Calculating the Parameters of the Thermal Equivalent Circuit A device in a TO-220 package is mounted with a 0.33-mm thick insulating foil on a small aluminum heat sink. This is shown in Figure 17.15a. The thermal resistance of the heat sink is Rth_KK = 25 K/W and its mass is msk = 2 g. The surface

Question

Calculating the Parameters of the Thermal Equivalent Circuit

A device in a TO-220 package is mounted with a 0.33-mm thick insulating foil on a small aluminum heat sink. This is shown in Figure 17.15a. The thermal resistance of the heat sink is [latex]R_{th–KK} = 25 K/W[/latex] and its mass is [latex]m_{sk} = 2 g[/latex]. The surface area of the TO-220 package is [latex]A_{sk} = 1 cm^2[/latex]. The surface area of the device chip is [latex]A_{cu} = 10 mm^2[/latex], the amount of copper around the pyramid stump is [latex]m_{cu} = 1 g[/latex], and the thickness of the copper is [latex]d_{cu }= 0.8 mm[/latex]. Find the parameters of the thermal equivalent circuit.

Accepted Answer

Because the heat sink is small and compact, there is no need to divide the structure into several RC elements. The first-order thermal equivalent circuit is shown in Figure 17.15b. [latex]m_{sk }= 2 g, R_{th–KK} = 25 K/W, d_{foil} = 0.3 mm, A_{foil} = 1 cm^2, A_{cu} = 10 mm^2, m_{cu} = 1 g[/latex], and [latex]d_{cu}= 0.8 mm[/latex].
From Table 17.2,

Table 17.2 Thermal Data for Common Materials [Ref. 1]

	[latex]ρ [g/cm^3][/latex]	[latex]\lambda _{th}[W/1mK2][/latex]	[latex]c [J/(gK)][/latex]
Silicon	2.4	140	0.7
Solder (Sn-Pb)	9	60	0.2
Cu	7.6 to 8.9	310 to 390	0.385 to 0.42
Al	2.7	170 to 230	0.9 to 0.95
[latex]Al_2 O_3[/latex]	3.8	24	0.8
FR4	—	24	—
Heat conductive paste	—	0.4 to 2.6	—
Insulating foil	—	0.9 to 2.7	—

the specific heat of the aluminum material is [latex]c_{sk} = 0.95 J/1gK^2[/latex]. Thus, the thermal capacitance of the heat sink becomes

[latex]C_{th-KK} = c_{sk} m_{sk}=0.95\frac{J}{gK}0.2\frac{J}{K}=1.9 g[/latex]

For insulating foil, Table 17.2 gives [latex]λ_{th-foil} = 1.1 W/mK[/latex]. Thus, the thermal resistance of the foil is

[latex]R_{th-foil}=\frac{d_{foil}}{\lambda_{th-foil} A_{foil}}=\frac{0.3 mm}{1.1\frac{W}{mK} imes 1 cm^2 }=2.7\frac{K}{W} [/latex]

For the copper chip, Table 17.2 gives [latex]c_{cu} = 0.39 J/(gK) [/latex].

The thermal capacitance of the chip is

[latex]C_{th7} = c_{cu} m_{cu} = 0.39\frac{J}{gK} imes 1 g=0.39\frac{J}{K}[/latex]

[latex]R_{th7} = \frac{d_{cu}}{\lambda _{th}A_{cu}}=\frac{0.8 mm}{390\frac{W}{mK} imes 10mm^2 }=0.205\frac{K}{W}[/latex]

Question 17.2: Calculating the Parameters of the Thermal Equivalent Circuit...

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