Materials Science and Engineering/Derivations/Semiconductor Devices

pn Junctino Electrostatics

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

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

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

$v_{average}=a\tau$

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

$J=N_{e}e{v}_D$

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

$J=\sigma E$

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

 $\sigma ={\mu }_e(N_{e}e)$

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

$-\underset{-x_p}{\overset{x_n}{\int }}Edx=\underset{V(-x_p)}{\overset{V(x_n)}{\int }}dV=V(x_n)-V(-x_p)=V_{bi}$

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

$J_N=q{\mu }_{N}nE+q{D}_N\frac{dn}{dx}=0$

Use the Einstein relationship:

$E=-\frac{D_n}{{\mu }_N}\frac{dn/dx}{n}=-\frac{kT}{q}\frac{dn/dx}{n}$

$V_{bi}=-\underset{-x_p}{\overset{x_n}{\int }}Edx=\frac{kT}{q}\underset{n(-x_p)}{\overset{n(x_n)}{\int }}\frac{dn}{n}=\frac{kT}{q}\ln \left[\frac{n(x_n)}{n(-x_p)}\right]$

$N_D$ 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

File:Pn junction electrostatics.png

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

$I=AJ$

$J=J_N(x)+J_p(x)$

$J_N=q{\mu }_{n}nE+q{D}_N\frac{dn}{dx}$

$J_p=q{\mu }_{p}pE-q{D}_p\frac{dp}{dx}$

Quasineutral Region Consideration

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

$0=D_N\frac{d^2\mathrm{\Delta }{n}_p}{d{x}^2}-\frac{\mathrm{\Delta }{n}_p}{{\tau }_n}$

$0=D_P\frac{d^2\mathrm{\Delta }{p}_n}{d{x}^2}-\frac{\mathrm{\Delta }{p}_n}{{\tau }_p}$

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

$J_N=q{D}_N\frac{d\mathrm{\Delta }{n}_p}{dx}$

$J_P=-q{D}_P\frac{d\mathrm{\Delta }{p}_n}{dx}$

Depletion Region

Continuity equations:

$0=\frac{1}{q}\frac{d{J}_N}{dx}+\frac{\partial n}{\partial t}{|}_{thermalR-G}$

$0=-\frac{1}{q}\frac{d{J}_P}{dx}+\frac{\partial p}{\partial t}{|}_{thermalR-G}$

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

$\mathrm{\Delta }{n}_p(x\to -\mathrm{\infty })=0$

$\mathrm{\Delta }{p}_n(x\to +\mathrm{\infty })=0$

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

$F_N-F_P\le E_{Fn}-E_{Fp}=q{V}_A$

"Law of Junction"

$np=n_i^2{e}^{q{V}_A/jT}$

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

$n(-x_p)p(-x_p)=n(-x_p)N_A=n_i^2{e}^{q{V}_A/kT}$

$n(-x_p)=\frac{n_i^2}{N_A}{e}^{q{V}_A/kT}$

 $\mathrm{\Delta }{n}_p(-x_p)=\frac{n_i^2}{N_A}{e}^{q{V}_A/kT}-1)$

$n(x_n)p(x_n)=p(x_n)N_D=n_i^2{e}^{q{V}_A/kT}$

$p(x_n)=\frac{n_i^2}{N_D}{e}^{q{V}_A/kT}$

 $\mathrm{\Delta }{p}_n(x_n)=\frac{n_i^2}{N_D}(e^{q{V}_A/kT}-1)$

Derivation

1. Solve minority carrier diffusion equations with regard to the boundary conditions to determine value of $\mathrm{\Delta }{n}_p$ and $\mathrm{\Delta }{p}_n$ 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.

$0=D_P\frac{d^2\mathrm{\Delta }{p}_n}{dx{\text{'}}^2}-\frac{\mathrm{\Delta }{p}_n}{{\tau }_p}$

$\mathrm{\Delta }{p}_n(x^{\prime }\to \mathrm{\infty })=0$

$\mathrm{\Delta }{p}_n(x^{\prime }=0)=\frac{n_i^2}{N_D}(e^{q{V}_A/kT}-1)$

General solution:

$\mathrm{\Delta }{p}_n(x^{\prime })=A_1{e}^{-x^{\prime }/L_P}+A_2{e}^{x^{\prime }/L_P}$

$L_P=\sqrt{D_P{\tau }_P}$

 $\mathrm{\Delta }{p}_n(x^{\prime })=\frac{n_i^2}{N_D}(e^{q{V}_A/kT}-1)e^{-x^{\prime }/L_P}$

 $J_P(x^{\prime })=-q{D}_P\frac{d\mathrm{\Delta }{p}_n}{d{x}^{\prime }}=q\frac{D_P}{L_P}\frac{n_i^2}{N_D}(e^{q{V}_A/kT}-1)e^{-x^{\prime }/L_P}$

 $\mathrm{\Delta }{n}_p(x^{″})=\frac{n_i^2}{N_A}(e^{q{V}_A/kT}-1)e^{-x^{″}/L_N}$

 $J_N(x^{″})=-q{D}_N\frac{d\mathrm{\Delta }{n}_p}{d{x}^{″}}=q\frac{D_N}{L_N}\frac{n_i^2}{N_A}(e^{q{V}_A/kT}-1)e^{-x^{\prime }/L_N}$

Evaluate at the depletion region edges

$J_N(x=-x_p)=J_N(x^{″}=0)=q\frac{D_N}{L_N}\frac{n_i^2}{N_A}(e^{q{V}_A/kT}-1)$

$J_P(x=x_n)=J_P(x^{\prime }=0)=q\frac{D_P}{L_P}\frac{n_i^2}{N_D}(e^{q{V}_A/kT}-1)$

Multiply the current density by the area:

$I=JA=qA(\frac{D_N}{L_N}\frac{n_i^2}{N_A}+\frac{D_P}{L_P}\frac{n_i^2}{N_D})(e^{q{V}_A/kT}-1)$

Ideal diode equation:

 $I=I_0(e^{q{V}_A/kT}-1)$
 $I_0=q_A(\frac{D_N}{L_N}\frac{n_i^2}{N_A}+\frac{D_P}{L_P}\frac{n_i^2}{N_D})$

The Effective Mass

Derivation 1

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

$v_g=\frac{1}{\hslash }\frac{\partial E}{\partial k}$

$dE=eE{v}_{g}dt$

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

Differentiate the equation of velocity

$\frac{d{v}_g}{dt}=\frac{1}{\hslash }\frac{d}{dt}\frac{\partial E}{\partial k}$

$\frac{d{v}_g}{dt}=\frac{1}{\hslash }\frac{{\partial }^{2}E}{\partial k^2}\frac{dk}{dt}$

$\frac{d{v}_g}{dt}=\frac{1}{{\hslash }^2}\frac{{\partial }^{2}E}{\partial k^2}eE$

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

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

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:

$v=\frac{d\omega }{dk}$ = $\frac{1}{\hslash }\cdot \frac{d\epsilon }{dk}$

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

$d\epsilon =\frac{d\epsilon }{dk}dk=-eEdx=-eEvdt=\frac{-eE}{\hslash }\frac{d\epsilon }{dk}dt$

Now we can say:

$\hslash \cdot \frac{dk}{dt}=\frac{dp}{dt}=m\cdot \frac{dv}{dt}$

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

$\frac{\hslash }{m}\cdot \frac{dk}{dt}=\frac{1}{\hslash }\cdot \frac{d}{dt}\frac{d\epsilon }{dk}=\frac{1}{\hslash }\cdot \frac{d^2\epsilon }{d{k}^2}\frac{dk}{dt}$

From this follows the definition of effective mass:

$\frac{1}{m}=\frac{1}{{\hslash }^2}\cdot \frac{d^2\epsilon }{d{k}^2}$

The Zimman Model

Wavefunction:

${\mathrm{\Psi }}_k=e^{ikx}$

Strong disturbance when individual reflections add in phase

$n\lambda =2a\sin \theta$

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

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

$\mathrm{\Psi }\pm =\frac{1}{\sqrt{2}}(e^{ikx}\pm e^{-ikx})=\sqrt{2}\left[\begin{array}{c}\cos kx\\ i\sin kx\end{array}\right]$

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

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

$V_{\pm }=\frac{1}{L}\int \left[\begin{array}{c}2{\cos }^{2}kx\\ 2{\sin }^{2}kx\end{array}\right]V(x)dx$

Average over one period

$V_{\pm }=\frac{1}{a}\underset{0}{\overset{a}{\int }}\left[\begin{array}{c}2{\cos }^{2}kx\\ 2{\sin }^{2}kx\end{array}\right]V(x)dx$

$V_{\pm }=\frac{1}{a}\underset{0}{\overset{a}{\int }}\left[\begin{array}{c}1+\cos 2kx\\ 1+\sin 2kx\end{array}\right]V(x)dx$

$\pm \frac{1}{a}\underset{0}{\overset{a}{\int }}\cos 2kxV(x)dx$

$V_{\pm }=\pm V_n$

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

$E=\frac{{\hslash }^2{k}^2}{2m}$

The total energy is the kinetic energy plus the potential energy

$E_{\pm }=\frac{{\hslash }^2{k}^2}{2m}\pm V_n$

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