 Update 14_doping_and_devices_solutions.md

parent 134d1c57
 ... ... @@ -13,12 +13,12 @@ $$n_e n_h = \frac{1}{2} \left(\frac{k_BT}{\pi\hbar^2}\right)^3 (m_e^{\ast}m_h^{\ast})^{3/2}e^{-\beta E_{gap}}$$ Charge balance condition: $$D = N_D - N_A - n_e + n_h$$ $$D = N_D - N_A - n_e + n_h = (doping)$$ ### Subquestion 2 $$n_e = \frac{1}{2}(\sqrt{D^2+4n_i^2}+D)$$ $$n_h = \frac{1}{2}(\sqrt{D^2+4n_i^2}-D)$$ $$n_e + n_A = \frac{1}{2}(\sqrt{D^2+4n_i^2}+D)$$ $$n_h + n_D = \frac{1}{2}(\sqrt{D^2+4n_i^2}-D)$$ where $n_i=n_{e,intrinsic}=n_{h,intrinsic}$. ### Subquestion 3 ... ... @@ -35,14 +35,22 @@ Check both cases with lecture notes approximated solutions by doing a Taylor exp ### Subquestion 1 If all the dopants are ionized, the Fermi level gets shifted up towards the conduction band. This result can be obtained when using results in Exercise 1 - Subquestion 2 and the following: ### Subquestion 2 $$n_D = \approx N_D$$ $$n_A = \approx N_A$$ $$n_e = n_h = n_i$$ ### Subquestion 2 Now, $$n_D^{\ast} = N_D (1-\frac{1}{e^({E_D-E_F})/k_BT+1})$$ $$n_A_{\ast} = N_A (1-\frac{1}{e^({E_F-E_A})/k_BT+1})$$ $\ast$ indicated non-ionized concentrations. ### Subquestion 3 Use Germianium Fermi Energy at room temperature and perform the [$\tebtit{key algorithm of describing the state of a semiconductor}$](13_semiconductors/#part-1-pristine-semiconductor) ## Exercise 3: Performance of a diode ... ...
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