Semiconductors

Abstract

In earlier sections we have seen that a perfect crystal will be (i) an insulator at $T=0~\mathrm K$ if there is a gap separating the filled and empty energy bands. (ii) a conductor at $T=0~ \mathrm K$ if the conduction band is only partially occupied.

Appendices

Problems

7.1

Intrinsic carrier concentration can be written

$$ n_i(T)=2.5\left( \frac{m_\mathrm{c}}{m}\frac{m_\mathrm{v}}{m}\right) ^{3/4} \left( \frac{k_\mathrm{B}T}{E_\mathrm{G}}\right) ^{3/2} \left( \frac{E_\mathrm{G} \text{ in } \text{ eV } }{\frac{1}{40} \, \text{ eV } }\right) ^{3/2} \mathrm e^{-\frac{E_\mathrm{G}}{2k_\mathrm{B}T}}\times 10^{19} / \mathrm cm^3. $$

Take $E_\mathrm{G}=1.5 ~\mathrm{eV}$, $m_\mathrm{h}=0.7 \,\mathrm{m}$, and $m_\mathrm{e}=0.06 \,\mathrm{m}$ roughly those of GaAs, and plot $\ln n_i$ vs T in the range $T=3 ~\mathrm K$ $\sim$ $300 ~\mathrm K$.

7.2

Plot the chemical potential $\zeta _i(T)$ vs T in the range $T=3 \,\mathrm K$ $\sim$ $300\, \mathrm K$ for values of $E_\mathrm{G}=1.5\, \mathrm{eV}$, $m_\mathrm{h}=0.7\, \mathrm{m}$, and $m_\mathrm{e}=0.06\, \mathrm{m}$.

7.3

For InSb, we have $E_\mathrm{G}\simeq 0.18 \,\mathrm{eV}$, $\epsilon _\mathrm{s} \simeq 17$, and $m_\mathrm{c^*}\simeq 0.014\, \mathrm{m}$.

• (a) Evaluate the binding energy of a donor.

• (b) Evaluate the orbit radius of a conduction electron in the ground state.

• (c) Evaluate the donor concentration at which overlap effects between neighboring impurities become significant.

• (d) If $N_\mathrm{d}=10^{14}\, \text{ cm } ^3$ in a sample of InSb, calculate $n_c$ at $T= 4$ K. (One can begin with the general charge neutrality condition in the low temperature region.)

• (e) Estimate the magnitude of the electric field needed to ionize the donor at zero temperature.

7.4

Let us consider a case that the work function of two metals differ by 2 eV; $E_{\mathrm{F}1} -E_{\mathrm{F}2} =2$ eV.

If the metals are brought into contact, electrons will flow from metal 1 to metal 2. Assume the transferred electrons are displaced by $3 \times 10^{-8}$ cm. How many electrons per $\mathrm cm^2$ are transferred?

7.5

Consider a semiconductor quantum well consisting of a very thin layer of narrow gap semiconductor of $E_\mathrm{G} = \varepsilon _\mathrm{c}-\varepsilon _\mathrm{v}$ contained in a wide band gap host material of $E_\mathrm{G} = \varepsilon _\mathrm{c}^\mathrm{H}-\varepsilon _\mathrm{v}^\mathrm{H}$ as shown in the figure below. The conduction and valence band edges are shown in the figure below. The dashed lines indicate the positions of energy levels associated with the quantized motion of electrons ($\varepsilon _0^\mathrm{c}$) and holes ($\varepsilon _0^\mathrm{v}$) in this quantum well. We can write the electron and hole energies, respectively, as $\varepsilon _\mathrm{c} (k) = \tilde{\varepsilon }_\mathrm{c} +\frac{\hbar ^2}{2m_\mathrm{c}}\left( k_x^2+k_y^2\right)$ and $\varepsilon _\mathrm{v} (k) = \tilde{\varepsilon }_\mathrm{v} -\frac{\hbar ^2}{2m_\mathrm{v}}\left( k_x^2+k_y^2\right)$, where $\tilde{\varepsilon }_\mathrm{c} =\varepsilon _\mathrm{c}+\varepsilon _0^\mathrm{c}$ and $\tilde{\varepsilon }_\mathrm{v} =\varepsilon _\mathrm{v}-\varepsilon _0^\mathrm{v}$.

• (a) Calculate the two-dimensional density of states for the electrons and holes assuming that other quantized levels can be ignored. Note that

  $$ L^2g_\mathrm{c}(\varepsilon ) d\varepsilon = \sum _{\begin{array}{c}\tiny k_x, k_y,\sigma \\ \varepsilon<\varepsilon _k <\varepsilon +d\varepsilon \end{array}} 1. $$

• (b) Determine $N_\mathrm{c}(T)$ and $P_\mathrm{v}(T)$ for this two-dimensional system. Remember that $N_\mathrm{c}(T) =\int _0^\infty d\varepsilon ~g_\mathrm{c}(\varepsilon )\mathrm e^{-{(\varepsilon -\tilde{\varepsilon }_\mathrm{c})}/{k_\mathrm{B}T}}.$

• (c) Determine $n_\mathrm{c}(T)$ and $p_\mathrm{v}(T)$ for the intrinsic case.

• (d) Determine the value of the chemical potential for this case.

7.6

Consider the metal–oxide–semiconductor structure with oxide layer width of a as shown below. We have assumed the semiconductor is p-type with $N_\mathrm{A}$ acceptors per unit volume and, therefore $E_\mathrm{F}$ located at the acceptor level.

We then apply a gate voltage $V_\mathrm{g}$, which lowers the Fermi level in the metal relative to that in the bulk of the semiconductor.

• (a) Sketch the resulting energy bands versus z if $V_\mathrm{g}$ is less than $E_\mathrm{G} (=\varepsilon _c-\varepsilon _v)$.

• (b) Where are the charges that give rise to the voltage drop across the oxide and the semiconductor depletion layer? Sketch the profile of the charge distribution across the oxide layer.

• (c) For $0<z<d$, the voltage drop across the depletion layer of width d is determined by

  $$ \frac{\partial ^2}{\partial z^2}V_\mathrm{d}(z)=-\frac{4\pi e^2 }{\epsilon _s}N_\mathrm{A}, $$

  where $\epsilon _s$ is the background dielectric constant of the semiconductor. Solve this equation for $V_\mathrm{d}(z)$ in the ‘standard’ depletion layer approximation.

• (d) Impose boundary conditions that (i) $V_\mathrm{d}(z)=V_\mathrm{d}(\infty )$ for $z \ge d$ and (ii) $\epsilon _o E_\mathrm{oxide} = \epsilon _s E_\mathrm{semiconductor}$ at $z=0$, where $\epsilon _o$ is the background dielectric constant of the oxide, and determine the voltage drop across the oxide and that across the depletion layer.

• (e) When the voltage drop across the depletion layer exceeds $E_\mathrm{G}$, electrons can transfer from the valence band into the potential well formed by the conduction band edge and the oxide band gap. Determine the value $V_\mathrm{threshold}$ of the gate voltage at which this occurs.

• (f) For $V_\mathrm{gate} > V_\mathrm{threshold}$, the depletion width remains essentially constant, and the conduction electrons in the ‘inversion layer’ produce a Hartree potential $V_\mathrm{H}(z)$ which satisfies

  $$ \frac{\partial ^2}{\partial z^2}V_\mathrm{H}(z)=-\frac{4\pi e^2 n_\mathrm{s}}{\epsilon _s}|\varPsi _0(z)|^2, $$

  where $n_\mathrm{s}$ is the number of conduction electrons per unit area and $\varPsi _0(z)$ is the solution of the differential equation

  $$ \left( -\frac{\hbar ^2}{2m}\frac{\partial ^2}{\partial z^2}+V_\mathrm{d}(z)+V_\mathrm{H}(z) -E_0\right) \varPsi _0(z)=0. $$

  Because $V_\mathrm{H}(z)$ depends on $n_\mathrm{s}$ and $E_\mathrm{F}-E_0= \frac{\hbar ^2 k_\mathrm{F}^2}{2m}=\frac{\pi \hbar ^2}{m} n_\mathrm{s}$, this must be done self-consistently. Determine $V_\mathrm{H}(z)$ and $E_0$ to obtain the average electronic energy in the system $\tilde{E}$. Hint: One can assume a variational function $\varPsi (z,\alpha )=N z\mathrm e^{-\alpha z}$ to evaluate $E_0(\alpha )$ and then minimize the average electronic energy in the system given by $\tilde{E} (\alpha )= E_0-\frac{1}{2}\langle V_\mathrm{H}\rangle +\frac{1}{2}\left( E_\mathrm{F}-E_0\right)$.

Summary

In this chapter we studied the physics of semiconducting material and artificial structures made of semiconductors. General properties of typical semiconductors are reviewed and temperature dependence of carrier concentration is considered for both intrinsic and doped cases. Then basic physics of p–n junctions is covered in equilibrium and the current-voltage characteristic of the junction is described. The characteristics of two-dimensional electrons are discussed for the electrons in surface space charge layers formed in metal-oxide-semiconductor structures, semiconductor superlattices, and quantum wells. The fundamentals of the quantum Hall effects and the effects of disorders and modulation doping are also discussed.

The densities of states in the conduction and valence bands are given by

$$ g_\mathrm{c}(\varepsilon ) = \frac{\sqrt{2}m_\mathrm{c}^{3/2}}{\pi ^2 \hbar ^3}\left( \varepsilon -\varepsilon _\mathrm{c}\right) ^{1/2}; g_\mathrm{v}(\varepsilon ) = \frac{\sqrt{2}m_\mathrm{v}^{3/2}}{\pi ^2 \hbar ^3} \left( \varepsilon _\mathrm{v}-\varepsilon \right) ^{1/2}. $$

In the case of nondegenerate regime, we have $\varepsilon _\mathrm{c} -\zeta \gg \varTheta$ and $\zeta -\varepsilon _\mathrm{v} \gg \varTheta$, where $\varTheta$ is $k_\mathrm{B}T$. Then the carrier concentrations become

$$ n_\mathrm{c}(T) = N_\mathrm{c}(T) \mathrm e^{-\frac{\varepsilon _\mathrm{c}-\zeta }{\varTheta }}; p_\mathrm{v}(T) = P_\mathrm{v}(T) \mathrm e^{-\frac{\zeta -\varepsilon _\mathrm{v}}{\varTheta }}, $$

where

$$ N_\mathrm{c}(T) =\int _{\varepsilon _\mathrm{c}}^\infty d\varepsilon g_\mathrm{c}(\varepsilon ) \mathrm e^{-\frac{\varepsilon -\varepsilon _\mathrm{c}}{\varTheta }} ; P_\mathrm{v}(T) =\int _{\infty }^{\varepsilon _\mathrm{v}} d\varepsilon g_\mathrm{v}(\varepsilon ). $$

The product $n_\mathrm{c}(T)p_\mathrm{v}(T)$ is independent of $\zeta$ such that

$$ n_\mathrm{c}(T)p_\mathrm{v}(T) = N_\mathrm{c}(T) P_\mathrm{v}(T) \mathrm e^{-E_\mathrm{G}/\varTheta }. $$

In the absence of impurities, $n_\mathrm{c}(T)=p_\mathrm{v}(T)$ and we have

$$ n_\mathrm{i}(T) = \left[ N_\mathrm{c}(T) P_\mathrm{v}(T)\right] ^{1/2} \mathrm e^{-E_\mathrm{G}/2\varTheta }. $$

The chemical potential now becomes

$$ \zeta _\mathrm{i}=\varepsilon _\mathrm{c}- \frac{1}{2}E_\mathrm{G} +\frac{3}{4}\varTheta \ln {\left( \frac{m_\mathrm{v}}{m_\mathrm{c}}\right) } ; \zeta _\mathrm{i}=\varepsilon _\mathrm{v}+ \frac{1}{2}E_\mathrm{G} +\frac{3}{4}\varTheta \ln {\left( \frac{m_\mathrm{v}}{m_\mathrm{c}}\right) }. $$

When donors are present, the chemical potential $\zeta$ will move from its intrinsic value $\zeta _\mathrm{i}$ to a value near the conduction band edge. If the concentration of donors is sufficiently small, the average occupancy of a single donor impurity state is given by

$$ \langle n_\mathrm{d} \rangle = \frac{1}{\frac{1}{2}\mathrm e^{\beta (\varepsilon _\mathrm{d}-\zeta )}+1}. $$

The numerical factor of $\frac{1}{2}$ in $\langle n_\mathrm{d} \rangle$ comes from the fact that either spin up or spin down states can be occupied but not both.

At a finite temperature, we have

$$ n_\mathrm{c}(T) = N_\mathrm{c}(T)\mathrm e^{-\beta (\varepsilon _\mathrm{c}-\zeta )}, p_\mathrm{v}(T) = P_\mathrm{v}(T)\mathrm e^{-\beta (\zeta -\varepsilon _\mathrm{v})}, $$

$$ n_\mathrm{d}(T) = \frac{N_\mathrm{d}}{\frac{1}{2}\mathrm e^{\beta (\varepsilon _\mathrm{d}-\zeta )}+1}, p_\mathrm{a}(T) = \frac{N_\mathrm{a}}{\frac{1}{2}\mathrm e^{\beta (\zeta -\varepsilon _\mathrm{a})}+1}. $$

In addition, we have charge neutrality condition given by

$$ n_\mathrm{c}+n_\mathrm{d}=N_\mathrm{d}-N_\mathrm{a}+p_\mathrm{v}+p_\mathrm{a}. $$

The set of these five equations should be solved numerically in order to have self consistent result for five unknowns.

The region of the p–n junction is a high resistance region and the electrical current density becomes

$$ j = e\left( J_\mathrm{h}^\mathrm{gen} +J_\mathrm{e}^\mathrm{gen}\right) \left( \mathrm e^{eV/\varTheta }-1\right) , $$

where $J_\mathrm{h}^\mathrm{gen}$ and $J_\mathrm{e}^\mathrm{gen}$ are hole and electron generation current densities, respectively.

Near the interface of metal-oxide-semiconductor structure under a strong enough gate voltage, the motion of the electrons is characterized by

$$ \varepsilon = \varepsilon _0 + \frac{\hbar ^2}{2m_\mathrm{c}^*} \left( k_x^2 + k_y^2\right) ; \varPsi _{n,k_x, k_y} =\frac{1}{L} \mathrm e^{i(k_xx+k_yy)}\xi _n(z). $$

Here $\xi _n(z)$ is the $n\mathrm{th}$ eigenfunction of a differential equation given by

$$ \left[ \frac{1}{2m_\mathrm{c}^*}\left( -i\hbar \frac{\partial }{\partial z}\right) ^2 + V_\mathrm{eff}(z) - \varepsilon _n\right] \xi _n(z) = 0. $$

If a quantum well is narrow, it leads to quantized motion and subbands:

$$ \varepsilon _n^{(\mathrm c)}(\mathbf k) = \varepsilon _n^{(\mathrm c)} + \frac{\hbar ^2}{2m_\mathrm{c}^*}\left( k_x^2 + k_y^2\right) . $$

In the presence of a dc magnetic field $\mathbf B$ applied normal to the plane of the 2DEG, the Hamiltonian of a single electron is written by

$$ H=\frac{1}{2m} \left( \mathbf p+\frac{e}{c}\mathbf A\right) ^2. $$

Here $\mathbf p=(p_x, p_y)$ and $\mathbf A(\mathbf r)$ is the vector potential whose curl gives $\mathbf B = (0,0,B)$. The electronic states are described by

$$ E_{nk}=\hbar \omega _\mathrm{c}\left( n+\frac{1}{2}\right) , \varPsi _{nk}(x,y, z)=\mathrm e^{iky}\ u_n\left( x+\frac{\hbar k}{m\omega _\mathrm{c}}\right) ; n=0, 1, 2, \ldots . $$

The density of states (per unit length) is given by $g(\varepsilon ) \propto \sum _n \delta \left( \varepsilon -\hbar \omega _\mathrm{c}(n+\frac{1}{2})\right)$. The total number of states per Landau level is equal to the magnetic flux through the sample divided by the flux quantum $\frac{hc}{e}$:

$$ N_\mathrm{L} = \frac{BL^2}{hc/e}. $$