Exercises — Ion implantation and diffusion

$J=-D\frac{\partial N}{\partial x}$ $J$ is the flux — how many atoms cross a unit area per second. The minus sign says atoms flow down the concentration slope: from crowded regions toward empty ones. If $N$ falls as $x$ grows, then $\partial N/\partial x<0$, and the minus flips it so $J>0$ (flow into the wafer). Look at figure s01: the red arrow (flux) points opposite to the direction in which the black concentration curve climbs.

• Higher voltage → deeper. More energy means the ion ploughs further before it runs out of energy and stops. Depth ($R_p$) is set by energy.
• Ion implantation gives precise dose, because dose is just counted from the beam charge ($Q=It/qA$) — you literally tally the ions. Diffusion's dose depends on temperature and time and is only moderate.

The exponent is zero (its largest possible value, since the exponent is never positive) exactly when $x=R_p$. There $e^0=1$, so $N=N_p$. The peak is buried at $x=R_p$, not at the surface. See figure s02.

Each ion carries charge $q$. Total charge delivered is $It$; number of ions is $It/q$; dose is that divided by area: $Q=\frac{It}{qA}=\frac{(10^{-5})(100)}{(1.6\times 10^{-19})(200)}=\frac{10^{-3}}{3.2\times 10^{-17}}\approx 3.13\times 10^{13}{\text{cm}}^{-2}.$

A Gaussian of height $N_p$ and width $\Delta R_p$ has area $N_p\Delta R_p\sqrt{2\pi }$. Setting that equal to the dose $Q$: $N_p=\frac{Q}{\sqrt{2\pi }\Delta R_p}=\frac{10^{14}}{(2.5066)(5\times 10^{-6})}\approx 7.98\times 10^{18}{\text{cm}}^{-3}.$ The whole dose is packed into a narrow spike, so the peak is huge.

$Dt=(10^{-14})(3600)=3.6\times 10^{-11}{\text{cm}}^2,\sqrt{Dt}=6.0\times 10^{-6}\text{cm}.$ $L=2\times 6.0\times 10^{-6}=1.2\times 10^{-5}\text{cm}=0.12\mu \text{m}.$ An hour at this temperature moves dopants only about a tenth of a micron — diffusion is a slow crawler.

Junction depth tracks the diffusion length: $x_j\propto \sqrt{Dt}$. Same $T$ means same $D$, so $x_j\propto \sqrt{t}$. $\frac{x_2}{x_1}=2=\sqrt{\frac{t_2}{t_1}}\Rightarrow \frac{t_2}{t_1}=4\Rightarrow t_2=4\times 30=120\text{min}.$ Four times the time for twice the depth. Depth is a slow function of time. See figure s03 — the $\sqrt{t}$ curve flattens.

$T_1=1273\text{K}$, $T_2=1373\text{K}$. The prefactor $D_0$ cancels in a ratio: $\frac{D_2}{D_1}=\exp \left[-\frac{E_a}{k}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\right].$ $\frac{1}{T_2}-\frac{1}{T_1}=\frac{1}{1373}-\frac{1}{1273}=-5.719\times 10^{-5}{\text{K}}^{-1}.$ $\frac{E_a}{k}=\frac{3.5}{8.617\times 10^{-5}}=4.062\times 10^4\text{K}.$ $\frac{D_2}{D_1}=\exp [-(4.062\times 10^4)(-5.719\times 10^{-5})]=e^{2.323}\approx \mathrm{10.2.}$ A 100 °C rise multiplies $D$ by roughly 10×. That is why temperature, not time, is the dominant control for diffusion.

$Dt=(2\times 10^{-14})(1800)=3.6\times 10^{-11}{\text{cm}}^2,\sqrt{Dt}=6.0\times 10^{-6}\text{cm}.$ $Q=\frac{2}{\sqrt{\pi }}(5\times 10^{20})(6.0\times 10^{-6})=(1.1284)(3.0\times 10^{15})\approx 3.39\times 10^{15}{\text{cm}}^{-2}.$ Unlike a fixed-dose drive-in, here dose grows with $\sqrt{t}$ because the surface keeps supplying atoms.

Work in cm. $\Delta R_p=5\times 10^{-6}\text{cm}$, so $\Delta R_p^2=2.5\times 10^{-11}{\text{cm}}^2$. $2Dt=2(4\times 10^{-15})(1200)=9.6\times 10^{-12}{\text{cm}}^2.$ $\Delta R_{\text{eff}}=\sqrt{2.5\times 10^{-11}+9.6\times 10^{-12}}=\sqrt{3.46\times 10^{-11}}=5.882\times 10^{-6}\text{cm}=0.0588\mu \text{m}.$ The anneal broadened the spike. New peak (dose is conserved — the anneal only spreads it): $N_p^{\prime }=\frac{Q}{\sqrt{2\pi }\Delta R_{\text{eff}}}=\frac{10^{14}}{(2.5066)(5.882\times 10^{-6})}\approx 6.78\times 10^{18}{\text{cm}}^{-3}.$ Broader profile → lower peak (same area). See figure s04: the annealed Gaussian is shorter and wider.

Set the Gaussian equal to $N_B$ and solve for $x$: $N_p\exp \left[-\frac{(x_j-R_p{)}^2}{2\Delta R_p^2}\right]=N_B.$ Take logs: $\frac{(x_j-R_p{)}^2}{2\Delta R_p^2}=\ln \frac{N_p}{N_B}=\ln \frac{5\times 10^{18}}{10^{16}}=\ln (500)=\mathrm{6.215.}$ $(x_j-R_p{)}^2=2(0.06{)}^2(6.215)=2(0.0036)(6.215)=0.04475\mu {\text{m}}^2.$ $x_j-R_p=\pm 0.2115\mu \text{m}.$ Two junctions form — one on each side of the buried peak. The deeper one: $x_j=0.30+0.2115=0.5115\mu \text{m}\approx 0.51\mu \text{m}.$ (The shallower junction sits at $0.30-0.21=0.09\mu \text{m}$.) Figure s05 shows the horizontal $N_B$ line cutting the Gaussian at two depths.

(a) Ion implantation. Only implantation gives precise dose (count the charge: $Q=It/qA$) and shallow, energy-set depth. Diffusion at ~1000 °C would spread too far and its dose is only moderate — wrong tool for a delicate threshold layer.

(b) Anneal is mandatory because implantation (i) smashes the crystal lattice (damage) and (ii) leaves most boron atoms in interstitial, electrically inactive sites. Heating repairs the lattice and moves boron onto substitutional lattice sites so it actually donates carriers. Without anneal, the dopants are physically present but do nothing electrically.

(c) $\Delta R_p=2\times 10^{-6}\text{cm}$, so $\Delta R_p^2=4\times 10^{-12}{\text{cm}}^2$. $2Dt=2(3\times 10^{-15})(600)=3.6\times 10^{-12}{\text{cm}}^2.$ $\Delta R_{\text{eff}}=\sqrt{4\times 10^{-12}+3.6\times 10^{-12}}=\sqrt{7.6\times 10^{-12}}=2.757\times 10^{-6}\text{cm}=0.0276\mu \text{m}.$ The anneal roughly doubled the width — a real design tension: you need the anneal to activate, but it also blurs your carefully-placed profile. Minimize $Dt$ (short, hot, "rapid thermal anneal") to activate while spreading as little as possible.

Same depth means same diffusion length: $\sqrt{D_A{t}_A}=\sqrt{D_B{t}_B}$, so $D_A{t}_A=D_B{t}_B$ (the product $Dt$ is what sets depth). $t_B=\frac{D_A{t}_A}{D_B}=\frac{D_A(60)}{5D_A}=12\text{min}.$ A 5× faster $D$ means 5× shorter time for the identical profile. Only the product $Dt$ matters — this is the single most important lever for planning any thermal step.