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17-Phys-A6 Solid State Physics · May 2013

Question 7 of 7: Point Defects — Donor Impurities, Vacancy Concentration, and Diffusion

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Notes on this paper

National Exams — Phys-A6: Solid State Physics — May 2013 (3 hours; closed book; useful equations and physical constants annexed to the paper). Any FIVE of the SEVEN questions constitute a complete exam paper (first five as they appear in the answer book are marked); all seven are solved below as a complete study resource.

Reference texts: C. Kittel, Introduction to Solid State Physics, 8th ed. (Ch. 1–3, 4, 6, 8, 14, 18); N. W. Ashcroft & N. D. Mermin, Solid State Physics, 1st ed. (Ch. 2, 4–7, 22, 28, 31–32).

Question 7: Point Defects — Donor Impurities, Vacancy Concentration, and Diffusion (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.

QuantityValue
Na density $\rho$$0.971\ \text{g/cm}^3$
Na atomic weight23 amu
Na vacancy formation energy $E_v$1.05 eV
Temperature (part b)300 K
Si–Al diffusion constants (from table)$D_0=8.0\ \text{cm}^2/\text{s}$, $E=3.47\ \text{eV}$
Temperature (part c)1200 K

Find. (a) why As-doping is useful; (b) equilibrium vacancy concentration in Na at 300 K; (c) diffusion coefficient of Al in Si at 1200 K.

Approach. (a) reason from donor doping and carrier control. (b) use the paper's own vacancy-equilibrium relation, eq. (18), with $n\ll N$. (c) use the Arrhenius diffusion law, eq. (19), reading $D_0,E$ for the Si–Al pair straight off the given table.

  1. Part (a) — usefulness of As impurities in Si. Arsenic is a Group-V element (5 valence electrons) substituting for a Group-IV Si atom (4 valence electrons); 4 of As's electrons bond into the lattice and the 5th is only weakly bound, occupying a shallow donor level just below the conduction-band edge. At room temperature this electron is thermally ionized into the conduction band, controllably increasing the free-electron concentration (and hence conductivity) by many orders of magnitude relative to intrinsic Si, while leaving a fixed, immobile positive donor ion behind. This controlled, reproducible n-type doping — paired with p-type (acceptor) doped regions — is what creates the p-n junctions underlying diodes, transistors, and essentially all planar silicon devices.
  2. Part (b) — vacancy concentration in Na at 300 K. Number density of Na sites: $$N=\frac{\rho N_A}{M}=\frac{(0.971\ \text{g/cm}^3)(6.02217\times10^{23}\ \text{mol}^{-1})}{23\ \text{g/mol}}=2.542\times10^{22}\ \text{cm}^{-3}$$ For $n\ll N$, eq. (18), $\dfrac{n}{N-n}=\exp\!\left(\dfrac{-E_v}{k_BT}\right)$, reduces to $n\approx N\exp(-E_v/k_BT)$. With $E_v=1.05\ \text{eV}=1.682\times10^{-12}\ \text{erg}$, $k_BT=(1.38062\times10^{-16})(300)=4.142\times10^{-14}\ \text{erg}$ (so $E_v/k_BT=40.62$): $$n=(2.542\times10^{22}\ \text{cm}^{-3})\,e^{-40.62}$$ $\boxed{n_{vac}\approx5.83\times10^{4}\ \text{cm}^{-3}}$ — an extremely dilute defect population, since $E_v=1.05\ \text{eV}$ is large compared with $k_BT\approx0.026\ \text{eV}$ at room temperature; the equilibrium vacancy fraction $n/N\sim2\times10^{-18}$ confirms $n\ll N$, justifying the approximation.
  3. Part (c) — Al diffusion coefficient in Si at 1200 K. Reading $D_0=8.0\ \text{cm}^2/\text{s}$ and $E=3.47\ \text{eV}=5.560\times10^{-12}\ \text{erg}$ for the Si–Al pair straight off the given defect table, and applying the Arrhenius law eq. (19), $D=D_0\exp(-E/k_BT)$, with $k_BT=(1.38062\times10^{-16})(1200)=1.657\times10^{-13}\ \text{erg}$ (so $E/k_BT=33.56$): $$D=(8.0\ \text{cm}^2/\text{s})\,e^{-33.56}$$ $\boxed{D\approx2.13\times10^{-14}\ \text{cm}^2/\text{s}}$ — this is the rate (diffusion coefficient) at which Al atoms migrate through the Si lattice at 1200 K; it lands within an order of magnitude of measured Al-in-Si diffusivities near this temperature, another useful internal check.
QuantityResult
(a) Role of As in Sishallow donor → controlled n-type conductivity (basis of p–n devices)
(b) Na vacancy concentration, 300 K$5.83\times10^4\ \text{cm}^{-3}$
(c) $D$(Al in Si), 1200 K$2.13\times10^{-14}\ \text{cm}^2/\text{s}$
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