Maxwell’s Equations

The Maxwell’s equations are:

\partial_\alpha F^{\beta\alpha} = \mu_0 j^\beta

\epsilon^{\alpha\beta\gamma\delta}\partial_\gamma F_{\alpha\beta} = 0

and the Lorentz force is:

{\d p_\alpha\over\d \tau} = q F_{\alpha\beta} u^\beta


j^\alpha = (c\rho, {\bf j})

F_{\alpha\beta} = \left(\begin{array}{cccc}
0 & {E_1\over c} & {E_2\over c} & {E_3\over c} \\
-{E_1\over c} & 0 & -B_3 & B_2 \\
-{E_2\over c} & B_3 & 0 & -B_1 \\
-{E_3\over c} & -B_2 & B_1 & 0 \\

This corresponds to:

\nabla\cdot{\bf E} = c^2\mu_0 \rho

\nabla\times{\bf B} = \mu_0 {\bf j} + {1\over c^2}{\partial{\bf E}
    \over \partial t}

\nabla\cdot{\bf B} = 0

\nabla\times{\bf E} = -{\partial{\bf B}\over\partial t}

Four Potential

The four potential is defined by:

A^\alpha = \left({\phi\over c}, {\bf A}\right)

F_{\alpha\beta} = \partial_\alpha A_\beta - \partial_\beta A_\alpha

this corresponds to:

{\bf E} = -\nabla\phi - {\partial {\bf A}\over\partial t}

{\bf B} = \nabla\times{\bf A}

The Maxwell’s equations can then be written as (note that the two eq. without sources are automatically satisfied by the four potential):

\partial_\alpha F^{\beta\alpha} =
    \partial_\alpha (\partial^\beta A^\alpha - \partial^\alpha A^\beta) =
    -\partial_\alpha \partial^\alpha A^\beta =
    \mu_0 j^\beta

where we have employed the Lorentz gauge \partial_\alpha A^\alpha=0.

Semiconductor Device Physics

In general, the task is to find the five quantities:

n({\bf x}, t),
p({\bf x}, t),
{\bf J}_n({\bf x}, t),
{\bf J}_h({\bf x}, t),
{\bf E}({\bf x}, t)

where n (p) is the electron (hole) concentration, {\bf J}_n ({\bf J}_p) is the electron (hole) current density, {\bf E} is the electric field.

And we have five equations that relate them. We start with the continuity equation:

\nabla\cdot{\bf J} +{\partial\rho\over\partial t} = 0

where the current density {\bf J} is composed of electron and hole current densities:

{\bf J} = {\bf J}_n + {\bf J}_p

and the charge density \rho is composed of mobile (electrons and holes) and fixed charges (ionized donors and acceptors):

\rho = q(p-n+C)

where n and p is the electron and hole concetration, C is the net doping concetration (C=p_D-n_A where p_D is the concentration of ionized donors, charged positive, and n_A is the concentration of ionized acceptors, charged negative) and q is the electron charge (positive). We get:

\nabla\cdot{\bf J}_n + \nabla\cdot{\bf J}_p + q\left(
    {\partial p\over\partial t}
    -{\partial n\over\partial t}
    +{\partial C\over\partial t}
    \right) = 0

Assuming the fixed charges C are time invariant, we get:

\nabla\cdot{\bf J}_n - q {\partial n\over\partial t} =
    -\left( \nabla\cdot{\bf J}_p + q{\partial p\over\partial t}
    \right) \equiv qR

where R is the net recombination rate for electrons and holes (a positive value means recombination, a negative value generation of carriers). We get the carrier continuity equations:

(1){\partial n\over\partial t} = -R + {1\over q} \nabla\cdot {\bf J}_n

{\partial p\over\partial t} = -R - {1\over q} \nabla\cdot {\bf J}_p

Then we need material relations that express how the current {\bf J} is generated using {\bf E} and n and p. A drift-diffusion model is to assume a drift current (q\mu_n n {\bf E}) and a diffusion (q D_n \nabla n), which gives:

(2){\bf J}_n = q\mu_n n {\bf E} + q D_n \nabla n

{\bf J}_p = q\mu_p p {\bf E} - q D_p \nabla p

where \mu_n, \mu_p, D_n, D_p are the carrier mobilities and diffusivities.

Final equation is the Gauss’s law:

\nabla\cdot (\varepsilon{\bf E}) = \rho

(3)\nabla\cdot(\varepsilon {\bf E}) = q(p-n + C)


Combining (2) and (1) we get the following three equations for three unknowns n, p and {\bf E}:

{\partial n\over\partial t} = -R + \nabla\cdot (\mu_n n {\bf E})
    +\nabla\cdot (D_n \nabla n)

{\partial p\over\partial t} = -R - \nabla\cdot (\mu_p p {\bf E})
    +\nabla\cdot (D_p \nabla p)

\nabla\cdot(\varepsilon {\bf E}) = q(p-n + C)

And it is usually assumed that the magnetic field is time independent, so {\bf E}=-\nabla\phi and we get:

(4){\partial n\over\partial t} = -R - \nabla\cdot (\mu_n n \nabla\phi)
    +\nabla\cdot (D_n \nabla n)

{\partial p\over\partial t} = -R + \nabla\cdot (\mu_p p \nabla\phi)
    +\nabla\cdot (D_p \nabla p)

\nabla\cdot(\varepsilon \nabla\phi) = -q(p-n + C)

These are three nonlinear (due to the terms \mu_n n \nabla\phi and \mu_p p \nabla\phi) equations for three unknown functions n, p and \phi.

Example 1

We can substract the first two equations and we get:

{\partial q(p-n)\over\partial t} = - q\nabla\cdot ((\mu_p p+\mu_n n){\bf E})
    +q\nabla\cdot(D_p \nabla p-D_n\nabla n)

\nabla\cdot(\varepsilon {\bf E}) = q(p-n+C)

and using \rho=q(p-n+C) and \sigma=q(\mu_p p+\mu_n n), we get:

{\partial \rho\over\partial t} -q{\partial C\over\partial t} =
    - \nabla\cdot (\sigma {\bf E})
    +q\nabla\cdot(D_p \nabla p-D_n\nabla n)

\nabla\cdot(\varepsilon {\bf E}) = \rho

So far we didn’t make any assumptions. Most of the times the net doping concetration C is time independent, which gives:

{\partial \rho\over\partial t} =
    - \nabla\cdot (\sigma {\bf E})
    +q\nabla\cdot(D_p \nabla p-D_n\nabla n)

\nabla\cdot(\varepsilon {\bf E}) = \rho

Assuming further D_p \nabla p-D_n\nabla n=0, we just get the equation of continuity and the Gauss law:

{\partial \rho\over\partial t} + \nabla\cdot (\sigma {\bf E}) = 0

\nabla\cdot(\varepsilon {\bf E}) = \rho

Finally, assuming also that that \rho doesn’t depend on time, we get:

\nabla\cdot (\sigma {\bf E}) = 0

\nabla\cdot(\varepsilon {\bf E}) = \rho

Example 2

As a simple model, assume D_n, D_p, \mu_n, \mu_p and \varepsilon are position independent and C=0, R=0:

{\partial n\over\partial t} =
    +\mu_n n \nabla\cdot {\bf E}
    +\mu_n {\bf E}\cdot\nabla n
    +D_n \nabla^2 n

{\partial p\over\partial t} =
    -\mu_p p \nabla\cdot {\bf E}
    -\mu_p {\bf E}\cdot\nabla p
    +D_p \nabla^2 p

\varepsilon\nabla\cdot {\bf E} = q(p-n)

Using {\bf E} = -\nabla \phi we get:

{\partial n\over\partial t} =
    -\mu_n n \nabla^2\phi
    -\mu_n \nabla\phi\cdot\nabla n
    +D_n \nabla^2 n

{\partial p\over\partial t} =
    +\mu_p p \nabla^2\phi
    +\mu_p \nabla\phi\cdot\nabla p
    +D_p \nabla^2 p

\varepsilon\nabla^2\phi = -q(p-n)

Example 3

Let’s calculate the 1D pn-junction. We take the equations (4) and write them in 1D for the stationary state ({\partial n\over\partial t}={\partial p\over\partial t}=0):

0 = -R - (\mu_n n \phi')' + (D_n n')'

0 = -R + (\mu_p p \phi')' + (D_p p')'

(\varepsilon \phi')' = -q(p-n + C)

We expand the derivatives and assume that \mu and D is constant:

0 = -R - \mu_n n' \phi' - \mu_n n \phi'' + D_n n''

0 = -R + \mu_p p' \phi' + \mu_p p \phi'' + D_p p''

\varepsilon \phi'' = -q(p-n + C)

and we put the second derivatives on the left hand side:

(5)n'' = {1\over D_n}(R + \mu_n n' \phi' + \mu_n n \phi'')

p'' = {1\over D_p}(R - \mu_p p' \phi' - \mu_p p \phi'')

\phi'' = -{q\over\varepsilon} (p-n + C)

now we introduce the variables y_i:

y_0 = n

y_1 = y_0' = n'

y_2 = p

y_3 = y_2' = p'

y_4 = \phi

y_5 = y_4' = \phi'

and rewrite (5):

y_1' = {1\over D_n}(R + \mu_n y_1 y_5 + \mu_n y_0 y_5')

y_3' = {1\over D_p}(R - \mu_p y_3 y_5 - \mu_p y_2 y_5')

y_5' = -{q\over\varepsilon} (y_2-y_0 + C)

So we are solving the following six nonlinear first order ODE:

(6)y_5' = -{q\over\varepsilon} (y_2-y_0 + C)

y_0' = y_1

y_1' = {1\over D_n}(R + \mu_n y_1 y_5 + \mu_n y_0 y_5')

y_2' = y_3

y_3' = {1\over D_p}(R - \mu_p y_3 y_5 - \mu_p y_2 y_5')

y_4' = y_5

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