A general metrology of stress on crystalline silicon with random crystal plane by using micro-Raman spectroscopy

Abstract

The requirement of stress analysis and measurement is increasing with the great development of heterogeneous structures and strain engineering in the field of semiconductors. Micro-Raman spectroscopy is an effective method for the measurement of intrinsic stress in semiconductor structures. However, most existing applications of Raman-stress measurement use the classical model established on the (001) crystal plane. A non-negligible error may be introduced when the Raman data are detected on surfaces/cross-sections of different crystal planes. Owing to crystal symmetry, the mechanical, physical and optical parameters of different crystal planes show obvious anisotropy, leading to the Raman-mechanical relationship dissimilarity on the different crystal planes. In this work, a general model of stress measurement on crystalline silicon with an arbitrary crystal plane was presented based on the elastic mechanics, the lattice dynamics and the Raman selection rule. The wavenumber-stress factor that is determined by the proposed method is suitable for the measured crystal plane. Detailed examples for some specific crystal planes were provided and the theoretical results were verified by experiments.

Appendix

For the Raman measurement on the (001) crystal plane, its SCS can be chosen as $\varvec {X}_{1}^{\prime}$ = [1 1 0] = [1 0 0]′, $\varvec {X}_{2}^{\prime}$ = [− 1 1 0] = [0 1 0]′, $\varvec {X}_{2}^{\prime}$ = [0 0 1] = [0 0 1]′. According to Eq. (1), the rotation matrix A is given by

$$ {\mathbf{A}} = \left( {\begin{array}{*{20}c} {\frac{1}{\sqrt 2 }} & {\frac{1}{\sqrt 2 }} & 0 \\ { - \frac{1}{\sqrt 2 }} & {\frac{1}{\sqrt 2 }} & 0 \\ 0 & 0 & 1 \\ \end{array} } \right). $$

By substituting matrix A into Eq. (6), the matrix H_ε and H_σ are achieved as follows,

$$ {\mathbf{H}}_{\varepsilon } = \left( {\begin{array}{*{20}c} {\frac{1}{2}} & {\frac{1}{2}} & 0 & {\frac{1}{2}} & 0 & 0 \\ {\frac{1}{2}} & {\frac{1}{2}} & 0 & { - \;\frac{1}{2}} & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ { - \;1} & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & {\frac{1}{\sqrt 2 }} & { - \frac{1}{\sqrt 2 }} \\ 0 & 0 & 0 & 0 & {\frac{1}{\sqrt 2 }} & {\frac{1}{\sqrt 2 }} \\ \end{array} } \right),\quad {\mathbf{H}}_{\sigma } = \left( {\begin{array}{*{20}c} {\frac{1}{2}} & {\frac{1}{2}} & 0 & 1 & 0 & 0 \\ {\frac{1}{2}} & {\frac{1}{2}} & 0 & { - \;1} & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ { - \;\frac{1}{2}} & {\frac{1}{2}} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & {\frac{1}{\sqrt 2 }} & { - \frac{1}{\sqrt 2 }} \\ 0 & 0 & 0 & 0 & {\frac{1}{\sqrt 2 }} & {\frac{1}{\sqrt 2 }} \\ \end{array} } \right) \, . $$

Using Eqs. (9) and (17), for the SCS of the C-Si, the compliance coefficient and the phonon deformation potential tensors can be written as

$$ {\mathbf{S}}^{\prime} = {\mathbf{H}}_{\varepsilon } {\mathbf{SH}}_{\varepsilon }^{\text{T}} = \left( {\begin{array}{*{20}c} \begin{aligned} \frac{1}{2}(s_{11} + s_{12} ) \hfill \\ + \;\frac{1}{4}s_{44} \hfill \\ \end{aligned} & \begin{aligned} \frac{1}{2}(s_{11} + s_{12} ) \hfill \\ - \frac{1}{4}s_{44} \hfill \\ \end{aligned} & {s_{12} } & 0 & 0 & 0 \\ \begin{aligned} \frac{1}{2}(s_{11} + s_{12} ) \hfill \\ - \;\frac{1}{4}s_{44} \hfill \\ \end{aligned} & \begin{aligned} \frac{1}{2}(s_{11} + s_{12} ) \hfill \\ + \;\frac{1}{4}s_{44} \hfill \\ \end{aligned} & {s_{12} } & 0 & 0 & 0 \\ {s_{12} } & {s_{12} } & {s_{11} } & 0 & 0 & 0 \\ 0 & 0 & 0 & {2(s_{11} - s_{12} )} & 0 & 0 \\ 0 & 0 & 0 & 0 & {s_{44} } & 0 \\ 0 & 0 & 0 & 0 & 0 & {s_{44} } \\ \end{array} } \right), $$

$$ {\mathbf{K}}^{\prime} = {\mathbf{H}}_{\sigma } {\mathbf{KH}}_{\sigma }^{\text{T}} = \left( {\begin{array}{*{20}c} \begin{aligned} \frac{1}{2}(k_{11} + k_{12} ) \hfill \\ + \;k_{44} \hfill \\ \end{aligned} & \begin{aligned} \frac{1}{2}(k_{11} + k_{12} ) \hfill \\ - \;k_{44} \hfill \\ \end{aligned} & {k_{12} } & 0 & 0 & 0 \\ \begin{aligned} \frac{1}{2}(k_{11} + k_{12} ) \hfill \\ - \;k_{44} \hfill \\ \end{aligned} & \begin{aligned} \frac{1}{2}(k_{11} + k_{12} ) \hfill \\ + \;k_{44} \hfill \\ \end{aligned} & {k_{12} } & 0 & 0 & 0 \\ {k_{12} } & {k_{12} } & {k_{11} } & 0 & 0 & 0 \\ 0 & 0 & 0 & {\frac{1}{2}(k_{11} - k_{12} )} & 0 & 0 \\ 0 & 0 & 0 & 0 & {k_{44} } & 0 \\ 0 & 0 & 0 & 0 & 0 & {k_{44} } \\ \end{array} } \right), $$

where k₁₁ = p, k₁₂ = q, k₄₄ = r. Suppose that p′ = (p + q)/2+ r and q′ = (p + q)/2− r, then ${k}_{11}^{\prime}$ = ${k}_{22}^{\prime}$  =  p′, ${k}_{33}^{\prime}$  =  p, ${k}_{12}^{\prime}$ =  q′, ${k}_{13}^{\prime}$ = ${k}_{23}^{\prime}$ =  q, ${k}_{44}^{\prime}$ = (p − q)/2, and ${k}_{55}^{\prime}$ = ${k}_{66}^{\prime}$ =  r.

We assume that the material is subjected to uniaxial stress along $\varvec {X}_{1}^{\prime}$, which is $\sigma_{1}^{\prime}$  =  σ′ ≠ 0. According to the generalized Hooke’s law, we obtain the strain in the SCS as follows,

$$ {{\mathbf{\varepsilon}}^{\prime}} = \left( {\begin{array}{*{20}c}{ {\varepsilon}^{\prime}_{1} } \\ {\varepsilon^{\prime}_{2} } \\ {\varepsilon^{\prime}_{3} } \\ {\varepsilon^{\prime}_{4} } \\ {\varepsilon^{\prime}_{5} } \\ {\varepsilon^{\prime}_{6} } \\ \end{array} } \right) = {\mathbf{S}}^{\prime}{\sigma}^{\prime} = \left( {\begin{array}{*{20}c} {\frac{1}{4}(2s_{11} + 2s_{12} + s_{44} )\sigma^{\prime}} \\ {\frac{1}{4}(2s_{11} + 2s_{12} - s_{44} )\sigma^{\prime}} \\ {s_{12} \sigma^{\prime}} \\ 0 \\ 0 \\ 0 \\ \end{array} } \right). $$

Substituting ε′ and K′ into Eq. (15), the secular matrix in the SCS then becomes

$$ \left| {\begin{array}{*{20}c} {p^{\prime}\varepsilon^{\prime}_{1} + q^{\prime}\varepsilon^{\prime}_{2} + q\varepsilon^{\prime}_{3} - \lambda^{\prime}} & 0 & 0 \\ 0 & {q^{\prime}\varepsilon^{\prime}_{1} + p^{\prime}\varepsilon^{\prime}_{2} + q\varepsilon^{\prime}_{3} - \lambda^{\prime}} & 0 \\ 0 & 0 & {q(\varepsilon^{\prime}_{1} + \varepsilon^{\prime}_{2} ) + p\varepsilon^{\prime}_{3} - \lambda^{\prime}} \\ \end{array} } \right| = 0, $$

and the eigenvalues and the corresponding eigenvectors can be easily obtained as follows

$$ \left\{ \begin{aligned} \lambda^{\prime}_{1} = p^{\prime}\varepsilon^{\prime}_{1} + q^{\prime}\varepsilon^{\prime}_{2} + q\varepsilon^{\prime}_{3} ,\quad {\mathbf{n}^{\prime}_{1}} = \left( {\begin{array}{*{20}c} 1 & 0 & 0 \\ \end{array} } \right)^{\prime } , \hfill \\ \lambda^{\prime}_{2} = q^{\prime}\varepsilon^{\prime}_{1} + p^{\prime}\varepsilon^{\prime}_{2} + q\varepsilon^{\prime}_{3} ,\quad {\mathbf{n}}^{\prime}_{2} = \left( {\begin{array}{*{20}c} 0 & 1 & 0 \\ \end{array} } \right)^{\prime } , \hfill \\ \lambda^{\prime}_{3} = q(\varepsilon^{\prime}_{1} + \varepsilon^{\prime}_{2} ) + p\varepsilon^{\prime}_{3} ,\quad\;\; {\mathbf{n}}^{\prime}_{3} = \left( {\begin{array}{*{20}c} 0 & 0 & 1 \\ \end{array} } \right)^{\prime } . \hfill \\ \end{aligned} \right. $$

By using Eq. (16),

$$ \left\{ \begin{aligned} & \Delta \omega_{1}^{\prime } = \frac{{\lambda_{1}^{\prime } }}{{2\omega_{0} }} = \frac{1}{2}\frac{{[(p + q)(s_{11} + s_{12} ) + rs_{44} ] + 2qs_{12} }}{{2\omega_{0} }}\sigma^{\prime}, \\ & \Delta \omega_{2}^{\prime } = \frac{{\lambda_{2}^{\prime } }}{{2\omega_{0} }} = \frac{1}{2}\frac{{[(p + q)(s_{11} + s_{12} ) - rs_{44} ] + 2qs_{12} }}{{2\omega_{0} }}\sigma^{\prime}, \\ & \Delta \omega_{3}^{\prime } = \frac{{\lambda_{3}^{\prime } }}{{2\omega_{0} }} = \frac{{q(s_{11} + s_{12} ) + ps_{12} }}{{2\omega_{0} }}\sigma^{\prime}, \\ \end{aligned} \right. $$

where ω₀ = 520 cm⁻¹, p = − 1.85ω ²₀ , q = − 2.31ω ²₀ , r = − 0.71ω ²₀ , s₁₁ = 7.68 × 10⁻¹² Pa⁻¹, s₁₂ = − 2.14 × 10⁻¹² Pa⁻¹, and s₄₄ = 12.6 × 10⁻¹² Pa⁻¹. Thus, the wavenumber increments Δω_j′ are given by

$$ \left\{ {\begin{array}{*{20}l} {\Delta \omega^{\prime}_{1} = \kappa_{1} \sigma^{\prime},\quad \kappa_{1} = - \;2.874, \, } \hfill \\ {\Delta \omega^{\prime}_{2} = \kappa_{2} \sigma^{\prime},\quad \kappa_{2} = - \;0.548, \, } \hfill \\ {\Delta \omega^{\prime}_{3} = \kappa_{3} \sigma^{\prime},\quad \kappa_{3} = - \;2.298. \, } \hfill \\ \end{array} } \right. $$

Hence, the Raman tensor R_i′ and its corresponding vector V_i′ in the SCS are given as follows

$$ \begin{array}{*{20}l} {{\mathbf{R}}^{\prime}_{1} = \frac{\sqrt 2 }{2}\left( {\begin{array}{*{20}c} 0 & 0 & d \\ 0 & 0 & { - \;d} \\ d & { - d} & 0 \\ \end{array} } \right),\quad {\mathbf{R}}^{\prime}_{2} = \frac{\sqrt 2 }{2}\left( {\begin{array}{*{20}c} 0 & 0 & d \\ 0 & 0 & d \\ d & d & 0 \\ \end{array} } \right),\quad {\mathbf{R}}^{\prime}_{3} = \left( {\begin{array}{*{20}c} d & 0 & 0 \\ 0 & { - \;d} & 0 \\ 0 & 0 & 0 \\ \end{array} } \right),} \\ {{\mathbf{V}}^{\prime}_{1} = \left( {\begin{array}{*{20}c} {\frac{1}{\sqrt 2 }} & { - \frac{1}{\sqrt 2 }} & 0 \\ \end{array} } \right)^{\prime } ,\quad {\mathbf{V}}^{\prime}_{2} = \left( {\begin{array}{*{20}c} {\frac{1}{\sqrt 2 }} & {\frac{1}{\sqrt 2 }} & 0 \\ \end{array} } \right)^{\prime } ,\quad {\mathbf{V}}^{\prime}_{3} = \left( {\begin{array}{*{20}c} 0 & 0 & 1 \\ \end{array} } \right)^{\prime } .} \\ \end{array} $$

Since V_i′ ≠ n′k, the Raman tensor R″k corresponding vector n′k is achieved by using Eq. (22) as follows

$$ {\mathbf{R}}^{\prime\prime}_{1} = \left( {\begin{array}{*{20}c} 0 & 0 & d \\ 0 & 0 & 0 \\ d & 0 & 0 \\ \end{array} } \right),\quad {\mathbf{R}}^{\prime\prime}_{2} = \left( {\begin{array}{*{20}c} 0 & 0 & 0 \\ 0 & 0 & d \\ 0 & d & 0 \\ \end{array} } \right),\quad {\mathbf{R}}^{\prime\prime}_{3} = \left( {\begin{array}{*{20}c} d & 0 & 0 \\ 0 & { - \;d} & 0 \\ 0 & 0 & 0 \\ \end{array} } \right). $$

Using Eq. (23), the Raman intensity of Δω_k′ in the SCS are

$$ \left\{ {\begin{array}{*{20}l} {I^{\prime}_{1} = Q\left| {{{\mathbf{e}}^{\prime}}_{i}^{\text{T}} {\mathbf{R}}^{\prime\prime}_{1} {\kern 1pt} {\mathbf{e}}^{\prime}_{s} } \right|^{2} = Qd^{2} ({e}^{\prime}_{i1} {e}^{\prime}_{s3} + e^{\prime}_{i3} e^{\prime}_{s1} )^{2} ,} \hfill \\ {I^{\prime}_{2} = Q\left| {{{\mathbf{e}}^{\prime}}_{i}^{\text{T}} {\mathbf{R}}^{\prime\prime}_{2} {\kern 1pt} {\mathbf{e}}^{\prime}_{s} } \right|^{2} = Qd^{2} (e^{\prime}_{i2} e^{\prime}_{s3} + e^{\prime}_{i3} e^{\prime}_{s2} )^{2} ,} \hfill \\ {I^{\prime}_{3} = Q\left| {{{\mathbf{e}}^{\prime}}_{i}^{\text{T}} {\mathbf{R}}^{\prime\prime}_{3} {\kern 1pt} {\mathbf{e}^{\prime}}_{s} } \right|^{2} = Qd^{2} (e^{\prime}_{i1} e^{\prime}_{s1} - e^{\prime}_{i2} e^{\prime}_{s2} )^{2} .} \hfill \\ \end{array} } \right. $$

For the backscattering from the plane surface of (001) crystal plane, the incident light and scattering light are all along the $\varvec {X}_{3}^{\prime}$ and all the polarization components in $\varvec {X}_{3}^{\prime}$ are zero. This means that e_i3′ = e_s3′ = 0. Hence, only I₃′ is visible (viz. nonzero) no matter the polarization configures of the incident light and scattering light. The Raman-mechanical relationship becomes into Eq. (29).