## Doping and remote doping ### Volume doping For volume doping, $d = N_D^{-\frac{1}{3}}, \, d = N_A^{-\frac{1}{3}}$, $d$ is the mean donor/acceptor separation. This value should be compared with the effective Bohr radius, $a_B^*$; - If $d \gg a_B^*$, then we may treat each dopant as an independent [[Envelope function and effective mass approximation#^fb8dcb|hydrogen-like impurity]], and states of neighboring dopants do not overlap. In this case, DOS should be like a delta peak. - If $d \sim a_B^*$, then overlap of states starts, and DOS peak gets broadened. If $d \ll a_B^*$, then the states of neighboring atoms strongly overlap, and states no longer bond to individual dopants. In this case, they form a spatially varying band edge energy, and shift the band edge into the band gap. ![[Drawing 2024-09-01 18.38.31.excalidraw.svg]] Above figure shows low doping, intermediate doping and high (degenerate) doping, from left to right. >[!Example] >In n doped GaAs with $a_{B}^{*}=100\ \mathrm{\mathring{A}}$, we have $d\approx a_{B}^{*}$ at donor concentration $N_{D}=10^{18}\ cm^{-3}$. This corresponds to a relative donor concentration of $1.6\times 10^{-5}$. ### Sheet (2D) doping For 2D, things are similar. By incorporating a plane, we'll get $\delta$-doping: $d = N_D^{-\frac{1}{2}}$. In GaAs, the characteristic sheet concentration is $N_D \approx {a_B^{*}}^{-2} = 10^{12} \, \text{cm}^{-2}$. One can definitely notice that, since we have a charged $\delta$ layer, an electric field would form due to the [[Poisson's equation and Green's function|Poisson equation]], $\nabla^{2}\phi = -\frac{\rho}{\varepsilon}$ Consider $z$ only, $\frac{\mathrm{d}^2 \phi}{\mathrm{d}z^2} = -\frac{Q}{\epsilon} = \frac{|e| N_D \delta(z)}{\varepsilon \varepsilon_{0}}$ Here this $Q$ is the charge density; $N_{D}$ is the number of electrons per unit area (cause it's a delta sheet). The solution of electrostatic potential would be, $\phi(z) = -\frac{|e| N_D}{\varepsilon \varepsilon_0} |z|,$and the potential $U(z)$ would be, $U(z) = -|e| \cdot \phi(z) = \frac{e^2 N_D}{\varepsilon \varepsilon_0} |z|$This triangular potential binds the donor electrons to the doping plane, as a heterostructure potential well creates bound states, binding electrons in the well. Although, in the real case, because of electron-electron interaction, it is not trivial, and solving the Schrödinger equation would be very difficult ([[Begin of DFT, Hohenberg-Kohn theorems|DFT]], self-consistency, etc.). >[!Note] >Actually, multiple self-consistency calculation methods would be introduced in another semiconductor related lecture "quantum transport". Besides solving the problem, $\delta$ doping also has a big issue, the motion of electrons is confined to the doping plane, fluctuating part of the donor potential is strong, and the electrons are strongly scattered, and electrical resistivity is high. But by combining $\delta$ doping and a heterointerface, we could avoid such drawback, namely **remote doping** ### Remote doping ![[Drawing 2024-09-01 22.57.23.excalidraw.svg]] Choose the [[Band engineering and heterointerface#Heterointerface|type I heterointerface]]. Due to the conduction/valence band offset, $e^-/\text{holes}$ would be favorable to move to the material with a smaller band gap (BG). However, due to the existence of the potential barrier, as well as the electrostatic potential, charges are kept near the interface, similar to a quantum well (QW). Quantum confined states exist along the growth direction. Therefore, we have formed an electric dipole by the plane with charged donors and a 2DEG, with a separation distance $d$ (quite large). So the fluctuating potential changes from $V_c \propto \frac{1}{r}$ to $\frac{1}{d}$. (Cause it's from Columbic potential, $r$ is dopant dependent ($N_{D}^{-1/2}$), and now we have the separation $d$ so it changes from $\frac{e^{2}}{4\pi\varepsilon\varepsilon_{0}r}$ to $\frac{e^{2}}{4\pi\varepsilon\varepsilon_{0}d}$.) Scattering is greatly reduced. >[!Info] >2DEG will be discussed in [[Electrostatics of GaAs-AlGaAs heterostructures]].