Commit f755a7f4 authored by Isidora Araya's avatar Isidora Araya
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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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