Failure Criterion for Moisture-Sensitive Plastic Packages of Integrated Circuit (IC) Devices: Application and Extension of the Theory of Thin Plates of Large Deflections

Abstract

The maximum value of the von Mises stress in the molding compound at the chip corner is suggested to be used as a suitable failure criterion for moisture-induced plastic packages of integrated circuit (IC) devices. This criterion is able to reflect the role of various geometric and material characteristics affecting the package propensity to moisture-induced failures during high-temperature reflow soldering. It is suggested also that the von Mises stress be determined from the constitutive equations which are a generalization of the von Karman equations for large deflections of plates with consideration of thermoelastic strains. The generalized von Karman equations are applied to the underchip layer of the molding compound and consider thermal effects associated with the thermal expansion (contraction) mismatch of the materials, in the package as well as with temperature gradients. The predicted stresses are in good agreement with experimental observations. The calculated von Mises stresses can be used particularly for the development of “figures of merit” that would enable one to separate packages that need to be “baked” and “bagged” (or “rebaked” and “rebagged”) from those that supposedly do not. The calculated stresses can be used also to judge whether the qualification test conditions for sufficiently reliable packages (say, thick packages with small chips) could be safely “derated” to an actual factory humidity profile. Finally the calculated von Mises stress can be helpful in the selection of the most feasible molding compound for the given package design. A more reliable (and more expensive) material might be needed in the case of a thin package with a large chip, while a low-cost compound can be successfully employed in the case of a thick package with a small chip.

Appendix. Clamped Plate of Finite Aspect Ratio Experiencing Large Deflections

If the temperature in a rectangular plate changes in the through-thickness direction only, then for an initially flat and stress flee plate, equations (10.16) can be used to evaluate the deflections w(x, y) and the stress function $\phi \left( {x,\,y} \right)$.

The functions w(x, y) and $\phi \left( {x,\,y} \right)$ can be sought in the form [8]:

$$w\left( {x,\,y} \right) = fw^ * \left( {x,\,y} \right), \quad \phi \left( {x,\,y} \right) = f^2 \phi ^ * \left( {x,\,y} \right),$$

((10.83))

where f is the maximum deflection at the center of the plate, the coordinate function $w^ * \left( {x,\,y} \right)$ is chosen in such a way that the boundary conditions for the deflection function at the plate’s contour be fulfilled, and the function $\phi ^ * \left( {x,\,y} \right)$ is to be determined. Substituting relationships (10.83) into the second equation in equation (10.16), we obtain the following equation for the function $\phi ^ * \left( {x,\,y} \right)$:

$$\nabla ^4 \phi ^ * \left( {x,\,y} \right) = - \frac{{E_c }}{2}L\left[ {w^ * \left( {x,\,y} \right), \,\,\, w^ * \left( {x,\,y} \right)} \right] = E_c \left[ {\left( {\frac{{\partial ^2 w^ * }}{{\partial x\partial y}}} \right)^2 - \frac{{\partial ^2 w^ * }}{{\partial x^2 }}\frac{{\partial ^2 w^ * }}{{\partial y^2 }}} \right],$$

((10.84))

where the equivalent modulus E _c is expressed by formula (10.10).

In an approximate analysis one can assume for a plate clamped at the support contour

$$w^ * \left( {x,\,y} \right) = \cos ^2 \frac{{\pi x}}{a}\cos ^2 \frac{{\pi y}}{b}.$$

((10.85))

Then equation (10.84) yields

$$\begin{array}{rcl} \nabla ^4 \phi ^ * \left( {x,\,y} \right) & \displaystyle = & \displaystyle - \frac{{\pi ^4 E_0 }}{{2a^2 b^2 }}\left( {\cos \frac{{2\pi x}}{a} + \cos \frac{{2\pi y}}{b} + \cos \frac{{4\pi x}}{a} + \cos \frac{{4\pi y}}{b}} \right. \\ && \displaystyle \left. { + \, 2\cos \frac{{2\pi x}}{a}\cos \frac{{2\pi y}}{b} + \cos \frac{{2\pi x}}{a}\cos \frac{{4\pi y}}{b} + \cos \frac{{4\pi x}}{a}\cos \frac{{2\pi y}}{b}} \right) \\ \end{array}.$$

((10.86))

This equation has the following solution:

$$\begin{array}{rcl} \phi ^ * \left( {x,\,y} \right) & \displaystyle = & \displaystyle Ax^2 + By^2 + D_0 \cos \frac{{2\pi x}}{a} + D_1 \cos \frac{{2\pi y}}{b} + D_2 \cos \frac{{4\pi x}}{a} + D_3 \cos \frac{{4\pi y}}{b} \\ && \displaystyle + \, D_4 \cos \frac{{2\pi x}}{a}\cos \frac{{2\pi y}}{b} + D_5 \cos \frac{{2\pi x}}{a}\cos \frac{{4\pi y}}{b} + D_6 \cos \frac{{4\pi x}}{a}\cos \frac{{2\pi y}}{b}\,\,. \\ \end{array}$$

((10.87))

Substituting expression (10.87) into equation (10.86) we obtain the formulae for the constants $D_0 \to D_6$. Introducing equation (10.87) into conditions (10.9) of the nondeformability of the contour, we determine the constants A and B. Then, formula (10.87) results in the following expression for the function φ ^*:

$$\begin{array}{l} \phi ^ * \left( {x,\,y} \right) \displaystyle = \displaystyle \frac{{E_0 }}{{32}}\left\{ {\frac{{3\pi ^2 }}{{2\left( {1 - \upsilon ^2 } \right)}}} \left[ {\left( {\frac{\upsilon }{{a^2 }} + \frac{1}{{b^2 }}} \right)x^2 + \left( {\frac{1}{{a^2 }} + \frac{\upsilon }{{b^2 }}} \right)y^2 } \right] - \frac{{a^2 }}{{b^2 }}\cos \frac{{\pi x}}{a} \right.\\ \qquad\qquad\;\, \displaystyle - \, \frac{{a^2 }}{{b^2 }}\cos \frac{{\pi y}}{b} - \left( {\frac{a}{{4b}}} \right)^2 \cos \frac{{4\pi x}}{a} - \left( {\frac{b}{{4a}}} \right)^2 \cos \frac{{4\pi y}}{b} \\ \qquad\qquad\;\, \displaystyle - \frac{{2a^2 b^2 }}{{\left( {a^2 + b^2 } \right)}}\cos \frac{{2\pi x}}{a}\cos \frac{{2\pi y}}{b} + \, \frac{{a^2 b^2 }}{{\left( {4a^2 + b^2 } \right)^2 }}\cos \frac{{2\pi x}}{a}\cos \frac{{4\pi y}}{b} \\ \qquad\qquad\;\, \displaystyle \left. - \frac{{a^2 b^2 }}{{\left( {a^2 + 4b^2 } \right)^2 }}\cos \frac{{4\pi x}}{a} {\cos \frac{{2\pi y}}{b}} \right\}. (10.88)\\ \end{array}$$

((10.88))

The in-plane stresses can then be evaluated as follows:

$$\begin{array}{l} \displaystyle \sigma _x^0 = \frac{{\partial ^2 \phi }}{{\partial y^2 }} = f^2 \frac{{\partial ^2 \phi ^ * }}{{\partial y^2 }}\, {=}\, \frac{{\pi ^2 E_0 f^2 }}{{32}}\left[ {\frac{3}{{1 - \upsilon ^2 }}} \right.\left( {\frac{1}{{a^2 }}\, {+}\, \frac{\upsilon }{{b^2 }}} \right)\, {+}\, \frac{1}{{a^2 }}\cos \frac{{\pi y}}{b}\, {+}\, \frac{1}{{a^2 }}\cos \frac{{4\pi y}}{b} \\ \qquad\;\;\displaystyle + \, \frac{{8a^2 }}{{\left( {a^2 + b^2 } \right)^2 }}\cos \frac{{2\pi x}}{a}\cos \frac{{2\pi y}}{b} + \frac{{16a^2 }}{{\left( {4a^2 + b^2 } \right)^2 }}\cos \frac{{2\pi x}}{a}\cos \frac{{4\pi y}}{b}\\\qquad\;\;\displaystyle + \frac{{4a^2 }}{{\left( {a^2 + 4b^2 } \right)^2 }}\cos \frac{{4\pi x}}{a}\left. {\cos \frac{{2\pi y}}{b}} \right], \\ \displaystyle \sigma _y^0\, {=}\, \frac{{\partial ^2 \phi }}{{\partial x^2 }} = f^2 \frac{{\partial ^2 \phi ^ * }}{{\partial x^2 }}\, {=}\, \frac{{\pi ^2 E_0 f^2 }}{{32}}\left[ {\frac{3}{{1 - \upsilon ^2 }}} \right.\left( {\frac{\upsilon }{{a^2 }}\, {+}\, \frac{1}{{b^2 }}} \right)\, {+}\, \frac{1}{{b^2 }}\cos \frac{{\pi x}}{a}\, {+}\, \frac{1}{{b^2 }}\cos \frac{{4\pi x}}{a} \\ \qquad\;\;\displaystyle + \, \frac{{8b^2 }}{{\left( {a^2 + b^2 } \right)^2 }}\cos \frac{{2\pi x}}{a}\cos \frac{{2\pi y}}{b} + \frac{{16a^2 }}{{\left( {a^2 + 4b^2 } \right)^2 }}\cos \frac{{4\pi x}}{a}\cos \frac{{2\pi y}}{b} \\\qquad\;\;\displaystyle + \, \frac{{4b^2 }}{{\left( {4a^2 + b^2 } \right)^2 }}\cos \frac{{2\pi x}}{a}\left. {\cos \frac{{4\pi y}}{b}} \right],\\ \end{array}$$

((10.89))

$$\begin{array}{l} \displaystyle \tau _{xy}^0 = - \frac{{\partial ^2 \phi }}{{\partial x\partial y}} = - f^2 \frac{{\partial ^2 \phi ^ * }}{{\partial x\partial y}} = \frac{{\pi ^2 E_0 abf^2 }}{4}\left[ \frac{1}{{\left( {a^2 + b^2 } \right)^2 }}\sin \frac{{2\pi x}}{a}\sin \frac{{2\pi y}}{b} \right. \\ \qquad\;\;\displaystyle \left. + \, \frac{1}{{\left( {4a^2 + b^2 } \right)^2 }}\sin \frac{{2\pi x}}{a}\sin \frac{{4\pi y}}{b} + {\frac{1}{{\left( {a^2 + 4b^2 } \right)^2 }}\sin \frac{{4\pi x}}{a}\sin \frac{{2\pi y}}{b}} \right]. \\ \end{array}$$

((10.89))

At the point x = a/2, y = 0, formulae (10.89) yield

$$\sigma _x^0 = \left[ {\sigma _x^0 } \right]_\infty \psi _x^0 \left( \gamma \right), \quad \sigma _y^0 = \left[ {\sigma _y^0 } \right]_\infty \psi _y^0 \left( \gamma \right)$$

((10.90))

where

$$\left. \begin{array}{l} \displaystyle \left[ {\sigma _x^0 } \right]_\infty = \frac{{5 - 2\upsilon ^2 }}{{1 - \upsilon ^2 }}\frac{{\pi ^2 E_0 f^2 }}{{32a^2 }}\,, \\ \displaystyle \left[ {\sigma _y^0 } \right]_\infty = \frac{{3\upsilon ^2 }}{{1 - \upsilon ^2 }}\frac{{\pi ^2 E_0 f^2 }}{{32a^2 }}\, \\ \end{array} \right\}$$

((10.91))

are the in-plane stresses, calculated for the case of an elongated plate ($b \to \infty ,\break \gamma = 0$), and the factors

$$\left. \begin{array}{l} \displaystyle \psi _x^0 \left( \gamma \right) = 1 + \frac{3}{{5 - 2\upsilon ^2 }}\gamma ^2 - \frac{{4\left( {1 - \upsilon ^2 } \right)}}{{5 - 2\upsilon ^2 }}\gamma ^4 \left[ {\frac{2}{{\left( {1 + \gamma ^2 } \right)^2 }} + \frac{4}{{\left( {1 + 4\gamma ^2 } \right)^2 }} - \frac{1}{{\left( {4 + \gamma ^2 } \right)^2 }}} \right], \\ \displaystyle \psi _y^0 \left( \gamma \right) = 1 + \frac{{2 + \upsilon ^2 }}{{3\upsilon }}\gamma ^2 - \frac{{4\left( {1 - \upsilon ^2 } \right)}}{{3\upsilon ^2 }}\gamma ^4 \left[ {\frac{2}{{\left( {1 + \gamma ^2 } \right)^2 }} + \frac{4}{{\left( {1 + 4\gamma ^2 } \right)^2 }} - \frac{1}{{\left( {4 + \gamma ^2 } \right)^2 }}} \right]. \\ \end{array} \right\}$$

((10.92))

(10.92)

reflect the effect of the finite aspect ratio $\gamma = a/b$. The shear stress at the point $x = a/2, \, y = 0$ is zero.

For an elongated plate ($\gamma = 0$), formulae (10.92) yield $\psi _x^0 (1) = \psi _y^0 (0) = 1$. For a square plate ($\gamma = 1$) these formulae result in the expressions

$$\psi _x^0 = \frac{{\left( {1 + \upsilon } \right)\left( {63 + 12\upsilon } \right)}}{{25\left( {5 - 2\upsilon ^2 } \right)}}, \qquad \psi _y^0 = 1 + \frac{{4\left( {3 + 10\upsilon ^2 } \right)}}{{75\upsilon }}.$$

((10.93))

If, for instance, v = 0.4, then the finite aspect ratio of the plate results in lower in-plane stresses in the x-direction and in higher stresses in the y-direction.

The first equation in equation (10.16) can be solved, using the Galerkin method (see, for instance, [7]). In accordance with the procedure of this method, we substitute formulae (10.83), with consideration of equation (10.84), into the first equation in equation (10.16), multiply the obtained expression by the coordinate function $w^ * \left( {x,\,y} \right)$, and integrate the result over the plate’s surface A. This leads to the following equation for the dimensionless deflection $\,\overline f = f/h$:

$$\bar f + \alpha \bar f^3 = \bar f_0,$$

((10.94))

where

$$\bar f_0 = \frac{1}{{Dh}}\frac{{\int_A {p\left( {x,\,y} \right)w^ * \left( {x,\,y} \right){\textrm{d}}A} }}{{\int_A {w^ * } \left( {x,\,y} \right)\nabla ^4 w^ * \left( {x,\,y} \right){\textrm{d}}A}}$$

((10.95))

is the dimensionless linear deflection ($\alpha = 0$) and

$$\alpha = - \frac{{h^3 }}{D}\frac{{\int_A {w^ * \left( {x,\,y} \right)L\left[ {w^ * \left( {x,\,y} \right) , \,\,w^ * \left( {x,\,y} \right)} \right]{\textrm{d}}A} }}{{\int_A {w^ * } \left( {x,\,y} \right)\nabla ^4 w^ * \left( {x,\,y} \right){\textrm{d}}A}}$$

((10.96))

is the parameter of nonlinearity. The solution to equation (10.94) can be written as

$$\bar f = \eta _f \bar f_0,$$

((10.97))

where the factor

$$\eta _f = \frac{1}{\sqrt[3]{\mu }}\left( {\sqrt[\lower 12pt\hbox{$^{^{3}}$}]{{1 + \sqrt {1 + \frac{8}{{27\mu }}} }} + \sqrt[\lower 12pt\hbox{$^{^{3}}$}]{{1 - \sqrt {1 + \frac{8}{{27\mu }}} }}} \right),\;\;\;\;\mu = 2\alpha \bar f_0^2$$

((10.98))

considers the effect of nonlinearity (“membrane” stresses).

Introducing equation (10.85) into formulae (10.95) and (10.96), we obtain

$$\bar f_0 = \frac{{pa^4 }}{{\pi ^4 D\left( {3 + 2\gamma ^2 + 3\gamma ^4 } \right)h}} = \frac{{12}}{{\pi ^4 }}\frac{p}{{E_D }}\frac{{1 - \upsilon ^2 }}{{3 + 2\gamma ^2 + 3\gamma ^4 }}\left( {\frac{a}{h}} \right)^4,$$

((10.99))

$$\begin{array}{l}\displaystyle \alpha = \frac{{3\left( {1 - \upsilon ^2 } \right)}}{{8\left( {3 + 2\gamma ^2 + 3\gamma ^3 } \right)}}\left[ \frac{9}{4}\frac{{1 + 2\upsilon \gamma ^2 + \gamma ^4 }}{{1 - \upsilon ^2 }} + \frac{{17}}{8}\left( {1 + \gamma ^4 } \right) + \frac{{12\gamma ^4 }}{{\left( {1 + \gamma ^2 } \right)^2 }} \right.\\\qquad\displaystyle \left. + \, \frac{{5\gamma ^4 }}{{\left( {1 + 4\gamma ^2 } \right)^2 }} + \frac{{5\gamma ^4 }}{{\left( {4 + \gamma ^2 } \right)^2 }} \right].\end{array}$$

((10.100))

Using formulae (10.94) for the bending moments acting in the plate’s cross sections, putting $M_T = 0$ and considering the first formula in equation (10.83), we have

$$\left. \begin{array}{rcl} M_x & \displaystyle = & \displaystyle - Df\left( {\frac{{\partial ^2 w^ * }}{{\partial x^2 }} + \upsilon \frac{{\partial ^2 w^ * }}{{\partial y^2 }}} \right), \\ M_y & = & \displaystyle - Df\left( {\frac{{\partial ^2 w^ * }}{{\partial y^2 }} + \upsilon \frac{{\partial ^2 w^ * }}{{\partial x^2 }}} \right), \\ M_{xy} & = & \displaystyle - D\left( {1 - \upsilon } \right)f\frac{{\partial ^2 w^ * }}{{\partial x\partial y}}\,. \\ \end{array} \right\}$$

((10.101))

Using equation (10.85), we obtain

$$\left. \begin{array}{rcl} M_x & \displaystyle = & \displaystyle 2\pi ^2 Df\left( {\frac{1}{{a^2 }}\cos \frac{{2\pi x}}{a}\cos \frac{{2\pi y}}{b} + \frac{\upsilon }{{b^2 }}\cos \frac{{2\pi y}}{b}\cos \frac{{2\pi x}}{a}} \right), \\ M_y & \displaystyle = & \displaystyle 2\pi ^2 Df\left( {\frac{1}{{b^2 }}\cos \frac{{2\pi y}}{b}\cos \frac{{2\pi x}}{a} + \frac{\upsilon }{{a^2 }}\cos \frac{{2\pi x}}{a}\cos \frac{{2\pi y}}{b}} \right), \\ M_{xy} & \displaystyle = & \displaystyle - D\left( {1 - \upsilon } \right)f\frac{{\pi ^2 }}{{ab}}\sin \frac{{2\pi x}}{a}\sin \frac{{2\pi y}}{b}\,. \\ \end{array} \right\}$$

At the point $x = a/2, \,y = 0$, these formulae yield

$$M_x = - \frac{{2\pi ^2 }}{{a^2 }}Df,\;\;\;\;M_y = - \frac{{2\pi ^2 }}{{a^2 }}\upsilon Df,\;\;\;\;M_{xy} = 0\,.$$

The bending stresses caused by these moments can be calculated as

$$\sigma _x = - \frac{{6M_x }}{{h^2 }} = \frac{{12\pi ^2 Df}}{{a^2\,h^2 }},\;\;\;\;\sigma _y = - \frac{{6M_y }}{{h^2 }} = \frac{{12\pi ^2 \upsilon Df}}{{a^2\,h^2 }}\,.$$

((10.102))

In the case of an elongated plate ($b \to \infty , \quad \gamma = 0$), formula (10.99) yields

$$\bar f_\infty = \bar f_0 = \frac{{pa^4 }}{{3\pi ^4 Dh}}\,.$$

Using this formula and formula (10.97), one can calculate the factor $\eta ^* = \bar f/f_\infty = \eta _f (\overline f _0 /f_\infty )$, which considers the effect of the finite aspect ratio of the plate on its maximum deflection.