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Dioxide film Essay

A set of windows are to be etched in a silicon dioxide film of thickness 6000 A. As patterned in the photoresist, the size of the windows is 6 um square. The question states that the etching process is ideal and isotropic, with no overetching or underetching. This implies that the time required to complete the etching process is exactly the time required for the etchant to etch to the interface between the silicon dioxide layer and the substrate.

I have also completed this question with the assumption that the etchant is a wet etchant. The slide entitled “Isotropic Wet Etching and Feature Size” in section 5 of the notes states the time required for a perfect etch using a wet etchant, with no overetching or underetching. This time is given in the following formula: where z is the thickness of the film, r is the etch rate of the etchant and ? is the time required for a perfect etch, with no overetching or underetching.

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Rearranging this equation to solve for z yields: The slide entitled “Isotropic Wet Etching and Feature Size” in section 5 of the notes states the amount of undercutting that would occur at the top of the silicon dioxide layer for a perfect etch, with no overetching or underetching. Since the etchant is isotropic, it must etch equally in all directions. Additionally, the etchant is always in contact with the top of the silicon dioxide layer during the etching process.

Therefore, it etches horizontally along the top of the silicon dioxide layer for the same amount of time that it etches vertically from the top of the silicon dioxide layer down to the substrate. Therefore, the length of the undercut that is generated at the top of the silicon dioxide layer is simply equal to the etch rate of the etchant multiplied by the time of the etching process. Mathematically, where xundercut is the length of the undercut that is generated at the top of the silicon dioxide layer. The above two equations can be combined to yield:

This result indicates that in an ideal etching process using a wet etchant, with no overetching or underetching, the length of the undercut that is generated at the top of the silicon dioxide layer is equal to the thickness of the silicon dioxide layer. According to the question, the silicon dioxide layer has a thickness of 6000 A. Therefore, z = 6000 A. Using the conversion factors 1 A = 10-10 m and 1 um = 10-6 m, z can be converted to um in the following manner: with significant figures applied We have just stated that xundercut must equal z.

Therefore, xundercut = 0. 6 um. Since the etchant is isotropic, it must etch equally in all directions. Therefore, each of the original sides of the windows must now be located a distance xundercut away from their initial positions, as defined by the patterning process. Therefore, the total distance between two opposite sides in the expanded window must be equal to with significant figures applied where dsides is the total distance between two opposite sides in the expanded window. Additionally, the corners of the original windows would also have etched isotropically.

These corners must link the four original sides of the windows at their new locations. Therefore, after the isotropic etching process, these corners should each be quarter-circles. Since these corners must also have etched a distance xundercut further into the silicon dioxide film, the radius of the quarter-circles must be equal to xundercut. Therefore, the final dimensions of the window, as measured at the top of the oxide layer, after ideal isotropic etching using a wet etchant is complete is represented in the following diagram: b).

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The question states that the etching process is ideal and isotropic, with no overetching or underetching. This implies that the time required to complete the etching process is exactly the time required for the etchant to etch to the interface between the silicon dioxide layer and the substrate. Therefore, the etching process should have ended at the exact moment the etchant touched the oxide-substrate interface. If the etching process ended at the exact moment the etchant touched the oxide-substrate interface, the etchant would simply have reproduced the dimensions of the original windows on the substrate surface.

It would not have had any additional time to expand these dimensions. This implies that the dimensions of the windows at the oxide-substrate interface should be identical to the dimensions of the original windows. In other words, the dimensions of the windows at the oxide-substrate interface should simply be 6 um square. Therefore, the final dimensions of the window, as measured at oxide-substrate interface, after ideal isotropic etching using a wet etchant is complete is represented in the following diagram: c).

I have completed this question with the assumption that the average slope of the window edge can be determined by drawing a line connecting the window edge at the top of the silicon dioxide film to the window edge at the oxide-substrate interface. The slope of this connecting line can then be determined and treated as the average slope at the window edge. I have also made the simplifying assumption that I can disregard the quarter-circular corners of the window at the top of the silicon dioxide layer when determining the slope.

Using this assumption, I need only to connect a point on a straight edge of the window at the top of the silicon dioxide layer to a point on a straight edge of the window at the oxide-substrate interface. I can then take the slope of this connecting line. Consider the diagram below of the cross-section of the two windows – one at the top of the silicon-dioxide film, the other at the bottom of the silicon-dioxide film. The diagram focusses on the amount that the window edges have expanded during the etching processes. The variable drun has been introduced in the above diagram to denote the horizontal component of the slope.

The variable drun has been introduced in the above diagram to denote the vertical component of the slope. From the above diagram, it is apparent that drise is equal to the thickness of the film, so drise = 0. 6um. Also from the above diagram, it is apparent that drun is equal to the amount of horizontal etching that was performed at the top of the silicon-dioxide film. Therefore, drun = 0. 6um. From the above diagram, we can see that we can determine the slope of the connecting line in the following manner: with significant figures

This slope is equal to the average slope at the window edge. d). I have denoted the total amount of time for which the etching process occurs by the symbol t. Overetching by 30% implies that the etchant is applied for 130% of the time it would take for a perfect etch to clear the entire thickness of the silicon dioxide film. Therefore, t can be expressed in terms of ? , the time required for a perfect etch: with significant figures Any number expressed as a percentage can equivalently be expressed as a decimal.

For example, the percentage value 130% can be equivalently expressed as 1.3. We may substitute the decimal value 1. 3 for the percentage value 130% in the above equation: with significant figures The above expression for ? can now be substituted into the above equation to yield: with significant figures Since the etchant is isotropic, it must etch equally in all directions. Additionally, the etchant is always in contact with the top of the silicon dioxide film during the etching process. Therefore, it etches horizontally along the top of the silicon dioxide film for the same amount of time that it etches vertically through the silicon dioxide film.

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Therefore, the length of the undercut that is generated at the top of the silicon dioxide film is simply equal to the etch rate of the etchant multiplied by the time of the etching process. Mathematically, where xundercut is the length of the undercut that is generated at the top of the silicon dioxide film. The previous two equations can now be combined to yield: with significant figures This result indicates that in an ideal etching process using a wet etchant, a 30% overetch, the length of the undercut that is generated at the top of the silicon dioxide film is equal to 1.3 times the thickness of the silicon dioxide film. It was shown in question 6-2-a that the silicon dioxide film is 0. 6 um thick. Therefore, z = 0. 6 um. Substituting this value for z into the above formula yields: with significant figures Since the etchant is isotropic, it must etch equally in all directions.

Therefore, each of the original sides of the windows must now be located a distance xundercut away from their initial positions, as defined by the patterning process. Therefore, the total distance between two opposite sides in the expanded window must be equal to with significant figures applied where dsides is the total distance between two opposite sides in the expanded window. This length dsides is equal to the dimension of the expanded window after the etching process is complete. Additionally, the corners of the original windows would also have etched isotropically. These corners must link the four original sides of the windows at their new locations. Therefore, after the isotropic etching process, these corners should each be quarter-circles.

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Dioxide film Essay
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Artscolumbia
A set of windows are to be etched in a silicon dioxide film of thickness 6000 A. As patterned in the photoresist, the size of the windows is 6 um square. The question states that the etching process is ideal and isotropic, with no overetching or underetching. This implies that the time required to complete the etching process is exactly the time required for the etchant to etch to the interface between the silicon dioxide layer and the substrate. I have also completed this question with the a
2018-01-07 18:44:48
Dioxide film Essay
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