Semiconductor Fabrication

Question 1

(a) $k_1 = \text{CD}\cdot\text{NA}/\lambda = (18)(1.35)/193 = 0.1259$. (2) Single-patterning has a hard physical floor of $k_1 = 0.25$ (interference of two-beam imaging; below this no aerial-image modulation exists). Since $0.126 < 0.25$, single ArF exposure is impossible. (3)

(b) At $k_1=0.30$ the smallest resolvable half-pitch on the ArF tool: $\text{CD}_{\min} = k_1\lambda/\text{NA} = 0.30\times193/1.35 = 42.9$ nm → resolvable pitch $\approx 85.8$ nm. (2) SAQP divides pitch by 4, so printed feature $= 42.9/2 \approx 21.4$ nm half-pitch region... more precisely the mandrel pitch must be $4\times$ the final pitch: final pitch $= 2\times18 = 36$ nm, so mandrel pitch $= 4\times36 = 144$ nm → mandrel half-pitch 72 nm, well above the 42.9 nm resolution limit. ✓ reaches 18 nm. (2) SAQP uses 2 spacer deposition/etch cycles. General rule: $n$ pitch-splitting (spacer) stages give pitch-division factor $2^n$ (SADP: $2$, SAQP: $4$, SAOP: $8$). (2)

(c) EUV: $k_1 = 18\times0.33/13.5 = 0.44$. (2) High-NA EUV: $k_1 = 18\times0.55/13.5 = 0.733$. (1) Both exceed $0.25$ so single-exposure works; high-NA gives huge $k_1$ margin (more manufacturable / lower LER risk / larger process window). (1)

(d) ArF immersion: $\text{DOF} = 0.5\times193/1.35^2 = 96.5/1.8225 = 52.9$ nm. (1) High-NA EUV: $\text{DOF} = 0.5\times13.5/0.55^2 = 6.75/0.3025 = 22.3$ nm. (1) High-NA EUV has ~2.4× smaller DOF → far tighter focus/CMP-planarity control needed; thinner resist stacks and better wafer flatness required. (2)

(e)

def min_exposures(cd_target, lam, na, k1_floor):
    # single exposure resolves cd = k1*lam/na; smallest cd at k1_floor:
    cd_single = k1_floor * lam / na
    n = 1                      # start: single patterning, division factor 2**0=1
    div = 1
    while cd_single / div > cd_target:
        n += 1
        div *= 2               # each extra stage halves effective pitch
    return n - 1               # number of spacer stages (0=single, 1=SADP, 2=SAQP)

Logic: each pitch-splitting stage doubles the density (div *= 2); loop until the achievable feature meets target. (5) Check: cd_single $=0.30\cdot193/1.35=42.9$; need $\le18$: $42.9/1=42.9>18$, $/2=21.4>18$, $/4=10.7\le18$ → 2 stages (SAQP). ✓

Question 2

(a) At steady state the three fluxes are equal: Gas-phase/interface: $F_1 = h(C^* - C_o)$; diffusion through oxide: $F_2 = D\,(C_o - C_i)/x_o$; reaction at Si interface: $F_3 = k_s C_i$. (3) Setting $F_2=F_3$ and $F_1=F_2$, solve for $C_i$: $C_i = \frac{C^*}{1 + k_s/h + k_s x_o/D}.$ Growth: $\dfrac{dx_o}{dt} = \dfrac{k_s C_i}{N_1}$ where $N_1$ = oxidant molecules per unit oxide volume. Integrating yields $x_o^2 + A x_o = B(t+\tau),\quad A = 2D\left(\tfrac1{k_s}+\tfrac1h\right),\ B=\tfrac{2DC^*}{N_1}.$ (2)

• Linear regime (thin oxide, $x_o \ll A$): $x_o \approx (B/A)t$ — reaction/interface-rate limited. (1)
• Parabolic regime (thick oxide): $x_o \approx \sqrt{Bt}$ — diffusion-limited (oxidant must diffuse through growing oxide). (1)

(b) $A = B/(B/A) = 0.287/1.63 = 0.1761\ \mu\text{m}$. Solve $x_o^2 + 0.1761 x_o - 0.287(1.5)=0 \Rightarrow x_o^2 + 0.1761 x_o - 0.4305 = 0.$ $x_o = \dfrac{-0.1761+\sqrt{0.1761^2 + 4(0.4305)}}{2} = \dfrac{-0.1761+\sqrt{0.03101+1.722}}{2}=\dfrac{-0.1761+1.3242}{2}=0.574\ \mu\text{m}.$ (5)

(c) $t = (x_o^2 + A x_o)/B = (0.25 + 0.1761\times0.5)/0.287 = (0.25+0.08805)/0.287 = 0.33805/0.287 = 1.178\ \text{hr}$ (≈70.7 min). (4)

(d) From $x_o^2+Ax_o=Bt$, differentiate: $\dfrac{dx_o}{dt}=\dfrac{B}{2x_o+A}$. For thin oxide $x_o\ll A$, $dx_o/dt \approx B/A$ (constant, linear). At 40 nm = 0.04 μm: rate $= 0.287/(0.08+0.1761)=0.287/0.2561 = 1.12\ \mu\text{m/hr}$ — nearly the linear value 1.63 and roughly constant, so a small time error gives a small, proportional thickness error. In the parabolic regime rate $\propto 1/x_o$ diverges control sensitivity... actually thickness grows as $\sqrt t$ so $d x_o/dt$ falls — but linear regime gives predictable, controllable thin films. (4)

Question 3

(a) $AD_0 = 1.1\times0.15 = 0.165$. Poisson: $Y = e^{-0.165} = 0.8479$ (84.8%). (2) Murphy: $Y = \left(\dfrac{1-e^{-0.165}}{0.165}\right)^2 = \left(\dfrac{0.15206}{0.165}\right)^2 = (0.9216)^2 = 0.8494$ (84.9%). (2) Murphy assumes defects cluster / density varies across the wafer (some regions defect-free), so more dies survive than the uniform-Poisson prediction; the gap widens as $AD_0$ grows. (2)

(b) Dies per wafer (gross) $= 650/1.1 = 590.9 \approx 590$. Good dies $= 590\times0.8479 = 500.3 \approx 500$ good dies. (4)

(c)

import math
def good_dies(wafer_area, die_area, D0):
    gross = int(wafer_area // die_area)
    Y = math.exp(-die_area * D0)
    good = int(round(gross * Y))
    premium = int(round(good * 0.30))
    return {"gross": gross, "yield": Y, "good": good,
            "bins": {"premium": premium, "standard": good - premium}}

Binning: dies that pass functional test still vary in max clock/leakage due to within-wafer parametric variation (Vt, channel-length, oxide-thickness gradients). Faster/lower-leakage dies are binned as premium; the split (top 30%) models this parametric distribution. (6)

[
 {"claim":"ArF single-exposure k1 for 18nm is ~0.126 (below 0.25 floor)","code":"k1=18*1.35/193; result = (abs(k1-0.1259)<0.005) and (k1<0.25)"},
 {"claim":"EUV k1=0.44 and high-NA EUV k1=0.733","code":"a=18*0.33/13.5; b=18*0.55/13.5; result = (abs(a-0.44)<0.01) and (abs(b-0.733)<0.01)"},
 {"claim":"DOF ratio ArF-immersion vs high-NA EUV ~2.37","code":"d1=0.5*193/1.35**2; d2=0.5*13.5/0.55**2; result = abs(d1/d2-2.37)<0.1"},
 {"claim":"Deal-Grove wet oxide 1.5hr gives ~0.574 um","code":"A=0.287/1.63; import sympy as sp; x=sp.symbols('x',positive=True); s=sp.solve(sp.Eq(x**2+A*x,0.287*1.5),x); result = abs(float(s[0])-0.574)<0.005"},
 {"claim":"Time for 0.5um oxide ~1.178 hr","code":"A=0.287/1.63; t=(0.5**2+A*0.5)/0.287; result = abs(t-1.178)<0.01"},
 {"claim":"Poisson yield 84.8% and ~500 good dies","code":"import math; Y=math.exp(-1.1*0.15); g=int(650//1.1)*Y; result = (abs(Y-0.8479)<0.001) and (abs(g-500)<3)"}
]