Materials Science and Engineering/Derivations/Semiconductor Devices

pn Junctino Electrostatics

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

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

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

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

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

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

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

$J=\sigma E\;$

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

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

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.

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

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

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

Depletion Width with Step Junction and VA not equal to zero

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

$I=AJ\;$

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

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

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

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.

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

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

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

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

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

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

Derivation

1. Solve minority carrier diffusion equations with regard to the boundary conditions to determine value of $\Delta n_{p}$ and $\Delta p_{n}$ in quasineutral region.
2. Determine the minority carrier current densities in quasineutral region
3. 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:

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

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

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

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

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

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

Evaluate at the depletion region edges

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

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

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:

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

The Effective Mass

Derivation 1

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

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

$dE=eEv_{g}dt\;$

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

Differentiate the equation of velocity

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

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

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

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

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

Derivation 2

In the free electron model, the electronic wave function can be in the form of $e^{ik\cdot z}$. 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.