To understand semiconductors one first has to understand the physics behind them. 0.0.1. Postulates of Quantum mechanicsUncertainty principle we can not now the position and the momentum at the same time:

    \[\Delta x \leq \hbar \frac{1}{\Delta p}\]

When we now the particle energy we can not now the particle postion in time.

    \[\Delta E \Delta t \leq \hbar\]

Particles evolve in time like waves. And the momentum of a particle is directly realted to its wavelength.

(1)   \begin{equation*}  \lambda=\frac{h}{p} \end{equation*}

Therefore electrons must be described with the wave function. The probability where the electron is at the moment can be described with \left.|\psi|\right|^2 0.0.2. How do we do quantum mechanicsWith the Schrödinger equations which is actually a wave equation.

(2)   \begin{equation*}  \frac{\partial^2 y}{\partial x^2}=-k^2 y \end{equation*}


(3)   \begin{equation*}  \frac{\partial^2 y}{\partial x^2}=k^2 y \end{equation*}

and the solution of this equations turns out to be y=\sin(k\cdot x)\cdot A_1 or y=\cos(k\cdot x)\cdot A_1 or when we write in in complex form y=e^{-k\cdot x}\cdot A_1 0.0.3. The Electron Wavenumber kIn equation 2 and 3 we used k, because the differential equation is easier to solve with the constant k instead of the term mentioned in equation 4.

(4)   \begin{equation*}  k=\sqrt{\frac{2 m E}{\hbar^2}} \end{equation*}

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In quantum mechanics, k does directly relate to the momentum of the particle p=\hbar  \cdot k. From standard physics we know equation 5

(5)   \begin{equation*}  E=\frac{p^2}{2m} \end{equation*}

When inserting equation 1 in equation 5 We get equation 6 and when we rearrange everything equation 7.

(6)   \begin{equation*}  E=\frac{\left(\frac{h}{\lambda}\right)^2}{2 m} \end{equation*}

(7)   \begin{equation*}  \lambda^2=\frac{h^2}{2 m E} \end{equation*}

Furthermore we know from the figure above that k=\frac{2 \cdot \pi}{\lambda} When we insert this in equation 7 we get after some rearinging equation 10 which is the same as equation 4. 0.0.4. Infinite Potential WellNow we want to find a solution to the differential equation. Let’s assume we have a cube of silicon. The electrons are for sure inside this cube/lattice. So the probability that they are outside of the latice is zero. This behaviour can also be described with the graphic below: C

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} For the condition of a solution of the differential equation one can say that the probability of finding an electron must be zero at both ends, therefore the only solution would be a sine (see the possible solutions for the differential equation).

    \[\Psi_0(x)=A_1\sin(k\cdot x)\]

Furthermore we now that A_1\sin(k\cdot L)=0 therefore k\cdot L = n \cdot \pi \Rightarrow k = \frac{n \cdot \pi}{L}. So therefore k can only take certain values, which means the energy is restricted to certain values. The last thing that remains now is to solve for A_1. We further know equation 8, which says integral of the probability of the particle must be one.

(8)   \begin{equation*}  \int_{-\infty}^{\infty}|\Psi|^2 d x=1 \end{equation*}

Therefore we end up with the following:

(9)   \begin{equation*} \begin{gathered} \int_0^L A_1^2 \sin ^2(k x) d x=1 \\ A_1=\sqrt{\frac{2}{L}} \end{gathered} \end{equation*}

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(10)   \begin{equation*}   k^2=\frac{2 m E}{\hbar^2} \end{equation*}

link to sphere

(11)   \begin{equation*} \begin{array}{r} n=\int \underbrace{g(E)}_{\text {density of states}} \cdot \underbrace{P(E)}_{\text { Probability state is occupied }} d E} \end{array} \end{equation*}

Now want to calculate g(E). To do that we have to calculate the total number of states divided by the total volume of the semiconductor.

(12)   \begin{equation*} g(E)=\frac{N}{L^3} \end{equation*}

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Assuming that the apples are equally spaced as in the plot above one can say that eacht apple is in a distance to the other with a distance of a. The density apple per volume is in this case \frac{1}{a^3} which is one apple per a^3. The total number of apples is know the volume divided by the apple density


The volume of a spherical shell is 4\cdot \pi \cdot r^2 \Delta \r. From before we know that the states of an electron are defined like the following: k=\frac{\pi}{L}\cdot n therefore the states are in a distance to each other of \frac{\pi}{L}. Futher more we can say that we have a state volume, therefore r^2 in this volume is just k^2 which is k_x^2+k_y^2+k_z^2 and therefore

    \[N=\frac{4 \cdot \pi \cdot k^2 \cdot \Delta k}{(\frac{\pi}{L})^3}\]

When we only take the positive k_x, k_y, k_z direction we also get a factor of \frac{1}{8}. The density of states in a certain volume is then given by the following formula:

    \[\frac{N}{L^3}=\frac{1}{8}\cdot\frac{4  \cdot k^2 \cdot \Delta k}{(\pi^2}\]

Now we want to substitute dk with dE. We already know:

    \[\begin{aligned} k &=\sqrt{\frac{2 m E}{\hbar^2}} \\ \frac{dk}{d E} &=\frac{1}{2} \sqrt{\frac{2 m}{\hbar^2 E}} \\ d k &=\frac{1}{2} \sqrt{\frac{2 m}{\hbar^2} \frac{1}{E}} d E \end{aligned}\]

With this information we can then replace k in the original equation which results in the following:

    \[\begin{aligned} &\frac{4 \cdot \frac{2 m E}{\hbar^2} \frac{1}{2} \sqrt{\frac{2 m}{\hbar^2} \frac{1}{E}} d E}{8 \pi^2} \\ &g(E) d E=\frac{2 \pi(2 m)^{3 / 2}}{n^3} \sqrt{E} d E \end{aligned}\]

And because an electron can have an up and down spin we get the following:

    \[&g(E) d E=2 \cdot \frac{2 \pi(2 m)^{3 / 2}}{n^3} \sqrt{E} d E\]

    \[\frac{-\hbar^2}{2 m} \frac{\partial^2 \Psi}{\partial x^2}=i \hbar \frac{\partial \Psi}{\partial t}\]

or 1. intrinsic & extrinsic semiconductors Where p-type and n-type materials meet is called a p-n junction. Electrons from the n-type material move to the p-type material and close the holes and vice versa. This is called \textbf{diffusion} and leads to a {depletion layer} In an n-channel mosfet the inversion is called inversion because the p-type material looks like n-type.

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