A 0.60 um film of silicon dioxide is to be etched with a buffered oxide etchant of etch rate 750 A min-1. Process data shows that the thickness may vary up to 10% and the etch rate may vary up to 15%.
Andy Somody
97300-6222
ENSC 495
Assignment #3
6-1).
a).
A 0.60 um film of silicon dioxide is to be etched with a buffered oxide etchant of etch rate 750 A min-1. Process data shows that the thickness may vary up to 10% and the etch rate may vary up to 15%. The maximum possible thickness of the silicon dioxide film is therefore 110% of its nominal value. Therefore, the maximum possible thickness of the silicon dioxide film can be determined through the following calculation:
where zmax is the maximum possible thickness of the silicon dioxide film and znominal is the nominal thickness of the silicon dioxide film. Therefore, znominal = 0.60 um. Any number expressed as a percentage can alternatively be expressed as a decimal. For example, 110% can be expressed as 1.1. Using this decimal format, the above formula can be rewritten in the following manner:
Substituting our previously determined value for znominal into the above formula yields:
with significant figures applied
Similarly, the minimum possible etch rate of the buffered oxide etchant is 85% of its nominal value. Therefore, the minimum possible etch rate of the buffered oxide etchant can be determined through the following calculation:
where rmin is the minimum possible etch rate of the buffered oxide etchant and rnominal is the minimum possible etch rate of the buffered oxide etchant. Therefore, rnominal = 750 A min-1. Using the conversion factors 1 A = 10-10 m and 1 um = 10-6 m, rnominal can be converted to um min-1 in the following manner:
with significant figures applied
As was demonstrated above, this percentage value can alternatively be expressed as a decimal. Therefore, 85% can be expressed as 0.85. Using this decimal format, the above formula can be rewritten in the following manner:
Substituting our previously determined value for rnominal into the above formula yields:
with significant figures applied
I have completed this question with the assumption that the etching process is perfect, with no overetching or underetching. This implies that the time required to complete the etching process is exactly the time required for the buffered oxide 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 buffered oxide 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 of the etchant and ? is the time required for a perfect etch, with no overetching or underetching. The thickness of our silicon dioxide film may vary up to 10% and the etch rate of our buffered oxide etchant may vary up to 15%. Therefore, the time required to complete the etching process may also vary. From the above equation for ?, we can see that the maximum possible time required to complete the etching process occurs when z is maximized and r is minimized. Therefore, we can slightly modify the above equation for ? to represent the maximum possible time required to complete the etching process:
where ?max is the maximum possible time required to complete the etching process, with no overetching or underetching. Substituting our previously determined values for zmax and rmin into the above formula yields:
with significant figures applied
Therefore, ?max represents the maximum possible time required to complete the etching process, with no overetching or underetching.
b).
I have completed this question with the assumption that the buffered oxide etchant is a wet etchant, and that it etches isotropically. For an isotropic wet etching process, undercutting will occur at the top of the silicon dioxide layer. 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 entire time for which the etching process occurs. 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 buffered oxide 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. I have completed question 6-1-a with the assumption that we are etching for the maximum possible time required to complete the etching process. As a result, whatever variations in film thickness or etch rate may occur, the film of silicon dioxide will be fully etched through. The maximum undercut will be generated if the buffered oxide etchant etches at its maximum possible rate. The maximum possible etch rate of the buffered oxide etchant is 115% of its nominal value. Therefore, the maximum possible etch rate of the buffered oxide etchant can be determined through the following calculation:
where rmax is the maximum possible etch rate of the buffered oxide etchant and rnominal is the nominal etch rate of the buffered oxide etchant. Therefore, rnominal = 750 A min-1. Using the conversion factors 1 A = 10-10 m and 1 um = 10-6 m, rnominal can be converted to um min-1 in the following manner:
with significant figures applied
As was demonstrated above, this percentage value can alternatively be expressed as a decimal. Therefore, 115% can be expressed as 1.15. Using this decimal format, the above formula can be rewritten in the following manner:
Substituting our previously determined value for rnominal into the above formula yields:
with significant figures applied
the above equation for xundercut can be modified slightly to yield the length of the undercut that is generated at the top of the silicon dioxide layer after the maximum possible etch time and with the maximum possible etch rate. Mathematically,
where xundercut_max is the length of the undercut that is generated at the top of the silicon dioxide layer after the maximum possible etch time and with the maximum possible etch rate. Substituting our previously determined values for rmax and ?max into the above equation yields:
with significant figures applied
The minimum undercut will be generated if the buffered oxide etchant etches at its minimum possible rate. The minimum possible etch rate of the buffered oxide etchant is 85% of its nominal value. Therefore, the minimum possible etch rate of the buffered oxide etchant can be determined through the following calculation:
where rmin is the minimum possible etch rate of the buffered oxide etchant and rnominal is the nominal etch rate of the buffered oxide etchant.
As was demonstrated above, this percentage value can alternatively be expressed as a decimal. Therefore, 85% can be expressed as 0.85. Using this decimal format, the above formula can be rewritten in the following manner:
Substituting our previously determined value for rnominal into the above formula yields:
with significant figures applied
the above equation for xundercut can be modified slightly to yield the length of the undercut that is generated at the top of the silicon dioxide layer after the maximum possible etch time and with the minimum possible etch rate. Mathematically,
where xundercut_min is the length of the undercut that is generated at the top of the silicon dioxide layer after the maximum possible etch time and with the minimum possible etch rate. Substituting our previously determined values for rmin and ?max into the above equation yields:
with significant figures applied
Therefore, the amount of undercut that will occur at the top of the film will at most be equal to xundercut_max and at least be equal to xundercut_min.
6-2).
a). 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. 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).
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. 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 ...
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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. 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:
e).
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.
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 figures
Similarly, 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. This has been done below:
Substituting our previously determined values for rCVD and ?Total into the above formula yields:
with significant figures applied
Using the conversion factors 1 A = 10-10 m and 1 um = 10-6 m, xundercut can be converted to um 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 CVD oxide 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 CVD oxide layer, after ideal isotropic etching using a wet etchant is complete is represented in the following diagram:
c).
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 bottom of the film. If the etching process ended at the exact moment the etchant touched the bottom of the film, 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 bottom of the film should be identical to the dimensions of the original windows. In other words, the dimensions of the windows at the bottom of the film should simply be 6 um square. Therefore, the final dimensions of the window, as measured at bottom of the film, after ideal isotropic etching using a wet etchant is complete is represented in the following diagram:
This question is slightly ambiguous. Therefore, I have solved the question through two different techniques. I first assumed that the question was asking for the average slope across both the window edges for both the fast and slow etching windows. I have denoted this solution to be Solution #1. I have later solved this same question in an alternative manner. I have denoted by second solution to be Solution #2.
Solution #1:
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 CVD oxide to the window edge at the bottom of the film. 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 CVD oxide 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 CVD oxide to a point on a straight edge of the window at the bottom of the film. I can then take the slope of this connecting line.
In question 6-3-a, we expressed the thickness of the silicon dioxide film, zSilicon_Dioxide, as zSilicon_Dioxide = 6000 A. Similarly, in question 6-3-a, we expressed the thickness of the CVD oxide, zCVD, as zCVD = 1000 A. Substituting these values into the above expression for zTotal yields:
with significant figures applied
Using the conversion factors 1 A = 10-10 m and 1 um = 10-6 m, zTotal can be converted to um in the following manner:
with significant figures applied
Consider the diagram below of the cross-section of the two windows - one at the top of the CVD oxide 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. The radius of the fast-etched CVD oxide and the radius of the slow-etched silicon-dioxide layers have both been shown in the above diagram, in accordance to Figure 6-2 in the "Etching Bilayer Resist" page of the notes. The rate of the fast-etched CVD oxide layer has been denoted Rf. From the above diagram, it is apparent that drise is equal to the total thickness of both the CVD oxide layer and the silicon-dioxide layer, so drise = zTotal = 0.7 um. Also from the above diagram, it is apparent that drun is equal to the distance that the top of the CVD oxide film was etched horizontally. Therefore, drun = 1.3um.
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.
Solution #2:
This question may be interpreted a slightly different way. It has asked to find the slope at the window edge. In section 5 of the notes, there is a sheet entitled "Etching Bilayer Film". Figure 6-2 on this sheet states that the slope of only the slow-etched portion of the window defines the slope of the window edge. According to Figure 6-2, we may assume that the slow-etched portion of the window etches slowly enough to have a constant slope, ?. I have therefore completed Solution #2 by solving for that slope ?.
Consider the diagram below of the cross-section of the two windows - one at the top of the CVD oxide 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 symbol ? has been used to denote the slope of the edge of the slow-etched window.
The radius of the fast-etched CVD oxide and the radius of the slow-etched silicon-dioxide layers have both been shown in the above diagram, in accordance to Figure 6-2 in the "Etching Bilayer Resist" page of the notes. The rate of the fast-etched CVD oxide layer has been denoted Rf. Note that the radius of the fast-etched CVD oxide etching profile is equal to the amount of undercut at the top of the CVD layer, 1.3 um. Note also that drun has been redefined to depict the new horizontal component of the slope, while drise has likewise been redefined to depict the new vertical component of the slope.
From the above diagram, it is apparent that drise is equal to the total thickness of both the silicon-dioxide layer, so drise = 0.6 um. Also from the above diagram, it is apparent that drun is equal to one side of a right triangle. The other sides of this triangle are equal to 1.3 um and 0.1 um. Therefore, the pythagorian theorem can be applied to solve for drun in the following manner:
with significant figures applied
From the above diagram, we can see that we can determine the slope of the window edge in the following manner:
with significant figures
This slope is equal to the slope ? in the above diagram for a bilayer film. This has been interpreted to be equal to the slope of the window edge in Solution #2.
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 film. Therefore, t can be expressed in terms of ?Total, 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
In question 6-3-a, we determined the time it would take for a perfect etch to clear the entire thickness of the film as ?Total = 6.5 min. Substituting this value into the above equation for t yields:
with significant figures applied
In question 6-3-b, an expression was determined for the quantity of the undercutting at the top of the fast etching layer of oxide, xundercut, in terms of the etch rate, rf, of the layer of oxide with a fast etch time, and the total etch time, t, required to clear both the upper and lower layers of oxide. This expression has been reproduced here:
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. 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. This has been done below:
Substituting our previously determined values for rCVD and t into the above formula yields:
with significant figures applied
Using the conversion factors 1 A = 10-10 m and 1 um = 10-6 m, xundercut can be converted to um 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 CVD oxide 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 CVD oxide layer, after ideal isotropic etching using a wet etchant is complete is represented in the following diagram:
e).
Before we can calculate the average slope of the window after a 30% overetch time, we must determine dimensions of the window as measured at the bottom of the film after a 30% overetch time.
This question is again slightly ambiguous. Therefore, I have solved the question through two different techniques. I first assumed that the question was asking for the average slope across both the window edges for both the fast and slow etching windows. I have denoted this solution to be Solution #1. I have later solved this same question in an alternative manner. I have denoted by second solution to be Solution #2.
Solution #1:
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 - as they would appear after a 30% overetching process. 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 total thickness of both the CVD oxide layer and the silicon-dioxide layer, so drise = zTotal = 0.7 um. The rate of the fast-etched CVD oxide layer has been denoted Rf, while the rate of the slow-etched silicon dioxide oxide layer has been denoted Rs.
The CVD oxide and SiO2 oxide layers etch at different rates, so the SiO2 only begins to etch uniformly in all directions with a constant radius once it clears the CVD oxide layer. It was shown in question 6-3-a that etching vertically down through the CVD layer requires a time ?CVD equal to:
with significant figures
It was also shown in question 6-3-d that the total amount of time spent etching, t, for a process with a 30% overetch time is:
with significant figures applied
Therefore, the maximum amount of time that the etchant can spend etching the silicon dioxide film is given by the following equation:
with significant figures
The etch rate of the silicon dioxide is given in the question to be rSilicon_Dioxide = 1000A/min. Additionally, we can express the radius of the silicon dioxide etch seen in the above diagram, RS, in the following manner:
Substituting our values for rSilicon_Dioxide and ?Silicon_Dioxide into the above formula for RSilicon_Dioxide yields:
with significant figures applied
Using the conversion factors 1 A = 10-10 m and 1 um = 10-6 m, RS can be converted to um in the following manner:
with significant figures applied
From the above diagram, we can see that RS forms a right triangle with xinterface and a side of length 0.6 um. We can apply the pythagorean theorem to this triangle to obtain a value for xinterface:
with significant figures applied
From the diagram, we determine the dimensions of the window at the oxide-substrate interface after the total etching time has elapsed. After the etching has completed, each side of the original 6 um square window will have been etched a horizontal distance xinterface further into the silicon dioxide layer. Therefore, the total distance between two opposite sides in the expanded window after the total etching time has elapsed must be equal to:
with significant figures applied
where dsides is the total distance between two opposite sides in the expanded window.
Observing the positions of xinterface and drun in the above diagram, we can also see that their sum is equal to s side of length 1.69 um. We can therefore solve for drun in the following manner:
with significant figures applied
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.
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:
Solution #2:
This question may be interpreted a slightly different way. It has asked to find the slope at the window edge. In section 5 of the notes, there is a sheet entitled "Etching Bilayer Film". Figure 6-2 on this sheet states that the slope of only the slow-etched portion of the window defines the slope of the window edge. According to Figure 6-2, we may assume that the slow-etched portion of the window etches slowly enough to have a constant slope, ?. I have therefore completed Solution #2 by solving for that slope ?.
Consider the diagram below of the cross-section of the two windows - one at the top of the CVD oxide 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 symbol ? has been used to denote the slope of the edge of the slow-etched window.
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 silicon-dioxide layer, so drise = 0.6 um.
The radius of the fast-etched CVD oxide and the radius of the slow-etched silicon-dioxide layers have both been shown in the above diagram, in accordance to Figure 6-2 in the "Etching Bilayer Resist" page of the notes. The rate of the fast-etched CVD oxide layer has been denoted Rf, while the rate of the slow-etched silicon-dioxide layer has been denoted Rs. Note that the radius of the fast-etched CVD oxide etching profile is equal to the amount of undercut at the top of the CVD layer, 1.69 um.
The CVD oxide and SiO2 oxide layers etch at different rates, so the SiO2 only begins to etch uniformly in all directions with a constant radius once it clears the CVD oxide layer. It was shown in solution #1 in this question that :
with significant figures applied
Additionally, we can see from the above diagram that Rs forms a right triangle with xinterface and a side of length 0.6 um. Therefore, we can apply the pythagorean theorem to this triangle to obtain a value for xinterface:
with significant figures applied
From the above diagram, we can see that the sum of xinterface and drun forms a right triangle with Rf and with a side of length 0.1 um. We can apply the pythagorean theorem to this triangle to obtain a value for xinterface + drun:
with significant figures applied
The last two equations can be combined to yield:
with significant figures applied
From the diagram, we determine the dimensions of the window at the oxide-substrate interface after the total etching time has elapsed. After the etching has completed, each side of the original 6 um square window will have been etched a horizontal distance xinterface further into the silicon dioxide layer. Therefore, the total distance between two opposite sides in the expanded window after the total etching time has elapsed 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 xinterface further into the oxide film, the radius of the quarter-circles must be equal to xinterface. 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:
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 slope ? in the above diagram for a bilayer film. This has been interpreted to be equal to the slope of the window edge in Solution #2.
6-4).
In a memory device, the cells are accessed using an array of parallel metal lines. The pitch of these lines limits the size of the device. We must determine the pitch of a set of parallel lines under the following conditions:
a). The lines are made of aluminum, 1 um thick
b). The minimum space-width resolvable with the photoresist technology in use is 2 um.
c). The lines are wet etched, isotropically, with a 50% overetch necessitated by uncertainties in etch rate and end-point detection.
d). The minimum line dimension, after etching, at the top of the line, is required by design considerations to be 4 um.
The pitch between the set of parallel lines is defined to be equal to the distance between a given point on one line and the same point on the adjacent line. Therefore, it is irrelevant whether we measure the pitch from the left sides of the lines, the right sides of the lines, or at any point in between. I will choose in completing this question to measure the pitch from the left side of one line to the left side on the adjacent line. All of the lines are wet etched, isotropically, with a 50% overetch necessitated by uncertainties in etch rate and end-point detection. Therefore, all of the lines are etched identically and are underetched in the same manner. As a result, it is irrelevant whether we measure the pitch of the lines at the top of the lines, the bottom of the lines, or at any point in between. I will choose in completing this question to measure the pitch from the top of the lines.
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.
The lines are wet etched, isotropically, with a 50% overetch necessitated by uncertainties in etch rate and end-point detection. I have denoted the total amount of time for which the etching process occurs by the symbol t. Overetching by 50% implies that the etchant is applied for 150% of the time it would take for a perfect etch to clear the entire thickness of the aluminum. 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 150% can be equivalently expressed as 1.5. We may substitute the decimal value 1.5 for the percentage value 150% 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 aluminum during the etching process. Therefore, it etches horizontally along the top of the aluminum for the same amount of time that it etches vertically through the aluminum. Therefore, the length of the undercut that is generated at the top of the aluminum 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 aluminum.
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 50% overetch, the length of the undercut that is generated at the top of the aluminum is equal to 1.5 times the thickness of the aluminum.
The question states that the aluminum lines are 1 um thick. Therefore, z = 1 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 aluminum lines must now be located a distance xundercut away from their initial positions, as defined by the patterning process. Therefore, the total distance between the left side and right side of any aluminum line must have been reduced by a distance 2*xundercut. In other words, the final width of the aluminum lines must have been reduced by 2*xundercut after the etching process. Therefore, the final width of the aluminum lines after etching, wfinal, can be related to the initial thickness of the aluminum lines before etching, winitial, by the formula:
Rearranging this equation to solve for winitial yields:
The question states that the minimum line dimension, after etching, at the top of the line, is required by design considerations to be 4 um. Therefore, wfinal = 4 um. Substituting this value for wfinal and our previously determined value for xundercut into the above formula yields:
with significant figures
The question also states that the minimum space-width resolvable with the photoresist technology in use is 2 um. This implies that, due to the limitations on the resolution of the photoresist technology, any structure created during photolithography could be offset by a maximum of ? 2 um. We must ensure that, during photolithography, no structure created is placed on top of another structure. Consider two lines, and designate the leftmost of these lines to be line 1, and the rightmost of these lines to be line 2. During photolithography, the right side of line 1 must be placed at least 2 um away from the left side of line 2 to ensure that, due to the limitations on the resolution of the photoresist technology, one of these lines is not patterned on top of the other line. We have shown previously that the initial width of each of the lines, before the etching process, is 7um. The total distance between the left side of line 1 and the left side of line 2 must be equal to the pitch between the lines. This width is given by:
where swresist is the minimum space-width resolvable with the photoresist technology in use. Substituting swresist = 2 um and winitial = 7 um into the above equation yields:
with significant figures
The following diagram displays the two adjacent lines and the pitch:
7-1).
a).
The slide entitled "Diffusion Solutions" in section 6 of the notes gives an equation relating the impurity concentration in a constant source diffusion as a function of the time required for the diffusion to occur, t, and the distance of a given concentration of dopant from the surface of the wafer, x. This equation is as follows:
where N0 is the concentration of the impurity at the surface of the wafer, D is the diffusion coefficient of the given diffusion, t is the time required for the diffusion to occur, x is the distance of a given concentration of dopant from the surface of the wafer, and N(x,t) is impurity concentration of the constant source diffusion as a function of x and t. The slide entitled "Useful erfc(x) Approximations" in section 6 of the notes defines erfc(x) as the Complementary Error Function. Mathematically, it is given as:
Combining the previous two equations yields:
The slide entitled "Diffusion Solutions" in section 6 of the notes also gives an equation relating the impurity concentration in a limited source diffusion as a function of the time required for the diffusion to occur, t, and the distance of a given concentration of dopant from the surface of the wafer, x. This equation is as follows:
where Q is the total impurity concentration present in the wafer, D is the diffusion coefficient of the given diffusion, t is the time required for the diffusion to occur, x is the distance of a given concentration of dopant from the surface of the wafer, and N(x,t) is impurity concentration of the limited source diffusion as a function of x and t.
According to an email received from Sun Djaja, the term concentration gradient refers to "how the concentration changes with the depth of the substrate". Therefore, the concentration gradients for the erfc and Gaussian distributions must be defined as the change in the concentration of a given dopant divided by the change in distance from the surface of the wafer. We have defined x as the distance a given concentration of dopant from the surface of the wafer and N(x,t) as the impurity concentration as a function of x and t. Mathematically, therefore, the concentration gradient must be defined as:
Therefore, we can determine the concentration gradients for the erfc and Gaussian distributions by evaluating the derivatives of the erfc and Gaussian distribution expressions for N(x,t) with respect to x. For the erfc distribution, this derivative has been performed below:
Since the derivative is being taken with respect to x, while the integral is with respect to s, the derivative of this integral can be easily evaluated:
Similarly, the derivative of the Gaussian distribution expression for N(x,t) with respect to x has been performed below:
After application of the chain rule, we obtain:
If the substrate doping density is Csub, we must derive expressions for the junction depths, xj, for the erfc and Gaussian distributions.
The slide entitled "Formation of PN Junction" in section 6 of the notes depicts the formation of a pn junction by a p-type Gaussian diffusion of boron into an n-type silicon substrate. Figure 4.7 on this slide illustrates that the impurity concentration of the p-type Gaussian diffusion of boron decreases as the distance from the surface of the silicon wafer is increased. The impurity concentration of the background n-type dopant remains constant at any distance from the surface of the silicon wafer. Figure 4.7 on this slide states that, "the metallurgical junction occurs at the point x = xj, where the net concentration is zero". This implies that xj occurs at the distance from the surface of the silicon wafer at which the concentration of the n-type dopant equals the concentration of the background p-type dopant. In this example provided on the slide, a p-type Gaussian diffusion of boron was made into an n-type substrate. Therefore, a p-type region is formed between the surface of the silicon wafer and xj. For distances deeper into the wafer than xj, an n-type region is formed. I have completed this question with the assumption that either a p-type diffusion is being made into a n-type substrate, or that an n-type diffusion is being made into a p-type substrate. This ensures that a pn junction will be formed at a distance xj from the surface of the wafer at which the impurity
concentration of the diffused dopant equals the background doping density of the substrate.
The above expression for N(x,t) in the erfc distribution can therefore be used to write an expression for N(xj,t) in the erfc distribution.
The question has stated that that the substrate doping density is Csub. Therefore, at a distance xj from the surface of the wafer, the impurity concentration of the diffused dopant must equal the background doping density of the substrate, implying N(xj,t) = Csub. Substituting this value for N(xj,t) into the above equation yields:
This equation can now be rearranged to solve for xj in the following manner:
The above expression has used the symbol N0 to represent the concentration of the impurity at the surface of the wafer. Question 7-1-b has used the symbol Cs to represent the concentration of the impurity at the surface of the wafer. Therefore, I will slightly modify the above expression for the concentration gradient in the erfc distribution to remain consistent with the notation provided in Question 7-1-b.
The above expression for N(x,t) in the Gaussian distribution can therefore be used to write an expression for N(xj,t) in the Gaussian distribution.
The question has stated that that the substrate doping density is Csub. Therefore, at a distance xj from the surface of the wafer, the impurity concentration of the diffused dopant must equal the background doping density of the substrate, implying N(xj,t) = Csub. Substituting this value for N(xj,t) into the above equation yields:
This equation can now be rearranged to solve for xj in the following manner in the Gaussian distribution:
b).
Assuming Cs = 1019 atoms/cm3 for an erfc distribution and QT = 1x1013 atoms/cm2 for a Gaussian distribution, Csub = 1015 atoms/cm3, and D = 1x10-15 cm2/sec (which is close to boron diffusivity at 900oC, we must calculate the junction depths and the concentration gradients for both distributions at diffusion times of 10, 30 and 60 min.
I have represented these times with the symbols t1 = 10 minutes, t2 = 30 minutes, t3 = 60 minutes. Using the conversion factor 1 minutes = 60 seconds, we can convert these diffusion times to seconds in the following manner:
with significant figures applied
with significant figures applied
with significant figures applied
In question 7-1-a, we derived an expression for the concentration gradient of the erfc distribution. That equation was given as:
The above expression has used the symbol N0 to represent the concentration of the impurity at the surface of the wafer. Question 7-1-b has used the symbol Cs to represent the concentration of the impurity at the surface of the wafer. Therefore, I will slightly modify the above expression for the concentration gradient to remain consistent with the notation provided in Question 7-1-b.
Substituting t = t1 = 600 sec, D = 1x10-15 cm2/sec, and Cs = 1019 atoms/cm3 into this expression, we can solve for the concentration gradient for the erfc distribution for a diffusion time of 10 minutes:
with significant figures applied
where d(N(x,t1)) is the concentration gradient for the erfc distribution after time t1 = 600 sec.
Substituting t = t2 = 1800 sec, D = 1x10-15 cm2/sec, and Cs = 1019 atoms/cm3 into this expression, we can solve for the concentration gradient for the erfc distribution for a diffusion time of 30 minutes:
with significant figures applied
where d(N(x,t2)) is the concentration gradient for the erfc distribution after time t2 = 1800 sec.
Substituting t = t3 = 3600 sec, D = 1x10-15 cm2/sec, and Cs = 1019 atoms/cm3 into this expression, we can solve for the concentration gradient for the erfc distribution for a diffusion time of 60 minutes:
with significant figures applied
where d(N(x,t3)) is the concentration gradient for the erfc distribution after time t3 = 3600 sec.
In question 7-1-a, we derived an expression for the concentration gradient of the Gaussian distribution. That equation was given as:
The expression derived in question 7-1-a for the concentration gradient of the Gaussian distribution used the symbol Q to represent the total impurity concentration. Question 7-1-b has used the symbol QT to represent the total impurity concentration. Therefore, I will slightly modify the expression derived in section 7-1-a for the concentration gradient to remain consistent with the notation provided in Question 7-1-b.
Substituting t = t1 = 600 sec, D = 1x10-15 cm2/sec, and QT = 1x1013 atoms/cm2 into this expression, we can solve for the concentration gradient for the Gaussian distribution for a diffusion time of 10 minutes:
with significant figures applied
where d(N(x,t1)) is the concentration gradient for the Gaussian distribution after time t1 = 600 sec.
Substituting t = t2 = 1800 sec, D = 1x10-15 cm2/sec, and QT = 1x1013 atoms/cm2 into this expression, we can solve for the concentration gradient for the Gaussian distribution for a diffusion time of 30 minutes:
with significant figures applied
where d(N(x,t2)) is the concentration gradient for the Gaussian distribution after time t2 = 1800 sec.
Substituting t = t3 = 3600 sec, D = 1x10-15 cm2/sec, and QT = 1x1013 atoms/cm2 into this expression, we can solve for the concentration gradient for the Gaussian distribution for a diffusion time of 60 minutes:
with significant figures applied
where d(N(x,t3)) is the concentration gradient for the Gaussian distribution after time t3 = 3600 sec.
The following expression for the junction depth of the erfc function was derived in question 7-1-a:
Figure 4.4 on the slide entitled "Comparison of Normalized and ERFC" in section 6 of the notes provides a graph comparing the Gaussian and complementary error profiles. This curve can be used to evaluate the value of erfc and its inverse. According to an email message I received from Dr. Chapman, "you do not need to solve the y= erfc(x) to find x for a y, just use the plot in the notes to extract the x (after all this is only 2 figure accuracy)"
Substituting t = t1 = 600 sec, D = 1x10-15 cm2/sec, Csub = 1015 atoms/cm3, and Cs = 1x1019 atoms/cm3 into this expression, we can solve for the junction depth for the erfc distribution for a diffusion time of 10 minutes:
Estimating the value the inverse of the erfc profile at the point 10^(-4) from the graph yields:
with significant figures
where xj1 is the junction depth for the erfc distribution after time t1 = 600 sec.
Substituting t = t2 = 1800 sec, D = 1x10-15 cm2/sec, Csub = 1015 atoms/cm3, and Cs = 1x1019 atoms/cm3 into this expression, we can solve for the junction depth for the erfc distribution for a diffusion time of 30 minutes:
Estimating the value the inverse of the erfc profile at the point 10^(-4) from the graph yields:
with significant figures
where xj2 is the junction depth for the erfc distribution after time t2 = 1800 sec.
Substituting t = t3 = 3600 sec, D = 1x10-15 cm2/sec, Csub = 1015 atoms/cm3, and Cs = 1x1019 atoms/cm3 into this expression, we can solve for the junction depth for the erfc distribution for a diffusion time of 60 minutes:
Estimating the value the inverse of the erfc profile at the point 10^(-4) from the graph yields:
with significant figures
where xj3 is the junction depth for the erfc distribution after time t3 = 3600 sec.
The expression derived in question 7-1-a for the junction depth of the Gaussian distribution used the symbol Q to represent the total impurity concentration. Question 7-1-b has used the symbol QT to represent the total impurity concentration. Therefore, I will slightly modify the expression derived in section 7-1-a for the junction depth to remain consistent with the notation provided in Question 7-1-b.
Substituting t = t1 = 600 sec, D = 1x10-15 cm2/sec, Csub = 1015 atoms/cm3, and QT = 1x1013 atoms/cm2 into this expression, we can solve for the junction depth for the Gaussian distribution for a diffusion time of 10 minutes:
with significant figures
where xj1 is the junction depth for the Gaussian distribution after time t1 = 600 sec.
Substituting t = t2 = 1800 sec, D = 1x10-15 cm2/sec, Csub = 1015 atoms/cm3, and QT = 1x1013 atoms/cm2 into this expression, we can solve for the junction depth for the Gaussian distribution for a diffusion time of 30 minutes:
with significant figures
where xj2 is the junction depth for the Gaussian distribution after time t2 = 1800 sec.
Substituting t = t3 = 3600 sec, D = 1x10-15 cm2/sec, Csub = 1015 atoms/cm3, and QT = 1x1013 atoms/cm2 into this expression, we can solve for the junction depth for the Gaussian distribution for a diffusion time of 60 minutes:
with significant figures
where xj3 is the junction depth for the Gaussian distribution after time t3 = 3600 sec.
The integrated dopants for the erfc distribution for diffusion times of 10, 30 and 60 minutes can be calculated.
The slide entitled "Diffusion Solutions" in section 6 of the notes gives an equation relating the total impurity concentration in a constant source diffusion as a function of the time required for the diffusion to occur, t, the diffusion coefficient of the given diffusion process, D, and the concentration of the impurity at the surface of the wafer, N0. This equation is given as follows:
The above expression has used the symbol N0 to represent the concentration of the impurity at the surface of the wafer. Question 7-1-b has used the symbol Cs to represent the concentration of the impurity at the surface of the wafer. Therefore, I will slightly modify the above expression for Q to remain consistent with the notation provided in Question 7-1-b.
We have been given in the question that Cs = 1019 atoms/cm3 for an erfc distribution and D = 1x10-15 cm2/sec.
Substituting t = t1 = 600 sec, D = 1x10-15 cm2/sec, and Cs = 1019 atoms/cm3 into this expression, we can solve for the total impurity concentration present in the wafer in the erfc distribution for a diffusion time of 10 minutes:
with significant figures applied
where Q1 is the total impurity concentration present in the wafer in the erfc distribution after time t1 = 600 sec.
Similarly, substituting t = t2 = 1800 sec, D = 1x10-15 cm2/sec, and Cs = 1019 atoms/cm3 into this expression, we can solve for the total impurity concentration present in the wafer in the erfc distribution for a diffusion time of 30 minutes:
with significant figures applied
where Q2 is the total impurity concentration present in the wafer after time t2 = 1800 sec.
Similarly, substituting t = t3 = 3600 sec, D = 1x10-15 cm2/sec, and Cs = 1019 atoms/cm3 into this expression, we can solve for the total impurity concentration present in the wafer in the erfc distribution for a diffusion time of 60 minutes:
with significant figures applied
where Q3 is the total impurity concentration present in the wafer after time t3 = 3600 sec.
The surface concentration for the Gaussian distribution for diffusion times of 10, 30 and 60 minutes can therefore also be calculated. The variable x is defined as the distance of a given concentration of dopant from the surface of the wafer. Therefore, at the surface of the wafer, by definition, x = 0 cm. We may use the expression introduced in question 7-1-a for N(x,t) in the Gaussian distribution. Again, the expression introduced in question 7-1-a for N(x,t) in the Gaussian distribution used the symbol Q to represent the total impurity concentration. Question 7-1-b has used the symbol QT to represent the total impurity concentration. Therefore, I will slightly modify the expression derived in section 7-1-a for N(x,t) to remain consistent with the notation provided in Question 7-1-b.
Substituting x = 0, N(x,t) = N(0,t1), t1 = 600 sec, D = 1x10-15 cm2/sec, and QT = 1x1013 atoms/cm2 into this expression, we can solve for N(0,t1) for the Gaussian distribution for a diffusion time of 10 minutes:
with significant figures applied
where N(0,t1) is the dopant concentration on the surface of the wafer for the Gaussian distribution after time t1 = 600 sec.
Similarly, substituting x = 0, N(x,t) = N(0,t2), t2 = 1800 sec, D = 1x10-15 cm2/sec, and QT = 1x1013 atoms/cm2 into this expression, we can solve for N(0,t2) for the Gaussian distribution for a diffusion time of 30 minutes:
with significant figures applied
where N(0,t2) is the dopant concentration on the surface of the wafer for the Gaussian distribution after time t2 = 1800 sec.
Similarly, substituting x = 0, N(x,t) = N(0,t3), t3 = 3600 sec, D = 1x10-15 cm2/sec, and QT = 1x1013 atoms/cm2 into this expression, we can solve for N(0,t3) for the Gaussian distribution for a diffusion time of 60 minutes:
with significant figures applied
where N(0,t3) is the dopant concentration on the surface of the wafer for the Gaussian distribution after time t3 = 3600 sec.
7-2).
a). Assume the measured phosphorous profile in a wafer can be represented by a Gaussian function with a diffusivity of D = 2.3x10-13 cm2/sec. The measured surface concentration is 1018 atoms/cm3, and the junction depth is 1 micron with a substrate acceptor concentration of 1015 atoms/cm3. Therefore, the diffusion is an n-type diffusion of phosphorous. I have assumed in solving this question that the substrate is a uniformly doped p-type substrate. The slide entitled "Formation of PN Junction" in section 6 of the notes depicts the formation of a pn junction by a p-type Gaussian diffusion of boron into an n-type silicon substrate. Figure 4.7 on this slide illustrates that the impurity concentration of the p-type Gaussian diffusion of boron decreases as the distance from the surface of the silicon wafer is increased. The impurity concentration of the background n-type dopant remains constant at any distance from the surface of the silicon wafer. Figure 4.7 on this slide states that, "the metallurgical junction occurs at the point x = xj, where the net concentration is zero". This implies that xj occurs at the distance from the surface of the silicon wafer at which the concentration of the n-type dopant equals the concentration of the background p-type dopant. In this example provided on the slide, a p-type Gaussian diffusion of boron was made into an n-type substrate. Therefore, a p-type region is formed between the surface of the silicon wafer and xj. For distances deeper into the wafer than xj, an n-type region is formed. In question 7-2, we are told that the measured phosphorous profile in the wafer can be represented by a Gaussian function. Therefore, the diffusion is an n-type diffusion of phosphorous. Since the n-type Gaussian diffusion of phosphorous is being made into a uniformly doped p-type substrate, a pn junction will be formed at a distance xj from the surface of the wafer at which the impurity concentration of the phosphorous equals the background impurity concentration of the substrate.
The slide entitled "Diffusion Solutions" in section 6 of the notes gives an equation relating the impurity concentration in a limited source diffusion as a function of the time required for the diffusion to occur, t, and the distance of a given concentration of dopant from the surface of the wafer, x. This equation is as follows:
where Q is the total impurity concentration present in the wafer, D is the diffusion coefficient of the given diffusion, t is the time required for the diffusion to occur, x is the distance of a given concentration of dopant from the surface of the wafer, and N(x,t) is impurity concentration of the limited source diffusion as a function of x and t. This equation can be rearranged to solve for Q in the following manner:
The question has stated that the measured surface concentration, which is the concentration of n-type phosphorous dopant present on the surface of the wafer, is 1018 atoms/cm3. The symbol N(0,t) can be used to denote the concentration of the dopant present on the surface of the wafer. Therefore, N(0,t) = 1018 atoms/cm3. The variable x is defined as the distance of a given concentration of dopant from the surface of the wafer. Therefore, at the surface of the wafer, by definition, x = 0 cm. Additionally, the question has stated that the measured phosphorous profile in the wafer can be represented by a Gaussian function with a diffusivity of D = 2.3x10-13 cm2/sec. Substituting these values for N(0,t) = 1018 atoms/cm3, x = 0 cm, and D = 2.3x10-13 cm2/sec into the above the above expression for Q yields:
with significant figures applied
The question has stated that the junction depth is 1 micron, implying xj = 1 um. Using the conversion factors 1 cm = 10-2 m and 1 um = 10-6 m, xj can be converted to cm in the following manner:
with significant figures applied
The question has also stated that the substrate acceptor concentration, which is the background p-type doping level of the substrate, is 1015 atoms/cm3. A pn junction will be formed at a distance xj from the surface of the wafer at which the phosphorous impurity concentration is equal to 1015 atoms/cm3. Therefore, for xj = 1x10-4 cm, N(x,t) = 1015 atoms/cm3. Additionally, the question has stated that the measured phosphorous profile in the wafer can be represented by a Gaussian function with a diffusivity of D = 2.3x10-13 cm2/sec.
The above expression for N(x,t) in the Gaussian distribution can be used to write an expression for N(xj,t) in the following manner:
We can now substitute our previously determined and given values for Q, N(xj,t), xj and D into the above equation for N(xj,t):
with significant figures applied
The above equation can now be rearranged to solve for the diffusion time, t, required to create the pn junction at xj = 1x10-4 cm in the diffusion process:
with significant figures applied
Substituting this diffusion time back into the above expression for Q allows us to solve for the total dopant in the diffused wafer:
with significant figures applied
b).
If the diffusion temperature was changed so that the D = 1.4x10-13 cm2/sec, we must now calculate the new junction depth. In question 7-1-a, we obtained an expression for the junction depth of a Gaussian distribution. This expression was given to be:
In question 7-1-a, we also introduced an equation for the impurity concentration in a Gaussian distribution:
The surface concentration, CS, is defined as being located directly on the surface of the wafer. Therefore, we may simultaneously substitute both x = 0 and N(0,t) = CS into the above expression for N(x,t) in a Gaussian distribution. Performing this substitution yields:
This expression can be simplified to the following equation:
It can be readily seen from the above equation that if we alter the diffusion coefficient, D, in a system, at least one of the factors CS, t or Q must change as well. However, question 7-2-b gives no information about which (if any) of the CS, t ,or Q variables must retain the same values they had in earlier portions of question 7. As a result, I have completed this question with the assumption that up to two of these variables may retain the same values they had in earlier portions of question 7, as long as all three are not forced to retain their old values. I have chosen to keep the values of Q and t at the same levels they were solved to be at in question 7-2-a. This implies that, when D is changed to its new value, the surface concentration CS must change to accommodate this variation in D. In section 6 of the notes, the slide entitled "Limited Source Diffusion Solutions" also supports the concept that, as the D*t product is altered, the surface concentration changes correspondingly.
I have also completed this question with the assumption that the substrate doping level will retain the same values that it had in question 7-1-b. Having made these assumptions, I can now substitute the previously determined values for Q, t and Csub into my above expression for xj. After performing these substitutions, I obtain the following result for xj:
with significant figures applied
