Commit 73408dde authored by Piotr's avatar Piotr
Browse files

Fixed some typo's.

parent 48ed4ffa
......@@ -111,7 +111,7 @@ $$ g(E) = (2m_e)^{3/2}\sqrt{E-E_G}/2\pi^2\hbar^3$$
Holes
$$ E = {p^2}/{2m_h}$$
$$ g(E) = (2m_h)^{3/2}\sqrt{E_h}/2\pi^2\hbar^3$$
$$ g(E) = (2m_h)^{3/2}\sqrt{-E}/2\pi^2\hbar^3$$
**The key algorithm of describing the state of a semiconductor:**
......@@ -180,7 +180,7 @@ $n_i$ is the **intrinsic carrier concentration**, and for a pristine semiconduct
Extra electron (or extra hole) is attracted to the extra charge of the nucleus.
In H the energy levels are:
$$ E_n = - \frac{me^4}{8\pi^2\hbar^3\varepsilon_0n^2} = -R_E /n^2= -\frac{13.6\text{eV}}{n^2}$$
$$ E_n = - \frac{me^4}{8\pi^2\hbar^3\varepsilon^2_0n^2} = -R_E /n^2= -\frac{13.6\text{eV}}{n^2}$$
Bohr radius (size of the ground state wave function): $4 \pi \varepsilon_0 \hbar^2/m_{\mathrm{e}} e^2$
......@@ -201,7 +201,7 @@ Likewise an acceptor adds an extra state at $E_A$ (close to the top of the valen
![](figures/doping.svg)
All donor/acceptor states at the same energy:
$$g_A = N_A \delta(E_A),\quad g_D = N_D \delta(E_G - E_D)$$
$$g_A(E) = N_A \delta(E-E_A),\quad g_D(E) = N_D \delta(E- (E_G - E_D))$$
How large can $N_D/N_A$ be? The distance between donors should be such that the states don't overlap, so the distance must be much larger than 4 nm. Therefore **maximal** concentration of donors before donor band merges with conduction band is $N_D \lesssim (1Å/4\textrm{nm})^3 \sim 10^{-5}\ll N_C$.
......
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