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:

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

where:

$j^\alpha = (c\rho, {\bf j}) F_{\alpha\beta} = \left(\begin{array}{cccc} 0 & {E_1\over c} & {E_2\over c} & {E_3\over c} \\ -{E_1\over c} & 0 & -B_3 & B_2 \\ -{E_2\over c} & B_3 & 0 & -B_1 \\ -{E_3\over c} & -B_2 & B_1 & 0 \\ \end{array}\right)$

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

Equations¶

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) $y_5' = -{q\over\varepsilon} (y_2-y_0 + C) y_0' = y_1 y_1' = {1\over D_n}(R + \mu_n y_1 y_5 + \mu_n y_0 y_5') y_2' = y_3 y_3' = {1\over D_p}(R - \mu_p y_3 y_5 - \mu_p y_2 y_5') y_4' = y_5$

Maxwell’s Equations¶

Four Potential¶

Semiconductor Device Physics¶

Equations¶

Example 1¶

Example 2¶

Example 3¶

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Maxwell’s Equations¶

Four Potential¶

Semiconductor Device Physics¶

Equations¶

Example 1¶

Example 2¶

Example 3¶

Table Of Contents

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Next topic

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