Silicon Processing, Powder Preparation and Sintering

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

1. Which sequence correctly describes the production of semiconductor-grade polysilicon by the Siemens route from metallurgical-grade silicon?

   1. React with HCl to give liquid SiHCl₃, purify it by fractional distillation, then reduce it with H₂ on heated silicon rods
   2. Reduce quartz with carbon at 1800 °C and then pull the product directly as a crystal without any further chemical purification
   3. Dissolve metallurgical silicon in molten aluminium, cool it slowly, and collect the silicon platelets that separate out
   4. Oxidise metallurgical silicon in steam to SiO₂ and then strip the oxide chemically to leave ultra-pure silicon behind
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   Answer: A — React with HCl to give liquid SiHCl₃, purify it by fractional distillation, then reduce it with H₂ on heated silicon rods

   The Siemens route turns silicon into a volatile liquid compound, SiHCl₃ (made with HCl near 300 °C), distils it repeatedly to remove the chlorides of Fe, B and P, and deposits silicon by hydrogen reduction on rods at about 1100 °C. Carbothermic reduction alone gives only 98 to 99 % Si, which is the tempting but insufficient route. Oxidation followed by stripping would remove silicon itself, not the impurities.

2. A silicon bar contains an impurity with segregation coefficient k = 0.05 (C_s/C_l at the interface). A molten zone is passed along the bar several times in one direction. After many passes, the impurity is most concentrated:

   1. at the end of the bar where the zone starts, because the first solid to freeze carries the most solute
   2. uniformly along the bar, because a zone only mixes the solute and cannot move it in any direction
   3. at the end of the bar where the zone finishes, because the liquid retains the solute that the freezing solid rejects
   4. in the middle of the bar, because the zone is at its longest there and holds the greatest quantity of solute
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   Answer: C — at the end of the bar where the zone finishes, because the liquid retains the solute that the freezing solid rejects

   For k < 1 the freezing solid takes only a fraction k of the liquid concentration, so the liquid zone becomes richer as it travels and deposits its load at the end where it freezes last. The first solid is the purest, since it takes k C₀. For k > 1 the result would be reversed.

3. Why does float-zone (FZ) silicon contain much less oxygen than Czochralski (CZ) silicon?

   1. The melt is argon-cooled so quickly that oxygen has no time to dissolve into it during growth
   2. The molten zone touches no crucible, whereas in CZ the fused-silica crucible dissolves slightly into the melt
   3. Oxygen has a segregation coefficient far below 1, so every zone pass sweeps nearly all of it to the far end of the rod
   4. The radio-frequency coil removes oxygen from the melt as gaseous SiO by heating it above its boiling point
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   Answer: B — The molten zone touches no crucible, whereas in CZ the fused-silica crucible dissolves slightly into the melt

   In CZ growth the silica crucible is the main oxygen source. In FZ the zone is held by surface tension, so there is no crucible and no oxygen pick-up. FZ is not a rapid-cooling process, and oxygen in silicon has k of about 1.25, so zone passes do not sweep it out, and the coil heats only a short zone near the melting point, not above the boiling point of silicon.

4. Silicon is oxidised so that the oxide is already thick and growth follows the parabolic Deal–Grove law, x ≈ √(Bt). The oxide is 0.30 µm thick after 3 h. Taking τ = 0 and no change in temperature, the thickness after a total of 12 h is:

   1. 0.45 µm, because the thickness grows with the square root of the time ratio multiplied by 1.5
   2. 0.90 µm, because the thickness is proportional to time and the time has increased by a factor of three
   3. 1.20 µm, because the thickness is proportional to the time and the time has increased by a factor of four, as in the linear regime
   4. 0.60 µm, because the thickness varies as the square root of time and the time has increased by a factor of four
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   Answer: D — 0.60 µm, because the thickness varies as the square root of time and the time has increased by a factor of four

   In the parabolic regime x² = Bt, so x₂/x₁ = √(t₂/t₁) = √4 = 2, giving 0.60 µm. The linear-in-time options, 0.90 and 1.20 µm, describe the thin-oxide linear regime, which does not apply here. Quadrupling the time doubles, not quadruples, the thickness.

5. Which statement correctly compares doping by ion implantation with doping by thermal diffusion?

   1. Implantation gives a profile that always peaks at the wafer surface, because ions are stopped immediately on entry
   2. Implantation leaves the lattice undamaged, so no anneal is required before the dopant becomes electrically active
   3. Implantation allows the dose to be set by beam current and time and the depth by ion energy, at low wafer temperature
   4. Diffusion places the dopant at a fixed depth below the surface and cannot spread sideways under a mask edge at all
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   Answer: C — Implantation allows the dose to be set by beam current and time and the depth by ion energy, at low wafer temperature

   Dose is Q = It/(qA) and depth follows the ion energy (projected range). The wafer need not be hot, so a thin resist or oxide can mask it. The profile peaks at R_p below the surface, and the damaged lattice must be annealed. Thermal diffusion is isotropic and does spread laterally.

6. In photolithography, a positive photoresist is one in which:

   1. the exposed regions crosslink and remain after development, so the pattern on the wafer is the reverse of the mask, as in a lift-off process
   2. the exposed regions become soluble in the developer and are removed, so the pattern on the wafer is the same as the mask
   3. the resist is deposited only where the mask is opaque, so no developer step is needed before the etch begins
   4. the resist is exposed by an ion beam instead of light, which gives a positive charge to the exposed region
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   Answer: B — the exposed regions become soluble in the developer and are removed, so the pattern on the wafer is the same as the mask

   A positive resist is made soluble by exposure, so developing removes the exposed areas and the pattern matches the mask. A negative resist behaves in the opposite way, crosslinking where exposed. The word positive has no connection with charge or with the use of ions.

7. What is the thermodynamic driving force for the sintering of a loose compact of metal powder?

   1. The reduction of the total surface free energy as solid–vapour surface is replaced by lower-energy grain boundary
   2. The melting of the particles at their contacts, which then freezes into a continuous solid on cooling
   3. The release of the elastic strain energy that was stored in the compact during die pressing at room temperature
   4. The latent heat of a polymorphic phase transformation that every powder undergoes at the sintering temperature
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   Answer: A — The reduction of the total surface free energy as solid–vapour surface is replaced by lower-energy grain boundary

   Powder has a very large surface area, and replacing surface by grain boundary lowers the free energy; atomic diffusion at 0.7 to 0.9 of the absolute melting temperature supplies the transport. Sintering is a solid-state process, so melting at contacts is not the mechanism. Stored strain relief is a separate effect, and not every powder has a polymorphic transformation.

8. Two identical compacts of the same powder are sintered. In one, the neck between particles grows and strength rises, but the density and the dimensions do not change at all. Which mechanism is mainly operating?

   1. Grain-boundary diffusion, because it carries atoms from the contact plane into the neck and so pulls the particle centres together
   2. Plastic flow by dislocation motion, because it removes material from the contact region and closes the pores quickly
   3. Lattice diffusion from the grain boundary, because the atoms removed from the boundary are deposited on the neck surface
   4. Surface diffusion, because it moves atoms from the particle surface to the neck without bringing the centres closer
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   Answer: D — Surface diffusion, because it moves atoms from the particle surface to the neck without bringing the centres closer

   Matter that is taken from the free surface and placed on the neck leaves the particle centres where they were, so the neck grows without shrinkage. Surface diffusion is one such non-densifying path (evaporation–condensation is another). Grain-boundary diffusion and lattice diffusion from the boundary both take matter from the contact region and so densify, and plastic flow also densifies.

9. The sintering time needed to reach a given neck-to-particle radius ratio by lattice diffusion scales with particle radius a as a³ (Herring’s scaling law, for particles of the same shape and material at the same temperature). A powder of 20 µm particles needs 16 h. The time for a powder of 10 µm particles is:

   1. 8 h, because halving the particle size halves the time
   2. 4 h, because the time scales as the square of the particle size
   3. 2 h, because the time scales as the cube of the particle size
   4. 32 h, because finer powder has more surface and so needs longer to sinter
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   Answer: C — 2 h, because the time scales as the cube of the particle size

   With t ∝ a³, halving a gives t = 16 × (1/2)³ = 2 h. The linear and square options use the wrong exponent, and the longer-time option reverses the trend: finer powder has a larger driving force per unit mass and shorter diffusion distances, so it sinters faster.

10. Compared with water-atomised metal powder, gas-atomised powder typically has:

    1. an irregular shape with higher oxygen content, which gives better green strength in die compaction
    2. a flake shape produced by repeated cold welding, which is the characteristic result of ball milling
    3. a spherical shape with lower oxygen content, because the melt is broken up by an inert-gas jet
    4. a sponge-like porous structure, because the metal oxide is reduced by hydrogen in the solid state
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    Answer: C — a spherical shape with lower oxygen content, because the melt is broken up by an inert-gas jet

    An inert-gas jet breaks the liquid stream into droplets that become spherical before freezing and that meet little oxygen. The irregular, oxide-rich shape belongs to water atomisation, the flake shape to milling of ductile metals and the sponge structure to solid-state oxide reduction.

11. Which of the following statements about segregation and crystal growth of silicon are correct? Take k = C_s/C_l.

    1. For an impurity with k < 1, the first part of a normally frozen bar to solidify is purer than the starting material
    2. In a Czochralski growth with a dopant of k < 1, the dopant concentration in the crystal rises from seed end to tail end
    3. Zone refining is most effective for impurities whose k is very close to 1, since they pass easily across the interface
    4. A neck is grown at the start of a Czochralski pull so that dislocations produced by thermal shock grow out of the crystal
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    Answer: A — For an impurity with k < 1, the first part of a normally frozen bar to solidify is purer than the starting material; B — In a Czochralski growth with a dopant of k < 1, the dopant concentration in the crystal rises from seed end to tail end; D — A neck is grown at the start of a Czochralski pull so that dislocations produced by thermal shock grow out of the crystal

    The first solid takes k C₀ < C₀, so it is purer; with k < 1 the melt is enriched as the crystal grows, so the crystal dopant level rises along its length (C_s = kC₀(1 − f)^(k−1)); and the thin neck lets dislocations escape through its sides. Zone refining relies on rejection of solute, so it fails when k is near 1, which is the reason the statement about k close to 1 is wrong.

12. Which of the following statements about thermal oxidation of silicon are correct?

    1. New oxide forms at the Si–SiO₂ interface, because the oxidant diffuses through the oxide already grown
    2. In the Deal–Grove model the linear regime applies to thick oxide and the parabolic regime to thin oxide
    3. An oxide of thickness x consumes about 0.44 x of silicon, because the oxide occupies roughly 2.27 times the volume of the silicon replaced
    4. At the same temperature, wet oxidation in steam is generally slower than dry oxidation in O₂, since water diffuses less readily in SiO₂
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    Answer: A — New oxide forms at the Si–SiO₂ interface, because the oxidant diffuses through the oxide already grown; C — An oxide of thickness x consumes about 0.44 x of silicon, because the oxide occupies roughly 2.27 times the volume of the silicon replaced

    Growth occurs at the inner interface, and the volume ratio gives 1/2.27 = 0.44 for the silicon consumed. The regimes are the other way round: thin oxide is reaction-limited (linear) and thick oxide diffusion-limited (parabolic). Wet oxidation is faster, not slower, because H₂O is much more soluble in SiO₂ than O₂ is.

13. Which of the following statements about sintering of metal powder compacts are correct?

    1. Raising the sintering temperature speeds every diffusion mechanism, because the diffusivity follows an Arrhenius law
    2. In the final stage the pores are isolated, and grain growth that sweeps past them can leave them trapped inside grains
    3. Sintering is performed above the melting temperature of the metal, so that the powder fuses into a liquid mass
    4. Surface-diffusion neck growth causes the compact to shrink faster than grain-boundary diffusion does
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    Answer: A — Raising the sintering temperature speeds every diffusion mechanism, because the diffusivity follows an Arrhenius law; B — In the final stage the pores are isolated, and grain growth that sweeps past them can leave them trapped inside grains

    Diffusion is thermally activated, so a hotter compact sinters faster; and grain growth can detach grain boundaries from pores, leaving the pores isolated in the grains and hard to remove. Conventional sintering stays below the melting temperature. Surface diffusion does not shrink the compact at all, so it cannot shrink it faster than grain-boundary diffusion.

14. A bar of silicon of uniform impurity concentration C₀ is zone refined in a single pass with a molten zone of length l. For the first pass, the solid concentration at distance x from the start is C(x) = C₀[1 − (1 − k) exp(−kx/l)]. The impurity has k = 0.2. At x = 4l the ratio C/C₀ is ____ (answer to two decimal places).

    Numerical answer — type the value.

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    Answer: 0.64

    kx/l = 0.2 × 4 = 0.8. exp(−0.8) = 0.4493. C/C₀ = 1 − 0.8 × 0.4493 = 1 − 0.3595 = 0.6405, which is 0.64. Check: at x = 0 the formula gives k = 0.2 and at large x it tends to 1, so 0.64 lies between, as it should.

15. Silicon is oxidised at a fixed temperature for which the Deal–Grove equation x² + Ax = B(t + τ) has A = 0.10 µm, B = 0.020 µm²/h and τ = 0. The oxide is grown from bare silicon for 5 h. Solve the quadratic exactly (do not use the parabolic approximation). The oxide thickness is ____ nm (answer as a whole number).

    Numerical answer — type the value.

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    Answer: 270

    x² + 0.10x − 0.10 = 0 (since B t = 0.020 × 5 = 0.10). x = [−0.10 + √(0.01 + 0.40)]/2 = (−0.10 + 0.6403)/2 = 0.2702 µm = 270 nm. The parabolic approximation √(Bt) = 0.316 µm = 316 nm would overestimate, because the oxide is not yet far beyond A/2.

16. Boron is predeposited into n-type silicon from a source that holds the surface concentration constant, so that C(x, t) = C_s erfc[x/(2√(Dt))]. The substrate concentration is C_b = 0.05 C_s, and erfc(z) = 0.05 at z = 1.39 (use this value). Take D = 4 × 10⁻¹⁴ cm²/s and t = 3600 s. The junction depth is ____ µm (answer to two decimal places).

    Numerical answer — type the value.

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    Answer: 0.33

    The junction is where C = C_b, i.e. erfc(z) = 0.05, so x_j = 2z√(Dt). Dt = 4 × 10⁻¹⁴ × 3600 = 1.44 × 10⁻¹⁰ cm², so √(Dt) = 1.2 × 10⁻⁵ cm = 0.12 µm. x_j = 2 × 1.39 × 0.12 = 0.3336 µm, which is 0.33 µm. Unit slip to avoid: 1 cm = 10⁴ µm.

17. Two copper spheres sinter by lattice diffusion, for which the neck radius x follows x⁵ ∝ t at fixed temperature and particle size. The neck radius is 3 µm after 2 h. Neglect any change of the geometry other than this scaling. The time at which the neck radius reaches 6 µm is ____ h (answer as a whole number).

    Numerical answer — type the value.

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    Answer: 64

    x⁵ ∝ t gives t₂/t₁ = (x₂/x₁)⁵ = 2⁵ = 32. So t₂ = 2 × 32 = 64 h. Doubling the neck radius needs a 32-fold longer time, which shows how strongly neck growth slows. Assuming x ∝ t would wrongly give 4 h.

18. A powder compact has a green relative density of 55 % of the theoretical density. On sintering it shrinks isotropically, with a linear shrinkage ΔL/L₀ = 8 %, and no mass is lost. The relative density after sintering, in per cent, is ____ (answer to one decimal place).

    Numerical answer — type the value.

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    Answer: 70.6

    Mass is conserved and volume falls by the cube of the linear factor: V/V₀ = (1 − 0.08)³ = 0.7787. Density therefore rises by 1/0.7787 = 1.2842, so the relative density is 55 × 1.2842 = 70.6 %. Using 8 % as a volume shrinkage would give 59.8 %, which is wrong.