6. Technology of realization of the current extractor device using the forces of Casimir in a vacuum

It can be seen in Figures 25 and 26 that if PZT is used as the piezoelectric material, then the peak current output goes from 2:10�<sup>8</sup> A to 6:10�<sup>8</sup> A, when the width of the piezoelectric layer bp changes from 50 μm to 150 μm (for a length of the piezoelectric layer lp ¼ 50μm, a thickness ap ¼ 10μm, a length of the Casimir electrode ls ¼ 200μm, starting interface z0 ¼ 200 A° and a Ratio of FCO=FCA ¼ 10).

As a result, it will be necessary, using micro-technology techniques on S.O.I silicon, to machine devices with on a high thickness while maintaining exceedingly small spaces between structures Casimir. The microelectronics laboratory of the ESIEE has acquired a great experience in the plasma etching of deep sub-micron structures by etching remarkably parallel layers of silicon of 100 mm separated by intervals of 0.8 mm and well parallel [6, 7] and Figure 53.

However, for the structures presented above, the space between the two surfaces of the reflectors must be of the order of 200 A°, almost 4000 times less wide … . which is not technologically feasible by engraving!

Yet it seems possible, to be able to be obtained this parallel space of the order of 200 A° between Casimir reflectors, not by etching layers but by making them thermally grow!

#### Figure 53.

Extract de [7]: High aspect ratio (HAR) structures manufactured using the Bosch process: (a) 800 nm-wide trenches with a depth of 99.5 lm (aspect ratio 124:1) and (b) 250 nm-wide trenches with a depth of 40 lm (aspect ratio 160:1). Some of the walls collapsed during the dicing procedure. © is a magnified view of the inset shown in (b) [8].

Indeed, the SS3 and SS2 surfaces of the Casimir reflector must;


For example, silicon has a molar mass of 28 g/mol and silicon dioxide SiO2 has a molar mass of 60 g/mol. However, it is well known that when we grow a silicon dioxide SiO2 of one unit we "attack" a silicon depth of the order of 28/60 = 46.6% (Figure 54).

The initial silicon layer reacts with the oxidizing element to form SiO2. We will thus "consume" Silicon. The Si=SiO2 interface will therefore end up "below" the initial surface. A simple calculation shows that the fraction of oxide thickness

Figure 54. Growth of SiO2 oxide on silicon.

Perspective Chapter: Device, Electronic,Technology for a M.E.M.S. Which Allow… DOI: http://dx.doi.org/10.5772/intechopen.105197

"below" the initial surface is 46% of the total oxide thickness; the fraction "above" therefore represents 54% according to S.M. Sze. We therefore moved the original silicon surface by 46% [9].

The same must happen, for example for thermal growth of alumina. As the molecular masses of Alumina and aluminum are MAl2O3 = 102 g/mole and MAl = 27 g/mole, we obtain an aluminum attack ratio of 27/102 = 26%, which implies that the original surface of this metal has shifted by 26% so that 74% of the alumina has grown out of the initial surface of the aluminum … .

Likewise, if titanium is used for thermal growth of TiO2, the molar mass ratio being MTiO2 = 79.9 g/mole and MTi = 47.8 g/mole we obtain a titanium attack ratio of 59.8% which implies that the original surface has moved by 59.8%.

So, this growth covers up the initial interface and it can be finely controlled! As a result, it should be possible to define finely the interface between the two Casimir reflectors using the oxidative growth conditions of a metal such as titanium or aluminum.

As regards the technological manufacture of electronics and structure, it therefore seems preferable:


We obtain: zod <sup>¼</sup> <sup>2</sup> <sup>∗</sup> <sup>ð</sup>zmd <sup>þ</sup> zof � zmaÞ þ z0 <sup>¼</sup> <sup>2</sup> <sup>∗</sup> <sup>ð</sup>zmd <sup>þ</sup> zofð Þ <sup>1</sup> � :<sup>26</sup> Þ þ z0. For example, if we start from an opening zod ¼ 3μm and deposit a metal layer of aluminum that is etched leaving a width zmd ¼ 1μm on each side of the reflector. Then an Alumina Al2O3 can grow, the thickness of which is precisely adjusted, simply by considerations of time, temperature, and pressure to increase a necessary thickness to have a desired interface z0.!

For example, if z0 ¼ 200A°, zod ¼ 3μm, zmd ¼ 1μm, then zof ¼ 0:662μm. So, we obtain a Casimir interface of 200 A°. The final remaining metal thickness will be zmf ¼ 0:338μm and will act as a conductor under the aluminum oxide.

Obviously, the growth of this metal oxide between the electrodes of the Casimir reflector modifies the composition of the dielectric present between these electrodes, therefore of the mean relative permittivity of the dielectric.

Let: ε<sup>0</sup> be the permittivity of vacuum and ε0. ε<sup>r</sup> the metal oxides one (ε<sup>r</sup> = relative permittivity = 8 in the case of Al2O3), zof the final oxide thickness on one of

Figure 55. Distribution of thickness.

the electrodes and z the thickness of the vacuum present between electrode (initially we want z ¼ z0).

Then the average permittivity e0m of the dielectric is:

$$\begin{split} \mathbf{e}\_{\mathbf{0}\ \mathbf{m}} &= \left( \mathbf{z}\_{\text{of}} \cdot \mathbf{e}\_{\mathbf{0}} \cdot \mathbf{e}\_{\mathbf{r}} + \mathbf{z} \cdot \mathbf{e}\_{\mathbf{0}} + \mathbf{z}\_{\text{of}} \cdot \mathbf{e}\_{\mathbf{0}} \cdot \mathbf{e}\_{\mathbf{r}} \right) / (2 \mathbf{z}\_{\text{of}} + \mathbf{z}) \mathbf{e}\_{\mathbf{0} \mathbf{m}} \\ &= \mathbf{e}\_{\mathbf{0}} \cdot (\mathbf{2} \cdot \mathbf{z}\_{\text{of}} \cdot \mathbf{e}\_{\mathbf{r}} + \mathbf{z}) / (2 \mathbf{z}\_{\text{of}} + \mathbf{z}) \cong \mathbf{e}\_{\mathbf{0}} \cdot \mathbf{e}\_{\mathbf{r}} \end{split}$$

because z is < <zof.!! For example, zof ¼ 6620 A° is large compared to z < ¼ 200A° therefore <sup>ε</sup>0m ffi <sup>8</sup><sup>∗</sup> <sup>ε</sup><sup>0</sup> in the case of Al2O3.

We have taken this change in permittivity into account in the preceding simulations.
