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

Step-by-step 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:

`(B)/(B/A) = D_o \ exp \Big [(-\varepsilon a)/(k_BT) \Big]`

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

Thus;

`(B)/(A) = 1.63 * 10^8 exp \Big [ (-2.05)/(8.617 * 10^(_-5)* 1173)\Big] \\ \\ = 0.2535 \ \ \mu m/hr`

`B= 386 \ exp \Big [-(0.78)/(8.617 * 10^(-3) * 1173) \Big] \\ \\ = 0.1719 \ \mu m^2/hr`

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

`t_(ox) = (x)/(B/A)+ (x^2)/(B) - t_o (initial)`

`where; \\ \\ t_(ox) = 2 \ hrs;\\ \\ x = 0.5 \\ \\ B/A = 0.2535 \\ \\ B = 0.1719`

∴

`2= (0.5)/(0.2535)+ (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 = (d_o)/(0.2535) + (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 = (-b \pm √(b^2-4ac))/(2a)`

`d_o = (-(0.6781) \pm √((0.6781)^2-4(1)(-0.245)))/(2(10))`

`d_o = (-(0.6781) \pm √(0.45981961+0.98))/(20)`

`d_o = (-(0.6781) \pm √(1.43981961))/(20)`

`d_o = (-(0.6781) + √(1.43981961))/(20) \ OR \ (-(0.6781) - √(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