Question 5 of 7: Reading the Diffusivity Chart; Phosphorus Diffusion Flux in Silicon

Question 5: Reading the Diffusivity Chart; Phosphorus Diffusion Flux in Silicon (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) The plotted $\log_{10}D$ vs. $1/T\times1000$ chart above. (b) 1 mm-thick Si wafer; initial P concentration $=1$ atom per $10^7$ Si atoms, to be raised by a factor of 500; $D_P=10.5\times10^{-10}$ m$^2$/s; Si is diamond cubic, 8 atoms/cell, $a_0=5.431$ Å.

Find. (a)(i)–(v) as stated. (b) Diffusion flux, expressed as P atoms passing through one unit cell of Si per minute.

Approach

Part (a) is entirely chart interpretation using the Arrhenius form of the diffusion coefficient, $D=D_0\exp(-Q_d/RT)$; part (b) is a steady-state Fick's-first-law flux calculation where the given "atoms per $10^7$/atoms per unit cell" concentrations must first be converted to a true volumetric concentration (atoms/m$^3$) before Fick's law can be applied, and the resulting SI flux then converted into the specific unit cell/minute units the question asks for.

1. (a)(i) Units of diffusivity. Fick's first law is $J=-D\,dC/dx$, where $J$ is a flux (amount per unit area per unit time, e.g. atoms/(m$^2$·s)) and $dC/dx$ is a concentration gradient (amount per unit volume per unit length, e.g. atoms/m$^4$). Solving for $D$, $$D=\frac{J}{dC/dx}=\frac{\text{atoms}/(\text{m}^2\cdot\text{s})}{\text{atoms}/\text{m}^4} =\boxed{\text{m}^2/\text{s}}.$$
2. (a)(ii) Straight lines with negative slope. Diffusivity follows the Arrhenius form $D=D_0e^{-Q_d/RT}$, so $$\ln D=\ln D_0-\frac{Q_d}{R}\cdot\frac{1}{T}.$$ Plotting $\ln D$ (equivalently $\log_{10}D$) against $1/T$ therefore gives a straight line with slope $-Q_d/R$ — negative because the activation energy $Q_d$ is always positive, i.e. diffusion is a thermally-activated process that slows (D falls) as $T$ falls ($1/T$ rises). The slope's magnitude is a direct, reproducible measure of the diffusing species' activation energy in that host lattice.
3. (a)(iii) Why D(O in MgO) < D(Mg in MgO) at a given T. Both species migrate via a cation/anion vacancy mechanism on their own sublattice, but the O$^{2-}$ anion ($r\approx1.40$ Å) is substantially larger than the Mg$^{2+}$ cation ($r\approx0.72$ Å). Moving the larger ion through the lattice requires more local lattice distortion to open a large-enough passage, which raises the migration activation energy — hence a lower $D$ at any given $T$ (visible on the chart as the O-in-MgO line sitting well below and to the right of Mg-in-MgO).
4. (a)(iv) Why D(Mg in MgO) is close to, but higher than, D(Ni in MgO). Ni$^{2+}$ substitutes for Mg$^{2+}$ on the same cation sublattice, with a similar ionic radius ($r_{\text{Ni}^{2+}}\approx0.69$ Å vs. $r_{\text{Mg}^{2+}}\approx0.72$ Å) and identical $+2$ charge, so both diffuse by the same cation-vacancy mechanism with similar activation energies — explaining why the two lines sit close together and roughly parallel. Mg is nonetheless a self-diffusing species in its own host lattice, with no size-mismatch strain energy to overcome, while Ni is a substitutional foreign ion that introduces a small local strain/binding penalty; this slightly raises Ni's effective activation energy, making Ni in MgO diffuse a little more slowly than Mg in MgO at the same temperature.
5. (a)(v) Comparing Si-in-Si and Ge-in-Ge. Reading two points off each line (approximate, to chart precision):

6. Interpreting the comparison. Both self-diffusion lines have a similar visual character (straight, negative slope), but Ge's line sits well to the right of Si's and its chart-read activation energy ($\approx$284 kJ/mol) is much lower than Si's ($\approx$495 kJ/mol) — in good agreement with literature values ($Q_d(\text{Si})\approx460$–500, $Q_d(\text{Ge})\approx290$–330 kJ/mol), which is a useful check on the chart reading.
   Check

   Chart-read activation energies carry the plotted line's own reading precision (± a fraction of a decade); they are reported here to 3 significant figures for traceability, not as exact literature values.
   Significance: Ge's weaker, longer covalent bonds (larger atomic radius, lower melting point 938°C vs. Si's 1414°C) make vacancy formation and migration easier, so Ge reaches a technologically useful diffusivity at a much lower processing temperature than Si does. This is exactly why early Ge-based devices could be processed at lower temperatures, while modern Si technology — which relies on Si's higher $Q_d$ for tighter, more thermally stable dopant-diffusion control — requires substantially higher processing temperatures for comparable diffusion depths.
7. (b) Convert given concentrations to atoms/m$^3$. Si is diamond cubic with 8 atoms/unit cell and $a_0=5.431\times10^{-10}$ m, so the Si atomic number density is $$n_{\text{Si}}=\frac{8}{a_0^3}=\frac{8}{(5.431\times10^{-10})^3}=5.00\times10^{28}\ \text{Si atoms/m}^3.$$ The initial P concentration (1 per $10^7$ Si atoms) is $C_1=n_{\text{Si}}/10^7$; raising it by a factor of 500 gives $C_2=500\,C_1$, so $$\Delta C=C_2-C_1=499\,C_1=499\times\frac{n_{\text{Si}}}{10^7}=2.495\times10^{24}\ \text{P atoms/m}^3.$$
8. Steady-state flux across the wafer. Treating the 1 mm wafer thickness as the diffusion distance in a steady-state, linear concentration profile, $$J=D_P\frac{\Delta C}{L}=10.5\times10^{-10}\times\frac{2.495\times10^{24}}{1.0\times10^{-3}} =\boxed{2.62\times10^{18}\ \text{atoms/(m}^2\cdot\text{s)}}.$$
9. Convert to atoms per unit cell per minute. The requested answer is not the raw SI flux but the number of P atoms crossing one unit-cell cross-section per minute — two separate unit conversions beyond $J$ itself: multiply by the unit cell's cross-sectional area $a_0^2$ (atoms/s through one cell), then by 60 (atoms/min): $$a_0^2=(5.431\times10^{-10})^2=2.949\times10^{-19}\ \text{m}^2,$$ $$\text{rate}=J\times a_0^2\times60=2.62\times10^{18}\times2.949\times10^{-19}\times60 =\boxed{46.3\ \text{P atoms per unit cell per minute}}.$$