Question

g An oxidized silicon (111) wafer has an initial field oxide thickness of d0. Wet oxidation at 950 °C is then used to grow a thin film gate of 500 nm in 50 minutes. What is the original field oxide thickness d0 (in nm)?

Answer 1

Answer:

Explanation:

From the information given:

oxidation of oxidized solution takes place at 950° C

For wet oxidation:

The linear and parabolic coefficient can be computed as:

$\dfrac{B}{B/A} = D_o \ exp \Big [\dfrac{-\varepsilon a}{k_BT} \Big]$

Using $D_o$ and $E_a$ values obtained from the graph:

Thus;

$\dfrac{B}{A} = 1.63 \times 10^8 exp \Big [ \dfrac{-2.05}{8.617 \times 10^{_-5}\times 1173}\Big] \\ \\ = 0.2535 \ \ \mu m/hr$

$B= 386 \ exp \Big [-\dfrac{0.78}{8.617 \times 10^{-3} \times 1173} \Big] \\ \\ = 0.1719 \ \mu m^2/hr$

So, the initial time required to grow oxidation is expressed as:

$t_{ox} = \dfrac{x}{B/A}+ \dfrac{x^2}{B} - t_o (initial)$

$where; \\ \\ t_{ox} = 2 \ hrs;\\ \\ x = 0.5 \\ \\ B/A = 0.2535 \\ \\ B = 0.1719$

∴

$2= \dfrac{0.5}{0.2535}+ \dfrac{0.5^2}{0.1719} - t_o (initial)$

$2 = 3.4267 - t_o (initial) \\ \\ t_o(initial) = 3.4267 - 2 \\ \\ t_o(initial) = 1.4267 \ hr$

NOW;

$1.4267 = \dfrac{d_o}{0.2535} + \dfrac{d_o^2}{0.1719} \\ \\ 1.4267 = 3.9448 \ d_o + 5.8173 \ d_o^2 \\ \\ d_o^2 + 0.6781 \ d_o = 0.2453$

$d_o = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

$d_o = \dfrac{-(0.6781) \pm \sqrt{(0.6781)^2-4(1)(-0.245)}}{2(10)}$

$d_o = \dfrac{-(0.6781) \pm \sqrt{0.45981961+0.98}}{20}$

$d_o = \dfrac{-(0.6781) \pm \sqrt{1.43981961}}{20}$

$d_o = \dfrac{-(0.6781) + \sqrt{1.43981961}}{20} \ OR \ \dfrac{-(0.6781) - \sqrt{1.43981961}}{20}$

$d_o =0.02609 \ OR \ -0.0939$

Thus; since we will consider the positive sign, the initial thickness $d_o$ is ;

≅ 0.261 μm