In a perfect etch using a wet etchant, with no overetching or underetching, the time required to complete the etching process is exactly the time required for the etchant to etch to the oxide-substrate interface.

Therefore, in a perfect etch, the etching process would have ended at the exact moment the etchant touched the oxide-substrate interface. If the etching process had 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 would have been identical to the dimensions of the original windows after the time ?. In other words, the dimensions of the windows at the oxide-substrate interface would simply be 6 um square after the time ?. This situation was shown in question 6-2-b above. Since the etchant is isotropic, it must etch equally in all directions. The etchant begins this isotropic etch the moment it comes into contact with the windows at the top of the silicon dioxide wafer.

At this point, the etchant begins to etch a radial path in all directions, starting from the edges of the windows. In a perfect etch, with no overetching or underetching, this radial path never touches the oxide-substrate interface. However, as soon as overetching is begun, the etchant that was etching vertically will reach the silicon-dioxide interface. It will be unable to continue etching vertically because of the presence of the substrate. Additionally, a portion of this radial path will reach the oxide-substrate interface. This causes the window dimension at the oxide-substrate interface to expand.

Etching occurs isotropically, so the horizontal expansion of the window at the top of the silicon dioxide film must be equal to the radial expansion of this radial path. In other words, the value of xundercut that was calculated previously must be equal to the length of the radial path. 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.

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In the above diagram, it can be readily seen that the horizontal etching distance at the top of the oxide layer is equal to the radial etching distance through the oxide layer. Also note in the above diagram that z is the thickness of the silicon dioxide layer. Additionally, I have introduced a new variable, xinterface, in the above diagram.

xinterface represents horizontal etching distance along the oxide-substrate interface. There is a right triangle in the above diagram formed by xundercut, z, and xinterface. Therefore, the pythagorean theorem can be used to solve for xinterface:This equation can then be rearranged to solve for xinterface: We have shown in question 2-2-a that z = 0.

6 um. Additionally, we have shown in 2-2-d that xundercut for a 30% overetching process = 0. 78 um. Therefore, we can solve for xinterface in the following manner: with significant figures applied 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. 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 bottom of the oxide layer, after ideal isotropic etching using a wet etchant is complete is represented in the following diagram: f). This question has been completed using the same technique that was performed in question 6-2-c.Again, 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.

drun can be calculated from the geometry of the above diagram in the following manner: with significant figures 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. 6-3). a). A gradual slope is desired at the edges of a set of windows.

Therefore, a 6000 A layer of silicon dioxide, of etch rate 1000 A min-1 is overlaid with a 1000 A layer of CVD oxide of etch rate 2000 A min-1. The window pattern in the photoresist measures 6 um on a side and is square. I have completed this question with the assumption that the etching process is perfect, with no overetching or underetching. The following formula was introduced in question 6-1-a for 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, and that it etches isotropically. 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 through the film, and ? is the time required for a perfect etch, with no overetching or underetching.

This formula can be modified slightly to represent the time required for a perfect etch through the 1000 A layer of CVD oxide using a wet etchant, with no overetching or underetching. This has been done below: where zCVD is the thickness of the CVD oxide layer, rCVD is the etch rate through the CVD oxide layer, and ? CVD is the time required for a perfect etch through the CVD oxide layer, with no overetching or underetching. From the values given in the question, we know that zCVD = 1000 A and rCVD = 2000 A min-1.

Substituting these values into the above equation for ? CVD yields: with significant figuresSimilarly, the formula for ? can be modified slightly to represent the time required for a perfect etch through the 6000 A layer of silicon dioxide using a wet etchant, with no overetching or underetching. This has been done below: where zSilicon_Dioxide is the thickness of the silicon dioxide layer, rSilicon_Dioxide is the etch rate through the silicon dioxide layer, and ? Silicon_Dioxide is the time required for a perfect etch through the silicon dioxide layer, with no overetching or underetching. From the values given in the question, we know that zSilicon_Dioxide = 6000 A and rSilicon_Dioxide = 1000 A min-1.Substituting these values into the above equation for ? Silicon_Dioxide yields: with significant figures The total etch time required to clear the film is the sum of the etch time required to etch the CVD oxide layer and the etch time required to etch the silicon dioxide layer. Therefore, the total etch time required to clear the film can be calculated in the following manner: where ? Total is the total etch time required to clear the film with a perfect isotropic etching process (no overcutting or undercutting).

Substituting the previously determined values for ? CVD and ? Silicon_Dioxide yields:With significant figures applied b). The slide entitled “Etching Bilayer Film” in section 5 of the notes describes a situation in which a lower layer of oxide with a slow etch time is overlaid by an upper layer of oxide with a fast etch time. This situation is analogous to the one presented in this question. Figure 6-2 on this slide indicates that the quantity of the undercutting caused by the etching process in this situation is completely determined by the etch rate of the fast etching film and the total etch time required to clear both the upper and lower layers of oxide.According to the slide, the radius of the undercutting generated by the layer of oxide with a fast etch time is given by the following formula: where Rf is the radius of the undercutting generated by the layer of oxide with a fast etch time, rf is the etch rate of the layer of oxide with a fast etch time, and t is the total etch time required to clear both the upper and lower layers of oxide. The etchant is in contact with the top of the fast etching layer of oxide for the entirety of the total etch time required to clear both the upper and lower layers of oxide.

Therefore, the quantity of the undercutting at the top of the fast etching layer of oxide must be equal to the radius of the undercutting generated by the fast etching layer of oxide. The above formula for Rf can therefore be modified slightly to represent the quantity of the undercutting at the top of the fast etching layer of oxide. This has been done below: where xundercut is the quantity of the undercutting at the top of the fast etching layer of oxide.In this question, the fast etching layer of oxide is the CVD oxide layer. The etch rate of the CVD oxide layer is represented by the symbol rCVD and the total etch time required to etch both the CVD oxide layer and the silicon dioxide layer with a perfect isotropic etching process (no overcutting or undercutting) is represented by the symbol ? Total. The above formula for xundercut can therefore be modified slightly to represent the quantity of the undercutting at the top of the CVD oxide layer in this particular question.