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Question : A 10- cm × 12- cm circuit board dissipating 24 W of heat is ...

A 10−cm×12−cm 10- cm \times 12- cm  circuit board dissipating 24 W of heat is to be conduction-cooled by a 1.2-mm-thick copper heat frame (k=386W/m.∘C) \left(k=386 W / m .^{\circ} C \right)  10cm×14cm10 cm \times 14 cm  in size. The epoxy laminate (k=0.26W/m.∘C) \left(k=0.26 W / m.^{\circ} C \right)  has a thickness of 0.8 mm and is attached to the heat frame with conductive epoxy adhesive (k=1.8W/m.∘C) (k= \left.1.8 W / m.^{\circ} C \right)  of 0.13- mm thickness, as shown in Fig. 15-31. The PCB is attached to a heat sink by clamping a 5-mm-wide portion of the edge to the heat sink from both ends. The temperature of the heat frame at this point is 20∘C 20^{\circ} C. Heat is uniformly generated on the PCB at a rate of 2 W per 1−cm×10−cm 1- cm \times 10- cm  strip. Considering only one-half of the PCB board because of symmetry, determine the maximum temperature on the PCB and the temperature distribution along the heat frame.

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SOLUTION A circuit board with uniform heat generation is to be conduction-cooled by a copper heat frame. Temperature distribution along the heat frame and the maximum temperature in the PCB are 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, and thus all the heat generated is conducted by the heat frame to the heat sink.

Analysis The PCB under consideration possesses thermal symmetry about the centerline. Therefore, the heat generated on the left half of the PCB is conducted to the left heat sink, and the heat generated on the right half is conducted to the right heat sink. Thus we need to consider only half of the board in the analysis.The maximum temperature will occur at a location furthest away from the heat sinks, which is the symmetry line. Therefore, the temperature of the electronic components located at the center of the PCB will be the highest, and their reliability will be the lowest.Heat generated in the components on each strip is conducted through the epoxy layer underneath. Heat is then conducted across the epoxy adhesive and to the middle of the copper heat frame. Finally, heat is conducted along the heat frame to the heat sink. The thermal resistance network associated with heat flow in the right half of the PCB is also shown in Fig. (15-31). Note that all vertical resistances are identical and are equal to the sum of the three resistances in series. Also note that heat conduction toward the heat sink is assumed to be predominantly along the heat frame, and conduction along the epoxy adhesive is considered to be negligible. This assumption is quite reasonable, since the conductivity-thickness product of the heat frame is much larger than those of the other two layers. The properties and dimensions of various sections of the PCB are summarized in this table.

 

Section and Material Thermal Conductivity, W/m. °C Thickness, mm Heat Transfer Surface Area
Epoxy board 0.26 0.8 10 mm x 100 mm
Epoxy adhesive 1.8 0.13 10 mm x 100 mm
Copper heat frame, _|_ (normal to frame) 386 0.6 10 mm x 100 mm
Copper heat frame,||(along the frame) 386 10 1.2 mm x 100 mm

Using the values in the table, the various thermal resistances are determined to be

Repoxy =(LkA)epoxy =0.8×10−3m(0.26W/m⋅∘C)(0.01m×0.1m)=3.077∘C/WR_{\text {epoxy }} =\left(\frac{L}{k A}\right)_{\text {epoxy }}=\frac{0.8 \times 10^{-3} m }{\left(0.26 W / m \cdot{ }^{\circ} C \right)(0.01 m \times 0.1 m )}=3.077^{\circ} C / W

 

Radhesive =(LkA)adhesive =0.13×10−3m(1.8W/m⋅∘C)(0.01m×0.1m)=0.072∘C/W R_{\text {adhesive }} =\left(\frac{L}{k A}\right)_{\text {adhesive }}=\frac{0.13 \times 10^{-3} m }{\left(1.8 W / m \cdot{ }^{\circ} C \right)(0.01 m \times 0.1 m )}=0.072^{\circ} C / W

 

Rcopper, ⊥=(LkA)copper, ⊥=0.6×10−3m(386W/m⋅∘C)(0.01m×0.1m)=0.002∘C/W R_{\text {copper, } \perp} =\left(\frac{L}{k A}\right)_{\text {copper, } \perp}=\frac{0.6 \times 10^{-3} m }{\left(386 W / m \cdot{ }^{\circ} C \right)(0.01 m \times 0.1 m )}=0.002^{\circ} C / W

 

Rframe =Rcopper, ∥=(LkA)copper, ∥=0.01m(386W/m⋅∘C)(0.0012m×0.1m) R_{\text {frame }} =R_{\text {copper, } \|}=\left(\frac{L}{k A}\right)_{\text {copper, } \|}=\frac{0.01 m }{\left(386 W / m \cdot{ }^{\circ} C \right)(0.0012 m \times 0.1 m )}

 

=0.216∘C/W =0.216^{\circ} C / W

 

The combined resistance between the electronic components on each strip and the heat frame can be determined, by adding the three resistances in series, to be

Rvertical =Repoxy +Radhesive +Rcopper, ⊥ R_{\text {vertical }} =R_{\text {epoxy }}+R_{\text {adhesive }}+R_{\text {copper, } \perp}

 

=(3.077+0.072+0.002)∘C/W =(3.077+0.072+0.002)^{\circ} C / W

 

=3.151∘C/W =3.151^{\circ} C / W

 

The various temperatures along the heat frame can be determined from the relation

ΔT=Thigh −T10w=Q˙R \Delta T=T_{\text {high }}-T_{10 w}=\dot{Q} R

where R is the thermal resistance between two specified points, Q˙ \dot{Q}  is the heat transfer rate through that resistance, and ΔT \Delta T  is the temperature difference across that resistance.The temperature at the location where the heat frame is clamped to the heat sink is given as T0=20∘C T_{0}=20^{\circ} C. Noting that the entire 12 W of heat generated on the right half of the PCB must pass through the last thermal resistance adjacent to the heat sink, the temperature T1 T_{1}  can be determined from

T1=T0+Q˙1−0R1−0=20∘C+(12W)(0.216∘C/W)=22.59∘C T_{1}=T_{0}+\dot{Q}_{1-0} R_{1-0}=20^{\circ} C +(12 W )\left(0.216^{\circ} C / W \right)=22.59^{\circ} C

Following the same line of reasoning, the temperatures at specified locations along the heat frame are determined to be

T2=T1+Q˙2−1R2−1=22.59∘C+(10W)(0.216∘C/W)=24.75∘C T_{2}=T_{1}+\dot{Q}_{2-1} R_{2-1}=22.59^{\circ} C +(10 W )\left(0.216^{\circ} C / W \right)=24.75^{\circ} C

 

T3=T2+Q˙3−2R3−2=24.75∘C+(8W)(0.216∘C/W)=26.48∘C T_{3}=T_{2}+\dot{Q}_{3-2} R_{3-2}=24.75^{\circ} C +(8 W )\left(0.216^{\circ} C / W \right)=26.48^{\circ} C

 

T4=T3+Q˙4−3R4−3=26.48∘C+(6W)(0.216∘C/W)=27.78∘C T_{4}=T_{3}+\dot{Q}_{4-3} R_{4-3}=26.48^{\circ} C +(6 W )\left(0.216^{\circ} C / W \right)=27.78^{\circ} C

 

T5=T4+Q˙5−4R5−4=27.78∘C+(4W)(0.216∘C/W)=28.64∘C T_{5}=T_{4}+\dot{Q}_{5-4} R_{5-4}=27.78^{\circ} C +(4 W )\left(0.216^{\circ} C / W \right)=28.64^{\circ} C

 

T6=T5+Q˙6−5R6−5=28.64∘C+(2W)(0.216∘C/W)=29.07∘C T_{6}=T_{5}+\dot{Q}_{6-5} R_{6-5}=28.64^{\circ} C +(2 W )\left(0.216^{\circ} C / W \right)=29.07^{\circ} C

 

Finally, T7 T_{7}, which is the maximum temperature on the PCB, is determined from

T7=T6+Q˙vertical Rvertical =29.07∘C+(2W)(3.151∘C/W)=35.37∘C T_{7}=T_{6}+\dot{Q}_{\text {vertical }} R_{\text {vertical }}=29.07^{\circ} C +(2 W )\left(3.151^{\circ} C / W \right)=35.37^{\circ} C

Discussion The maximum temperature difference between the PCB and the heat sink is only 15.37∘C 15.37^{\circ} C, which is very impressive considering that the PCB has no direct contact with the cooling medium. The junction temperatures in this case can be determined by calculating the temperature difference between the junction and the leads of the chip carrier at the point of contact to the PCB and adding 35.37∘C 35.37^{\circ} C  to it. The maximum temperature rise of 15.37∘C 15.37^{\circ} C can be reduced, if necessary, by using a thicker heat frame.