Semiconductor Doping — Diffusion
Question
- Describe the two basic diffusion mechanisms for dopant atoms in a semiconductor lattice. Which has the higher activation energy and why?
Answer
Vacancy (substitutional) diffusion: a lattice atom acquires enough thermal energy to leave its site, creating a vacancy. A neighbouring dopant atom then migrates into that vacancy. Typical for dopant atoms larger than the host. Activation energy is 2–3 eV because energy is needed both to create the vacancy and to move the dopant atom into it.
Interstitial diffusion: the dopant atom moves through the spaces between lattice sites without occupying a lattice position. It hops from one interstitial site to another by overcoming potential barriers. Activation energy is 0.5–1.5 eV — lower than vacancy diffusion because no vacancy needs to be created first.
Question
- Write down Fick’s first and second laws of diffusion. Derive the second law from the first by considering the flux into and out of a thin slab of thickness dx.
Answer
Fick’s first law: F = −D(∂C/∂x). The flux of atoms is proportional to the concentration gradient, directed from high to low concentration.
Consider a slab between x and x+dx. Flux in: F₁ = −D(∂C/∂x). Flux out: F₂ = F₁ + dF, where by Taylor expansion F₂ = −D(∂C/∂x + (∂²C/∂x²)dx).
The rate of change of concentration in the slab: (∂C/∂t)dx = F₁ − F₂ = D(∂²C/∂x²)dx.
Fick’s second law (the diffusion equation): ∂C/∂t = D(∂²C/∂x²).
This assumes D is constant (independent of concentration).
Question
- For the constant surface concentration (predeposition) case, state the boundary conditions, the solution to the diffusion equation, and the expression for total dose Q(t).
Answer
Boundary conditions: at t=0, C(x,0) = 0 (no dopant initially). At x=0, C(0,t) = Cₛ (surface held at solid solubility limit, ≈5×10²⁰ cm⁻³ for B, P, As in Si). At x=∞, C(∞,t) = 0.
Solution: C(x,t) = Cₛ · erfc(x / 2√(Dt))
where erfc(z) = 1 − erf(z) and erf(z) = (2/√π) ∫₀ᶻ exp(−y²) dy.
Total dose: Q(t) = (2/√π) · Cₛ · √(Dt) ≈ 1.13 · Cₛ · √(Dt)
√(Dt) is called the diffusion length.
Question
- For the constant total dopant (drive-in) case, state the boundary conditions, the solution, and how the surface concentration varies with time.
Answer
Boundary conditions: at x=0, ∂C/∂x|ₓ₌₀ = 0 (no flux through the surface — the surface is sealed, e.g. by an oxide cap). At x=∞, C(∞,t) = 0. The initial dopant distribution is approximated as a delta function at x=0 with total dose Q.
Solution (Gaussian): C(x,t) = (Q / √(πDt)) · exp(−x² / 4Dt)
Surface concentration: Cₛ(t) = Q / √(πDt)
The surface concentration decreases as 1/√t while the dopant penetrates deeper. Total dose Q is conserved throughout.
Question
- A two-step doping process uses a predeposition followed by a drive-in. Explain why this sequence is used and how you would determine the required predeposition time and drive-in conditions.
Answer
The predeposition step (constant surface concentration, erfc profile) introduces a controlled dose Q of dopant near the surface. The surface concentration is fixed at the solid solubility limit — you can’t choose it independently. By controlling time and temperature, you control Q = 1.13·Cₛ·√(D₁t₁).
The drive-in step (constant total dopant, Gaussian profile) then redistributes this dose deeper into the wafer with no additional dopant added. The surface is typically sealed with an oxide. You now have independent control of the final surface concentration Cₛ = Q/√(πD₂t₂) and junction depth.
This two-step process decouples dose control from profile shaping — predeposition sets Q, drive-in sets the depth and surface concentration.
Question
- The diffusion coefficient follows D = D₀·exp(−Eₐ/kT). For boron in silicon, D₀ = 10.5 cm²/s and Eₐ = 3.69 eV. Calculate D at 1000°C and at 1100°C. Comment on the sensitivity of D to temperature.
Answer
T₁ = 1000°C = 1273 K. kT₁ = 8.617×10⁻⁵ × 1273 = 0.1097 eV. D₁ = 10.5 · exp(−3.69/0.1097) = 10.5 · exp(−33.64) = 10.5 × 2.44×10⁻¹⁵ = 2.56×10⁻¹⁴ cm²/s.
T₂ = 1100°C = 1373 K. kT₂ = 8.617×10⁻⁵ × 1373 = 0.1183 eV. D₂ = 10.5 · exp(−3.69/0.1183) = 10.5 · exp(−31.19) = 10.5 × 2.80×10⁻¹⁴ = 2.94×10⁻¹³ cm²/s.
A 100°C increase raised D by over an order of magnitude (~11.5×). This extreme sensitivity is why furnace temperature must be precisely controlled during diffusion — even a few degrees of error significantly affects the doping profile.
Question
- What is dopant segregation at the Si/SiO₂ interface during thermal oxidation? Explain the effect for boron (m < 1) and phosphorus (m > 1).
Answer
The segregation coefficient m = [impurity concentration in Si] / [impurity concentration in SiO₂] describes how dopant redistributes between Si and its growing oxide.
For boron (m < 1): the oxide preferentially takes up boron from the Si. Boron near the interface is depleted from the Si side and accumulates in the oxide. This reduces the surface doping concentration in the Si — problematic for p-type regions that need to maintain their doping level.
For phosphorus (m > 1): the oxide rejects phosphorus. Phosphorus is pushed back into the Si, causing a pile-up at the Si surface near the interface. Surface concentration increases.
This is an important consideration when thermal oxidation is performed on already-doped structures.
Question
- Under what conditions does the simple constant-D diffusion model break down? What form does the diffusion equation take in this case?
Answer
The model breaks down when the doping concentration Cₛ exceeds the intrinsic carrier concentration nᵢ at the diffusion temperature. At this point, the number of vacancies (and hence D) becomes concentration-dependent — higher doping creates more charged vacancies through Fermi-level effects.
The diffusion equation becomes: ∂C/∂t = ∂/∂x [D(C) · ∂C/∂x]
where D is now a function of C rather than a constant. This nonlinear PDE has no closed-form solution and must be solved numerically. The resulting profiles deviate from the ideal erfc or Gaussian shapes — typically showing a “box-like” profile at high concentrations with a sharp drop-off at the diffusion front.
Semiconductor Doping — Ion Implantation
Question
- Describe the ion implantation process and the key components of an ion implanter. What are the advantages of implantation over furnace diffusion?
Dopant atoms (B, P, As) are ionised, accelerated to 10–500 keV (up to >1 MeV), mass-selected by a bending magnet + slit (only the desired isotope/charge state passes through), and scanned across the wafer surface.
Advantages over diffusion: (1) precise dose control — the ion current is measured directly, giving accurate dose (Q = It/eA); (2) low processing temperature — implantation occurs near room temperature, avoiding unwanted diffusion of existing dopant profiles; (3) depth control via energy — peak concentration can be placed below the surface, not just at it; (4) capable of very shallow or very deep profiles by adjusting energy.
Disadvantage: high capital cost (though high throughput compensates on a per-wafer basis).
Question
- The implanted dopant distribution is approximately Gaussian. Write down the expression for C(x) in terms of dose Q, projected range Rₚ, and straggle ΔRₚ. How do Rₚ and ΔRₚ depend on ion energy and mass?
C(x) = (Q / (√(2π)·ΔRₚ)) · exp(−½·((x − Rₚ)/ΔRₚ)²)
where Rₚ is the projected range (depth of the concentration peak) and ΔRₚ is the standard deviation (straggle).
Both Rₚ and ΔRₚ increase approximately linearly with ion energy. For a given energy, lighter ions (lower atomic number) have larger Rₚ and ΔRₚ because they lose energy more slowly per collision (less nuclear stopping). For B, P, and As in Si at 10–1000 keV, Rₚ ranges from ~10 to ~100 nm.
Question
- What is channelling in ion implantation? How does it affect the dopant profile, and describe three methods to prevent it.
Answer
Channelling occurs when incident ions align with a major crystallographic direction of the single-crystal substrate. The ions are steered between rows of atoms through open “channels”, experiencing less nuclear stopping and penetrating much deeper than predicted by the Gaussian model. The resulting profile has a long tail extending well beyond the expected range.
Prevention methods: (1) Tilt the wafer 7–10° off the major crystal axis so ions don’t align with any channel. (2) Deposit an amorphous layer (e.g. thin SiO₂) on the surface to randomise the ion trajectories before they enter the crystal. (3) Pre-amorphise the surface by implanting an electrically inert species (e.g. nitrogen or silicon) to destroy the crystal order in the near-surface region.
Question
- Why must implanted wafers be annealed after implantation? What is the trade-off, and how is rapid thermal annealing used to address it?
Answer
Implantation causes extensive lattice damage — each ion can displace hundreds of lattice atoms through nuclear collisions. In the damaged region, carrier mobility and lifetime are severely degraded. Furthermore, the implanted dopant atoms are mostly interstitial and not electrically active until they occupy substitutional lattice sites.
Annealing at ~1000°C repairs the crystal damage and activates the dopants. However, this high temperature also causes diffusion, broadening the carefully controlled implant profile.
Rapid thermal annealing (RTA) solves this by heating the wafer extremely quickly (using high-intensity lamps, electron beams, or pulsed lasers) for very short durations. The high temperature activates dopants and repairs damage, but the brief duration minimises diffusion. Examples: pulsed laser (~10⁻¹⁰ s, melting/recrystallisation), electron beam (~10⁻⁶ s, solid-phase epitaxy), or lamp-based (~10⁻⁴ s, fast heating/cooling).
Question
- List four important applications of ion implantation in semiconductor device fabrication.
Answer
(1) Threshold voltage adjustment in enhancement-mode MOSFETs — a shallow, precisely controlled dose shifts Vth to the desired value. (2) Source/drain formation in submicron MOSFETs — shallow, heavily doped p-n junctions with abrupt profiles. (3) Well formation in CMOS — deep implants (e.g. n-well for PMOS) followed by drive-in diffusion. (4) Buried SiO₂ layers for silicon-on-insulator (SOI) technology — high-dose oxygen implantation (SIMOX process).
28. An implanter operates at 100 keV with a beam current of 10 µA, uniformly scanning a 150 mm diameter wafer. Calculate the time required to achieve a dose of 10¹⁵ ions/cm². Assume singly charged ions.
Wafer area: A = π(7.5)² = 176.7 cm².
Total charge needed: dose × area × charge per ion = 10¹⁵ × 176.7 × 1.6×10⁻¹⁹ C = 176.7 × 1.6×10⁻⁴ = 0.02827 C.
Time = total charge / beam current = 0.02827 / 10×10⁻⁶ = 2827 s ≈ 47 minutes.
This illustrates why high-dose implants (>10¹⁵ cm⁻²) are time-consuming — high-current implanters are needed for production throughput.