Advanced Semiconductor Physics

An n-channel MOSFET has $\mu_n = 450$ cm$^2$/(V$\cdot$S), $C_{ox} = 34.5$ nF/cm$^2$ ($t_{ox} = 10$ nm SiO$_2$), $W = 10$ $\mu$M, $L = 1$ $\mu$M, $V_T = 0.7$ V.

For $V_G = 2$ V, $V_D = 0.5$ V (linear region since $V_D < V_G - V_T = 1.3$ V):

$I_D = 450 \times 34.5 \times 10^{-9} \times \frac{10 \times 10^{-4}}{1 \times 10^{-4}}\left[1.3 \times 0.5 - \frac{0.25}{2}\right]$

$= 450 \times 3.45 \times 10^{-7} \times 10 \times [0.65 - 0.125]$

$= 1.55 \times 10^{-3} \times 0.525 = 8.14 \times 10^{-4}\ \text{A} = 0.814\ \text{mA}$

At saturation ($V_D = 2$ V):

$I_D = \frac{1}{2} \times 450 \times 34.5 \times 10^{-9} \times 10 \times (1.3)^2 = 1.30\ \text{mA}$

A Type II superconductor has $\xi_0 = 5$ nm and $\lambda_0 = 50$ nm at $T = 0$. At $T = 0.9\,T_c$:

(a) Calculate $\xi(T)$, $\lambda(T)$And $\kappa(T)$.

(b) Calculate $B_{c1}$ and $B_{c2}$ at $T = 0.9\,T_c$.

(c) How many flux quanta per unit area are present at $B = B_{c2}/2$?

Solution:

(a) At $T = 0.9\,T_c$: $1 - T/T_c = 0.1$.

$\xi = \frac{\xi_0}{\sqrt{0.1}} = \frac{5}{0.316} = 15.8\ \text{nm}, \quad \lambda = \frac{50}{\sqrt{0.1}} = 158\ \text{nm}$

$\kappa = \lambda/\xi = 10 \quad \text{(independent of T\text{)}}$

(b) $B_{c2} = \frac{\Phi_0}{2\pi\xi^2} = \frac{2.07 \times 10^{-15}}{2\pi \times (15.8 \times 10^{-9})^2} = \frac{2.07 \times 10^{-15}}{1.57 \times 10^{-15}} = 1.32\ \text{T}$

$B_{c1} = \frac{\Phi_0}{4\pi\lambda^2}\ln\kappa = \frac{2.07 \times 10^{-15}}{4\pi \times (158 \times 10^{-9})^2}\ln 10 = \frac{2.07 \times 10^{-15}}{3.14 \times 10^{-13}} \times 2.303 = 1.52 \times 10^{-3}\ \text{T} = 1.52\ \text{mT}$

(c) At $B = B_{c2}/2 = 0.66$ T: number of flux quanta per m$^2$ = $B/\Phi_0 = 0.66/(2.07 \times 10^{-15}) = 3.19 \times 10^{14}\ \text{m}^{-2}$.

Average spacing between vortices: $a \approx (2\Phi_0/(\sqrt{3}B))^{1/2} = (2 \times 2.07 \times 10^{-15}/(1.73 \times 0.66))^{1/2} = 60\ \text{nm}$.

Consider a spinless particle on a 1D lattice with Hamiltonian:

$\hat{H} = \sum_n \left(t\,e^{i\phi}\,\hat{c}_n^\dagger\hat{c}_{n+1} + t\,e^{-i\phi}\,\hat{c}_{n+1}^\dagger\hat{c}_n\right)$

(a) Show that the dispersion is $\varepsilon(k) = -2t\cos(k + \phi)$.

(b) Calculate the Berry phase for an electron traversing the Brillouin zone $-\pi/a \to \pi/a$.

(c) Show that the Berry phase is $\gamma = 2\pi\phi/(2\pi/a)$ and interpret physically.

Solution:

(a) Substituting $\psi_k(n) = e^{ikna}/\sqrt{N}$:

$\varepsilon(k)\psi_k(n) = t\,e^{i\phi}\,e^{ika}\psi_k(n) + t\,e^{-i\phi}\,e^{-ika}\psi_k(n)$

$\varepsilon(k) = t\,e^{i(k+\phi)a} + t\,e^{-i(k+\phi)a} = -2t\cos(k + \phi)$

(b) The Bloch function is $u_k(n) = e^{i\phi n}$ (up to normalisation). The Berry connection:

$A(k) = \langle u_k | i\partial_k | u_k \rangle = i \cdot i\phi = -\phi$

Wait, more carefully. In a continuum formulation:

$A(k) = \langle u_k | \frac{\partial}{\partial k} | u_k \rangle = \frac{\partial}{\partial k}(\arg u_k) = \frac{\partial}{\partial k}(0) = 0$

Since $u_k(x) = e^{ikx}$ has $\partial_k \ln u_k = ix$ and $\langle u_k|ix|u_k\rangle$ averages to zero.

Actually, for this model the Berry phase arises from the $\phi$-dependent phase winding. Let us use the proper formulation. The wavefunction $\psi_k(x) = e^{ikx}u_k(x)$ where $u_k$ has the periodicity of the lattice. With the flux $\phi$The Berry connection picks up an extra term. The Berry phase for one circuit of the BZ is:

$\gamma = \oint A(k)\,dk = 2\pi\phi$

(c) The Berry phase $\gamma = 2\pi\phi$ is directly proportional to the flux $\phi$ per unit cell. This is the Aharonov—Bohm effect in a lattice: the flux threading each plaquette shifts the band minimum and modifies the group velocity. For $\phi = \pi$The band is inverted ($\varepsilon = 2t\cos k$), which is the basis for the Rice—Mele model of topological insulators.

A silicon p-n junction has $N_A = 10^{24}$ m$^{-3}$ and $N_D = 10^{22}$ m$^{-3}$ at $T = 300$ K.

(a) Calculate the built-in potential $V_0$.

(b) Calculate the depletion width $W$ and the depletion capacitance per unit area at zero bias.

(c) Under forward bias $V = 0.5$ V, calculate the current density. Assume $I_0/A = 10^{-12}$ A/m$^2$.

(d) What is the breakdown voltage if the critical field for Zener breakdown in Si is $E_{\text{crit} \approx 3 \times 10^8}$ V/m?

Solution:

(a) $V_0 = \frac{k_BT}{e}\ln\frac{N_A N_D}{n_i^2} = 0.02585 \times \ln\frac{10^{24} \times 10^{22}}{(1.5 \times 10^{16})^2}$

$= 0.02585 \times \ln\frac{10^{46}}{2.25 \times 10^{32}} = 0.02585 \times \ln(4.44 \times 10^{13})$

$= 0.02585 \times 31.43 = 0.812\ \text{V}$

(b) $W = \sqrt{\frac{2\varepsilon_s V_0}{e}\left(\frac{1}{N_A} + \frac{1}{N_D}\right)}$

Since $N_D \ll N_A$The depletion region extends mainly into the n-side:

$W \approx \sqrt{\frac{2 \times 11.7 \times 8.85 \times 10^{-12} \times 0.812}{1.6 \times 10^{-19} \times 10^{22}}} = \sqrt{\frac{1.68 \times 10^{-10}}{1.6 \times 10^{-3}}} = \sqrt{1.05 \times 10^{-7}} = 3.24 \times 10^{-4}\ \text{m} = 0.324\ \text{mm}$

This is a wide depletion region because of the asymmetric doping.

$C/A = \frac{\varepsilon_s}{W} = \frac{11.7 \times 8.85 \times 10^{-12}}{3.24 \times 10^{-4}} = 3.19 \times 10^{-7}\ \text{F}/m^2 = 319\ \text{nF}/m^2$

(c) $J = J_0(e^{eV/(k_BT)} - 1) = 10^{-12}(e^{0.5/0.02585} - 1) = 10^{-12}(e^{19.34} - 1) = 10^{-12} \times 2.48 \times 10^8 = 2.48 \times 10^{-4}\ \text{A}/m^2 = 0.248\ \text{mA}/m^2$

(d) The maximum field occurs at the metallurgical junction and for a one-sided junction is approximately $E_{\max} = 2V_0/W$. For breakdown: $V_{BD} \approx E_{\text{crit} \cdot W_{BD}/2}$.

With $W_{BD} = \sqrt{2\varepsilon_s V_{BD}/(eN_D)}$ (one-sided):

V_{BD} = \frac{\varepsilon_s E_{\text{crit}^2}{2eN_D} = \frac{11.7 \times 8.85 \times 10^{-12} \times (3 \times 10^8)^2}{2 \times 1.6 \times 10^{-19} \times 10^{22}} = \frac{9.29 \times 10^{-1}}{3.2 \times 10^{-3}} = 290\ \text{V}}