What is the derivation of the formula for the probability of transmission through a barrier? This site  has a formula for the transmission probability over a barrier (in transistors):
$$ T \propto \exp [-2(2m^*/\hbar ^2)^{1/2}(q\phi)^{1/2} d ]. $$
Where $T$ is the transmission probability over barrier, $m^*$ is the effective mass of silicon, $\phi$ is the barrier height and $d$ = barrier width. I'm assuming $q$ is charge (of an electron), but it is not specified.
The article does not state its sources, so I'm wondering how this formula is obtained, can anyone explain this equation or give a source?
 A: The formula given is a little bit cavalier, but they're not trying to nail down the exact form, just get some idea of the functional dependence of the behavior. Note also that they aren't providing the exact form of the barrier to begin with, so it's also probably the best they can do. All of this is OK, and it's why they give such a simplistic result.
That said, here's how to get it: First, assume that the electron propagates as a plane wave (as it would in free space): $\Psi(x) \propto e^{i k x / \hbar}$. Assume that within the barrier, the electron also propagates according to this formula. Then assume that the "barrier height" is the positive value $q\phi = $(potential energy inside the barrier) $-$ (energy of the electron) ; this is equal to the charge $q$ times a potential $\phi$ because the barrier is electrostatic. The effective momentum of the electron in the barrier is given by
$$\frac{\hbar^2 k^2}{2m^*} = -q \phi \Rightarrow k = \sqrt{ \frac{ -2m^*q\phi }{\hbar^2}} = i \sqrt{\frac{2m^*q\phi}{\hbar^2}}$$
Putting this into the expression for the wavefunction of a plane wave, the exponent becomes real and negative, so the wavefunction decays exponentially.
$$\Psi(x) \propto e^{-\sqrt{\frac{2m^* |q\phi|}{\hbar^2}}x}$$
Letting $x$ be the barrier width is really taking this ratio, where I am letting the near edge of the barrier be at position 0:
$$\frac{\Psi(d)}{\Psi(0)} \propto \frac{e^{-\sqrt{\frac{2m^* |q\phi|}{\hbar^2}}d}}{e^{-\sqrt{\frac{2m^* |q\phi|}{\hbar^2}}0}} = e^{-\sqrt{\frac{2m^* |q\phi|}{\hbar^2}}d}$$
Now comes the question of how to normalize this. The idea of "the probability of being in a given place when in a given state" doesn't really make sense for a current, so we normalize according to the probability current $J$ of the electrons (which, multiplied by their charge, just gives the electric current). Let $J_I$ be the incident current and $J_T$ be the transmitted current. Then
$$ \frac{J_T}{J_I} = \left ( \frac{\Psi(d)}{\Psi(0)} \right )^2 \propto  e^{-2\sqrt{\frac{2m^* |q\phi|}{\hbar^2}}d}$$
(The square is because the amplitude of the wavefunction is proportional to the square root of the probability rather than the probability itself.) Of course, the fraction $J_T/J_I$ is the number of transmitted electrons over the number of incident electrons, i.e., the transmission probability $T$. Therefore 
$$ T \propto e^{-2\sqrt{\frac{2m^* |q\phi|}{\hbar^2}}d}$$
Q.E.D.
