Thermo-Mechanical Analysis of a Surface-Mounted Resistor

The drive for miniaturizing electronic devices has resulted in today’s extensive use of surface-mount electronic components. An important aspect in electronics design and the choice of materials is a product’s durability and lifetime. For surface-mount resistors and other components producing heat it is a well-known problem that temperature cycling can lead to cracks propagating through the solder joints, resulting in premature failure (Ref. 1). For electronics in general there is a strong interest in changing the soldering material from lead- or tin-based solder alloys to other mixtures.

The following multiphysics example models the heat transport and structural stresses and deformations resulting from the temperature distribution using the Heat Transfer in Solids and Solid Mechanics interfaces.

This application requires either the Structural Mechanics Module or the MEMS Module.

Model Definition

Figure 1 shows a photograph of a surface-mount resistor together with a diagram of it on a printed circuit board (PCB).

Figure 1: A photo and diagram of a typical surface-mounted resistor soldered to a PCB.

Table 1 shows the dimensions of the resistor and other key components in the model including the PCB.

Table 1: Component Dimensions.
Component	Length	Width	Height
Resistor (Alumina)	6 mm	3 mm	0.5 mm
PCB (FR4)	16 mm	8 mm	1.6 mm
Cu pad	2 mm	3 mm	35 μm
Termination (Silver)	0.5 mm	3 mm	25 μm
Stand-off (gap to PCB)	-	-	105 μm

The simulation makes use of the symmetry so that it needs to include only half of the component (Figure 2). The modeling of the PCB is terminated a distance away from the resistor, in order to reduce effects of the boundary conditions.

Figure 2: The simulation models only one half of the resistor.

In operation, the resistor dissipates 0.2 W of power as heat. Conduction to the PCB and convection to the surrounding air provide cooling. In this model, the heat transfer occurs through conduction in the subdomains. The model simplifies the surface cooling and describes it using a heat transfer coefficient, h, in this case set to 10 W/(m2·K); the surrounding air temperature, Tinf, is at 300 K. The resulting heat transfer equation and boundary condition (included in the model using the Heat Transfer interface) are

where k is the thermal conductivity, and Q is the heating power per unit volume of the resistor (equal to 22.2 MW/m3 corresponding to 0.2 W in total).

The model handles thermal expansion using a static structural analysis using the Structural Mechanics interface (a description of the corresponding equations is available in the Structural Mechanics Module User’s Guide). The thermal and mechanical material properties in this model are taken from the material library. The data for the solid materials are temperature independent, and the reference values are shown in the table below.

Table 2: Material properties.
Material	E (GPa)	ν	α (ppm)	k (W/(m·K))	ρ (kg/m3)	Cp (J/(kg·K))
Silver	83	0.37	18.9	420	10500	230
Alumina	300	0.222	8.0	27	3900	900
Cu	110	0.35	17	400	8700	385
Fr4	22	0.28	18	0.3	1900	1369
60Sn-40Pb	10	0.4	21	50	9000	150

Air has temperature-dependent and pressure-dependent properties in the built-in material library. Because the temperature is a variable of the problem, it is automatically used. The pressure is by default set to 1 atm.

The stresses are zero at room temperature, 293 K. The boundary condition for the Solid Mechanics interface is that the cuts in the PCB do not rotate and have no net force normal to the cut.

Results and Discussion

The isosurfaces in Figure 3 show the temperature distribution at steady state. The highest temperature occurs in the center of the resistor. The circuit board also heats up significantly.

Figure 3: Temperature distribution in the resistor and the circuit board at steady state.

Thermal stresses appear as a result of the temperature increase; they arise from the materials’ different expansion coefficients and from the bending of the PCB. Figure 4 plots the equivalent stress (von Mises) together with the resulting deformation of the assembly.

Figure 4: The thermally induced distribution of von Mises stress together with the deformation (magnified).

The highest stresses seem to occur in the termination material. It is interesting to compare these equivalent stresses to the yield stress and thereby investigate whether or not the material is irreversibly deformed. In that case the solder is the weak point. Figure 5 the stress in the solder alone.

Figure 5: Close-up of the von Mises stress in the solder joints.

The yield stress for solder is approximately 220 MPa. The highest equivalent stress appears to be about 40 percent of this value. The structure does not get permanently deformed directly when heated. However, it is possible that the solder displays creep strains over time because of the combination of fairly high stress levels and elevated temperatures.

Notes About the COMSOL Implementation

In this example, the Thermal Stress interface automatically adds and couples the Solid Mechanics interface and the Heat Transfer interface. This is done by the predefined multiphysics feature .

Build the geometry as an assembly of the bottom plate and the resistor to make it possible to mesh the parts independently. Use continuity conditions for the temperature and the displacements to connect the top and bottom parts of the geometry.

It is assumed that the temperature gradient in the normal direction on the cuts in the PCB is zero. The mechanical boundary conditions on the cuts must not impose a general state of compressive stress due to thermal expansion. At the same time, the restraint from the part of the PCB that is not modeled means that there is no rotation of the cross section. To obtain this effect, the entire cut must have the same (but unknown) displacement in the direction normal to the cut. This can be achieved by using a special setting in the condition which allows a translation normal to the symmetry plane.

You solve the problem sequentially using two stationary study steps. The heat transfer problem is nonlinear because the air has temperature-dependent properties. The structural problem, on the other hand, is linear. For the structural analysis, use a memory-efficient iterative solver to make it possible to solve the problem also on computers with limited memory.

References

1. H. Lu, C. Bailey, M. Dusek, C. Hunt, and J. Nottay, “Modeling the Fatigue Life of Solder Joints of Surface Mount Resistors,” EMAP, 2000.

2. J.M. Coulson and J.F. Richardson, Chemical Engineering, vol. 1, Pergamon Press, appendix, 1990.

Heat_Transfer_Module/Thermal_Stress/surface_resistor

Modeling Instructions

From the menu, choose .

New

In the window, click

Model Wizard

In the window, click

In the tree, select .

Click .

Click

In the tree, select .

Click

Global Definitions

Parameters 1

In the window, under click .

In the window for , locate the section.

In the table, enter the following settings:


Name	Expression	Value	Description
T_air	300[K]	300 K	Air temperature
h_air	10[W/(m^2*K)]	10 W/(m²·K)	Heat transfer coefficient
Psource	0.2[W]/2	0.1 W	Heat dissipated by the resistor on the half geometry
p0	1[atm]	1.0133E5 Pa	Air pressure
T0	80[degC]	353.15 K	Initial temperature guess

Geometry 1

In the window, under click .

In the window for , locate the section.

From the list, choose .

Create the geometry. To simplify this step, insert a prepared 2D geometry sequence. On the toolbar, click . Browse to the model’s Application Library folder and double-click the file surface_resistor.mph.

Work Plane 1 (wp1)>Union 1 (uni1)

The 2D geometry should now look as in the figure below.

Extrude 1 (ext1)

In the toolbar, click

Select the object only.

In the window for , locate the section.

In the table, enter the following settings:


Distances (mm)
1.5

Click

Block 1 (blk1)

In the toolbar, click

In the window for , locate the section.

In the text field, type 4.

In the text field, type 16.

In the text field, type 1.6.

Locate the section. In the text field, type -4.

In the text field, type -1.6.

Click

Click the

button in the toolbar.

Now, create an imprint of the resistor’s bottom boundary on the printed circuit board to make a pair with matching parts.

Form Union (fin)

In the window, under click .

In the window for , locate the section.

From the list, choose .

Select the check box.

In the toolbar, click

The completed geometry is shown in Figure 2.

Add Material

In the toolbar, click

to open the window.

Go to the window.

In the tree, select .

Click in the window toolbar.

In the tree, select .

Click in the window toolbar.

In the tree, select .

Click in the window toolbar.

In the tree, select .

Click in the window toolbar.

In the tree, select .

Click in the window toolbar.

In the tree, select .

Click in the window toolbar.

In the toolbar, click

to close the window.

Materials

FR4 (Circuit Board) (mat1)

In the window, under click .

In the window for , locate the section.

From the list, choose .

Click

Select Domain 1 only.

Alumina (mat2)

In the window, click .

Select Domain 5 only.

Copper (mat3)

In the window, click .

Select Domains 2 and 7 only.

Silver [solid] (mat4)

In the window, click .

Select Domains 4 and 9 only.

Solder, 60Sn-40Pb (mat5)

In the window, click .

Select Domains 3 and 8 only.

Air (mat6)

In the window, click .

Select Domain 6 only.

Definitions

Symmetry Plane

In the toolbar, click

In the window for , type Symmetry Plane in the text field.

Locate the section. From the list, choose .

Locate the section. In the text field, type 0.1.

Locate the section. From the list, choose .

Solid Mechanics (solid)

In the window, under click .

Select Domains 1–5 and 7–9 only.

Symmetry 1

In the toolbar, click

and choose .

In the window for , locate the section.

From the list, choose .

Continuity 1

In the toolbar, click

and choose .

In the window for , locate the section.

Under , click

In the dialog box, select in the list.

Click .

Roller 1

In the toolbar, click

and choose .

Select Boundary 8 only.

Symmetry 2

In the toolbar, click

and choose .

Select Boundary 2 only.

In the window for , click to expand the section.

From the list, choose .

Symmetry 3

In the toolbar, click

and choose .

Select Boundary 9 only.

In the window for , locate the section.

From the list, choose .

Prescribed Displacement 1

In the toolbar, click

and choose .

Select Point 1 only.

In the window for , locate the section.

Select the check box.

Heat Transfer in Solids (ht)

In the window, under click .

Fluid 1

In the toolbar, click

and choose .

Select Domain 6 only.

Heat Source 1

In the toolbar, click

and choose .

Select Domain 5 only.

In the window for , locate the section.

Click the button.

In the P0 text field, type Psource.

Heat Flux 1

In the toolbar, click

and choose .

Select Boundaries 3, 4, 11, 14, 19, 29, 30, 44, 46, and 49–58 only.

In the window for , locate the section.

Click the button.

In the h text field, type h_air.

In the Text text field, type T_air.

Next, add a continuity condition to the identity pair to couple the domains together.

Continuity 1

In the toolbar, click

and choose .

In the window for , locate the section.

Under , click

In the dialog box, select in the list.

Click .

Because the material properties are temperature dependent, the solution converges better if you supply an initial guess of the temperature.

Initial Values 1

In the window, click .

In the window for , locate the section.

In the T text field, type T0.

Mesh 1

Free Triangular 1

In the toolbar, click

and choose .

Select Boundaries 10, 13, 16, 20, 24, 33, 37, and 40 only.

Size 1

Right-click and choose .

In the window for , locate the section.

From the list, choose .

Click the button.

Locate the section. Select the check box.

In the associated text field, type 0.1.

Click

Swept 1

In the toolbar, click

In the window for , locate the section.

From the list, choose .

Select Domains 2–9 only.

Distribution 1

In the toolbar, click

In the window for , locate the section.

In the text field, type 10.

Click

Free Tetrahedral 1

In the toolbar, click

Size 1

Right-click and choose .

In the window for , locate the section.

From the list, choose .

Select Boundaries 5–7 only.

Locate the section. From the list, choose .

Click

Because the heat transfer problem is independent of the displacements, use the first stationary study step to find the temperature distribution and the second stationary step to solve for the displacements.

Study 1

Step 1: Stationary

In the window, under click .

In the window for , locate the section.

In the table, clear the check box for .

Stationary 2

In the toolbar, click

and choose .

In the window for , locate the section.

In the table, clear the check box for .

Click to expand the section. Find the subsection. From the list, choose .

From the list, choose .

In the toolbar, click

Results

Surface 1

In the window, expand the node, then click .

In the window for , locate the section.

From the list, choose .

Hold down the left mouse button and drag in the window to rotate the geometry so that you see the opposite side of the resistor, which is where the largest stresses occur, as shown in Figure 4. Similarly, use the right mouse button to translate the geometry and the middle button to zoom.

Now, study the stresses in the solder.

Study 1/Solution 1 (3) (sol1)

In the toolbar, click