Energy Dispersion of the Conduction Band Minimum of Strained Si: Method 2

The above formulae have been derived, assuming that the conduction band minima are located at ${\ensuremath{\mathitbf{k}}}_{\mathrm{min}} = 2\pi / a_0 (0,0,\pm0.85)$, which is a good approximation only for small shear strain. As can be seen from Figure 3.12 the minimum of the conduction band is expected to move towards the $X$ point as the strain-induced splitting between the conduction band becomes larger. As a direct consequence, the conduction bands shape is deformed and the previous assumption ${\ensuremath{\mathitbf{k}}}_{\mathrm{min}} = 2\pi / a_0 (0,0,0.85)$ is not satisfied.

Thus, a more general model for the effect of shear strain on the effective masses needs to be developed, which takes the movement of the conduction band minimum as a function of strain into account. The effective masses are subsequently evaluated at the position of the conduction band minimum ${\ensuremath{\mathitbf{k}}}_{\mathrm{min}}({\ensuremath{\varepsilon_{xy}}})$.

The position of the minimum can be found from (3.73) by setting $k_x = k_y = 0$:

$$E_- = \frac{\hbar^2}{2 \ensuremath{m_\mathrm{l}}} k_z^2 - \sqrt{\... ...th{m_\mathrm{l}}^2}k_z^2 + (2 \Xi_{u'})^2 {\ensuremath{\varepsilon_{xy}}}^2}\ .$$ (3.90)

Here, relations (3.75) and (3.77) are used to replace $A_4$ and $A_1$, and $k_0$ denotes the position of the unstrained conduction band minimum measured from the zone boundary $X$, $k_0=0.15\cdot2\pi/a_0$. Setting

$$\frac{\partial E_-}{\partial k_z} = 0\ ,$$ (3.91)

the position of the conduction band minimum $k_{z,\mathrm{min}}$ can be obtained

$$k_{z,\mathrm{min}} = \left\{ \begin{array}{ll} k_0 \sqrt{1 - \kap... ...d $\vert{\ensuremath{\varepsilon_{xy}}}\vert>1/\kappa$} \end{array} \right. \ .$$ (3.92)

It can be seen that for strain smaller in magnitude than $1/\kappa$, the minimum position is shifted towards the $X$ point. For $\vert{\ensuremath{\varepsilon_{xy}}}\vert=1/\kappa$, the position of the minimum is located at the $X$ point, thus $k_{z,\mathrm{min}}=0$. The position is fixed, even when ${\ensuremath {\varepsilon _{xy}}}$ is further increased. The changing position of $k_{z,\mathrm{min}}$ is visualized in Figure 3.12, where the impact of shear strain on the shape of the conduction bands $\Delta _1$ and $\Delta _{2'}$ is plotted.

The strain dependent longitudinal mass $\ensuremath{m_\mathrm{l}}({\ensuremath{\varepsilon_{xy}}})$ can be obtained from (3.73) by calculating

$$\frac{1}{\ensuremath{m_\mathrm{l}}({\ensuremath{\varepsilon_{xy}}... ... k_z^2} {\Big\vert}_{{\ensuremath{\mathitbf{k}}} = (0,0,k_{z,\mathrm{min}})}\ .$$ (3.93)

The following expressions can be derived after some algebraic manipulations

$$\ensuremath{m_\mathrm{l}}({\ensuremath{\varepsilon_{xy}}}) = \left\{ \begin{array}{ll} \ensuremath{m_\mathrm{l}}\Bigl (1-\ka... ...uad $\vert{\ensuremath{\varepsilon_{xy}}}\vert>1/\kappa$} \end{array} \right ..$$ (3.94)

Similar to (3.92), the dependence of the longitudinal mass on strain is described by two expressions, depending on whether the magnitude of strain is smaller or bigger than $1/\kappa$. Note that in the approximation of Section 3.7.2 the effect of shear strain was modeled by the expansion (3.76) about the conduction band minimum of the unstrained lattice, such that the longitudinal mass $\ensuremath{m_\mathrm{l}}$ is not influenced by strain.

To derive the transverse effective masses that include the dependence on $k_{z,\mathrm{min}}$, first (3.73) is transformed to the rotated coordinate system introduced in (3.85).

$$E_{\pm}(\hat{{\ensuremath{\varepsilon_{}}}},{\ensuremath{\mathitb... ...} {\ensuremath{\varepsilon_{xy}}} + \frac{A_3}{2} (k_x^2 - k_y^2)\right )^2}\ .$$ (3.95)

The effective mass in the $x'=[110]$ and $y'=[1\bar{1}0]$ direction can be obtained from

$$\frac{1}{m_{\mathrm{t},x'}({\ensuremath{\varepsilon_{xy}}})} = \h... ... k_{x'}^2} {\Big\vert}_{{\ensuremath{\mathitbf{k}}} = (0,0,k_{z,\mathrm{min}})}$$ (3.96)

and

$$\frac{1}{m_{\mathrm{t},y'}({\ensuremath{\varepsilon_{xy}}})} = \h... ...{y'}^2} {\Big\vert}_{{\ensuremath{\mathitbf{k}}} = (0,0,k_{z,\mathrm{min}})}\ .$$ (3.97)

The strain dependence of the transverse masses is given by

$$m_{\mathrm{t},x'}({\ensuremath{\varepsilon_{xy}}}) = \left\{ \beg... ...quad $\vert{\ensuremath{\varepsilon_{xy}}}\vert>1/\kappa$} \end{array} \right .$$ (3.98)

for the [110] direction and by

$$m_{\mathrm{t},y'}({\ensuremath{\varepsilon_{xy}}}) = \left\{ \beg... ...quad $\vert{\ensuremath{\varepsilon_{xy}}}\vert>1/\kappa$} \end{array} \right..$$ (3.99)

for the $[1\bar{1}0]$ direction. Here, $\mathrm{sgn}$ denotes the signum function. For $\vert{\ensuremath{\varepsilon_{xy}}}\vert<1/\kappa$ the effective masses derived in this way are consistent with equations (3.87) and (3.88). However, for $\vert{\ensuremath{\varepsilon_{xy}}}\vert>1/\kappa$ the transverse masses are constant and depend on the sign of the strain only.

Finally, an analytical expression for the valley shift induced by shear strain $\varepsilon_{xy}$, which was given in Section 3.6.2, is calculated. According to (3.92), equation (3.90) has to be evaluated at $k_{z,\mathrm{min}} = k_0 \sqrt{1 - \kappa^2 {\ensuremath{\varepsilon_{xy}}}^2}$ for $\vert{\ensuremath{\varepsilon_{xy}}}\vert<1/\kappa$. For $\vert{\ensuremath{\varepsilon_{xy}}}\vert>1/\kappa$ the energy shift at the $X$ point determines the overall valley shift. The shift between the valley pair along [001] and the valley pairs along [100] or [010] is obtained from

$$\delta E_{\mathrm{shear}} = E({\ensuremath{\varepsilon_{xy}}},{\e... ...ad $\vert{\ensuremath{\varepsilon_{xy}}}\vert>1/\kappa$} \end{array} \right.\ .$$ (3.100)

This derivation shows that the valley splitting cannot be obtained if the spectrum of the strained crystal is expanded around the conduction band minimum $k_0$ of the unstrained crystal, as it was assumed in Section 3.7.2.

Figure 3.13: Constant energy ellipsoids for the Si conduction band with ${\ensuremath {\varepsilon _{xy}}}=0$ (left side) and ${\ensuremath{\varepsilon_{xy}}}\neq 0$ (right side). In unstrained Si there are six equivalent valleys along the three principal axes. In the deformed lattice the valleys along ${\ensuremath{\mathitbf{k}}}_z$ go down in energy, hence the surface of constant energy is bigger. The constant energy lines in the ${\ensuremath{\mathitbf{k}}}_x{\ensuremath{\mathitbf{k}}}_y$-plane are characterized by two masses, $m_{\mathrm{t},x'}$ and $m_{\mathrm{t},y'}$.

E. Ungersboeck: Advanced Modelling Aspects of Modern Strained CMOS Technology