal energy generation"k_c=121 [W/m-K] "thermal conductivity of chip"W_s=1.5 [cm]*convert(cm,m) "spreader width"th_s=7.5 [mm]*convert(mm,m) "spreader thickness"k_s=85 [W/m-K] "spreader conductivity"th_v=1.2 [mm]*convert(mm,m) "via thickness"

D_v=0.5 [mm]*convert(mm,m) "via diameter"vd=100 [1/cm^2]*convert(1/cm^2,1/m^2) "via density"k_v=25.3 [W/m-K] "via conductivity"k_uf=2.1 [W/m-K] "underfill conductivity"th_d=0.35 [mm]*convert(mm,m) "dielectric thickness"k_d=0.85 [W/m-K] "dielectric conductivity"R``_c=8.2e-5 [K-m^2/W] "dielectric to spreader contact resistance"h_bar_a=37.2 [W/m^2-K] "air heat transfer coefficient"e=0.9 [-] "emissivity"h_bar_w=458 [W/m^2-K] "water heat transfer coefficient"T_w=converttemp(C,K,20 [C]) "water temperature"T_a=converttemp(C,K,30 [C]) "air temperature"

The isothermal model is used to obtain a lower value of each resistance and therefore a lowervalue of the chip temperature. The isothermal resistance network is shown in Figure 2.

5 Wg =

Tus

Rc

= 0.26 K/W

Rd

= 4.12 K/W

Rcontact = 0.82 K/W

Rs,iso = 0.39 K/W

Rconv,w,iso = 9.7 K/W

Tw

upper,isoq

Rv = 2.42 K/W

Ruf

= 7.11 K/W Rconv,a,iso

= 268.8 K/W

Ta

Rrad,iso

= 1303 K/W

Figure 2: Isothermal resistance network.

The resistance of the chip and dielectric material are computed according to:

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2

cc

c c

thR

W k= (1)

2

dd

c d

thR

W k= (2)

The contact resistance between the dielectric and the spreader is calculated according to:

2

ccontact

c

RR

W

= (3)

"Isothermal model"R_c=th_c/(W_c^2*k_c) "chip resistance"R_d=th_d/(W_c^2*k_d) "dielectric resistance"R_contact=R``_c/W_c^2 "contact resistance"

The conduction resistance of the spreader and convection resistance to the water, estimated usingthe isothermal model (in which the entire spreader area can be used for conduction and

convection), are:

, 2

ss iso

s s

thR

W k= (4)

, , 2

1conv w iso

s w

RW h

= (5)

R_s_iso=th_s/(W_s^2*k_s) "isothermal spreader resistance"R_conv_w_iso=1/(h_bar_w*W_s^2) "isothermal water convection resistance"

The number of vias is estimated according to:

2

v c N W vd = (6)

The area of each via is:

2

4

vv

DA = (7)

The resistance to conduction through all of the vias is:

vv

v v v

thR

k N A= (8)

The resistance to conduction through the underfill surrounding the vias is:

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( )2v

uf

uf c v v

thR

k W N A=

(9)

The resistance to convection from the air for the isothermal model is:

, , 2

1conv a iso

c a

RW h

= (10)

N_v=W_c^2*vd "number of vias"A_v=pi*D_v^2/4 "area of each via"R_v=th_v/(k_v*N_v*A_v) "via resistance"R_uf=th_v/(k_uf*(W_c^2-N_v*A_v)) "resistance of underfill"R_conv_a_iso=1/(h_bar_a*W_c^2) "isothermal air convection resistance"

The radiation resistance from the top surface depends on the temperature of the top surface, Tus,iso

in Figure 2. This temperature is guessed initially and used to compute the resistance to radiation:

( ) ( ), 2 2 2, ,1

rad iso

c us iso a us iso a

RW T T T T

=+ +

(11)

T_us_iso=T_a "guess for upper surface temperature"R_rad_iso=1/(e*sigma#*W_c^2*(T_us_iso+T_a)*(T_us_iso^2+T_a^2)) "radiation resistance"

The resistance from the chip to the water is:

, , , ,lower iso c d contact s iso conv w iso R R R R R R= + + + + (12)

The resistance from the chip to the air is:

1 1

,

, , ,

1 1 1 1upper iso

v uf conv a iso rad iso

R R R R R

= + + +

(13)

The chip temperature is computed according to:

( ) ( ), ,, ,

c iso w c iso a

lower iso upper iso

T T T T g

R R

= + (14)

R_lower_iso=R_c+R_d+R_contact+R_s_iso+R_conv_w_iso "resistance from generation to water"R_upper_iso=(1/R_v+1/R_uf)^(-1)+(1/R_conv_a_iso+1/R_rad_iso)^(-1)"resistance from generation to air"g_dot=(T_c_iso-T_w)/R_lower_iso+(T_c_iso-T_a)/R_upper_iso "chip temperature"T_c_iso_C=converttemp(K,C,T_c_iso) "in C"

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The problem is solved and the guess values updated (select Update Guess Values from the

Calculate menu). The guess for the upper surface temperature is commented out:

{T_us_iso=T_a} "guess for upper surface temperature"

The heat transfer rate to the air ( ,upper isoq

in Figure 2) is:

( ),,

,

c iso a

upper iso

upper iso

T Tq

R

= (15)

The upper surface temperature is computed according to:

1

, , ,

1 1us iso c iso upper iso

v uf

T T qR R

= +

(16)

q_dot_upper_iso=(T_c_iso-T_a)/R_upper_iso "heat transfer to air"T_us_iso=T_c_iso-q_dot_upper_iso*(1/R_v+1/R_uf)^(-1) "upper surface temperature"

which leads to Tus,iso = 365.4 K (92.25C). The values of each resistance are labeled in Figure 2.

b.) Obtain an upper bound for the chip temperature using a resistance model. Your solutionshould include a resistance network that is clearly labeled with the numerical value of each

resistance for the conditions shown in Figure P1.2-28.

The adiabatic model is used to obtain an upper bound on the chip temperature. The adiabaticresistance network is shown in Figure 3.

5 Wg =

Tus,v,ad

Rc = 0.26 K/W

Rd = 4.12 K/W

Rcontact = 0.82 K/W

Rs,ad = 0.88 K/W

Rconv,w,ad = 21.8 K/W

Tw

Rv = 2.42 K/W

Ruf = 7.11 K/W

Ta

Rrad,v,ad = 5196 K/W

Ta

Tc,ad

Rconv,a,v,ad = 1369 K/W

Tus,uf,adTa

Rrad,uf,ad = 1285 K/W

Ta

Rconv,a,uf,ad= 334.5 K/W

Figure 3: Adiabatic resistance network.

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The resistance of the chip and dielectric material and the contact resistance do not change. The

resistance of the spreader and convection to the water do change because heat is not allowed to

spread in order to take advantage of the entire area of the spreader in the isothermal model.

, 2

ss ad

c s

thR

W k

= (17)

, , 2

1conv w ad

c w

RW h

= (18)

"Adiabatic model"R_conv_w_ad=1/(h_bar_w*W_c^2) "adiabatic water convection resistance"R_s_ad=th_s/(W_c^2*k_s) "adiabatic spreader resistance"

In the adiabatic model, the heat is not allowed to spread as it leaves the vias; therefore, theconvection resistance from the surface above the vias and the convection resistance from the

surface above the underfill must be computed separately:

, , ,

1conv a v ad

a v v

Rh N A

= (19)

( ), , , 2

1conv a uf ad

a c v v

Rh W N A

=

(20)

The surface temperature above the vias and underfill (Tus,v,ad and Tus,uf,ad) are assumed in order to

calculate the radiation resistances:

( ) ( ), , 2 2, , , ,1

rad v ad

v v us v ad a us v ad a

R N A T T T T

=+ +

(21)

( ) ( ) ( ), , 2 2 2, , , ,1

rad uf ad

c v v us uf ad a us uf ad a

RW N A T T T T

= + +

(22)

T_us_v_ad=T_a "guess for upper surface temperature over vias"T_us_uf_ad=T_a "guess for upper surface temperature over underfill"R_rad_v_ad=1/(e*sigma#*N_v*A_v*(T_us_v_ad+T_a)*(T_us_v_ad^2+T_a^2))

"radiation resistance over vias"R_rad_uf_ad=1/(e*sigma#*(W_c^2-N_v*A_v)*(T_us_uf_ad+T_a)*(T_us_uf_ad^2+T_a^2))"radiation resistance over underfill"

The resistance from the chip to the water is:

, , , ,lower ad c d contact s ad conv w ad R R R R R R= + + + + (23)

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The resistance from the chip to the air is:

1

,

, ,

1 1upper ad

upper v upper uf

RR R

= +

(24)

where

1

,

, , , , ,

1 1upper v v

conv a v ad rad v ad

R RR R

= + +

(25)

1

,

, , , , ,

1 1upper uf uf

conv a uf ad rad uf ad

R RR R

= + +

(26)

The chip temperature is computed according to:

( ) ( ), ,, ,

c ad w c ad a

lower ad upper ad

T T T T g

R R

= + (27)

R_lower_ad=R_c+R_d+R_contact+R_s_ad+R_conv_w_ad "resistance from generation to water"R_upper_v=R_v+(1/R_conv_a_v_ad+1/R_rad_v_ad)^(-1) "upper resistance through vias"R_upper_uf=R_uf+(1/R_conv_a_uf_ad+1/R_rad_uf_ad)^(-1) "upper resistance through underfill"R_upper_ad=(1/R_upper_v+1/R_upper_uf)^(-1) "resistance from generation to air"g_dot=(T_c_ad-T_w)/R_lower_ad+(T_c_ad-T_a)/R_upper_ad "chip temperature"

T_c_ad_C=converttemp(K,C,T_c_ad) "in C"

The problem is solved and the guess values updated (select Update Guess Values from the

Calculate menu). The guess for the upper surface temperature is commented out:

{T_us_v_ad=T_a "guess for upper surface temperature over vias"T_us_uf_ad=T_a "guess for upper surface temperature over underfill"}

The heat transfer rates to the air through the vias (, ,upper v adq in Figure 3) and through the underfill

(, ,upper uf adq ) are calculated:

( ),, ,

,

c ad a

upper v ad

upper v

T Tq

R= (28)

( ),, ,

,

c ad a

upper uf ad

upper uf

T Tq

R

= (29)

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The upper surface temperatures over the via (Tus,v,ad) and over the underfill (Tus,v,uf) are computedaccording to:

, , , , ,us v ad c ad upper v ad vT T q R= (30)

, , , , ,us uf ad c ad upper uf ad uf T T q R= (31)

q_dot_upper_v_ad=(T_c_ad-T_a)/R_upper_v "heat transfer to air through vias"q_dot_upper_uf_ad=(T_c_ad-T_a)/R_upper_uf "heat transfer to air through underfill"T_us_v_ad=T_c_ad-q_dot_upper_v_ad*R_v "upper surface over vias"T_us_uf_ad=T_c_ad-q_dot_upper_uf_ad*R_uf "upper surface over underfill"

which leads to Tus,ad = 418 K (144.9C). The values of each resistance are labeled in Figure 3.

c.) You need to reduce the chip temperature (or increase the power level of the chip). Based on

your resistance network from (a), provide two areas of research which would provide the

most benefit and two areas that are not likely to provide much benefit. For example,"improving convection from the air to the upper surface" would be an area of research.

Clearly justify your answers.

The majority of the heat travels from the chip surface to the water; this is evident because the left

side of the resistance network in Figure 2 is much smaller than the right side. The largest

resistances in the series on the left correspond to the convection with the water and conductionacross the dielectric. Therefore, the two areas of research that would provide the most benefit

are:

improve the convection coefficient with the water improve the conductivity or reduce the thickness of the dielectric

Two resistances that are not very important according to Figure 2 are the resistance to radiation

and the resistance of the chip material. Therefore, two areas of research that would provide very

little benefit are:

increase the emissivity of the surface increase the conductivity of the chip material

d.) You have decided to aggressively cool the top surface using impinging jets to increase the

value ofah and want to understand the potential benefit of this approach. Plot the upper and

lower bounds on the chip temperature as a function ofah for 1 W/m

2-K