New improved capacitance–voltage model for symmetrical step junction: a way to a unified model for realistic junctions

Abstract

Based on the mass-action law, a new capacitance–voltage (C–V) model for symmetrical step junction is presented. Furthermore, we propose a unified C–V model for realistic junction and for any applied-voltage.

Appendices

Appendix 1: Capacitance-peak

The capacitance-peak can be obtained from

$$\begin{aligned} \left. {\frac{\hbox {dC}\left( {\hbox {V}_{\mathrm{app}} } \right) }{\hbox {dV}_{\mathrm{app}} }} \right| _{\mathrm{V}_{\mathrm{app}} =\mathrm{V}_{\mathrm{peak}} } =0 \end{aligned}$$

(65)

By including Eq. (36) into Eq. (65) and solving it for $\hbox {K}_{\mathrm{peak}}$, we get

$$\begin{aligned} \,\hbox {K}_{\mathrm{peak}} =2\sinh ^{-1}\left( {\hbox {K}_{\mathrm{peak}} } \right) -\frac{\hbox {K}_{\mathrm{peak}} }{1+\sqrt{1+\hbox {K}_{\mathrm{peak}}^2 }} \end{aligned}$$

(66)

Solving the above expressions gives

$$\begin{aligned} \hbox {K}_{\mathrm{peak}}&= 2.80421655769\approx 2.8 \nonumber \\ \Rightarrow \,\hbox {V}_{\mathrm{peak}}&= \hbox {V}_\mathrm{d} -3.4485\,\hbox {u}_\mathrm{T} \end{aligned}$$

(67)

Including Eq. (66) into Eq. (36) yields

$$\begin{aligned} \,\hbox {C}_{\mathrm{peak}} =\frac{\upvarepsilon _{\mathrm{sc}} }{\hbox {L}_{\mathrm{Di}} }\sqrt{\frac{2\hbox {K}_0\,\hbox {K}_{\mathrm{peak}} }{1+\hbox {K}_{\mathrm{peak}}^2 +\sqrt{1+\hbox {K}_{\mathrm{peak}}^2 }}}\approx 0.2433\frac{\upvarepsilon _{\mathrm{sc}} }{\hbox {L}_{\mathrm{De}} }\nonumber \\ \end{aligned}$$

(68)

Appendix 2: C–V inflection points

The inflection points of the C–V characteristic for any SSJ are determined by the following equation

$$\begin{aligned} \frac{\hbox {d}^{2}\hbox {C}\left( {\hbox {V}_{\mathrm{app}} } \right) }{\hbox {dV}_{\mathrm{app}}^2 }=0 \end{aligned}$$

(69)

Including Eq. (36) into Eq. (69) and solving, we obtain after some manipulations the analytic equation relating to the inflection points

$$\begin{aligned}&4\left( {\frac{\hbox {sinh}^{-1}\left( \,\hbox {K} \right) }{\hbox {K}}} \right) ^{2}\times \left( {\begin{array}{l} 64+48\hbox {K}^{2}-24\hbox {K}^{4}-17\hbox {K}^{6}-\hbox {K}^{8} \\ +\frac{32\hbox {K}^{2}+16\hbox {K}^{4}-14\hbox {K}^{6} -5\hbox {K}^{8}}{1+\sqrt{1+\hbox {K}^{2}}} \\ \end{array}} \right) \nonumber \\&\qquad +\left( {\begin{array}{l} 208+292\hbox {K}^{2}+125\hbox {K}^{4}+17\hbox {K}^{6}+ \\ \frac{104\hbox {K}^{2}+120\hbox {K}^{4}+39\hbox {K}^{6} +3\hbox {K}^{8}}{1+\sqrt{1+\hbox {K}^{2}}} \\ \end{array}} \right) \nonumber \\&\quad =2\frac{\hbox {sinh}^{-1}\left( \,\hbox {K} \right) }{\hbox {K}}\times \left( {\begin{array}{l} 224+256\hbox {K}^{2}+58\hbox {K}^{4}-2\hbox {K}^{6} \\ +\frac{112\hbox {K}^{2}+100\hbox {K}^{4}+11\hbox {K}^{6} -\;\hbox {K}^{8}}{1+\sqrt{1+\hbox {K}^{2}}}\;\\ \end{array}} \right) \end{aligned}$$

(70)

Solving the above expression gives

$$\begin{aligned} \left\{ {{\begin{array}{l} {\hbox {K}_1 = 12.4573105953\approx 12.5} \\ {\hbox {K}_2 =0.539083653266\approx 0.54} \\ \end{array} }} \right. \end{aligned}$$

(71)

The corresponding applied-voltages $\hbox {V}_{1}$ and $\hbox {V}_{2}$ are given, respectively, by:

$$\begin{aligned} \left\{ {\begin{array}{l} \hbox {V}_1 =\hbox {V}_\mathrm{d} -6.431\,\hbox {u}_\mathrm{T} \equiv \,\hbox {V}_{\mathrm{peak}} -2.982\,\hbox {u}_\mathrm{T} \\ \hbox {V}_2 =\hbox {V}_\mathrm{d} -0.151\,\hbox {u}_\mathrm{T} \equiv \,\hbox {V}_{\mathrm{peak}} +3.298\,\hbox {u}_\mathrm{T} \cong \,\hbox {V}_\mathrm{d} \\ \end{array}} \right. \end{aligned}$$

(72)

At these voltages, the capacitances $\hbox {C}(\hbox {V}_{1})$ and $\hbox {C}(\hbox {V}_{2})$ are given, respectively, by

$$\begin{aligned} \left\{ {\begin{array}{l}\,\hbox {C}\left( {\hbox {V}_1 :\hbox {K}\equiv \,\hbox {K}_1 } \right) =0.215\frac{\upvarepsilon _{\mathrm{sc}}}{\hbox {L}_{\mathrm{Di}}} \sqrt{\hbox {K}_0 }\approx \frac{\upvarepsilon _{\mathrm{sc}} }{4.5\,\hbox {L}_{\mathrm{De}} } \\ \hbox {C}\left( {\hbox {V}_2 :\hbox {K}\equiv \,\hbox {K}_2 } \right) =0.174\frac{\upvarepsilon _{\mathrm{sc}}}{\hbox {L}_{\mathrm{Di}}} \sqrt{\hbox {K}_0 }\approx \frac{\upvarepsilon _{\mathrm{sc}} }{6.0\hbox {L}_{\mathrm{De}} } \\ \end{array}} \right. \end{aligned}$$

(73)