Commit 42e3f98c authored by Isidora Araya's avatar Isidora Araya
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Update 14_doping_and_devices_solutions.md

parent bbb91eba
......@@ -55,14 +55,35 @@ $$ I_s(T) \propto e^{-E_{gap}/k_BT}$$
### Subquestion 1
### Subquestion 2
This a "particle in a box" problem.
$$\frac{-\hbar^2}{2m_e^{\*}} \frac{\partial^2 \Phi_e(z)}{\partial z^2} = (E-U_0)\Phi_e $$
$$\frac{-\hbar^2}{2m_h^{\*}} \frac{\partial^2 \Phi_h(z)}{\partial z^2} = (E-U_0)\Phi_h $$
### Subquestion 3
$$E_e = E_c + \frac{\hbar^2 (k_x^2+k_y^2)}{2m_e^{\*}}$$
$$E_h = E_v - \frac{\hbar^2 (k_x^2+k_y^2)}{2m_h^{\*}}$$
### Subquestion 4
This is a 2D electron/hole gas. Apply 2D density of states.
$$g_e = \frac{4\pi m^_e{\*}}{\hbar^2}$$
$$g_h = \frac{4\pi m^_h{\*}}{\hbar^2}$$
### Subquestion 5
Setting $$ E_e - E_h - E_c + E_v = 1 eV = 2\frac{\hbar^2 (k_x^2+k_y^2)}{2m_e^{\*}}$$
L can be found here for $k_x$ and $k_y$.
### Subquestion 6
For a laser one wants to fix the emission wavelength to a certain value. With
this setup the band gap is "easy" to design (set by L, which is fixed).
### Subquestion 7
If donor impurities are put outside of the well (on both sides, for example)
the donated electrons can reduce their energies by falling into the well,
but the ionized dopants remain behind. This gives an advanttage because an
extremely high mobility for electrons can be obtained within the quantum well
(there are no ionized dopants in the well to scatter off of).
This is called modulation doping.
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