2.1.2 Approximation for the Space Charge Region

In this section the interaction of the beam of injected electrons and the semiconductor is investigated for the space charge region under idealized conditions where the influence of the surface potential barrier is neglected. The effects caused by a surface potential barrier are discussed in Section 2.2.

Under these idealized conditions and for small injection currents the energy shift of the emitted secondary electrons is determined by the doping. For small beam currents the carrier concentration caused by the electron beam is several orders of magnitude smaller than the equilibrium carrier concentration. Therefore the change in the surface potential $\Delta\ensuremath{\varphi}$ is small compared to the equilibrium potential $\ensuremath{\varphi}$.

$$\Delta \varphi \ll \varphi$$ (2.20)

The pn-junction is located where the net doping equals zero.

N_A^- + N_D⁺ = 0 (2.21)

For conditions near to thermal equilibrium the carrier concentration in the space charge region around the pn-junction is very small and therefore the net-charge is determined mainly by the ionized dopants. This is illustrated in the following by the example of a p⁺n-diode where total ionization is assumed. The acceptor concentration is given by

$$\left\{\vphantom{\begin{array}{ll}\mathrm{N_{A0}}\cdot \left(1 - \exp\left(\di... ...\frac{x}{\mathrm{d}}}\right)\right) & x \leq 0 \\ 0 & 0 < x\end{array}}\right. \begin{array}{ll}\mathrm{N_{A0}}\cdot \left(1 - \exp\left(\displaystyle{\frac{x}{\mathrm{d}}}\right)\right) & x \leq 0 \\ 0 & 0 < x\end{array} \left.\vphantom{\begin{array}{ll}\mathrm{N_{A0}}\cdot \left(1 - \exp\left(\dis... ...\frac{x}{\mathrm{d}}}\right)\right) & x \leq 0 \\ 0 & 0 < x\end{array}}\right.$$ (2.22)

The donor concentration N_D is assumed to be constant and several orders of magnitude smaller than the maximum acceptor concentration N_A0. Fig. 2.5 shows a plot of the doping concentration.

The Poisson equation for this doping distribution is then given by

$${\frac{\partial ^{2} \varphi(x)}{\partial x^{2}}} \left\{\vphantom{\begin{array}{ll}0 & x < -\mathrm{d_{p}}\\ [2mm]\displaystyl... ... & 0 \leq x \leq\mathrm{d_{n}}\\ [3mm]0 & \mathrm{d_{n}}< x\end{array}}\right. \begin{array}{ll}0 & x < -\mathrm{d_{p}}\\ [2mm]\displaystyle{-\frac{\mathrm{... ...N_{D}}}} & 0 \leq x \leq\mathrm{d_{n}}\\ [3mm]0 & \mathrm{d_{n}}< x\end{array} \left.\vphantom{\begin{array}{ll}0 & x < -\mathrm{d_{p}}\\ [2mm]\displaystyle... ... & 0 \leq x \leq\mathrm{d_{n}}\\ [3mm]0 & \mathrm{d_{n}}< x\end{array}}\right.$$ (2.23)

In this equation d_p and d_n are the depletion widths of the p-doped and n-doped region, respectively. The electric field and the potential can be calculated by integrating (2.23) with the following boundary conditions.

$${\frac{\partial \varphi(-\mathrm{d_{p}})}{\partial x}} {\frac{\partial \varphi(\mathrm{d_{n}})}{\partial x}}$$ (2.24)

$${\frac{\partial \varphi(0-)}{\partial x}} {\frac{\partial \varphi(0+)}{\partial x}}$$ (2.25)

$$\varphi \infty \varphi$$ (2.26)

$$\varphi \varphi$$ (2.27)

$$\varphi \varphi \infty$$ (2.28)

Figure 2.5: Doping concentration of a p⁺n diode with N _A0 = 10¹⁶ cm^-3, N_D = 10¹⁴ cm^-3, and d= 10^-8 cm.

Neglecting terms of order exp$\left(\vphantom{\frac{\mathrm{d_{p}}}{\mathrm{d}}}\right.$${\frac{\mathrm{d_{p}}}{\mathrm{d}}}$ $\left.\vphantom{\frac{\mathrm{d_{p}}}{\mathrm{d}}}\right)$ the electric field and the potential are given by

$$\left\{\vphantom{\begin{array}{ll}0 & x \leq -\mathrm{d_{p}}\\ [2mm]\displays... ... \leq x \leq \mathrm{d_{n}}\\ [3mm]0 & \mathrm{d_{n}}\leq x\end{array}}\right. \begin{array}{ll}0 & x \leq -\mathrm{d_{p}}\\ [2mm]\displaystyle{\frac{\mathr... ...ght)} &0 \leq x \leq \mathrm{d_{n}}\\ [3mm]0 & \mathrm{d_{n}}\leq x\end{array} \left.\vphantom{\begin{array}{ll}0 & x \leq -\mathrm{d_{p}}\\ [2mm]\displayst... ... \leq x \leq \mathrm{d_{n}}\\ [3mm]0 & \mathrm{d_{n}}\leq x\end{array}}\right.$$ (2.29)

and

$$\varphi \left\{\vphantom{\begin{array}{ll}0 & x \leq -\mathrm{d_{p}}\\ [2mm]\displays... ...mathrm{d_{n}}\\ [3mm]\varphi(\infty) & \mathrm{d_{n}}\leq x\end{array}}\right. \begin{array}{ll}0 & x \leq -\mathrm{d_{p}}\\ [2mm]\displaystyle{-\frac{\math... ...x \leq \mathrm{d_{n}}\\ [3mm]\varphi(\infty) & \mathrm{d_{n}}\leq x\end{array} \left.\vphantom{\begin{array}{ll}0 & x \leq -\mathrm{d_{p}}\\ [2mm]\displayst... ...mathrm{d_{n}}\\ [3mm]\varphi(\infty) & \mathrm{d_{n}}\leq x\end{array}}\right.$$ (2.30)

The potential $\varphi$($\infty$) is given by

$$\varphi \infty {\frac{\ensuremath{E_{\mathrm{g}}}}{\mathrm{q}}} {\frac{\ensuremath{\mathrm{k_{B}}}\cdot T}{\mathrm{q}}} \left(\vphantom{\frac{\mathrm{N_{A0}}\cdot\mathrm{N_{D}}}{N_{\mathrm{C}}\cdot N_{\mathrm{V}}}}\right. {\frac{\mathrm{N_{A0}}\cdot\mathrm{N_{D}}}{N_{\mathrm{C}}\cdot N_{\mathrm{V}}}} \left.\vphantom{\frac{\mathrm{N_{A0}}\cdot\mathrm{N_{D}}}{N_{\mathrm{C}}\cdot N_{\mathrm{V}}}}\right)$$ (2.31)

where $\ensuremath{E_{\mathrm{g}}}$ is the band gap energy and N_C and N_V are the effective carrier density of states for the conduction and valence band, respectively.

$$\left(\vphantom{\frac{T}{\ensuremath{\mathrm{300~K}}}}\right. {\frac{T}{\ensuremath{\mathrm{300~K}}}} \left.\vphantom{\frac{T}{\ensuremath{\mathrm{300~K}}}}\right)^{1.5}_{}$$ N_C (2.32)

$$N_{\mathrm{V}} = 1.8\cdot10^{19}\cdot\left(\frac{T}{\ensuremath{\mathrm{300~K}}}\right)^{1.5} \ensuremath{\mathrm{cm^{-3}}}$$ (2.33)

The extension of the space charge region is given by

$$\sqrt{\frac{\mathrm{N_{A0}}\cdot\mathrm{N_{D}}\cdot\mathrm{d}^{2} + 2\cdot\mat... ...\cdot\varphi(\infty)}{\mathrm{N_{A0}}^{2} -\mathrm{N_{A0}}\cdot\mathrm{N_{D}}}}$$ (2.34)

and

$${\frac{\mathrm{N_{A0}}\cdot\mathrm{d}- (\mathrm{N_{A0}}- \mathrm{N_{D}})\cdot\mathrm{d_{p}}}{\mathrm{N_{D}}}}$$ (2.35)

The pn-junction is located at

$$\left(\vphantom{1 + \displaystyle{\frac{\mathrm{N_{D}}}{\mathrm{N_{A0}}}}}\right. {\frac{\mathrm{N_{D}}}{\mathrm{N_{A0}}}} \left.\vphantom{1 + \displaystyle{\frac{\mathrm{N_{D}}}{\mathrm{N_{A0}}}}}\right)$$ (2.36)

For N_A0 = 10¹⁶ cm^-3, N_D = 10¹⁴ cm^-3, $T = \ensuremath{\mathrm{300~K}}$, and d = 1 nm the following values can be calculated from (2.31), (2.34), (2.35), (2.36), and (2.28).

$$\varphi \infty = \ensuremath{\mathrm{0.892~V}}$$ (2.37)

$$\mathrm{d_{p}} = \ensuremath{\mathrm{35.3~nm}}$$ (2.38)

$$\mathrm{d_{n}} = \ensuremath{\mathrm{3.39~\mu m}}$$ (2.39)

$$x_{0} = \ensuremath{\mathrm{-9.95\cdot 10^{-12}~m}}$$ (2.40)

$$E(x_{0}) = \ensuremath{\mathrm{-0.52\cdot 10^{6}~Vm^{-1}}}$$ (2.41)

The electric field and the potential in the space charge region are shown in Fig. 2.6 and Fig. 2.7, respectively.

When the electron beam is positioned in such a way that the secondary electron-hole pairs are generated in the space charge region the generated electrons and holes are separated immediately by the strong electric field. Because of the long diffusion lengths compared to the extension of the space charge region nearly no recombination occurs in the space charge region. In a one-dimensional approximation the current density for electrons and holes caused by the injection is constant from the location of the injection to the n-doped and p-doped end of the space charge region, respectively.

Figure 2.6: Electric field in the space charge region.

Figure 2.7: Potential in the space charge region.

For beam currents of approximately 1 nA and primary electron energies of several keV the current density at the surface is of the order of 10⁷ Am^{- 2} which is several orders of magnitude more than the equilibrium drift and diffusion currents. The electron current density caused by injection is given by

$${\frac{\partial n}{\partial x}} \mu_{\mathrm{n}}^{}$$ (2.42)

the hole current density caused by injection is given by

$${\frac{\partial p}{\partial x}} \mu_{\mathrm{p}}^{}$$ (2.43)

For small beam currents the electric field E(x) can be approximated by the equilibrium electric field. The differential equation (2.42) can be rewritten as

$${\frac{\mu_{\mathrm{n}}}{\mathrm{D}_{\mathrm{n}}}} {\frac{J_{\mathrm{n}}^{\mathrm{inj}}}{\mathrm{q}\cdot\mathrm{D}_{\mathrm{n}}}}$$ (2.44)

To find the general solution n(x) of this differential equation we have to find the solution of the corresponding homogeneous differential equation and a particular solution of the inhomogeneous equation. The homogeneous differential equation is given by

$${\frac{\mu_{\mathrm{n}}}{\mathrm{D}_{\mathrm{n}}}}$$ (2.45)

The solution of this homogeneous differential equation is

$$\left(\vphantom{\frac{\mu_{\mathrm{n}}}{\mathrm{D}_{\mathrm{n}}}\cdot\left(\ensuremath{\varphi}(x)- \ensuremath{\varphi}(\xi) \right)}\right. {\frac{\mu_{\mathrm{n}}}{\mathrm{D}_{\mathrm{n}}}} \left(\vphantom{\ensuremath{\varphi}(x)- \ensuremath{\varphi}(\xi) }\right. \varphi \varphi \xi \left.\vphantom{\ensuremath{\varphi}(x)- \ensuremath{\varphi}(\xi) }\right) \left.\vphantom{\frac{\mu_{\mathrm{n}}}{\mathrm{D}_{\mathrm{n}}}\cdot\left(\ensuremath{\varphi}(x)- \ensuremath{\varphi}(\xi) \right)}\right)$$ (2.46)

A particular solution to the inhomogeneous equation can be found by the method of variation of the parameters.

$${\frac{J_{\mathrm{n}}^{\mathrm{inj}}}{\mathrm{q}\cdot\mathrm{D}_{\mathrm{n}}}} \left(\vphantom{\frac{\mu_{\mathrm{n}}}{\mathrm{D}_{\mathrm{n}}}\cdot\ensuremath{\varphi}(x)}\right. {\frac{\mu_{\mathrm{n}}}{\mathrm{D}_{\mathrm{n}}}} \varphi \left.\vphantom{\frac{\mu_{\mathrm{n}}}{\mathrm{D}_{\mathrm{n}}}\cdot\ensuremath{\varphi}(x)}\right) \int_{\xi}^{x} \left(\vphantom{-\frac{\mu_{\mathrm{n}}}{\mathrm{D}_{\mathrm{n}}}\cdot\ensuremath{\varphi}(\zeta)}\right. {\frac{\mu_{\mathrm{n}}}{\mathrm{D}_{\mathrm{n}}}} \varphi \zeta \left.\vphantom{-\frac{\mu_{\mathrm{n}}}{\mathrm{D}_{\mathrm{n}}}\cdot\ensuremath{\varphi}(\zeta)}\right) \zeta$$ (2.47)

The general solution of (2.44) is the sum of (2.46) and (2.47).

The constant C can be determined by calculating n(d_n) and solving for C.

$$\left(\vphantom{-\frac{\mu_{\mathrm{n}}}{\mathrm{D}_{\mathrm{n}}}\cdot\ensuremath{\varphi}(\xi)}\right. {\frac{\mu_{\mathrm{n}}}{\mathrm{D}_{\mathrm{n}}}} \varphi \xi \left.\vphantom{-\frac{\mu_{\mathrm{n}}}{\mathrm{D}_{\mathrm{n}}}\cdot\ensuremath{\varphi}(\xi)}\right) \left(\vphantom{-\frac{\mu_{\mathrm{n}}}{\mathrm{D}_{\mathrm{n}}}\cdot\ensuremath{\varphi}(\mathrm{d_{n}})}\right. {\frac{\mu_{\mathrm{n}}}{\mathrm{D}_{\mathrm{n}}}} \varphi \left.\vphantom{-\frac{\mu_{\mathrm{n}}}{\mathrm{D}_{\mathrm{n}}}\cdot\ensuremath{\varphi}(\mathrm{d_{n}})}\right) {\frac{J_{\mathrm{n}}^{\mathrm{inj}}}{\mathrm{q}\cdot\mathrm{D}_{\mathrm{n}}}} \int_{\xi}^{\mathrm{d_{n}}} \left(\vphantom{-\frac{\mu_{\mathrm{n}}}{\mathrm{D}_{\mathrm{n}}}\cdot\ensuremath{\varphi}(\zeta)}\right. {\frac{\mu_{\mathrm{n}}}{\mathrm{D}_{\mathrm{n}}}} \varphi \zeta \left.\vphantom{-\frac{\mu_{\mathrm{n}}}{\mathrm{D}_{\mathrm{n}}}\cdot\ensuremath{\varphi}(\zeta)}\right) \zeta$$ (2.49)

Inserting this expression into (2.48) the general solution finally can be written as

An analogous calculation for p(x) starting with (2.43) gives

The charge density caused by the injection of electrons is proportional to the sum of (2.50) and (2.51).

$$\rho_{\mathrm{inj}}^{} \left(\vphantom{p(x) - n(x)}\right. \left.\vphantom{p(x) - n(x)}\right)$$ (2.52)

At the pn-junction $\rho_{\mathrm{inj}}^{}$ is very flat and therefore $\Delta$$\rho_{\mathrm{inj}}^{}$ will be small at the junction.