**3. Passive heat exchanger design**

The simplest design technique uses the thermal resistance model of the heat exchanger. It supposes that the coolant (air for example) temperature,*Ta*, is

constant in contact with the heatsink. Using this model, the power of heat dissipated by a heatsink is written as follows:

$$Q = \frac{T\_f - T\_a}{R\_{th}} \tag{10}$$

Where:

*Q* is the heat flux from the component to ambient air (Watts), *TJ* and *Ta* are, respectively, junction and air temperatures (°C), and *R*th is the thermal resistance of the heat exchanger (°C/W). Three design problems can generally be encountered:


#### **3.1 Heat exchanger sizing**

In this case, it is desired to find a commercially available heatsink, capable of dispersing the heat generated by the operation, in surroundings of temperature *Ta*, of an electronic component with a maximum power generation *Qmax* and a maximum junction operating temperature,*Top*.

The sizing procedure is carried out in two steps.

STEP 1: From component, thermal interface, and heatsink datasheets select those which satisfy the space constraint. Get the values of *Qmax*,*Top*, *RJC*, *RCS*, and *RSA*.

STEP 2: Write Eq. (10) for *Qmax* and determine the thermal resistance of the heatsink which will insure operating temperature not exceed *Top*.

Written for *Qmax* and *Top* Eq. (10) gives the following:

$$Q\_{\text{max}} = \frac{T\_{op} - T\_a}{R\_{th}} \tag{11}$$

Rearranging, we obtain:

$$R\_{th} = \frac{T\_{op} - T\_a}{Q\_{max}} \tag{12}$$

*Illustration 1*

*An electronic chip is required to operate at a surrounding temperature of 45°C. This chip is mounted with a casing having a thermal resistance RJC, and with thermal interface materials having a global resistance RCS. The data sheet of the chip shows that it can generate up to Qmax at Top.*

*Supposing there are no space or geometric constraints, what are the heat sinks that can be used among those of Table 1.*

*Heat Exchangers for Electronic Equipment Cooling DOI: http://dx.doi.org/10.5772/intechopen.100732*

*Data: RJC* ¼ 0*:*3°*C=W RCS* ¼ 0*:*4°*C=W:* Q max *=10 W Top = 85°C. Solution Using Eq. (12), the thermal resistance at maximum power is: Rth* <sup>¼</sup> <sup>85</sup>�<sup>45</sup> <sup>10</sup> ¼ 4°*C=W The adequate heatsink resistance is obtained using Eq. (9) as follows:*

$$R\_{\rm SA} = R\_{th} - \left(R\_{\rm JC} + R\_{\rm CS}\right)$$

$$R\_{\rm JC} + R\_{\rm CS} = \mathbf{0}.7 \, \text{\textdegree C} \prime\_{\rm W} \Longrightarrow R\_{\rm SA} = \mathbf{3.3 \, \text{\textdegree C}} \prime\_{\rm W}$$

*This means that only heat sinks with thermal resistances lower or equal to 3.3°C/W will be able to give heat exchangers with Rth* ≤ 4°*C=W, thus ensuring safe operation of this chip.*

*This is the case for references 1 and 6 only.*
