Level 3 — ProductionSemiconductor Fabrication

Semiconductor Fabrication

45 minutes60 marksprintable — key stays hidden on paper

Chapter: 4.3 Semiconductor Fabrication Level: 3 — From-scratch derivations, code-from-memory, explain-out-loud Time limit: 45 minutes Total marks: 60

Instructions: Show all derivations. For "explain-out-loud" prompts, write as if teaching a peer — reasoning must be explicit. Code may be pseudocode or Python.


Q1. Resolution & the lithography wavelength ladder (12 marks)

(a) State the Rayleigh criterion for the minimum resolvable half-pitch in projection lithography, defining every symbol. (3)

(b) A DUV scanner uses an ArF source (λ=193nm\lambda = 193\,\text{nm}), numerical aperture NA=1.35NA = 1.35 (immersion), and k1=0.28k_1 = 0.28. Compute the minimum half-pitch. (3)

(c) EUV uses λ=13.5nm\lambda = 13.5\,\text{nm} with NA=0.33NA = 0.33 and k1=0.4k_1 = 0.4. Compute its half-pitch and compare with (b), explaining out loud why EUV lets you skip multi-patterning that DUV would need for the same node. (4)

(d) Derive the depth of focus (DOF) expression DOF=k2λ/NA2DOF = k_2\,\lambda/NA^2 trend and explain why higher NANA is a double-edged sword. (2)


Q2. Thermal oxidation — Deal–Grove from scratch (12 marks)

(a) Derive the Deal–Grove relation xo2+Axo=B(t+τ)x_o^2 + A x_o = B(t + \tau) from the three fluxes (gas-phase, diffusion through oxide, interface reaction) in steady state. Define AA, BB, τ\tau. (7)

(b) Show that for long times oxide growth is parabolic (xoBtx_o \approx \sqrt{Bt}) and for short times it is linear (xo(B/A)(t+τ)x_o \approx (B/A)(t+\tau)). (3)

(c) Explain out loud why growing 100 nm of oxide consumes silicon, and compute how much Si thickness is consumed (given oxide-to-silicon thickness ratio 0.44\approx 0.44). (2)


Q3. Yield, defect density, and binning (10 marks)

(a) Derive/justify the Poisson yield model Y=eD0AY = e^{-D_0 A}, where D0D_0 is defect density and AA is die area. (2)

(b) A wafer process has D0=0.15defects/cm2D_0 = 0.15\,\text{defects/cm}^2. Compute the yield for a die of area A=2cm2A = 2\,\text{cm}^2 and for A=4cm2A = 4\,\text{cm}^2. (3)

(c) Explain out loud why doubling die area more than doubles the yield loss, and how chiplets/binning respond to this. (3)

(d) A 300 mm wafer (radius 150 mm) holds usable dies of 2cm22\,\text{cm}^2; ignoring edge loss, estimate gross die count and good die count using (b). (2)


Q4. Damascene / metallization — explain and sequence (8 marks)

(a) Write the ordered step sequence of the copper dual-damascene process (trench + via), and state why Cu cannot be patterned by conventional plasma etch (motivating damascene). (5)

(b) Explain the role of the barrier/liner (e.g. Ta/TaN) and of CMP in this flow. (3)


Q5. Transistor scaling: FinFET → GAA (10 marks)

(a) Explain out loud the electrostatic problem in planar MOSFETs at short channel that motivated FinFETs. (3)

(b) Compare FinFET vs Gate-All-Around nanosheet on: gate control geometry, effective width tunability, and drive current per footprint. (4)

(c) For a FinFET, derive the effective channel width for nn fins of height HfinH_{fin} and width WfinW_{fin} (double-gate approximation), and compute WeffW_{eff} for n=3n=3, Hfin=50nmH_{fin}=50\,\text{nm}, Wfin=7nmW_{fin}=7\,\text{nm}. (3)


Q6. Code-from-memory: ALD cycle counter (8 marks)

Write a function (Python/pseudocode) ald_thickness(gpc_nm, target_nm) that returns the number of ALD cycles needed to reach a target thickness given growth-per-cycle (GPC), rounding up. Then extend it to ald_time(gpc_nm, target_nm, t_cycle_s) returning total process time. Explain out loud why ALD is self-limiting and gives this linear thickness-vs-cycles behaviour. (8)


Answer keyMark scheme & solutions

Q1 (12)

(a) Rayleigh: half-pitch=k1λNA\text{half-pitch} = k_1 \dfrac{\lambda}{NA}. (1)

  • λ\lambda = exposure wavelength; NANA = numerical aperture of projection lens (=nsinθ= n\sin\theta); k1k_1 = process-dependent constant (theoretical floor 0.25 for single exposure). (2)

(b) HP=0.28×1931.35=0.28×142.96=40.0nmHP = 0.28 \times \dfrac{193}{1.35} = 0.28 \times 142.96 = 40.0\,\text{nm}. (3) (accept ~40 nm)

(c) HP=0.4×13.50.33=0.4×40.91=16.4nmHP = 0.4 \times \dfrac{13.5}{0.33} = 0.4 \times 40.91 = 16.4\,\text{nm}. (2) Explain: EUV's ~14× shorter wavelength gives a far smaller λ/NA\lambda/NA even at higher k1k_1 and modest NANA, so a single EUV exposure resolves features that DUV can only reach by splitting the pattern across multiple masks/etch steps (multi-patterning) — cutting mask count, cost, and overlay error. (2)

(d) DOF=k2λ/NA2DOF = k_2 \lambda / NA^2: resolution scales as 1/NA1/NA but DOF scales as 1/NA21/NA^2. (1) So raising NANA sharpens resolution but shrinks the usable focus window quadratically, tightening flatness/CMP and wafer-stage tolerances. (1)

Q2 (12)

(a) Three fluxes in steady state (F1=F2=F3=FF_1=F_2=F_3=F): (1 setup)

  • Gas→surface: F1=h(CC0)F_1 = h(C^* - C_0)
  • Diffusion through oxide: F2=DC0CixoF_2 = D\dfrac{C_0 - C_i}{x_o}
  • Interface reaction: F3=ksCiF_3 = k_s C_i (2)

Set equal and solve: from F=ksCiF=k_s C_i and F=D(C0Ci)/xoF=D(C_0-C_i)/x_o and F=h(CC0)F=h(C^*-C_0), eliminate C0,CiC_0,C_i: F=C1ks+1h+xoDF = \frac{C^*}{\frac{1}{k_s} + \frac{1}{h} + \frac{x_o}{D}} (2) Growth rate dxodt=F/N\dfrac{dx_o}{dt} = F/N (NN = oxidant atoms per unit oxide volume). Integrate: xo2+Axo=B(t+τ)x_o^2 + A x_o = B(t+\tau) (1) with A=2D(1ks+1h)A = 2D\left(\tfrac1{k_s}+\tfrac1h\right), B=2DCNB = \dfrac{2DC^*}{N}, τ=xi2+AxiB\tau = \dfrac{x_i^2 + A x_i}{B} (accounts for initial oxide). (1)

(b) Long tt: xo2Axoxo2BtxoBtx_o^2 \gg A x_o \Rightarrow x_o^2 \approx Bt \Rightarrow x_o\approx\sqrt{Bt} (parabolic, diffusion-limited). (1.5) Short tt: Axoxo2xoBA(t+τ)A x_o \gg x_o^2 \Rightarrow x_o \approx \dfrac{B}{A}(t+\tau) (linear, reaction-limited). B/AB/A = linear rate constant. (1.5)

(c) Oxygen must reach the Si interface and react, converting Si into SiO₂, so the interface moves into the wafer. (1) Si consumed =0.44×100=44nm= 0.44 \times 100 = 44\,\text{nm}. (1)

Q3 (10)

(a) Defects fall randomly and independently; number on a die of area AA is Poisson with mean λ=D0A\lambda = D_0 A. Yield = P(0 killer defects) =eλ=eD0A= e^{-\lambda} = e^{-D_0 A}. (2)

(b) Y(2)=e0.15×2=e0.3=0.741Y(2) = e^{-0.15\times2} = e^{-0.3} = 0.741 (74.1%). (1.5) Y(4)=e0.15×4=e0.6=0.549Y(4) = e^{-0.15\times4} = e^{-0.6} = 0.549 (54.9%). (1.5)

(c) Yield decays exponentially with area, so loss (1Y)(1-Y) grows faster than linearly for large AA (compounding). Doubling area squares the survival probability: Y(2A)=Y(A)2Y(2A)=Y(A)^2, so more than double the loss fraction. (2) Chiplets split a big SoC into small high-yield dies; binning sells partially-defective dies at lower spec (disable a core/lower clock) rather than scrapping. (1)

(d) Wafer area =π(15cm)2=706.86cm2= \pi (15\,\text{cm})^2 = 706.86\,\text{cm}^2. Gross dies =706.86/2353= 706.86/2 \approx 353. (1) Good dies 353×0.741262\approx 353 \times 0.741 \approx 262. (1)

Q4 (8)

(a) Dual-damascene sequence: (3, deduct 0.5 per misordering)

  1. Deposit ILD (dielectric, e.g. low-k oxide).
  2. Pattern & etch via holes (lithography + dry etch).
  3. Pattern & etch trenches (metal-line pattern).
  4. Deposit barrier/liner (TaN/Ta).
  5. Deposit Cu seed (PVD) then electroplate Cu to overfill trench+via.
  6. Anneal.
  7. CMP to remove overburden, leaving inlaid Cu lines/vias. Cu has no volatile etch byproduct at reasonable temperature (halides non-volatile), so subtractive plasma etch of Cu is impractical → deposit-then-polish (damascene) instead. (2)

(b) Barrier/liner (Ta/TaN) blocks Cu diffusion into the dielectric (Cu is a fast diffuser and poisons Si/oxide) and improves adhesion. (1.5) CMP removes the overfilled Cu/barrier overburden and planarizes each layer so the next lithography level has a flat, in-focus surface. (1.5)

Q5 (10)

(a) In short-channel planar MOSFETs the drain field increasingly controls the channel (DIBL, punch-through, poor subthreshold slope); single-side gate loses electrostatic control → high leakage. (3)

(b) (4, 1.33 each)

  • Geometry: FinFET gate wraps 3 sides of a fin; GAA wraps all 4 sides of stacked nanosheets → better electrostatic control, lower leakage.
  • Effective width: FinFET width is quantized (add whole fins); GAA tunes width by nanosheet width → finer design granularity.
  • Drive current/footprint: GAA stacks multiple sheets vertically → higher WeffW_{eff}/drive per footprint at advanced nodes.

(c) Double-gate fin: each fin contributes conduction on two sidewalls plus top (approx): Weffn(2Hfin+Wfin)W_{eff}\approx n(2H_{fin}+W_{fin}). (1) Weff=3(2×50+7)=3×107=321nmW_{eff} = 3(2\times50 + 7) = 3\times107 = 321\,\text{nm}. (2) (Accept Weff=n2Hfin=300W_{eff}=n\cdot2H_{fin}=300 nm for pure double-gate.)

Q6 (8)

import math
def ald_thickness(gpc_nm, target_nm):
    return math.ceil(target_nm / gpc_nm)          # cycles, rounded up
 
def ald_time(gpc_nm, target_nm, t_cycle_s):
    return ald_thickness(gpc_nm, target_nm) * t_cycle_s

(4 for correct code incl. ceil, 1 for time extension)

Explain: each ALD cycle = self-limiting half-reactions (precursor dose saturates surface, then co-reactant). Once all surface sites react, no further growth that pulse — so exactly one monolayer-fraction (GPC) deposits per cycle regardless of dose. Hence thickness =GPC×cycles= GPC \times \text{cycles} (linear), giving atomic-level, conformal control. (3)

[
  {"claim":"DUV half-pitch ~40nm", "code":"hp=0.28*193/1.35; result = abs(hp-40.0)<0.5"},
  {"claim":"EUV half-pitch ~16.4nm", "code":"hp=0.4*13.5/0.33; result = abs(hp-16.36)<0.2"},
  {"claim":"Yields 74.1% and 54.9%", "code":"from sympy import exp,N; y2=N(exp(-0.3)); y4=N(exp(-0.6)); result = abs(y2-0.7408)<0.001 and abs(y4-0.5488)<0.001"},
  {"claim":"Wafer gross die ~353 and good ~262", "code":"from sympy import pi,exp,N; area=N(pi*15**2); gross=int(area/2); good=int(gross*N(exp(-0.3))); result = gross==353 and 260<=good<=263"},
  {"claim":"FinFET Weff=321nm", "code":"n=3;H=50;W=7; result = n*(2*H+W)==321"}
]