# Materials Science and Engineering/Derivations/Semiconductor Devices
## pn Junctino Electrostatics
### Effect of an Electric Field - Conductivity and Ohm's Law

$a={\frac {e}{m}}E$
$v_{average}=a\tau \;$
$v_{D}=\left({\frac {e}{m}}\tau \right)E$
$J=N_{e}ev_{D}\;$
$J={\frac {N_{e}e^{2}\tau }{m}}E$
$J=\sigma E\;$
$\sigma =\left({\frac {e}{m}}\tau \right)(N_{e}e)$

### The Built-in Potential (Vbi)

The electric field is the derivative of the potential with position
$E=-{\frac {dV}{dx}}$
Integrate across the depletion region
$-\int _{-x_{p}}^{x_{n}}Edx=\int _{V(-x_{p})}^{V(x_{n})}dV=V(x_{n})-V(-x_{p})=V_{bi}$
The sum of the drift and the diffusion at equilibrium is equal to zero
$J_{N}=q\mu _{N}nE+qD_{N}{\frac {dn}{dx}}=0$
Use the Einstein relationship:
$E=-{\frac {D_{n}}{\mu _{N}}}{\frac {dn/dx}{n}}=-{\frac {kT}{q}}{\frac {dn/dx}{n}}$
$V_{bi}=-\int _{-x_{p}}^{x_{n}}Edx={\frac {kT}{q}}\int _{n(-x_{p})}^{n(x_{n})}{\frac {dn}{n}}={\frac {kT}{q}}\ln \left[{\frac {n(x_{n})}{n(-x_{p})}}\right]$
$N_{D}$ and $N_{A}$ are the n- and p-side doping concentrations.
$n(x_{n})=N_{D}\;$
$n(-x_{p})={\frac {n_{i}^{2}}{N_{A}}}$

### Depletion Width with Step Junction and VA not equal to zero
## pn Junction Diode: I-V Characteristics
### Assumptions

* Diode operated in steady state
* Doping profile modeled by nondegenerately doped step junction
* Diode is one-dimensional
* In quasineutral region the low-level injection prevails
* The only processes are drift, diffusion, and thermal recombination-generation

### General Relationships

$I=AJ\;$
$J=J_{N}(x)+J_{p}(x)\;$
$J_{N}=q\mu _{n}nE+qD_{N}{\frac {dn}{dx}}$
$J_{p}=q\mu _{p}pE-qD_{p}{\frac {dp}{dx}}$

### Quasineutral Region Consideration

The quasineutral p-region and n-region are adjacent to the depletion region
$0=D_{N}{\frac {d^{2}\Delta n_{p}}{dx^{2}}}-{\frac {\Delta n_{p}}{\tau _{n}}}$
$0=D_{P}{\frac {d^{2}\Delta p_{n}}{dx^{2}}}-{\frac {\Delta p_{n}}{\tau _{p}}}$
The electric field is zero and the derivative of the electron and hole concentration is zero in the quasineutral regions.
$J_{N}=qD_{N}{\frac {d\Delta n_{p}}{dx}}$
$J_{P}=-qD_{P}{\frac {d\Delta p_{n}}{dx}}$

### Depletion Region

Continuity equations:
$0={\frac {1}{q}}{\frac {dJ_{N}}{dx}}+{\frac {\partial n}{\partial t}}|_{thermalR-G}$
$0=-{\frac {1}{q}}{\frac {dJ_{P}}{dx}}+{\frac {\partial p}{\partial t}}|_{thermalR-G}$
Assume that the thermal recombination-generation is zero throughout depletion region. Sum the $J_{N}$ and $J_{P}$ solutions.
$J=J_{N}(-x_{p})+J_{P}(x_{n})$
Strategy to find the minority carrier current density in the quasineutral regions:
* Evaluate current densities at the depletion region edges
* Add edge current densities
* Multiply by A to find the current

### Boundary Conditions
#### Ohmic Contacts

$\Delta n_{p}(x\rightarrow -\infty )=0$
$\Delta p_{n}(x\rightarrow +\infty )=0$

#### Depletion Region Edge

Establish boundary conditions at the edges of the depletion region.
Multiply defining equations of the electron quasi-Fermi level, $F_{N}$, and the hole quasi-Fermi level, $F_{P}$.
$np=n_{i}^{2}e^{(F_{N}-F_{P})/kT}$
Monotonic variation in levels
$F_{N}-F_{P}\leq E_{Fn}-E_{Fp}=qV_{A}$
"Law of Junction"
$np=n_{i}^{2}e^{qV_{A}/jT}$
Evaluate the "law of junction" at the depletion region edges to find the boundary conditions.
$n(-x_{p})p(-x_{p})=n(-x_{p})N_{A}=n_{i}^{2}e^{qV_{A}/kT}$
$n(-x_{p})={\frac {n_{i}^{2}}{N_{A}}}e^{qV_{A}/kT}$
$n(x_{n})p(x_{n})=p(x_{n})N_{D}=n_{i}^{2}e^{qV_{A}/kT}$
$p(x_{n})={\frac {n_{i}^{2}}{N_{D}}}e^{qV_{A}/kT}$

### Derivation

* Solve minority carrier diffusion equations with regard to the boundary conditions to determine value of $\Delta n_{p}$ and $\Delta p_{n}$ in quasineutral region.
* Determine the minority carrier current densities in quasineutral region
* Evaluate quasineutral region solutions of $J_{N}(x)$ and $J_{P}(x)$. Multiply result by area
Solve the equation below with regard to two boundary conditions.
$0=D_{P}{\frac {d^{2}\Delta p_{n}}{dx'^{2}}}-{\frac {\Delta p_{n}}{\tau _{p}}}$
$\Delta p_{n}(x'\rightarrow \infty )=0$
$\Delta p_{n}(x'=0)={\frac {n_{i}^{2}}{N_{D}}}(e^{qV_{A}/kT}-1)$
General solution:
$\Delta p_{n}(x')=A_{1}e^{-x'/L_{P}}+A_{2}e^{x'/L_{P}}$
$L_{P}={\sqrt {D_{P}\tau _{P}}}$
Evaluate at the depletion region edges
$J_{N}(x=-x_{p})=J_{N}(x''=0)=q{\frac {D_{N}}{L_{N}}}{\frac {n_{i}^{2}}{N_{A}}}(e^{qV_{A}/kT}-1)$
$J_{P}(x=x_{n})=J_{P}(x'=0)=q{\frac {D_{P}}{L_{P}}}{\frac {n_{i}^{2}}{N_{D}}}(e^{qV_{A}/kT}-1)$
Multiply the current density by the area:
$I=JA=qA\left({\frac {D_{N}}{L_{N}}}{\frac {n_{i}^{2}}{N_{A}}}+{\frac {D_{P}}{L_{P}}}{\frac {n_{i}^{2}}{N_{D}}}\right)(e^{qV_{A}/kT}-1)$
Ideal diode equation:

## The Effective Mass
### Derivation 1

The velocity of an electron in a one-dimensional lattice is in terms of the group velocity
$v_{g}={\frac {1}{\hbar }}{\frac {\partial E}{\partial k}}$
$dE=eEv_{g}dt\;$
$dE=eE{\frac {1}{\hbar }}{\frac {\partial E}{\partial k}}dt$
Differentiate the equation of velocity
${\frac {dv_{g}}{dt}}={\frac {1}{\hbar }}{\frac {d}{dt}}{\frac {\partial E}{\partial k}}$
${\frac {dv_{g}}{dt}}={\frac {1}{\hbar }}{\frac {\partial ^{2}E}{\partial k^{2}}}{\frac {dk}{dt}}$
${\frac {dv_{g}}{dt}}={\frac {1}{\hbar ^{2}}}{\frac {\partial ^{2}E}{\partial k^{2}}}eE$
$m{\frac {dv}{dt}}=eE$

### Derivation 2

In the free electron model, the electronic wave function can be in the form of $e^{ik\cdot z}$. For a wave packet the group velocity is given by:
$v={{d\omega } \over {dk}}$ = ${{1} \over {\hbar }}\cdot {{d\varepsilon } \over {dk}}$
In presence of an electric field E, the energy change is:
$d\varepsilon ={{d\varepsilon } \over {dk}}{dk}=-eE{dx}=-eEv{dt}={-eE \over {\hbar }}{{d\varepsilon } \over {dk}}{dt}$
Now we can say: $\hbar \cdot {{dk} \over {dt}}={{dp} \over {dt}}=m\cdot {{dv} \over {dt}}$
where p is the electron's momentum. Just put previous results in this last equation and we get:
${{\hbar } \over {m}}\cdot {{dk} \over {dt}}={{1} \over {\hbar }}\cdot {{d} \over {dt}}{{d\varepsilon } \over {dk}}={{1} \over {\hbar }}\cdot {{d^{2}\varepsilon } \over {dk^{2}}}{{dk} \over {dt}}$ From this follows the definition of effective mass:
${{1} \over {m}}={{1} \over {\hbar ^{2}}}\cdot {{d^{2}\varepsilon } \over {dk^{2}}}$

## The Zimman Model

Wavefunction:
$\Psi _{k}=e^{ikx}\;$
Strong disturbance when individual reflections add in phase
$n\lambda =2a\sin \theta \;$
$k={\frac {n\pi }{a}}$
$\Psi _{-k}=e^{-ikx}\;$
$\Psi {\pm }={\frac {1}{\sqrt {2}}}(e^{ikx}\pm e^{-ikx})={\sqrt {2}}{\begin{bmatrix}\cos kx\\i\sin kx\end{bmatrix}}$
The potential energy is from the actual potential $V(x)$ weighted by the probability function $|\Psi _{\pm }|^{2}$
$V_{\pm }={\frac {1}{L}}\int |\Psi _{\pm }|^{2}V(x)dx$
$V_{\pm }={\frac {1}{L}}\int {\begin{bmatrix}2\cos ^{2}kx\\2\sin ^{2}kx\end{bmatrix}}V(x)dx$
Average over one period
$V_{\pm }={\frac {1}{a}}\int _{0}^{a}{\begin{bmatrix}2\cos ^{2}kx\\2\sin ^{2}kx\end{bmatrix}}V(x)dx$
$V_{\pm }={\frac {1}{a}}\int _{0}^{a}{\begin{bmatrix}1+\cos 2kx\\1+\sin 2kx\end{bmatrix}}V(x)dx$
$\pm {\frac {1}{a}}\int _{0}^{a}\cos 2kxV(x)dx$
$V_{\pm }=\pm V_{n}$
The kinetic energy is the same in the case of both wave functions
$E={\frac {\hbar ^{2}k^{2}}{2m}}$
The total energy is the kinetic energy plus the potential energy
$E_{\pm }={\frac {\hbar ^{2}k^{2}}{2m}}\pm V_{n}$
The energy of an electron cannot be between the lower and higher value.