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Jung Nguyen
lectures
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823c5ab4
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823c5ab4
authored
Apr 04, 2020
by
Isidora Araya
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Update 14_doping_and_devices_solutions.md
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@@ -20,6 +20,8 @@ $$ n_{e} = \frac{1}{2}(\sqrt{D^2+4n_i^2}+D)$$
$$ n_{h} =
\f
rac{1}{2}(
\s
qrt{D^2+4n_i^2}-D)$$
where $D = N_D - N_A$ and $n_i=n_{e,intrinsic}=n_{h,intrinsic}$.
Note that for N_D = N_A results for intrinsic semiconductors are recovered.
### Subquestion 3
If $D<<n_i$, then the doping is not important and results of intrinsic are
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@@ -33,22 +35,34 @@ 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.
If all the dopants are ionized ($n_D=0$), 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:
$$ n_e - n_h = N_D - N_A $$
??? "hint?"
Solve now for $E_F$ using $n_e$ and $n_h$ expressions obtained in the
[
previous lecture
](
/13_semiconductors/#part-1-pristine-semiconductor
)
.
### Subquestion 2
Now,
$$ n_D = N_D
\f
rac{1}{e^({E_D-E_F})/k_BT+1} $$
### Subquestion 3
??? hint "how?"
Use Germianium Fermi Energy at room temperature and solve E_F
via using the n_e solution in Exercise 1 and by applying the definition of n_e.
Check
[
key algorithm of describing the state of a semiconductor
](
/13_semiconductors/#part-1-pristine-semiconductor
)
Because not all dopants are ionized, the charge conservation eq. becomes:
$$ n_e - n_h + n_D = N_D - N_A $$
Doing the same as in subquestion 1, an expression for $E_F$ can be found.
??? hint "Values"
For Germanium at $T=300K$, $n_i = 2.4
\t
imes 10^{13} cm^{-3}$
This can be obtained when plugging $m_e^{
\a
st}=0.55 m_e$, $m_h^{
\a
st}=0.37 m_e$ and $T=300K$ in
[
the results here
](
/13_semiconductors/#semiconductor-density-of-states-and-fermi-level
)
.
???
## Exercise 3: Performance of a diode
### Subquestion 1
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