## Electrostatics of GaAs-AlGaAs heterostructures The sketches show how the heterostructures looks like, and a vertical cut labelling the potentials/energies. ![[Drawing 2024-09-03 15.41.11.excalidraw.svg]] ![[Drawing 2024-09-03 15.56.20.excalidraw.svg]] Thanks to the fact that GaAs and AlGaAs have the same lattice constant, we do not have mismatch for such heterostructures. And this is a typical model for **2D electron gas**, or 2DEG. The doping layer follows the distribution ($z = 0$ at 2DEG), $N_d(\mathbf{r}, z) = \sum_i \delta(\mathbf{r} - \mathbf{r}_i) \delta(z + s)= N_d \, \delta(z + s) + C(\mathbf{r}) \delta(z + s)$$C(\mathbf{r})$ term is the fluctuating part, we neglect it here. So we have the electric field being, $E = \frac{-|e| n_s}{\varepsilon \varepsilon_0}$ here we assume $\varepsilon$ for GaAs, AlGaAs being the same. Then, $\phi(z) = -\frac{|e| n_s}{\varepsilon \varepsilon_0} \, z, \quad \text{for} \; -s < z < 0$ select $z = 0$ as zero potential. Now include the doping layer, $E = \frac{|e| (n_s - N_d)}{\varepsilon \varepsilon_0}.$ Electrostatic potential becomes, $\phi(z) = \frac{|e|n_s}{\varepsilon \varepsilon_0} S - \frac{|e|(n_s - N_d)}{\varepsilon \varepsilon_0} (z + S), \quad \text{for } -s - d < z < -s$ and at the metal interface, we'll have $\phi(-s - d - c) = \frac{|e|n_s}{\varepsilon \varepsilon_0} S + \frac{|e|(n_s - N_d)}{\epsilon \epsilon_0} (d + c)$ And if we check the potential difference, $\begin{aligned} \Delta \Phi_s &= -\frac{e^2 n_s}{\varepsilon \varepsilon_0}s = -\frac{e n_s}{C_s}, \\ \Delta \Phi_d &= -\frac{e^2 (n_s - N_d) }{\varepsilon \varepsilon_0}d = -\frac{e^2 (n_s - N_d)}{C_d}, \\ \Delta \Phi_c &= -\frac{e^2 (n_s - N_d) }{\varepsilon \varepsilon_0}c = - \frac{e^2 (n_s - N_d)}{C_c} \end{aligned}$ >[!Notice] >The expression $\left[ \frac{\varepsilon\varepsilon_{0}}{s} \right]$ is a capacitance per unit area, so we can write $\Delta \Phi$ as $\frac{e^{2}n}{C}$. Now check the potential energy difference at both sides, $\mu_G - \mu_{2DEG} = \left( -\sum_{d,s,c} \Phi(n_s) - \Phi_b \right) - (E_0(n_s) + E_F(n_s))= -|e| U_G$ >[!Note] >We are not at equilibrium, $\mu$ difference is caused by external voltage. Calculate the capacitance per unit area, $\frac{1}{C/A} = -\frac{\mathrm{d}(-|e|V_G)}{e^2 \, \mathrm{d}n_s} = \frac{1}{\varepsilon \varepsilon_0}\left( s + d+ \frac{\varepsilon \varepsilon_0}{e^2}\frac{\mathrm{d}E_F}{\mathrm{d}n_s} + \frac{\varepsilon \varepsilon_0}{e^2} \frac{\mathrm{d}E_0(n_s)}{\mathrm{d}n_s} \right)$ Notice there are 3 parts, $\sum \frac{1}{C_s}$ are the geometry capacitance, $C_{geo}$; $\frac{\mathrm{d}E_0(n_s)}{e^{2}\mathrm{d}n_s}$ is Fermi energy related term, one over the DOS, (i.e., $D(E_{F})=\frac{\mathrm{d}n_{s}}{\mathrm{d}E_{F}}$), and this part is called quantum capacitance, $\frac{1}{C_q} = \frac{\mathrm{d} E_F}{e^2\mathrm{d}n_{s}} = \frac{1}{e^2 D_{2D}}$, so $C_{q}=e^{2}D_{2D}$, $D_{2D}=\frac{m^{*}}{\pi \hbar^{2}}$, and the length scale is $\frac{\varepsilon \varepsilon_{0}}{e^{2}D_{2D}}=\frac{a_{B}^{*}}{4}$; the last term, $\frac{\mathrm{d}E_{0}}{e^{2}\mathrm{d}n_{s}}$ is called stand-off capacitance, which will be calculated using variational method, we name it $C_{so}$. The method will be briefly talked in [[Fang-Howard variational approach]]. The combined form of capacitance would be, $\frac{1}{C_{\text{tot}}} = \frac{1}{C_{\text{geo}}} + \frac{1}{C_q} + \frac{1}{C_{so}}$