Question 7 of 7: Reading the Diffusivity–Temperature Chart; Phosphorus Diffusion into a Silicon Wafer

Question 7: Reading the Diffusivity–Temperature Chart; Phosphorus Diffusion into a Silicon Wafer (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. (a) Fig. 1: $\log_{10}D$ vs. $1/T\times1000$ for several oxide and semiconductor diffusion systems (straight lines, negative slope). (b) 1 mm-thick pure-Si wafer; background phosphorus concentration $=1$ P atom per $10^7$ Si atoms; surface concentration raised by a factor of $500$; $D_{P\ \text{in}\ Si}=10.5\times10^{-10}$ m$^2$/s; Si is diamond cubic, $8$ atoms/unit cell, $a_0=0.5431$ nm.

Find. (a)(i) Derive the SI units of $D$. (ii) Interpret the straight-line, negative-slope form of the data. (iii)–(v) Explain three specific comparisons read off the chart. (b) The steady-state diffusion flux, expressed as phosphorus atoms crossing one silicon unit cell per minute.

Approach

(a) follows directly from Fick's first law and the Arrhenius temperature dependence of $D$. (b) treats the wafer as a steady-state Fick's-first-law problem: convert the two given atomic-ratio concentrations into a true volumetric concentration using silicon's own atomic density (from its diamond-cubic unit cell), compute the areal flux, then convert to an atoms-per-unit-cell-per-minute rate using the unit-cell face area.

1. (i) Units of $D$. Fick's first law is $J=-D\,dc/dx$, with flux $J$ in atoms/(m$^2$·s), concentration $c$ in atoms/m$^3$, and position $x$ in m, so $dc/dx$ has units atoms/m$^4$: $$D=\frac{J}{dc/dx}\ \Rightarrow\ [D]=\frac{\text{atoms}/(\text{m}^2\,\text{s})}{\text{atoms}/\text{m}^4}=\text{m}^2/\text{s}$$ $$\boxed{[D]=\text{m}^2/\text{s}}$$
2. (ii) Nature of the straight, negative-sloped lines. Diffusivity follows the Arrhenius relation $D=D_0\exp(-Q/RT)$; taking natural logs, $\ln D=\ln D_0-(Q/R)(1/T)$, which is linear in $1/T$ with slope $-Q/R$. A straight line on a $\log D$ vs. $1/T$ plot therefore signals a single, essentially temperature-independent activation energy $Q$ over the plotted range; the negative slope simply means $D$ rises (steeply, exponentially) as temperature rises ($1/T$ falls) — the hallmark of a thermally activated, diffusion-controlled process. A steeper line corresponds to a larger $Q$.
3. (iii) Oxygen vs. magnesium diffusion in MgO. O$^{2-}$ is a much larger ion ($r\approx0.140$ nm) than the Mg$^{2+}$ cation ($r\approx0.072$ nm). A larger ion must force more lattice distortion to squeeze through the saddle point between adjacent sites, so anion diffusion carries a substantially higher activation energy (a steeper Arrhenius line) than the much smaller, more mobile cation. This is the general rule for oxide ceramics: cation diffusion on the (usually more open) cation sublattice is almost always faster than anion diffusion, so $D_{O\ \text{in}\ MgO}<D_{Mg\ \text{in}\ MgO}$ at any given temperature.
4. (iv) Mg self-diffusion vs. Ni impurity diffusion in MgO. Ni$^{2+}$ ($r\approx0.069$ nm) is a similarly small cation occupying the same cation sublattice as Mg$^{2+}$, using essentially the same cation-vacancy diffusion mechanism — hence the two Arrhenius lines lie close together, with comparable (not wildly different) activation energies. Mg diffusion here is a true self-diffusion process using the host's native vacancy population, whereas Ni is diffusing as a substitutional impurity; impurity diffusion typically requires a small additional activation energy for local lattice relaxation around the foreign ion (solute–vacancy binding effects), so it trails self-diffusion slightly — consistent with the Mg line sitting just above (higher $D$ than) the Ni line at the same temperature.
5. (v) Silicon vs. germanium diffusion data. On the chart, the Ge-in-Ge self-diffusion line sits above the Si-in-Si line (higher $D$, shallower slope/lower activation energy) at comparable $1/T$ — diffusion is faster, and requires less thermal activation, in germanium than in silicon at a given absolute temperature. This tracks germanium's lower melting point ($T_{m,Ge}\approx938^\circ$C vs. $T_{m,Si}\approx1414^\circ$C) and its weaker Ge–Ge covalent bond compared with Si–Si — a general trend across the diamond-cubic semiconductor family, where diffusion activation energy scales with bond strength/melting point. Significance: germanium-based processing (dopant diffusion, annealing) can be carried out at correspondingly lower temperatures than silicon processing to reach a comparable doping profile, but germanium devices are also more prone to unwanted dopant spreading during any later high-temperature step.
6. (b) Silicon's atomic (Si-atom) density. Diamond cubic, $8$ atoms/cell, $a_0=5.431\times10^{-10}$ m $=5.431\times10^{-8}$ cm, so $a_0^3=1.601\times10^{-22}$ cm$^3$: $$N_{Si}=\frac{8}{a_0^3}\approx4.99\times10^{22}\ \text{Si atoms/cm}^3$$ (This matches the well-known handbook silicon atomic density, a good check on the unit-cell data.)
7. (b) Phosphorus concentrations. Background (bulk) ratio $=1$ P atom per $10^7$ Si atoms; surface ratio raised by a factor of $500$: $$c_{bulk}=\frac{N_{Si}}{10^7}\approx4.994\times10^{15}=0.00499\times10^{18}\ \text{atoms/cm}^3,\qquad c_{surf}=500\,c_{bulk}\approx2.497\times10^{18}\ \text{atoms/cm}^3$$
8. (b) Steady-state flux. Fick's first law across the $1$ mm ($=0.1$ cm) wafer thickness, with $D=10.5\times10^{-10}\,\text{m}^2/\text{s}=10.5\times10^{-6}\,\text{cm}^2/\text{s}$: $$J=D\,\frac{c_{surf}-c_{bulk}}{\Delta x}=(10.5\times10^{-6})\frac{(2.497-0.00499)\times10^{18}}{0.1} =(10.5\times10^{-6})\frac{2.492\times10^{18}}{0.1}$$ $$J\approx2.62\times10^{14}\ \text{atoms/(cm}^2\text{s)}$$
9. (b) Atoms per unit cell per minute. Multiply by the unit-cell face area $a_0^2=(5.431\times10^{-8})^2=2.950\times10^{-15}$ cm$^2$ and convert seconds to minutes ($\times60$): $$\text{rate}=J\,a_0^2\times60=(2.62\times10^{14})(2.950\times10^{-15})(60)$$ $$\boxed{\text{rate}\approx46.3\ \text{P atoms per unit cell per minute}}$$

[Figure not reproduced: Fig. Q7(a) — the five lines of the printed Fig. 1 used in parts (iii)–(v), redrawn from the printed chart. The Mg and Ni lines overlap between $1/T\times1000=0.53$ and $0.59$, where Mg sits only $0.2$–$0.6$ of a decade above Ni — the "close to, but higher than" pairing o. See the official exam paper.]