Materials Science and Engineering/Derivations/Semiconductor Devices

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

$$- \int_{-x_p}^{x_n} E dx = \int_{V(-x_p)}^{V(x_n)} dV = V(x_n) - V(-x_p) = V_{bi}$$

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

$$J_N = q \mu_N n E + q D_N \frac{dn}{dx} = 0$$

Use the Einstein relationship:

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

$$V_{bi} = - \int_{-x_p}^{x_n} E dx = \frac{kT}{q} \int_{n(-x_p)}^{n(x_n)} \frac{dn}{n} = \frac{kT}{q} \ln \left [ \frac{n(x_n)}{n(-x_p)} \right ]$$

$$N_D$$ and $$N_A$$ are the n- and p-side doping concentrations.

$$n(x_n) = N_D\;$$

$$n(-x_p) = \frac{n_i^2}{N_A}$$

$$V_{bi} = \frac{kT}{q} \ln \left ( \frac{N_A N_D}{n_i^2} \right )$$

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
$$I = AJ\;$$

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

$$J_N = q \mu_n n E + q D_N \frac{dn}{dx}$$

$$J_p = q \mu_p p E - q D_p \frac{dp}{dx}$$

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

$$0 = D_N \frac{d^2 \Delta n_p}{dx^2} - \frac{\Delta n_p}{\tau_n}$$

$$0 = D_P \frac{d^2 \Delta p_n}{dx^2} - \frac{\Delta p_n}{\tau_p}$$

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

$$J_N = q D_N \frac{d \Delta n_p}{dx}$$

$$J_P = - qD_P \frac{d \Delta p_n}{dx}$$

Depletion Region
Continuity equations:

$$0 = \frac{1}{q} \frac{dJ_N}{dx} + \frac{ \partial n}{\partial t} |_{thermal R-G}$$

$$0 = -\frac{1}{q} \frac{dJ_P}{dx} + \frac{ \partial p}{\partial t} |_{thermal R-G}$$

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

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

$$F_N - F_P \le E_{Fn} - E_{Fp} = qV_A$$

"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.

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

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

$$\Delta n_p (-x_p) = \frac{n_i^2}{N_A} e^{q V_A / kT} - 1 )$$

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

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

$$\Delta p_n (x_n) = \frac{n_i^2}{N_D} (e^{qV_A/kT} - 1)$$

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:

$$\Delta p_n(x') = A_1 e^{-x'/L_P} + A_2 e^{x'/L_P}$$

$$L_P = \sqrt{D_P \tau_P}$$

$$\Delta p_n(x') = \frac{n_i^2}{N_D} (e^{q V_A / kT} - 1) e^{-x'/L_P}$$

$$J_P(x') = -qD_P \frac{d \Delta p_n}{dx'} = q \frac{D_P}{L_P} \frac{n_i^2}{N_D} (e^{qV_A / kT} - 1) e^{-x'/L_P}$$

$$\Delta n_p(x) = \frac{n_i^2}{N_A} (e^{q V_A / kT} - 1) e^{-x/L_N}$$

$$J_N(x) = -qD_N \frac{d \Delta n_p}{dx} = q \frac{D_N}{L_N} \frac{n_i^2}{N_A} (e^{qV_A / kT} - 1) e^{-x'/L_N}$$

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:

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

Derivation 1
The velocity of an electron in a one-dimensional lattice is in terms of the group velocity

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

$$dE = e E v_g dt \;$$

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

Differentiate the equation of velocity

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

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

$$\frac{dv_g}{dt} = \frac{1}{\hbar^2} \frac{\partial^2 E}{\partial k^2} e E$$

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

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

Derivation 2
In the free electron model, the electronic wave function can be in the form of $$e^{i k \cdot z}$$. For a wave packet the group velocity is given by:

$$ v = {{d \omega} \over {d k}}$$ = $${{1} \over {\hbar}} \cdot {{d \varepsilon} \over {d k}} $$ In presence of an electric field E, the energy change is: $$ d \varepsilon = {{d \varepsilon} \over {d k}} {d k} = -eE{d x} = -eEv{d t} = {-eE \over {\hbar}} {{d \varepsilon} \over {d k}}{dt} $$

Now we can say: $$ \hbar \cdot {{d k} \over {d t}} = {{d p} \over {d t}} = m \cdot {{d v} \over {d t}}$$

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

$$ {{\hbar} \over {m}} \cdot {{d k} \over {d t}} = {{1} \over {\hbar}} \cdot {{d} \over {d t}}{{d \varepsilon} \over {d k}} ={{1} \over {\hbar}} \cdot {{d^{2} \varepsilon} \over {dk^{2}}}{{d k}\over {d t}}$$ From this follows the definition of effective mass:

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

The Zimman Model
Wavefunction:

$$\Psi_k = e^{ikx} \;$$

Strong disturbance when individual reflections add in phase

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

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

$$\Psi_{-k} = e^{-ikx}\;$$

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

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

$$V_{\pm} = \frac{1}{a} \int_0^a \begin{bmatrix} 2 \cos^2 kx\\ 2 \sin^2 kx \end{bmatrix} V(x) dx$$

$$V_{\pm} = \frac{1}{a} \int_0^a \begin{bmatrix} 1 + \cos 2kx\\ 1 + \sin 2kx \end{bmatrix} V(x) dx$$

$$\pm \frac{1}{a} \int_0^a \cos 2kx V(x) dx$$

$$V_{\pm} = \pm V_n$$

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.