Abrupt pn junctions in the depletion approximation

In an abrupt pn junction, the doping changes abruptly from p to n. It is common to solve for the band bending, the local electric field, the carrier concentration profiles, and the local conductivity in the depletion approximation. In this approximation, it is assumed that there is a depletion width W around the transition from p to n where the charge carrier densities are negligible. Outside of the depletion width, the charge carrier densities are equal to the doping concentrations. The semiconductor is electrically neutral outside of the depletion width. Using this depletion approximation it is possible to calculate important properties of pn junctions. The charge density distribution for an abrupt junction is,

$$\rho(x) = \begin{cases} 0 & \mbox{for } x\lt x_p, \\ -eN_A & \mbox{for } x_p\lt x\lt 0, \\ eN_D & \mbox{for } 0 \lt x\lt x_n, \\ 0 & \mbox{for } x_n\lt x.\end{cases}$$

The charge density can be integrated to determine the electric field $E = \int\frac{\rho}{\epsilon}dx$ ,

$$E(x) = \begin{cases} 0 & \mbox{for } x\lt x_p, \\ -\frac{eN_A}{\epsilon}(x-x_p) & \mbox{for } x_p\lt x\lt 0, \\ \frac{eN_D}{\epsilon}(x-x_n) & \mbox{for } 0 \lt x\lt x_n, \\ 0 & \mbox{for } x_n\lt x.\end{cases}$$

At the transition from p-doped to n-doped at $x=0$, there is a maximum in the electric field, $E = \frac{eN_Ax_p}{\epsilon} = -\frac{eN_Dx_n}{\epsilon}$. Due to the charge neutrality condition, $|N_Ax_p|=|N_Dx_n|$. Further $x_p < 0 < x_n$ so the electric field is pointed in the negative $x$-direction, from n towards p. The electric field can be integrated to determine the electrostatic potential $ \phi=-\int E dx $,

$$\phi(x) = \begin{cases} -\frac{eN_Ax_p^2}{2\epsilon} & \mbox{for } x\lt x_p, \\ \frac{eN_A}{\epsilon}\left(\frac{x^2}{2}-x_px\right) & \mbox{for } x_p\lt x\lt 0, \\ -\frac{eN_D}{\epsilon}\left(\frac{x^2}{2}-x_nx\right) & \mbox{for } 0 \lt x\lt x_n, \\ \frac{eN_Dx_n^2}{\epsilon} & \mbox{for } x_n\lt x.\end{cases}$$

The voltage across the junction is the difference in the electrostatic potential $V_{bi} - V=\phi (x_n)-\phi(x_p)$,

$$V_{bi} - V= \frac{eN_D x_n^2}{2 \epsilon} + \frac{eN_A x_p^2}{2 \epsilon},$$

where $V_{bi}$ is the built-in voltage. Using the charge neutrality condition $ \left( N_A|x_p|=N_D|x_n|=N_A (W-|x_n|)=N_D (W-|x_p|) \right)$, expressions for $|x_p|$ and $|x_n|$ can be obtained,

$$|x_p| = \frac{N_DW}{N_A+N_D}, \\ |x_n| = \frac{N_AW}{N_A+N_D}.$$

Substituting these expressions into the formula for $V_{bi} - V$ and solving for $W$ yields,

$$W= \sqrt{\frac{2\epsilon(N_D+N_A)(V_{bi}-V)}{eN_DN_A}}.$$

$N_A$ = 

 1/cm³

$N_D$ = 

 1/cm³ $E_g$ =  eV

$N_v(300)$ = 

 1/cm³

$N_c(300)$ = 

 1/cm³

$\epsilon_r$ = 

$T$ = 

K

$\mu_p$ = 

 cm²/V s

$\mu_n$ = 

 cm²/V s

$\tau_p$ = 

s 

$\tau_n$ = 

s

$V$ =  V
V 

$E_g=$  eV  $W=$  μm  $x_p=$  μm  $x_n=$  μm  $V_{bi}=$  V  $C_j=$  nF/cm²

$n_i=$  1/cm³  $D_p=$  cm²/s  $D_n=$  cm²/s  $L_p=$  μm  $L_n=$  μm

Band diagram

 [eV]

x [μm]

Current-Voltage Characteristics

 $\frac{I}{10^5 I_S}$

$V$ [V]

Charge density

ρ [C/m³]

x [μm]

Electric field

E [V/m]

x [μm]

 

Carrier Densities

$n/N_A$
$p/N_A$

x [μm]

log(Carrier Densities)

[1/cm³]

x [μm]

Capacitance - Voltage

$\frac{1}{C_j^2}$
$\left[\frac{\text{cm}^4}{\text{nF}^2}\right]$

V [V]

Current densities

$j$
$\left[\text{A/cm}^2\right]$

x [μm]

$\vec{j}_{n,\text{drift}}= ne\mu_n\vec{E}$,   $\vec{j}_{p,\text{drift}}= pe\mu_p\vec{E}$,
$\vec{j}_{n,\text{diffusion}}= eD_n\frac{dn}{dx}$, and $\vec{j}_{p,\text{diffusion}}= -eD_p\frac{dp}{dx}$