Deal Grove Example Question

Question

1. A ⟨100⟩ silicon wafer is oxidised in dry O₂ at 1000°C for 2 hours. Calculate the oxide thickness grown. Use Table 1 and the Deal-Grove model.

Step 1: Extract A and B at 1000°C dry, ⟨100⟩

From Table 1, dry O₂, ⟨100⟩ (X_i = 25 nm initial oxide assumed):

• B/A: D₀ = 3.71×10⁶ μm/hr, E_A = 2.00 eV
• B: D₀ = 772 μm²/hr, E_A = 1.23 eV

At T = 1273 K, k_BT = 0.1097 eV:

$B/A=3.71\times 10^6\times \exp !\left(\frac{-2.00}{0.1097}\right)=3.71\times 10^6\times e^{-18.23}$

$e^{-18.23}=1.20\times 10^{-8}$

$B/A=3.71\times 10^6\times 1.20\times 10^{-8}=0.04452\mu \text{m/hr}$

$B=772\times \exp !\left(\frac{-1.23}{0.1097}\right)=772\times e^{-11.21}=772\times 1.353\times 10^{-5}=0.01045\mu {\text{m}}^2/\text{hr}$

$A=\frac{B}{B/A}=\frac{0.01045}{0.04452}=0.2348\mu \text{m}$

Step 2: Account for initial oxide X_i = 25 nm = 0.025 μm

$\tau =\frac{X_i^2+A{X}_i}{B}=\frac{(0.025{)}^2+0.2348\times 0.025}{0.01045}=\frac{6.25\times 10^{-4}+5.87\times 10^{-3}}{0.01045}=\frac{6.495\times 10^{-3}}{0.01045}=0.6216\text{hr}$

Step 3: Solve for X at t = 2 hr

$X^2+AX=B(t+\tau )=0.01045\times (2+0.6216)=0.01045\times 2.6216=0.02740\mu {\text{m}}^2$

Quadratic in X:

$X^2+0.2348X-0.02740=0$

$X=\frac{-0.2348+\sqrt{(0.2348{)}^2+4\times 0.02740}}{2}=\frac{-0.2348+\sqrt{0.05513+0.10960}}{2}$

$=\frac{-0.2348+\sqrt{0.16473}}{2}=\frac{-0.2348+0.4059}{2}=\frac{0.1711}{2}$

$\boxed{X\approx 0.0855\mu \text{m}=85.5\text{nm}}$

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2. A ⟨111⟩ silicon wafer already has 150 nm of SiO₂ from a previous process. It is placed in a wet O₂ furnace at 1200°C. How long to grow a further 400 nm?

Step 1: A and B at 1200°C wet, ⟨111⟩

At T = 1473 K, k_BT = (8.617×10⁻⁵)(1473) = 0.1269 eV

$B/A=1.63\times 10^8\times \exp !\left(\frac{-2.05}{0.1269}\right)=1.63\times 10^8\times e^{-16.15}$

$e^{-16.15}=9.61\times 10^{-8}$

$B/A=1.63\times 10^8\times 9.61\times 10^{-8}=15.66\mu \text{m/hr}$

$B=386\times \exp !\left(\frac{-0.78}{0.1269}\right)=386\times e^{-6.147}=386\times 2.133\times 10^{-3}=0.8233\mu {\text{m}}^2/\text{hr}$

$A=\frac{0.8233}{15.66}=0.05257\mu \text{m}$

Step 2: τ for X_i = 0.150 μm

$\tau =\frac{(0.150{)}^2+0.05257\times 0.150}{0.8233}=\frac{0.02250+7.886\times 10^{-3}}{0.8233}=\frac{0.03039}{0.8233}=0.03692\text{hr}$

Step 3: t + τ for X_f = 0.550 μm

$t+\tau =\frac{(0.550{)}^2+0.05257\times 0.550}{0.8233}=\frac{0.3025+0.02891}{0.8233}=\frac{0.33141}{0.8233}=0.4026\text{hr}$

$t=0.4026-0.03692=0.3657\text{hr}$

$\boxed{t\approx 21.9\text{minutes}}$

High temperature wet oxidation is fast — makes sense physically.

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3. A process requires exactly 500 nm of gate oxide on ⟨100⟩ silicon. The engineer proposes a two-step process: first dry O₂ at 1100°C, then wet O₂ at 1000°C. The dry step runs for 3 hours. How long is the wet step?

Step 1: Oxide grown in dry step — A, B at 1100°C dry ⟨100⟩

T = 1373 K, k_BT = (8.617×10⁻⁵)(1373) = 0.1183 eV

$B/A=3.71\times 10^6\times \exp !\left(\frac{-2.00}{0.1183}\right)=3.71\times 10^6\times e^{-16.91}=3.71\times 10^6\times 4.50\times 10^{-8}=0.1670\mu \text{m/hr}$

$B=772\times \exp !\left(\frac{-1.23}{0.1183}\right)=772\times e^{-10.40}=772\times 3.035\times 10^{-5}=0.02343\mu {\text{m}}^2/\text{hr}$

$A=\frac{0.02343}{0.1670}=0.1403\mu \text{m}$

Initial oxide X_i = 25 nm = 0.025 μm:

$\tau =\frac{(0.025{)}^2+0.1403\times 0.025}{0.02343}=\frac{6.25\times 10^{-4}+3.508\times 10^{-3}}{0.02343}=\frac{4.133\times 10^{-3}}{0.02343}=0.1764\text{hr}$

Solve for X after t = 3 hr:

$X^2+0.1403X=0.02343\times (3+0.1764)=0.02343\times 3.1764=0.07442$

$X^2+0.1403X-0.07442=0$

$X=\frac{-0.1403+\sqrt{(0.1403{)}^2+4\times 0.07442}}{2}=\frac{-0.1403+\sqrt{0.01968+0.29768}}{2}=\frac{-0.1403+\sqrt{0.31736}}{2}$

$=\frac{-0.1403+0.5633}{2}=\frac{0.4230}{2}=0.2115\mu \text{m}$

Oxide after dry step: 211.5 nm

Step 2: Wet step at 1000°C ⟨100⟩ — A, B already calculated in Q6(c) above

A = 0.424 μm, B = 0.3142 μm²/hr

τ for X_i = 0.2115 μm:

$\tau =\frac{(0.2115{)}^2+0.424\times 0.2115}{0.3142}=\frac{0.04473+0.08968}{0.3142}=\frac{0.13441}{0.3142}=0.4278\text{hr}$

t + τ for X_f = 0.500 μm:

$t+\tau =\frac{(0.500{)}^2+0.424\times 0.500}{0.3142}=\frac{0.250+0.212}{0.3142}=\frac{0.462}{0.3142}=1.471\text{hr}$

$t=1.471-0.4278=1.043\text{hr}$

$\boxed{t\approx 62.6\text{minutes}}$

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4. Two identical ⟨100⟩ wafers start bare. Wafer 1 is oxidised in wet O₂ at 900°C for 10 hours. Wafer 2 is oxidised in dry O₂ at 1100°C for 10 hours. Which has more oxide, and by how much?

Wafer 1: wet O₂ at 900°C ⟨100⟩

T = 1173 K, k_BT = 0.1011 eV

$B/A=9.70\times 10^7\times e^{-2.05/0.1011}=9.70\times 10^7\times e^{-20.28}=9.70\times 10^7\times 1.553\times 10^{-9}=0.1506\mu \text{m/hr}$

$B=386\times e^{-0.78/0.1011}=386\times e^{-7.715}=386\times 4.46\times 10^{-4}=0.1722\mu {\text{m}}^2/\text{hr}$

$A=\frac{0.1722}{0.1506}=1.1434\mu \text{m}$

X_i = 0 nm (bare), so τ = 0.

$X^2+1.1434X=0.1722\times 10=1.722$

$X^2+1.1434X-1.722=0$

$X=\frac{-1.1434+\sqrt{(1.1434{)}^2+4\times 1.722}}{2}=\frac{-1.1434+\sqrt{1.3074+6.888}}{2}=\frac{-1.1434+\sqrt{8.195}}{2}$

$=\frac{-1.1434+2.8627}{2}=\frac{1.7193}{2}=0.8597\mu \text{m}$

Wafer 2: dry O₂ at 1100°C — from Q3 above

A = 0.1403 μm, B = 0.02343 μm²/hr, τ = 0.1764 hr (for X_i = 25 nm)

$X^2+0.1403X=0.02343\times (10+0.1764)=0.02343\times 10.1764=0.2384$

$X^2+0.1403X-0.2384=0$

$X=\frac{-0.1403+\sqrt{(0.1403{)}^2+4\times 0.2384}}{2}=\frac{-0.1403+\sqrt{0.01968+0.9536}}{2}=\frac{-0.1403+\sqrt{0.9733}}{2}$

$=\frac{-0.1403+0.9866}{2}=\frac{0.8463}{2}=0.4231\mu \text{m}$

Comparison:

Wafer	Conditions	Oxide thickness
1	Wet 900°C, 10 hr	860 nm
2	Dry 1100°C, 10 hr	423 nm

Wafer 1 has ~437 nm more oxide — despite being at 300°C lower temperature. This illustrates that wet oxidation is dramatically faster than dry due to the much higher diffusivity and solubility of H₂O in SiO₂ compared to O₂, which dominates over the temperature disadvantage.