Effect of Band Nonparabolicity on the Inter Band Tunneling in Semiconductors

Abstract

A simple yet generalized theory is developed to study inter band tunneling property of narrow band gap III–V compound semiconductors. The band structures of these low band gap semiconductors with sufficiently separated split-off valance band are usually described by the three energy band model of Kane, so this has been adopted here for the analysis of interband tunneling property in the case of InAs, InSb, and In_1-xGa_xAs_yP_1-y lattice matched to InP as representative direct band gap semiconductors having varied split-off valence band compared to their bulk state band gap energy. It has been found that the magnitude of tunneling rate from heavy hole decreases with increasing band nonparabolicity and the impact is more significant at high electric field in the three-band model of Kane than those with simple parabolic energy band approximations reflecting the direct influence of energy band parameters on inter band tunneling transitions. With proper consideration of band nonparabolicity, the results of the analysis of tunneling rate of these narrow gap materials show significant deviations from the results when simple parabolic band approximation is considered. The exact physical basis of the sources of deviation in the nonparabolic case from the corresponding parabolic band approximations is discussed in association to band coupling effect, transverse energy dependence, and the interplay between them. Moreover, under certain limiting conditions, our results reduce to the well-known results of parabolic band approximation and thus providing an indirect test to the accuracy of our generalized formulations.

Appendices

Derivation of Interband Transition Matrix Element

The spin vector is represented by

$$ \overrightarrow{S}=\frac{\mathrm{\hslash}}{2}\overrightarrow{\sigma} $$

(A.1)

where the Pauli spin matrices are

$$ {\sigma}_x=\left(\begin{array}{cc}0& 1\\ {}1& 0\end{array}\right);{\sigma}_y=\left(\begin{array}{cc}0& -i\\ {}i& 0\end{array}\right);{\sigma}_z=\left(\begin{array}{cc}1& 0\\ {}0& -1\end{array}\right) $$

(A.2)

The interband transition matrix element (IME) can be expressed as

$$ {\displaystyle \begin{array}{l}{Z}_{cv}(K)=i\left\langle {u}_1\left(\overrightarrow{k},\overrightarrow{r}\right)\left|\left(\partial /{k}_z\right){u}_2\left(\overrightarrow{k},\overrightarrow{r}\right)\right.\right\rangle \\ {}=i\left\{\left[-1\left\langle \left.(S)\right|(S)\right\rangle \left\langle \left.\downarrow {}^{\prime}\right|\uparrow {}^{\prime}\right\rangle \right]{a}_{k+}\frac{\partial {a}_{k-}}{\partial {k}_z}\right.\end{array}} $$

(A.3)

$$ \left.+\left[\left\langle \left(\frac{X^{\prime }-{iY}^{\prime }}{\sqrt{2}}\right)\left|\left(-\frac{X^{\prime }-{iY}^{\prime }}{\sqrt{2}}\right)\right.\right\rangle \left\langle \uparrow {}^{\prime}\left|\downarrow {}^{\prime}\right.\right\rangle \right]{b}_{k+}\frac{\partial {b}_{k-}}{\partial {k}_z}+\left[\left\langle \left.{Z}^{\prime}\right|{Z}^{\prime}\right\rangle \left\langle \uparrow {}^{\prime}\left|\downarrow {}^{\prime}\right.\right\rangle \right]{c}_{k+}\frac{\partial {c}_{k-}}{\partial {k}_z}\right\} $$

(A.4)

Expanding Eq. (22), |X^′〉, |Y^′〉, and |Z^′〉 can be expressed in terms of |X〉, |Y〉, and |Z〉 as|X^′〉 = cos θ cos ϕ|X〉 + cos θ sin ϕ|Y〉 − sin θ|Z〉, |Y^′〉 =  − sin ϕ|X〉 + cos ϕ|Y〉and

$$ \left|{Z}^{\prime}\right\rangle =\sin \theta \cos \phi \left|X\right\rangle +\sin \theta \sin \phi \left|Y\right\rangle +\cos \theta \left|Z\right\rangle $$

(A.5)

Thus, using Eqs. (20), (21) and (A.5), it can be inferred that

$$ \left\langle {X}^{\prime }|{X}^{\prime}\right\rangle =\left\langle {Y}^{\prime }|{Y}^{\prime}\right\rangle =\left\langle {Z}^{\prime }|{Z}^{\prime}\right\rangle =1 $$

(A.6)

and 〈X^′| Y^′〉 = 〈Y^′| Z^′〉 = 〈Z^′| X^′〉 = 0,〈S| X^′〉 = 〈X^′| S〉 = 0

$$ \left\langle S|{Y}^{\prime}\right\rangle =\left\langle {Y}^{\prime }|S\right\rangle =0,\kern1em \left\langle S|{Z}^{\prime}\right\rangle =\left\langle {Z}^{\prime }|S\right\rangle =0 $$

(A.7)

Now, using Eqs. (A.3), (A.6), and (A.7), we obtained

$$ {Z}_{cv}\left(\mathrm{k}\right)=i\left\{\left({a}_{k+}\frac{\partial {a}_{k-}}{\partial {k}_z}\right)\right.\left.-\left({c}_{k+}\frac{\partial {c}_{k-}}{\partial {k}_z}\right)\right\}\left\langle {\downarrow}^{\prime }|{\uparrow}^{\prime}\right\rangle $$

(A.8)

which is rewritten as

$$ {Z}_{cv}\left(\mathrm{k}\right)=i\left\{X(k)\right\}\left\langle {\downarrow}^{\prime }|{\uparrow}^{\prime}\right\rangle $$

(A.9)

where

$$ X(k)=\left\{\left({a}_{k+}\frac{\partial {a}_{k-}}{\partial {k}_z}\right)\right.\left.-\left({c}_{k+}\frac{\partial {c}_{k-}}{\partial {k}_z}\right)\right\} $$

(A.10)

This equation suggests that the carriers in the z-orbitals of the VB are only involved in interband tunneling process. By suitably tailoring the energy band constants by external effect (like strain etc.), the interband tunneling rate may be modified. This is a new observation made in this paper. Besides, the heavy-hole band being made up of x and y-type atomic orbitals cannot take part in the interband tunneling transition process as is noted in previous works [16].

Using Eqs. (11)–(16), and after tedious algebraic manipulation we obtain,

$$ {\displaystyle \begin{array}{l}X(k)=\frac{1}{4{\left(\zeta \left(\overrightarrow{k}\right)+{\delta}^{\prime}\right)}^2}\frac{d\zeta \left(\overrightarrow{k}\right)}{{d k}_z}\left\{{\beta}^2{\left(\frac{\left(\zeta \left(\overrightarrow{k}\right)+{E}_g\right)}{\left(\zeta \left(\overrightarrow{k}\right)-{E}_g+4{\delta}^{\prime }/{\beta}^2\right)}\right)}^{1/2}\left({E}_g-{\delta}^{\prime }+4{\delta}^{\prime }/{\beta}^2\right)\right.\\ {}\left.+{t}^2\left({\left(\frac{\left(\zeta \left(\overrightarrow{k}\right)-{E}_g\right)}{\left(\zeta \left(\overrightarrow{k}\right)+{E}_g\right)}\right)}^{1/2}\left({E}_g-{\delta}^{\prime}\right)\right)\right\}\end{array}} $$

(A.11)

From Eqs. (12) and (15)

${\beta}^2=\left(\frac{\chi +\Delta {E}_g}{\chi}\right)$,${E}_g-4{\delta}^{\prime }/{\beta}^2={E}_g\left(\frac{\chi -3\Delta {E}_g}{\chi +\Delta {E}_g}\right)$, ${t}^2=\left(\frac{\chi -\Delta {E}_g-\left(4{\Delta}^2/3\right)}{\chi}\right)$

$$ {E}_g-{\delta}^{\prime }={E}_g\left(1-\frac{\Delta {E}_g}{\chi}\right) $$

(A.12)

Using Eqs. (A.11) and (A.12), we may approximate X(k) as

$$ X(k)=\frac{1}{4{\left(\zeta \left(\overrightarrow{k}\right)+{\delta}^{\prime}\right)}^2}\frac{d\zeta \left(\overrightarrow{k}\right)}{{d k}_z}{E}_g\left(1-{\delta}^{\prime }/{E}_g\right)\left(2-{\delta}^{\prime }/{E}_g-{\left({\delta}^{\prime }/{E}_g\right)}^2-4{\Delta}^2/3\chi \right)\left(\frac{\zeta \left(\overrightarrow{k}\right)}{{\left(\zeta {\left(\overrightarrow{k}\right)}^2-{E_g}^2\right)}^{1/2}}\right) $$

(A.13)

From Eqs. (17) and (A.13), one can write

$$ X(k)=\frac{1}{4}\left(\frac{\mathrm{\hslash}{E_g}^{3/2}}{{\left(\zeta \left(\overrightarrow{k}\right)+{\delta}^{\prime}\right)}^2{\left({m}_r\right)}^{1/2}}\right)\left(1-{\delta}^{\prime }/{E}_g\right)\left(2-{\delta}^{\prime }/{E}_g+2{\left({\delta}^{\prime }/{E}_g\right)}^2-4{\Delta}^2/3\chi \right) $$

(A.13a)

On transforming the spin vectors, one may express

|↑^′〉 = e^−iϕ/2 cos(θ/2)|↑〉 + e^iϕ/2 sin(θ/2)|↓〉 (and)

$$ \left|{\downarrow}^{\prime}\right\rangle =-{e}^{- i\phi /2}\sin \left(\theta /2\right)\left|\uparrow \right\rangle +{e}^{i\phi /2}\cos \left(\theta /2\right)\left|\downarrow \right\rangle $$

Then,

$$ \left\langle {\downarrow}^{\prime }|{\uparrow}^{\prime}\right\rangle ={\left(-{e}^{- i\phi /2}\sin \left(\theta /2\right)\left|\uparrow \right\rangle +{e}^{i\phi /2}\cos \left(\theta /2\right)\left|\downarrow \right\rangle \right)}^{\ast } $$

$$ \times \left({e}^{- i\phi /2}\cos \left(\theta /2\right)\left|\uparrow \right\rangle +{e}^{i\phi /2}\sin \left(\theta /2\right)\left|\downarrow \right\rangle \right) $$

$$ {\displaystyle \begin{array}{l}=\left(-{e}^{i\phi /2}\sin \left(\theta /2\right)\left\langle \uparrow \right|+{e}^{- i\phi /2}\cos \left(\theta /2\right)\left\langle \downarrow \right|\right)\\ {}\times \left({e}^{- i\phi /2}\cos \left(\theta /2\right)\left|\uparrow \right\rangle +{e}^{i\phi /2}\sin \left(\theta /2\right)\left|\downarrow \right\rangle \right)\end{array}} $$

or

$$ \left\langle {\downarrow}^{\prime }|{\uparrow}^{\prime}\right\rangle =-\sin \left(\theta /2\right)\cos \left(\theta /2\right)\left\langle \uparrow |\uparrow \right\rangle +{e}^{- i\phi /2}{\cos}^2\left(\theta /2\right)\left\langle \downarrow |\uparrow \right\rangle $$

$$ -{e}^{- i\phi /2}{\sin}^2\left(\theta /2\right)\left\langle \uparrow |\downarrow \right\rangle +\sin \left(\theta /2\right)\cos \left(\theta /2\right)\left\langle \downarrow |\downarrow \right\rangle $$

The x-component of the above expression is written as,

$$ {\left\langle {\downarrow}^{\prime }|{\uparrow}^{\prime}\right\rangle}_x=-\sin \left(\theta /2\right)\cos \left(\theta /2\right){\left\langle \uparrow |\uparrow \right\rangle}_x+{e}^{- i\phi /2}{\cos}^2\left(\theta /2\right){\left\langle \downarrow |\uparrow \right\rangle}_x $$

$$ -{e}^{- i\phi /2}{\sin}^2\left(\theta /2\right){\left\langle \uparrow |\downarrow \right\rangle}_x+\sin \left(\theta /2\right)\cos \left(\theta /2\right){\left\langle \downarrow |\downarrow \right\rangle}_x $$

Using the Eqs. (A.1) and (A.2) and noting the spinor relations

$$ {\left\langle \uparrow |\uparrow \right\rangle}_x=0,{\left\langle \downarrow |\uparrow \right\rangle}_x=\frac{1}{2}={\left\langle \uparrow |\downarrow \right\rangle}_x,{\left\langle \downarrow |\downarrow \right\rangle}_x=0, $$

one may write

$$ {\left\langle {\downarrow}^{\prime }|{\uparrow}^{\prime}\right\rangle}_x=\frac{1}{2}\left(\cos \phi \cos \theta -i\sin \phi \right) $$

Proceeding in similar manner,

$$ {\left\langle {\downarrow}^{\prime }|{\uparrow}^{\prime}\right\rangle}_y=\frac{1}{2}\left(i\cos \phi +\cos \theta \sin \phi \right) $$

and

$$ {\left\langle {\downarrow}^{\prime }|{\uparrow}^{\prime}\right\rangle}_z=\frac{1}{2}\left(-\sin \theta \right) $$

In this context, we define the three mutual orthogonal vectors

$$ {\widehat{r}}_1=\cos \theta \cos \phi \widehat{i}+\cos \theta \sin \phi \widehat{j}-\sin \theta \widehat{k} $$

(A.14)

$$ {\widehat{r}}_2=\sin \phi \widehat{i}+\cos \phi \widehat{j} $$

(A.15)

and

$$ {\widehat{r}}_3=\sin \theta \cos \phi \widehat{i}+\sin \theta \sin \phi \widehat{j}+\cos \theta \widehat{k} $$

(A.16)

Therefore,

$$ {\displaystyle \begin{array}{l}\left\langle {\downarrow}^{\prime }|{\uparrow}^{\prime}\right\rangle =\widehat{i}{\left\langle {\downarrow}^{\prime }|{\uparrow}^{\prime}\right\rangle}_x+\widehat{j}{\left\langle {\downarrow}^{\prime }|{\uparrow}^{\prime}\right\rangle}_y+\widehat{k}{\left\langle {\downarrow}^{\prime }|{\uparrow}^{\prime}\right\rangle}_z\\ {}=\frac{1}{2}\left[\left(\cos \phi \cos \theta -i\sin \phi \right)\widehat{i}+\left(i\cos \phi +\cos \theta \sin \phi \right)\widehat{j}+\left(-\sin \theta \right)\widehat{k}\right]\\ {}=\frac{1}{2}\left[\left({\widehat{r}}_1+i{\widehat{r}}_2\right)\right]=-\frac{1}{2}\left[\left(i{\widehat{r}}_1-{\widehat{r}}_2\right)\right]\end{array}} $$

(A.17)

It may be noted that the magnitude of vector $\frac{1}{2}\left[\left({\widehat{r}}_1+i{\widehat{r}}_2\right)\right]$ is always unity and is independent of the polar angles θ andϕ.

By similar reasoning, one can write

$$ {\displaystyle \begin{array}{l}\left\langle {\uparrow}^{\prime }|{\uparrow}^{\prime}\right\rangle =\frac{1}{2}\left[\left(\cos \phi \sin \theta \right)\widehat{i}+\left(\sin \phi \sin \theta \right)\widehat{j}+\left(\cos \theta \right)\widehat{k}\right]\\ {}=\frac{1}{2}{\widehat{r}}_3\ \mathrm{and}\ \left\langle {\downarrow}^{\prime }|{\downarrow}^{\prime}\right\rangle =-\frac{1}{2}{\widehat{r}}_3,\left\langle {\uparrow}^{\prime }|{\downarrow}^{\prime}\right\rangle =\frac{1}{2}\left({\widehat{r}}_1-i{\widehat{r}}_2\right)\end{array}} $$

(A.18)

Thus, using Eqs. (A.13a) and (A.17), one can have a simplified form of IME as below

$$ {\displaystyle \begin{array}{l}{Z}_{cv}\left(\mathrm{k}\right)=i\left\{\frac{1}{4}\left(\frac{\mathrm{\hslash}{E_g}^{3/2}}{{\left(\zeta \left(\overrightarrow{k}\right)+{\delta}^{\prime}\right)}^2{\left({m}_r\right)}^{1/2}}\right)\right.\\ {}\times \left.\left(1-{\delta}^{\prime }/{E}_g\right)\left(2-{\delta}^{\prime }/{E}_g+2{\left({\delta}^{\prime }/{E}_g\right)}^2-4{\Delta}^2/3\chi \right)\right\}\frac{1}{2}\left[\left({\widehat{r}}_1+i{\widehat{r}}_2\right)\right]\end{array}} $$

(A.19)

Derivation of ${M}_{n{n}^{\prime }}\left({k}_{\perp}\right)$

The point σ which has the greatest contribution to the integration in Eq. (7) is the point where conduction and valence bands meet in the complex plane. Therefore, at this pointk _z  = σ,

$$ {E}_c\left(\sigma \right)={E}_v\left(\sigma \right) $$

(B.1)

Using Eqs. (17) and (A.18), we obtain

$$ \sigma =\pm i\sqrt{\eta^2+{k_{\perp}}^2} $$

(B.2)

where $\eta =\sqrt{m_r{E}_g/{\mathrm{\hslash}}^2}$.

From Eqs. (7), (17), and (B.2)

$$ {M}_{cv}\left({k}_{\perp}\right)=\frac{F}{G}\underset{-G/2}{\overset{G/2}{\int }}{Z}_{cv}\left(\mathrm{k}\right)\exp \left(\frac{i}{F}\underset{0}{\overset{\sigma }{\int }}\zeta \left(\overrightarrow{k}\right){dk}_{z^{\prime }}\right){dk}_z $$

(B.3)

It is noted that the δ^′ in the denominator of can be neglected relative to magnitude of $\zeta \left(\overrightarrow{k}\right)$ for evaluating the integration. Then, in the vicinity of the point σ at which $\zeta \left(\overrightarrow{k}\right)$ = 0 contributes maximum to the integral. The integration may be evaluated similar to [14] by expanding $\zeta \left(\overrightarrow{k}\right)$ as a function of k _z and ρ. Using Eq. (17) and (B.2), $\zeta \left(\overrightarrow{k}\right)$ can be expressed as

$$ \zeta \left(\overrightarrow{k}\right)=\frac{\mathrm{\hslash}{E_g}^{1/2}}{{m_r}^{1/2}}\sqrt{{k_z}^2-{\sigma}^2}=\frac{\mathrm{\hslash}{E_g}^{1/2}}{{m_r}^{1/2}}\sqrt{k_z+\sigma}\sqrt{k_z-\sigma } $$

which simplifies to

$$ \zeta \left(\overrightarrow{k}\right)=\frac{\sqrt{2}\mathrm{\hslash}{E_g}^{1/2}}{{m_r}^{1/2}}{\sigma}^{1/2}{\rho}^{1/2} $$

(B.4)

for

$$ {k}_z-\sigma =\rho $$

(B.5)

leading to the value of the integral inside the exponent of Eq. (B.3) as

$$ \underset{0}{\overset{\sigma }{\int }}\zeta \left(\overrightarrow{k}\right){dk}_z=\frac{2\sqrt{2}\mathrm{\hslash}{E_g}^{1/2}}{3{m_r}^{1/2}}\left|{\sigma}^2\right| $$

(B.6)

atk _z  = 0. Next, substituting this value and extending the limits of integration to ±∞ in Eq. (B.3), one can obtain

$$ {M}_{cv}\left({k}_{\perp}\right)=\frac{iF{\Upsilon}_{cv}}{G}\times \frac{1}{2}\left[\left({\widehat{r}}_1+i{\widehat{r}}_2\right)\right]\left(\frac{\mathrm{\hslash}{E_g}^{3/2}}{{\left({m}_r\right)}^{1/2}}\right)\underset{0}{\overset{\infty }{\int }}\left(\frac{dk_z}{\left({\eta}^2+{k_{\perp}}^2\right)}\right)\exp \left(\frac{i}{F}\underset{0}{\overset{\sigma }{\int }}\zeta \left(\overrightarrow{k}\right){dk}_{z^{\prime }}\right) $$

which can be further manipulated to evaluate for k _z  = σ as

$$ {M}_{cv}\left({k}_{\perp}\right)=\frac{\pi F{\Upsilon}_{cv}{\left({m}_r{E}_g\right)}^{1/2}}{\left(3/2\right)\mathrm{\hslash}\sigma G}\times \frac{1}{2}\left[\left({\widehat{r}}_1+i{\widehat{r}}_2\right)\right]\exp \left(-\frac{\pi \mathrm{\hslash}{\sigma}^2{E_g}^{1/2}}{4F{\left({m}_r\right)}^{1/2}}\right) $$

(B.7)