Act like a helpful tutor and exlain step-by-step solution with explanation and final answer:5. (10pts) Aluminium is to be diffused into a silicon wafer via the common two-step process. The
pre-deposi

Question

Act like a helpful tutor and exlain step-by-step solution with explanation and final answer:5. (10pts) Aluminium is to be diffused into a silicon wafer via the common two-step process. The
pre-deposition diffusion is conducted at 975°C for 1.25 h with a surface concentration of 4 x 102®
Al atoms/m?3. The drive-in diffusion is conducted at 1050°C for 0.95 h. The bulk concentration of
Alin Si is 1.75 x 10° atoms/m?. Please calculate: 1) the value of @,, which is total aluminium
concentration introduced by pre-deposition process; 2) the value of x;, which is the junction
depth after the drive-in diffusion process.

For this system, values of Qsand Do are 3.41 eV/atom and 1.38 x 10~* m?/s, respectively.

VIPSolutions · Accepted Answer

Let's solve the problem step-by-step.

---

## **Step 1: Pre-deposition Diffusion (Constant Surface Concentration)**

Given:
- Surface concentration, $ C_s = 4 \times 10^{26} $ atoms/m³
- Pre-deposition temperature, $ T_1 = 975^\circ\text{C} = 1248\,\text{K} $
- Time, $ t_1 = 1.25\,\text{h} = 450\,\text{s} $
- Diffusion parameters: $ Q_d = 3.41\,\text{eV/atom} $, $ D_ = 1.38 \times 10^{-4}\,\text{m}^2/\text{s} $

First, convert $ Q_d $ to Joules (1 eV = $ 1.602 \times 10^{-19} $ J):
$$
Q_d = 3.41\,\text{eV} \times 1.602 \times 10^{-19}\,\text{J/eV} = 5.463 \times 10^{-19}\,\text{J}
$$

Calculate diffusion coefficient at 975°C:
$$
D_1 = D_ \exp\left(-\frac{Q_d}{kT_1}\right)
$$
Where Boltzmann constant $ k = 1.38 \times 10^{-23} $ J/K.

$$
D_1 = 1.38 \times 10^{-4} \exp\left( - \frac{5.463 \times 10^{-19}}{1.38 \times 10^{-23} \times 1248} \right)
$$

Calculate the exponent:
$$
\frac{5.463 \times 10^{-19}}{1.38 \times 10^{-23} \times 1248} = \frac{5.463 \times 10^{-19}}{1.722 \times 10^{-20}} \approx 31.73
$$
So,
$$
D_1 = 1.38 \times 10^{-4} \exp(-31.73) \approx 1.38 \times 10^{-4} \times (1.7 \times 10^{-14}) \approx 2.35 \times 10^{-18}\,\text{m}^2/\text{s}
$$

---

## **Step 2: Total Dose $ Q_ $ Calculation (Pre-deposition Total Al)**

For constant source diffusion:
$$
Q_ = 2 C_s \sqrt{\frac{D_1 t_1}{\pi}}
$$
Plug in the numbers:
$$
Q_ = 2 \times 4 \times 10^{26} \sqrt{ \frac{2.35 \times 10^{-18} \times 450}{\pi} }
$$

Calculate inside the sqrt:
$$
2.35 \times 10^{-18} \times 450 = 1.0575 \times 10^{-14}
$$
$$
\frac{1.0575 \times 10^{-14}}{\pi} = 3.37 \times 10^{-15}
$$
$$
\sqrt{3.37 \times 10^{-15}} = 5.81 \times 10^{-8}
$$

Now,
$$
Q_ = 8 \times 10^{26} \times 5.81 \times 10^{-8} = 4.65 \times 10^{19}\,\text{atoms/m}^2
$$

---

## **Step 3: Drive-In Diffusion**

Given:
- $ T_2 = 105^\circ\text{C} = 1323\,\text{K} $
- $ t_2 = .95\,\text{h} = 342\,\text{s} $

Calculate $ D_2 $:
$$
D_2 = D_ \exp\left(-\frac{Q_d}{kT_2}\right)
$$
$$
\frac{5.463 \times 10^{-19}}{1.38 \times 10^{-23} \times 1323} = \frac{5.463 \times 10^{-19}}{1.82574 \times 10^{-20}} \approx 29.93
$$
$$
D_2 = 1.38 \times 10^{-4} \exp(-29.93) \approx 1.38 \times 10^{-4} \times 1.02 \times 10^{-13} = 1.41 \times 10^{-17}\,\text{m}^2/\text{s}
$$

---

## **Step 4: Junction Depth $ x_j $ Calculation**

For drive-in diffusion (Gaussian profile):
$$
C(x, t_2) = \frac{Q_}{\sqrt{\pi D_2 t_2}} \exp\left(-\frac{x^2}{4 D_2 t_2}\right)
$$
Set $ C(x_j, t_2) = C_B = 1.75 \times 10^{19} $ atoms/m³:

$$
C_B = \frac{Q_}{\sqrt{\pi D_2 t_2}} \exp\left(-\frac{x_j^2}{4 D_2 t_2}\right)
$$

Solve for $ x_j $:

$$
\exp\left(-\frac{x_j^2}{4 D_2 t_2}\right) = \frac{C_B \sqrt{\pi D_2 t_2}}{Q_}
$$
$$
-\frac{x_j^2}{4 D_2 t_2} = \ln\left( \frac{C_B \sqrt{\pi D_2 t_2}}{Q_} \right)
$$
$$
x_j^2 = -4 D_2 t_2 \ln\left( \frac{C_B \sqrt{\pi D_2 t_2}}{Q_} \right)
$$
$$
x_j = \sqrt{ -4 D_2 t_2 \ln\left( \frac{C_B \sqrt{\pi D_2 t_2}}{Q_} \right) }
$$

Calculate $ \sqrt{\pi D_2 t_2} $:

$$
D_2 t_2 = 1.41 \times 10^{-17} \times 342 = 4.822 \times 10^{-14}
$$
$$
\sqrt{ \pi \times 4.822 \times 10^{-14} } = \sqrt{ 1.515 \times 10^{-13} } = 1.23 \times 10^{-6.5} \approx 1.23 \times 10^{-6}
$$

Now plug into the ratio:

$$
\frac{C_B \sqrt{\pi D_2 t_2}}{Q_} = \frac{1.75 \times 10^{19} \times 1.23 \times 10^{-6}}{4.65 \times 10^{19}}
= \frac{2.15 \times 10^{13}}{4.65 \times 10^{19}} = 4.62 \times 10^{-7}
$$

Now the log:

$$
\ln(4.62 \times 10^{-7}) = \ln(4.62) + \ln(10^{-7}) = 1.53 - 16.12 = -14.59
$$

So,

$$
x_j = \sqrt{ -4 \times 4.822 \times 10^{-14} \times (-14.59) }
= \sqrt{ 4 \times 4.822 \times 10^{-14} \times 14.59 }
= \sqrt{ 2.813 \times 10^{-12} }
= 1.68 \times 10^{-6} \text{ m } = 1.68\,\mu\text{m}
$$

---

# **Final Answers**

### **1. Total Aluminum Dose ($ Q_ $)**
$$
Q_ = \boxed{4.65 \times 10^{19} \text{ atoms/m}^2}
$$

### **2. Junction Depth ($ x_j $)**
$$
x_j = \boxed{1.68\,\mu\text{m}}
$$

---

If you need any clarification or more steps explained, let me know!

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