Materials Science and Engineering/Derivations/Semiconductor Devices

pn Junctino Electrostatics

Effect of an Electric Field - Conductivity and Ohm's Law

 ${\displaystyle E={\frac {U}{L}}}$


${\displaystyle a={\frac {e}{m}}E}$

${\displaystyle v_{average}=a\tau \;}$

${\displaystyle v_{D}=\left({\frac {e}{m}}\tau \right)E}$

${\displaystyle J=N_{e}ev_{D}\;}$

${\displaystyle J={\frac {N_{e}e^{2}\tau }{m}}E}$

${\displaystyle J=\sigma E\;}$

${\displaystyle \sigma =\left({\frac {e}{m}}\tau \right)(N_{e}e)}$

 ${\displaystyle \sigma =\mu _{e}(N_{e}e)\;}$


The Built-in Potential (Vbi)

The electric field is the derivative of the potential with position

${\displaystyle E=-{\frac {dV}{dx}}}$

Integrate across the depletion region

${\displaystyle -\int _{-x_{p}}^{x_{n}}Edx=\int _{V(-x_{p})}^{V(x_{n})}dV=V(x_{n})-V(-x_{p})=V_{bi}}$

The sum of the drift and the diffusion at equilibrium is equal to zero

${\displaystyle J_{N}=q\mu _{N}nE+qD_{N}{\frac {dn}{dx}}=0}$

Use the Einstein relationship:

${\displaystyle E=-{\frac {D_{n}}{\mu _{N}}}{\frac {dn/dx}{n}}=-{\frac {kT}{q}}{\frac {dn/dx}{n}}}$

${\displaystyle V_{bi}=-\int _{-x_{p}}^{x_{n}}Edx={\frac {kT}{q}}\int _{n(-x_{p})}^{n(x_{n})}{\frac {dn}{n}}={\frac {kT}{q}}\ln \left[{\frac {n(x_{n})}{n(-x_{p})}}\right]}$

${\displaystyle N_{D}}$ and ${\displaystyle N_{A}}$ are the n- and p-side doping concentrations.

${\displaystyle n(x_{n})=N_{D}\;}$

${\displaystyle n(-x_{p})={\frac {n_{i}^{2}}{N_{A}}}}$

 ${\displaystyle V_{bi}={\frac {kT}{q}}\ln \left({\frac {N_{A}N_{D}}{n_{i}^{2}}}\right)}$


pn Junction Diode: I-V Characteristics

Assumptions

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

General Relationships

${\displaystyle I=AJ\;}$

${\displaystyle J=J_{N}(x)+J_{p}(x)\;}$

${\displaystyle J_{N}=q\mu _{n}nE+qD_{N}{\frac {dn}{dx}}}$

${\displaystyle J_{p}=q\mu _{p}pE-qD_{p}{\frac {dp}{dx}}}$

Quasineutral Region Consideration

The quasineutral p-region and n-region are adjacent to the depletion region

${\displaystyle 0=D_{N}{\frac {d^{2}\Delta n_{p}}{dx^{2}}}-{\frac {\Delta n_{p}}{\tau _{n}}}}$

${\displaystyle 0=D_{P}{\frac {d^{2}\Delta p_{n}}{dx^{2}}}-{\frac {\Delta p_{n}}{\tau _{p}}}}$

The electric field is zero and the derivative of the electron and hole concentration is zero in the quasineutral regions.

${\displaystyle J_{N}=qD_{N}{\frac {d\Delta n_{p}}{dx}}}$

${\displaystyle J_{P}=-qD_{P}{\frac {d\Delta p_{n}}{dx}}}$

Depletion Region

Continuity equations:

${\displaystyle 0={\frac {1}{q}}{\frac {dJ_{N}}{dx}}+{\frac {\partial n}{\partial t}}|_{thermalR-G}}$

${\displaystyle 0=-{\frac {1}{q}}{\frac {dJ_{P}}{dx}}+{\frac {\partial p}{\partial t}}|_{thermalR-G}}$

Assume that the thermal recombination-generation is zero throughout depletion region. Sum the ${\displaystyle J_{N}}$ and ${\displaystyle J_{P}}$ solutions.

${\displaystyle J=J_{N}(-x_{p})+J_{P}(x_{n})}$

Strategy to find the minority carrier current density in the quasineutral regions:

• Evaluate current densities at the depletion region edges
• Multiply by A to find the current

Boundary Conditions

Ohmic Contacts

${\displaystyle \Delta n_{p}(x\rightarrow -\infty )=0}$

${\displaystyle \Delta p_{n}(x\rightarrow +\infty )=0}$

Depletion Region Edge

Establish boundary conditions at the edges of the depletion region.

Multiply defining equations of the electron quasi-Fermi level, ${\displaystyle F_{N}}$, and the hole quasi-Fermi level, ${\displaystyle F_{P}}$.

${\displaystyle np=n_{i}^{2}e^{(F_{N}-F_{P})/kT}}$

Monotonic variation in levels

${\displaystyle F_{N}-F_{P}\leq E_{Fn}-E_{Fp}=qV_{A}}$

"Law of Junction"

${\displaystyle np=n_{i}^{2}e^{qV_{A}/jT}}$

Evaluate the "law of junction" at the depletion region edges to find the boundary conditions.

${\displaystyle n(-x_{p})p(-x_{p})=n(-x_{p})N_{A}=n_{i}^{2}e^{qV_{A}/kT}}$

${\displaystyle n(-x_{p})={\frac {n_{i}^{2}}{N_{A}}}e^{qV_{A}/kT}}$

 ${\displaystyle \Delta n_{p}(-x_{p})={\frac {n_{i}^{2}}{N_{A}}}e^{qV_{A}/kT}-1)}$


${\displaystyle n(x_{n})p(x_{n})=p(x_{n})N_{D}=n_{i}^{2}e^{qV_{A}/kT}}$

${\displaystyle p(x_{n})={\frac {n_{i}^{2}}{N_{D}}}e^{qV_{A}/kT}}$

 ${\displaystyle \Delta p_{n}(x_{n})={\frac {n_{i}^{2}}{N_{D}}}(e^{qV_{A}/kT}-1)}$


Derivation

1. Solve minority carrier diffusion equations with regard to the boundary conditions to determine value of ${\displaystyle \Delta n_{p}}$ and ${\displaystyle \Delta p_{n}}$ in quasineutral region.
2. Determine the minority carrier current densities in quasineutral region
3. Evaluate quasineutral region solutions of ${\displaystyle J_{N}(x)}$ and ${\displaystyle J_{P}(x)}$. Multiply result by area

Solve the equation below with regard to two boundary conditions.

${\displaystyle 0=D_{P}{\frac {d^{2}\Delta p_{n}}{dx'^{2}}}-{\frac {\Delta p_{n}}{\tau _{p}}}}$

${\displaystyle \Delta p_{n}(x'\rightarrow \infty )=0}$

${\displaystyle \Delta p_{n}(x'=0)={\frac {n_{i}^{2}}{N_{D}}}(e^{qV_{A}/kT}-1)}$

General solution:

${\displaystyle \Delta p_{n}(x')=A_{1}e^{-x'/L_{P}}+A_{2}e^{x'/L_{P}}}$

${\displaystyle L_{P}={\sqrt {D_{P}\tau _{P}}}}$

 ${\displaystyle \Delta p_{n}(x')={\frac {n_{i}^{2}}{N_{D}}}(e^{qV_{A}/kT}-1)e^{-x'/L_{P}}}$

 ${\displaystyle J_{P}(x')=-qD_{P}{\frac {d\Delta p_{n}}{dx'}}=q{\frac {D_{P}}{L_{P}}}{\frac {n_{i}^{2}}{N_{D}}}(e^{qV_{A}/kT}-1)e^{-x'/L_{P}}}$

 ${\displaystyle \Delta n_{p}(x'')={\frac {n_{i}^{2}}{N_{A}}}(e^{qV_{A}/kT}-1)e^{-x''/L_{N}}}$

 ${\displaystyle J_{N}(x'')=-qD_{N}{\frac {d\Delta n_{p}}{dx''}}=q{\frac {D_{N}}{L_{N}}}{\frac {n_{i}^{2}}{N_{A}}}(e^{qV_{A}/kT}-1)e^{-x'/L_{N}}}$


Evaluate at the depletion region edges

${\displaystyle J_{N}(x=-x_{p})=J_{N}(x''=0)=q{\frac {D_{N}}{L_{N}}}{\frac {n_{i}^{2}}{N_{A}}}(e^{qV_{A}/kT}-1)}$

${\displaystyle J_{P}(x=x_{n})=J_{P}(x'=0)=q{\frac {D_{P}}{L_{P}}}{\frac {n_{i}^{2}}{N_{D}}}(e^{qV_{A}/kT}-1)}$

Multiply the current density by the area:

${\displaystyle I=JA=qA\left({\frac {D_{N}}{L_{N}}}{\frac {n_{i}^{2}}{N_{A}}}+{\frac {D_{P}}{L_{P}}}{\frac {n_{i}^{2}}{N_{D}}}\right)(e^{qV_{A}/kT}-1)}$

Ideal diode equation:

 ${\displaystyle I=I_{0}(e^{qV_{A}/kT}-1)}$
${\displaystyle I_{0}=q_{A}\left({\frac {D_{N}}{L_{N}}}{\frac {n_{i}^{2}}{N_{A}}}+{\frac {D_{P}}{L_{P}}}{\frac {n_{i}^{2}}{N_{D}}}\right)}$


The Effective Mass

Derivation 1

The velocity of an electron in a one-dimensional lattice is in terms of the group velocity

${\displaystyle v_{g}={\frac {1}{\hbar }}{\frac {\partial E}{\partial k}}}$

${\displaystyle dE=eEv_{g}dt\;}$

${\displaystyle dE=eE{\frac {1}{\hbar }}{\frac {\partial E}{\partial k}}dt}$

Differentiate the equation of velocity

${\displaystyle {\frac {dv_{g}}{dt}}={\frac {1}{\hbar }}{\frac {d}{dt}}{\frac {\partial E}{\partial k}}}$

${\displaystyle {\frac {dv_{g}}{dt}}={\frac {1}{\hbar }}{\frac {\partial ^{2}E}{\partial k^{2}}}{\frac {dk}{dt}}}$

${\displaystyle {\frac {dv_{g}}{dt}}={\frac {1}{\hbar ^{2}}}{\frac {\partial ^{2}E}{\partial k^{2}}}eE}$

${\displaystyle m{\frac {dv}{dt}}=eE}$

 ${\displaystyle m^{*}=\hbar ^{2}\left({\frac {\partial ^{2}E}{\partial k^{2}}}\right)^{-1}}$


Derivation 2

In the free electron model, the electronic wave function can be in the form of ${\displaystyle e^{ik\cdot z}}$. For a wave packet the group velocity is given by:

${\displaystyle v={{d\omega } \over {dk}}}$ = ${\displaystyle {{1} \over {\hbar }}\cdot {{d\varepsilon } \over {dk}}}$

In presence of an electric field E, the energy change is:

${\displaystyle d\varepsilon ={{d\varepsilon } \over {dk}}{dk}=-eE{dx}=-eEv{dt}={-eE \over {\hbar }}{{d\varepsilon } \over {dk}}{dt}}$

Now we can say:

${\displaystyle \hbar \cdot {{dk} \over {dt}}={{dp} \over {dt}}=m\cdot {{dv} \over {dt}}}$

where p is the electron's momentum. Just put previous results in this last equation and we get:

${\displaystyle {{\hbar } \over {m}}\cdot {{dk} \over {dt}}={{1} \over {\hbar }}\cdot {{d} \over {dt}}{{d\varepsilon } \over {dk}}={{1} \over {\hbar }}\cdot {{d^{2}\varepsilon } \over {dk^{2}}}{{dk} \over {dt}}}$

From this follows the definition of effective mass:

${\displaystyle {{1} \over {m}}={{1} \over {\hbar ^{2}}}\cdot {{d^{2}\varepsilon } \over {dk^{2}}}}$

The Zimman Model

Wavefunction:

${\displaystyle \Psi _{k}=e^{ikx}\;}$

Strong disturbance when individual reflections add in phase

${\displaystyle n\lambda =2a\sin \theta \;}$

${\displaystyle k={\frac {n\pi }{a}}}$

${\displaystyle \Psi _{-k}=e^{-ikx}\;}$

${\displaystyle \Psi {\pm }={\frac {1}{\sqrt {2}}}(e^{ikx}\pm e^{-ikx})={\sqrt {2}}{\begin{bmatrix}\cos kx\\i\sin kx\end{bmatrix}}}$

The potential energy is from the actual potential ${\displaystyle V(x)}$ weighted by the probability function ${\displaystyle |\Psi _{\pm }|^{2}}$

${\displaystyle V_{\pm }={\frac {1}{L}}\int |\Psi _{\pm }|^{2}V(x)dx}$

${\displaystyle V_{\pm }={\frac {1}{L}}\int {\begin{bmatrix}2\cos ^{2}kx\\2\sin ^{2}kx\end{bmatrix}}V(x)dx}$

Average over one period

${\displaystyle V_{\pm }={\frac {1}{a}}\int _{0}^{a}{\begin{bmatrix}2\cos ^{2}kx\\2\sin ^{2}kx\end{bmatrix}}V(x)dx}$

${\displaystyle V_{\pm }={\frac {1}{a}}\int _{0}^{a}{\begin{bmatrix}1+\cos 2kx\\1+\sin 2kx\end{bmatrix}}V(x)dx}$

${\displaystyle \pm {\frac {1}{a}}\int _{0}^{a}\cos 2kxV(x)dx}$

${\displaystyle V_{\pm }=\pm V_{n}}$

The kinetic energy is the same in the case of both wave functions

${\displaystyle E={\frac {\hbar ^{2}k^{2}}{2m}}}$

The total energy is the kinetic energy plus the potential energy

${\displaystyle E_{\pm }={\frac {\hbar ^{2}k^{2}}{2m}}\pm V_{n}}$

The energy of an electron cannot be between the lower and higher value.