15 Chapter Cooling of Electronic Equipment 15-55 A plastic DIP with 16 leads is cooled by forced air. Using data supplied by the manufacturer, the junction temperature is to be determined. Assumptions Steady operating conditions exist. Analysis The junction-to-ambient thermal resistance of the device with 16 leads corresponding to an air velocity of 300 m/min is determined from Fig.15-23 to be
R junction
ambient
Air 25 C 300 m/min
50 C / W
2W
Then the junction temperature becomes
Q T junction T junction Tambient R junction Tambient
ambient
QR junction
ambient
25 C + (2 W)(50 C/W) = 125 C
When the fan fails the total thermal resistance is determined from Fig.15-23 by reading the value for zero air velocity (the intersection point of the curve with the vertical axis) to be
R junction
ambient
70 C / W
which yields
Q T junction T junction Tambient R junction Tambient
ambient
QR junction
ambient
25 C + (2 W)(70 C/W) = 165 C
15-56 A PCB with copper cladding is given. The percentages of heat conduction along the copper and epoxy layers as well as the effective thermal conductivity of the PCB are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat conduction along the PCB is one-dimensional since heat transfer from side surfaces is negligible. 3 The thermal properties of epoxy and copper layers are constant. Analysis Heat conduction along a layer is proportional to the thermal conductivity-thickness product (kt) which is determined for each layer and the entire PCB to be
( kt ) copper ( kt ) epoxy ( kt ) PCB (386 W / m. C)(0.06 10 m) (0.26 W / m. C)(0.5 10-3 m) ( kt ) copper ( kt ) epoxy
-3
PCB 12 cm 12 cm
0.02316 W/ C 0.00013 W/ C 0.02329 W/ C
Q
0.02316 0.00013
Therefore the percentages of heat conduction along the epoxy board are
f epoxy (kt) epoxy (kt) PCB 0.00013 W/ C 0.02316 W/ C 0.0056 0.6%
Copper t = 0.06 mm
Epoxy t = 0.5 mm
and
fcopper
(100 0.6)% 99.4%
Then the effective thermal conductivity becomes
k eff (kt) epoxy t epoxy (kt) copper t copper (0.02316 0.00013) W/ C (0.06 + 0.5) 10 -3 m 41.6 W/m. C
15-16
Chapter 15 Cooling of Electronic Equipment 15-57 "!PROBLEM 15-057" "GIVEN" length=0.12 "[m]" width=0.12 "[m]" "t_copper=0.06 [mm], parameter to be varied" t_epoxy=0.5 "[mm]" k_copper=386 "[W/m-C]" k_epoxy=0.26 "[W/m-C]" "ANALYSIS" kt_copper=k_copper*t_copper*Convert(mm, m) kt_epoxy=k_epoxy*t_epoxy*Convert(mm, m) kt_PCB=kt_copper+kt_epoxy f_copper=kt_copper/kt_PCB*Convert(, %) f_epoxy=100-f_copper k_eff=(kt_epoxy+kt_copper)/((t_epoxy+t_copper)*Convert(mm, m))
Tcopper [mm] 0.02 0.025 0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065 0.07 0.075 0.08 0.085 0.09 0.095 0.1
fcopper [%] 98.34 98.67 98.89 99.05 99.17 99.26 99.33 99.39 99.44 99.48 99.52 99.55 99.58 99.61 99.63 99.65 99.66
keff [W/m-C] 15.1 18.63 22.09 25.5 28.83 32.11 35.33 38.49 41.59 44.64 47.63 50.57 53.47 56.31 59.1 61.85 64.55
15-17
Chapter 15 Cooling of Electronic Equipment
15-18
Chapter 15 Cooling of Electronic Equipment 15-58 The heat generated in a silicon chip is conducted to a ceramic substrate to which it is attached. The Q temperature difference between the front and back surfaces of the chip is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat conduction along the chip is onedimensional. Analysis The thermal resistance of silicon chip is
Rchip L kA 0.5 10-3 m (130 W / m. C)(0.006 0.006)m2
6 6
3W Chip 0.5 mm
01068 C / W .
Ceramic substrate
Then the temperature difference across the chip becomes
T QRchip (3 W)(0.1068 C/W) = 0.32 C
15-59E The dimensions of an epoxy glass laminate are given. The thermal resistances for heat flow along the layers and across the thickness are to be determined. Assumptions 1 Heat conduction in the laminate is one-dimensional in either case. 2 Thermal properties of the laminate are constant. Analysis The thermal resistances of the PCB along the 7 in long side and across its thickness are
Ralong
length
Qlength
6 in
7 in
Qthickness
L kA (7/12) ft (0.15 Btu/h.ft. F)(6/12 ft)(0.05/12 ft) 1867 h. F/Btu
(a)
Racross
(b)
thickness
L kA (0.05/12) ft (0.15 Btu/h.ft. F)(7/12 ft)(6/12 ft) 0.095 h. F/Btu
0.05 in
15-19
Chapter 15 Cooling of Electronic Equipment 15-60 Cylindrical copper fillings are planted throughout an epoxy glass board. The thermal resistance of the board across its thickness is to be determined. Assumptions 1 Heat conduction along the board is one-dimensional. 2 Thermal properties of the board are constant. Analysis The number of copper fillings on the board is
n Area of board Area of one square (150 mm)(180 mm) (3 mm)(3 mm)
3 mm
Copper filing
1 mm 3 mm
Epoxy board
3000
The surface areas of the copper fillings and the remaining part of the epoxy layer are
Acopper Atotal Aepoxy D2 (0.001 m) 2 (3000) 0.002356 m2 4 4 (length)( width) (015 m)(0.18 m) = 0.027 m2 . n Atotal Acopper 0.027 0.002356 0.024644 m2
The thermal resistance of each material is
Rcopper Repoxy L kA L kA 0.0014 m (386 W / m. C)(0.002356 m2 ) 0.0014 m (0.26 W / m. C)(0.024644 m2 ) 0.00154 C / W 0.2185 C / W
Since these two resistances are in parallel, the equivalent thermal resistance of the entire board is
1 Rboard 1 Repoxy 1 Rcopper 1 0.2185 C/W 1 0.00154 C/W Rboard 0.00153 C/W
15-20
Chapter 15 Cooling of Electronic Equipment 15-61 "!PROBLEM 15-061" "GIVEN" length=0.18 "[m]" width=0.15 "[m]" k_epoxy=0.26 "[W/m-C]" t_board=1.4/1000 "[m]" k_filling=386 "[W/m-C], parameter to be varied" "D_filling=1 [mm], parameter to be varied" s=3/1000 "[m]" "ANALYSIS" A_board=length*width n_filling=A_board/s^2 A_filling=n_filling*pi*(D_filling*Convert(mm, m))^2/4 A_epoxy=A_board-A_filling R_filling=t_board/(k_filling*A_filling) R_epoxy=t_board/(k_epoxy*A_epoxy) 1/R_board=1/R_epoxy+1/R_filling kfilling [W/m-C] 10 29.5 49 68.5 88 107.5 127 146.5 166 185.5 205 224.5 244 263.5 283 302.5 322 341.5 361 380.5 400 Dfilling [mm] 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 Rboard [C/W] 0.04671 0.01844 0.01149 0.008343 0.00655 0.005391 0.00458 0.003982 0.003522 0.003157 0.00286 0.002615 0.002408 0.002232 0.00208 0.001947 0.00183 0.001726 0.001634 0.00155 0.001475 Rboard [C/W] 0.005977 0.004189 0.003095 0.002378 0.001884 0.001529 0.001265 0.001064 0.0009073 0.0007828 0.0006823
15-21
Chapter 15 Cooling of Electronic Equipment 1.6 1.7 1.8 1.9 2 0.0005999 0.0005316 0.0004743 0.0004258 0.0003843
15-22
Chapter 15 Cooling of Electronic Equipment 15-62 A circuit board with uniform heat generation is to be conduction cooled by a copper heat frame. Temperature distribution along the heat PCB frame and the maximum temperature in the PCB are to be determined. 15 cm 18 cm Assumptions 1 Steady operating conditions exist 2 Thermal properties are constant. 3 There is no direct heat dissipation from the surface of the PCB, and thus all the heat generated is conducted by the heat frame to the heat sink. Heat frame Epoxy Analysis The properties and dimensions of adhesive Cold plate various section of the PCB are summarized below as Section and material Thermal Thickness Heat transfer surface conductivity area Epoxy board 2 mm 0.26 W/m. C 10 mm 120 mm Epoxy adhesive 0.12 mm 1.8 W/m. C 10 mm 120 mm Copper heat frame 1.5 mm 386 W/m. C 10 mm 120 mm (normal to frame) Copper heat frame 10 mm 386 W/m. C 15 mm 120 mm (along the frame) Using the values in the table, the various thermal resistances are determined to be L 0.002 m Repoxy 6.41 C / W kA (0.26 W / m. C)(0.01 m 012 m) . L 0.00012 m Radhesive 0.056 C / W kA (1.8 W / m. C)(0.01 m 0.12 m) L 0.0015 m Rcopper , 0.0032 C / W kA (386 W / m. C)(0.01 m 0.12 m) L 0.01 m R frame Rcopper , parallel 0144 C / W . kA (386 W / m. C)(0.0015 0.12 m) The combined resistance between the electronic components on each strip and the heat frame can be determined by adding the three thermal resistances in series to be Rvertical Repoxy Radhesive Rcopper, 6.41 0.056 0.0032 6.469 C / W The temperatures along the heat frame can be determined from the relation
T1 T2 T3 T4 T5 T6 T7 T8 T0 T1 T2 Q1 0 R1 Q2 1 R2 Q R
3 2 0 1
T
Thigh
Tlow
QR . Then,
30 C + (22.5 W)(0.144 C / W) = 33.24 C 33.24 C + (19.5 W)(0.144 C / W) = 36.05 C 36.05 C + (16.5 W)(0.144 C / W) = 38.42 C 38.42 C + (13.5 W)(0.144 C / W) = 40.36 C 40.36 C + (10.5 W)(0.144 C / W) = 41.87 C 4187 C + (7.5 W)(0.144 C / W) = 42.95 C . 42.95 C + (4.5 W)(0.144 C / W) = 43.60 C 4388 C + (1.5 W)(0.144 C / W) = 43.81 C .
T9
3 2 3 4
T3 Q4 3 R4 T4 Q5 4 R5 T Q R
5
6 5 6 5 6 7
3W
Repox
y
T6 T7
Q7 6 R7 Q8 7 R8
The maximum surface temperature on the PCB is Tmax T9 T8 Qvertical Rvertical 4381 C + (3 W)(6.469 C / W) = 63.2 C .
22.5 W T0 T1 19.5 W T2 16.5 W T3 13.5 W T4 10.5 W T5 7.5 W T6 4.5 W T7 1.5 W T8
Radhesive
Rcopper
15-23
Chapter 15 Cooling of Electronic Equipment 15-63 A circuit board with uniform heat generation is to be conduction cooled by aluminum wires inserted into it. The magnitude and location of the maximum temperature in the PCB is to be determined. Assumptions 1 Steady conditions operating exist 2 Thermal properties are constant. 3 There is no direct heat dissipation from the surface of the PCB. Analysis The number of wires in the board is
n 150 mm 2 mm 75
Double sided PCB 12 cm 15 cm Aluminum wire, D = 1 mm
The surface areas of the aluminum wires and the remaining part of the epoxy layer are
Aalu min um Atotal Aepoxy D (0.001 m) (75) 0.0000589 m2 4 4 (length)( width) (0.003 m)(0.15 m) = 0.00045 m2 n Atotal Aalu min um 0.00045 0.0000589 0.0003911 m2
2 2
2 mm
Considering only half of the circuit board because of symmetry, the thermal resistance of each material per 1-cm length is determined to be
Ralu min um Repoxy L kA L kA 0.01 m (237 W / m. C)(0.0000589 m 2 ) 0.01 m (0.26 W / m. C)(0.0003911 m 2 ) 0.716 C / W 98.34 C / W
3 mm
Since these two resistances are in parallel, the equivalent thermal resistance per cm is determined from
1 Rboard 1 Repoxy 1 Ralu min um 1 0.716 C / W 1 98.34 C / W Rboard 0.711 C / W
Maximum temperature occurs in the middle of the plate along the 20 cm length, which is determined to be
Tmax Tend Tboard ,total Tend Qi Rboard ,1
cm
Tend
Rboard ,1
cm
Qi
30 C + (0.711 C / W)(15 +13.5 +12 +10.5 + 9 + 7.5 + 6 + 4.5 + 3 +1.5)W = 88.7 C
15 W
13.5 W
12 W
10.5 W
9W
7.5 W
6W
4.5 W
3W
1.5 W
30 C
1 cm
Tmax Rboard
15-24
Chapter 15 Cooling of Electronic Equipment 15-64 A circuit board with uniform heat generation is to be conduction cooled by copper wires inserted in it. The magnitude and location of the maximum temperature in the PCB is to be determined. Assumptions 1 Steady operating conditions exist 2 Thermal properties are constant. 3 There is no direct heat dissipation from the surface of the PCB. Analysis The number of wires in the circuit board is
n 150 mm 2 mm 75
Double sided PCB 12 cm 15 cm Copper wire, D = 1 mm
The surface areas of the copper wires and the remaining part of the epoxy layer are
Acopper Atotal Aepoxy D2 (0.001 m) 2 n (75) 0.0000589 m2 4 4 (length)( width) (0.003 m)(0.15 m) = 0.00045 m2 Atotal Acopper 0.00045 0.0000589 0.0003911 m2
2 mm
Considering only half of the circuit board because of symmetry, the thermal resistance of each material per 1-cm length is determined to be
Rcopper Repoxy L kA L kA 0.01 m (386 W / m. C)(0.0000589 m2 ) 0.01 m (0.26 W / m. C)(0.0003911 m2 ) 0.440 C / W 98.34 C / W
3 mm
Since these two resistances are in parallel, the equivalent thermal resistance is determined from
1 Rboard 1 Repoxy 1 Rcopper 1 0.440 C / W 1 98.34 C / W Rboard 0.438 C / W
Maximum temperature occurs in the middle of the plate along the 20 cm length which is determined to be
Tmax Tend Tboard ,total Tend Qi Rboard ,1
cm
Tend
Rboard ,1
cm
Qi
30 C + (0.438 C / W)(15 +13.5 +12 +10.5 + 9 + 7.5 + 6 + 4.5 + 3 +1.5)W = 66.1 C
15 W
13.5 W
12 W
10.5 W
9W
7.5 W
6W
4.5 W
3W
1.5 W
30 C
1 cm
Tmax Rboard
15-25
Chapter 15 Cooling of Electronic Equipment 15-65 A circuit board with uniform heat generation is to be conduction cooled by aluminum wires inserted into it. The magnitude and location of the maximum temperature in the PCB is to be determined. Assumptions 1 Steady operating conditions exist 2 Thermal properties are constant. 3 There is no direct heat dissipation from the surface of the PCB. Analysis The number of wires in the board is
n 150 mm 4 mm 37
Double sided PCB 12 cm 15 cm Aluminum wire, D = 1 mm
The surface areas of the aluminum wires and the remaining part of the epoxy layer are
Aalu min um Atotal Aepoxy D2 (0.001 m) 2 n (37) 0.000029 m2 4 4 (length)( width) (0.003 m)(0.15 m) = 0.00045 m2 Atotal Aalu min um 0.00045 0.000029 0.000421 m2
4 mm
Considering only half of the circuit board because of symmetry, the thermal resistance of each material per 1-cm length is determined to be
Ralu min um Repoxy L kA L kA 0.01 m (237 W / m. C)(0.000029 m2 ) 0.01 m (0.26 W / m. C)(0.000421 m2 ) 1455 C / W . 9136 C / W .
3 mm
Since these two resistances are in parallel, the equivalent thermal resistance is determined from
1 Rboard 1 Repoxy 1 Ralu min um 1 1455 C / W . 1 9136 C / W . Rboard 1432 C / W .
Maximum temperature occurs in the middle of the plate along the 20 cm length which is determined to be
Tmax Tend Tboard ,total Tend Qi Rboard ,1
cm
Tend
Rboard ,1
cm
Qi
30 C + (1.432 C / W)(15 +13.5 +12 +10.5 + 9 + 7.5 + 6 + 4.5 + 3 +1.5)W = 148.1 C
15 W
13.5 W
12 W
10.5 W
9W
7.5 W
6W
4.5 W
3W
1.5 W
30 C
1 cm
Tmax Rboard
15-26
Chapter 15 Cooling of Electronic Equipment 15-66 A thermal conduction module with 80 chips is cooled by water. The junction temperature of the chip is to be determined. Assumptions 1 Steady operating conditions exist 2 Heat transfer through various components is one-dimensional. Analysis The total thermal resistance between the junction and cooling water is
Rtotal R junction
water
Junction
Rchip
4W
Rchip
Rint ernal
Rexternal
12 9 7 17.2 C .
Rinternal
Then the junction temperature becomes
T junction Twater QR junction
water
18 C + (4 W)(17.2 C/W) = 86.8 C
Rexternal
Cooling water
15-67 A layer of copper is attached to the back surface of an epoxy board. The effective thermal conductivity of the board and the fraction of heat conducted through copper are to be determined. Assumptions 1 Steady operating conditions exist 2 Heat transfer is one-dimensional. Analysis Heat conduction along a layer is proportional to the thermal conductivity-thickness product (kt) which is determined for each layer and the entire PCB to be
(kt) copper (kt) epoxy (kt) PCB (386 W/m. C)(0.0001 m) (0.26 W/m. C)(0.0003 m) (kt) copper (kt) epoxy 0.0386 W/ C 0.000078 W/ C 0.038678 W/ C
Copper, t = 0.1 mm
PCB 15 cm 20 cm
Q
0.0386 0.000078
Epoxy, t = 0.3 mm
The effective thermal conductivity can be determined from
k eff
(kt) epoxy t epoxy
(kt) copper t copper
(0.0386 0.000078) W/ C (0.0003 m + 0.0001 m)
96.7 W/m. C
Then the fraction of the heat conducted along the copper becomes
f (kt) copper (kt) PCB 0.0386 W/ C 0.038678 W/ C 0.998 99.8%
Discussion Note that heat is transferred almost entirely through the copper layer.
15-27
Chapter 15 Cooling of Electronic Equipment 15-68 A copper plate is sandwiched between two epoxy boards. The effective thermal conductivity of the board and the fraction of heat conducted through copper are to be determined. Assumptions 1 Steady operating conditions exist 2 Heat transfer is one-dimensional. Analysis Heat conduction along a layer is proportional to the thermal conductivity-thickness product (kt) which is determined for each layer and the entire PCB to be
( kt ) copper ( kt ) epoxy ( kt ) PCB (386 W / m. C)(0.0005 m) (2)(0.26 W / m. C)(0.003 m) ( kt ) copper ( kt ) epoxy 0193 W/ C . 0.00156 W/ C 019456 W/ C .
Copper, t = 0.5 mm 12 cm
0193 0.00156 .
Q
The effective thermal conductivity can be determined from
k eff (kt) epoxy t epoxy (kt) copper t copper (0.00156 0.193) W/ C (2 0.003 m) + 0.0005 m 29.9 W/m. C
Then the fraction of the heat conducted along the copper becomes
f (kt) copper (kt) PCB 0.193 W/ C 0.19456 W/ C 0.992 99.2%
Epoxy, t = 3 mm
15-69E A copper heat frame is used to conduct heat generated in a PCB. The temperature difference between the mid section and either end of the heat frame is to be determined. Assumptions 1 Steady operating conditions exist 2 Heat transfer is one-dimensional. Analysis We assume heat is generated uniformly on the 6 in 8 in board, and all the heat generated is conducted by the heat frame along the 8-in side. Noting that the rate of heat transfer along the heat frame is variable, we consider 1 in 8 in strips of the board. The rate of heat generation in each strip is (20 W)/8 = 2.5 W, and the thermal resistance along each strip of the heat frame is
R frame L kA (1/12) ft (223 Btu/h.ft. F)(6/12 ft)(0.06/12 ft) 0.149 h. F/Btu
Tend
1 in 10 W PCB 6 in 8 in
8 in Heat frame Cold plate
7.5 W
5W
2.5 W
Tmid Rboard
Maximum temperature occurs in the middle of the plate along the 20 cm length. Then the temperature difference between the mid section and either end of the heat frame becomes
Tmax Tmid section - edge of frame Qi R frame,1
in
R frame,1
in
Qi
(0.149 F.h/Btu)(10 + 7.5 + 5 + 2.5 W)(3.4121 Btu/h.W) = 12.8 F
15-28
Chapter 15 Cooling of Electronic Equipment 15-70 A power transistor is cooled by mounting it on an aluminum bracket that is attached to a liquidcooled plate. The temperature of the transistor case is to be determined. Assumptions 1 Steady operating conditions exist 2 Conduction heat transfer is one-dimensional. Analysis The rate of heat transfer by conduction is Liquid channels
Transistor
2 cm 2 cm
Qconduction
(080)(12 W) = 9.6 W .
The thermal resistance of aluminum bracket and epoxy adhesive are
Ralu min um Repoxy L kA L kA 0.01 m 0.703 C / W (237 W / m. C)(0.003 m)(0.02 m) 0.0002 m 1852 C / W . (1.8 W / m. C)(0.003 m)(0.02 m)
Aluminum bracket
The total thermal resistance between the transistor and the cold plate is
Rtotal Rcase
cold plate
R plastic
Repoxy
Ralu min um
2.5 1852 0.703 5.055 C / W .
Then the temperature of the transistor case is determined from
Tcase Tcold
plate
QRcase
cold plate
50 C + (9.6 W)(5.055 C/W) = 98.5 C
15-29