Mixing
how many squares can go in circle?
I am reading a book on computer architecture. The author discusses how processor chips are made from wafers. Dies ( the heart of the chip, that does calculations) are cut from circular wafer plates.
Because dies are square (or rectangular I should say) and the wafer is a circle, not all of the area of the wafer is used for dies.
To calculate how many dies can go into a single wafer, the author uses this equation:
The second part of the equation compensates for the problem of squares in a circle. I am interested in the derivation of the second part. Can you help me out?
geometry
asked Nov 21 '18 at 10:18
potato
1013
add a comment |
I am reading a book on computer architecture. The author discusses how processor chips are made from wafers. Dies ( the heart of the chip, that does calculations) are cut from circular wafer plates.
Because dies are square (or rectangular I should say) and the wafer is a circle, not all of the area of the wafer is used for dies.
To calculate how many dies can go into a single wafer, the author uses this equation:
The second part of the equation compensates for the problem of squares in a circle. I am interested in the derivation of the second part. Can you help me out?
geometry
asked Nov 21 '18 at 10:18
potato
1013
add a comment |
I am reading a book on computer architecture. The author discusses how processor chips are made from wafers. Dies ( the heart of the chip, that does calculations) are cut from circular wafer plates.
Because dies are square (or rectangular I should say) and the wafer is a circle, not all of the area of the wafer is used for dies.
To calculate how many dies can go into a single wafer, the author uses this equation:
The second part of the equation compensates for the problem of squares in a circle. I am interested in the derivation of the second part. Can you help me out?
geometry
asked Nov 21 '18 at 10:18
potato
1013
I am reading a book on computer architecture. The author discusses how processor chips are made from wafers. Dies ( the heart of the chip, that does calculations) are cut from circular wafer plates.
Because dies are square (or rectangular I should say) and the wafer is a circle, not all of the area of the wafer is used for dies.
To calculate how many dies can go into a single wafer, the author uses this equation:
The second part of the equation compensates for the problem of squares in a circle. I am interested in the derivation of the second part. Can you help me out?
geometry
geometry
asked Nov 21 '18 at 10:18
potato
1013
asked Nov 21 '18 at 10:18
potato
1013
asked Nov 21 '18 at 10:18
potato
1013
asked Nov 21 '18 at 10:18
potato
1013
asked Nov 21 '18 at 10:18
potato
1013
1013
add a comment |
add a comment |
1 Answer
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oldest
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$$text{Dies per wafer} = frac{ pi times (text{Wafer diameter} / 2)^2}{text{Die area}} - frac{pi times (text{Wafer diameter})}{sqrt{2 times text{Die area}}}$$
Because $text{Wafer diameter} / 2 = text{Wafer radius}$, $pi times text{Wafer diameter} = text{Wafer circumference}$, and $sqrt{text{Die area}} = text{Square die edge length}$, the above is equivalent to
$$text{Dies per wafer} = frac{text{Wafer area}}{text{Die area}} - sqrt{frac{1}{2}} frac{text{Wafer circumference}}{text{Square die edge length}}$$
The left side is the number of dies you could get, if the wafer had a shape that could be exactly packed with dies. The right side is the estimated number of dies "lost" because of the disc shape.
The way you can derive this estimate is simple. Let's assume $r$ is the wafer radius, and $L$ is the edge length for each square die, and that the area unusable for dies corresponds to the area within $L/sqrt{2}$ of the perimeter; within half the diagonal of each die. That gives us the estimate on the number of dies $N$ we can fit on the wafer as
$$begin{aligned}
N(r, L) &approx frac{pi (r - sqrt{1/2}L)^2}{L^2} \
; &= frac{pi r^2}{L^2} - frac{2 pi sqrt{1/2} r L}{L^2} + frac{pi L^2}{2 L^2} \
; &= pi left(frac{r}{L}right)^2 - frac{2 pi r}{sqrt{2} L} + frac{pi}{2} \
end{aligned}$$
The third term $pi/2 approx 1.57$, and is obviously insignificant for the estimate, so we can drop it. Our estimate then becomes
$$N(r, L) = frac{pi r^2}{L^2} - frac{2 pi r}{sqrt{2} L}$$
If we write $r$ as $text{Wafer diameter} / 2$, and $L^2$ as $text{Die area}$, it becomes
$$text{Dies per wafer} = frac{pi times (text{Wafer diameter} / 2)^2}{text{Die area}} - frac{pi times (text{Wafer diameter})}{sqrt{2 times (text{Die area})}}$$
which is exactly the estimate at hand.
Of course, we don't know if the original authors derived it this way, but this is definitely a sensible way one can derive the estimate.
Comparing to bounds estimates shown here, $lambda = text{Wafer radius} / text{Square die edge length}$. The above estimate of the number of dies then equates to
$$N(lambda) = pi lambda^2 - sqrt{frac{1}{2}} 2 pi lambda = pi lambda left ( lambda - sqrt{2} right )$$
which is smaller than the upper bounds derived in that other answer. In other words, the $text{Dies per wafer}$ estimate is quite conservative.
answered Nov 21 '18 at 14:02
Nominal Animal
6,7952517
Thanks @Nominal Animal for taking your time. I have a subquestion. Can you further explain this sentance: "area unusable for dies corresponds to the area within L/2‾√ of the perimeter". How did you get that L/2‾√ relation?
– potato
Nov 21 '18 at 22:37
@potato: $L / sqrt{2}$ is half the length of the diagonal of a square with edges of length $L$. I actually derived the logic "backwards", writing it as $N(r, L) = frac{pi r^2}{L^2} - frac{2 pi r}{sqrt{2} L}$ first. The right side is the ratio of the wafer perimeter to the die edge length, divided by $sqrt{2}$. Complete to a square (with an additional small constant not depending on $r$ or $L$), and you get to my original description. (I did not know the definition before I got there, but I knew I could work it out when I saw the formula.)
– Nominal Animal
Nov 22 '18 at 1:39
Thanks again for your time. I was hoping for a "true" derivation without the equation to start with. I mean, would you get the same result, if you didn't know the original equation?
– potato
Nov 22 '18 at 11:37
@potato: I'm not a telepath, so I can only show you an example of how such an approximation can be derived. If we assume dies can be at the very edge of the wafer, the estimate $maxleft(N_0(r/L), N_1(r/L)right)$ I derived here says we can fit many more dies on each wafer than the estimate you are asking about. [...]
– Nominal Animal
Nov 22 '18 at 16:47
[...] If, however, there was a reason to estimate that near the wafer perimeter, an area equal to an annulus with thickness (difference in radii) half the diagonal of the die is unusable, then you'd get the approximation/estimate you are talking about (ignoring the irrelevant small constant number of dies). Is there a reason for that? I don't know. You need to ask about that in the physics or possibly electrical engineering subforums, because it really depends on what kind of limitations the manufacturing method has.
– Nominal Animal
Nov 22 '18 at 16:52
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$$text{Dies per wafer} = frac{ pi times (text{Wafer diameter} / 2)^2}{text{Die area}} - frac{pi times (text{Wafer diameter})}{sqrt{2 times text{Die area}}}$$
Because $text{Wafer diameter} / 2 = text{Wafer radius}$, $pi times text{Wafer diameter} = text{Wafer circumference}$, and $sqrt{text{Die area}} = text{Square die edge length}$, the above is equivalent to
$$text{Dies per wafer} = frac{text{Wafer area}}{text{Die area}} - sqrt{frac{1}{2}} frac{text{Wafer circumference}}{text{Square die edge length}}$$
The left side is the number of dies you could get, if the wafer had a shape that could be exactly packed with dies. The right side is the estimated number of dies "lost" because of the disc shape.
The way you can derive this estimate is simple. Let's assume $r$ is the wafer radius, and $L$ is the edge length for each square die, and that the area unusable for dies corresponds to the area within $L/sqrt{2}$ of the perimeter; within half the diagonal of each die. That gives us the estimate on the number of dies $N$ we can fit on the wafer as
$$begin{aligned}
N(r, L) &approx frac{pi (r - sqrt{1/2}L)^2}{L^2} \
; &= frac{pi r^2}{L^2} - frac{2 pi sqrt{1/2} r L}{L^2} + frac{pi L^2}{2 L^2} \
; &= pi left(frac{r}{L}right)^2 - frac{2 pi r}{sqrt{2} L} + frac{pi}{2} \
end{aligned}$$
The third term $pi/2 approx 1.57$, and is obviously insignificant for the estimate, so we can drop it. Our estimate then becomes
$$N(r, L) = frac{pi r^2}{L^2} - frac{2 pi r}{sqrt{2} L}$$
If we write $r$ as $text{Wafer diameter} / 2$, and $L^2$ as $text{Die area}$, it becomes
$$text{Dies per wafer} = frac{pi times (text{Wafer diameter} / 2)^2}{text{Die area}} - frac{pi times (text{Wafer diameter})}{sqrt{2 times (text{Die area})}}$$
which is exactly the estimate at hand.
Of course, we don't know if the original authors derived it this way, but this is definitely a sensible way one can derive the estimate.
Comparing to bounds estimates shown here, $lambda = text{Wafer radius} / text{Square die edge length}$. The above estimate of the number of dies then equates to
$$N(lambda) = pi lambda^2 - sqrt{frac{1}{2}} 2 pi lambda = pi lambda left ( lambda - sqrt{2} right )$$
which is smaller than the upper bounds derived in that other answer. In other words, the $text{Dies per wafer}$ estimate is quite conservative.
answered Nov 21 '18 at 14:02
Nominal Animal
6,7952517
Thanks @Nominal Animal for taking your time. I have a subquestion. Can you further explain this sentance: "area unusable for dies corresponds to the area within L/2‾√ of the perimeter". How did you get that L/2‾√ relation?
– potato
Nov 21 '18 at 22:37
@potato: $L / sqrt{2}$ is half the length of the diagonal of a square with edges of length $L$. I actually derived the logic "backwards", writing it as $N(r, L) = frac{pi r^2}{L^2} - frac{2 pi r}{sqrt{2} L}$ first. The right side is the ratio of the wafer perimeter to the die edge length, divided by $sqrt{2}$. Complete to a square (with an additional small constant not depending on $r$ or $L$), and you get to my original description. (I did not know the definition before I got there, but I knew I could work it out when I saw the formula.)
– Nominal Animal
Nov 22 '18 at 1:39
Thanks again for your time. I was hoping for a "true" derivation without the equation to start with. I mean, would you get the same result, if you didn't know the original equation?
– potato
Nov 22 '18 at 11:37
@potato: I'm not a telepath, so I can only show you an example of how such an approximation can be derived. If we assume dies can be at the very edge of the wafer, the estimate $maxleft(N_0(r/L), N_1(r/L)right)$ I derived here says we can fit many more dies on each wafer than the estimate you are asking about. [...]
– Nominal Animal
Nov 22 '18 at 16:47
[...] If, however, there was a reason to estimate that near the wafer perimeter, an area equal to an annulus with thickness (difference in radii) half the diagonal of the die is unusable, then you'd get the approximation/estimate you are talking about (ignoring the irrelevant small constant number of dies). Is there a reason for that? I don't know. You need to ask about that in the physics or possibly electrical engineering subforums, because it really depends on what kind of limitations the manufacturing method has.
– Nominal Animal
Nov 22 '18 at 16:52
add a comment |
$$text{Dies per wafer} = frac{ pi times (text{Wafer diameter} / 2)^2}{text{Die area}} - frac{pi times (text{Wafer diameter})}{sqrt{2 times text{Die area}}}$$
Because $text{Wafer diameter} / 2 = text{Wafer radius}$, $pi times text{Wafer diameter} = text{Wafer circumference}$, and $sqrt{text{Die area}} = text{Square die edge length}$, the above is equivalent to
$$text{Dies per wafer} = frac{text{Wafer area}}{text{Die area}} - sqrt{frac{1}{2}} frac{text{Wafer circumference}}{text{Square die edge length}}$$
The left side is the number of dies you could get, if the wafer had a shape that could be exactly packed with dies. The right side is the estimated number of dies "lost" because of the disc shape.
The way you can derive this estimate is simple. Let's assume $r$ is the wafer radius, and $L$ is the edge length for each square die, and that the area unusable for dies corresponds to the area within $L/sqrt{2}$ of the perimeter; within half the diagonal of each die. That gives us the estimate on the number of dies $N$ we can fit on the wafer as
$$begin{aligned}
N(r, L) &approx frac{pi (r - sqrt{1/2}L)^2}{L^2} \
; &= frac{pi r^2}{L^2} - frac{2 pi sqrt{1/2} r L}{L^2} + frac{pi L^2}{2 L^2} \
; &= pi left(frac{r}{L}right)^2 - frac{2 pi r}{sqrt{2} L} + frac{pi}{2} \
end{aligned}$$
The third term $pi/2 approx 1.57$, and is obviously insignificant for the estimate, so we can drop it. Our estimate then becomes
$$N(r, L) = frac{pi r^2}{L^2} - frac{2 pi r}{sqrt{2} L}$$
If we write $r$ as $text{Wafer diameter} / 2$, and $L^2$ as $text{Die area}$, it becomes
$$text{Dies per wafer} = frac{pi times (text{Wafer diameter} / 2)^2}{text{Die area}} - frac{pi times (text{Wafer diameter})}{sqrt{2 times (text{Die area})}}$$
which is exactly the estimate at hand.
Of course, we don't know if the original authors derived it this way, but this is definitely a sensible way one can derive the estimate.
Comparing to bounds estimates shown here, $lambda = text{Wafer radius} / text{Square die edge length}$. The above estimate of the number of dies then equates to
$$N(lambda) = pi lambda^2 - sqrt{frac{1}{2}} 2 pi lambda = pi lambda left ( lambda - sqrt{2} right )$$
which is smaller than the upper bounds derived in that other answer. In other words, the $text{Dies per wafer}$ estimate is quite conservative.
answered Nov 21 '18 at 14:02
Nominal Animal
6,7952517
Thanks @Nominal Animal for taking your time. I have a subquestion. Can you further explain this sentance: "area unusable for dies corresponds to the area within L/2‾√ of the perimeter". How did you get that L/2‾√ relation?
– potato
Nov 21 '18 at 22:37
@potato: $L / sqrt{2}$ is half the length of the diagonal of a square with edges of length $L$. I actually derived the logic "backwards", writing it as $N(r, L) = frac{pi r^2}{L^2} - frac{2 pi r}{sqrt{2} L}$ first. The right side is the ratio of the wafer perimeter to the die edge length, divided by $sqrt{2}$. Complete to a square (with an additional small constant not depending on $r$ or $L$), and you get to my original description. (I did not know the definition before I got there, but I knew I could work it out when I saw the formula.)
– Nominal Animal
Nov 22 '18 at 1:39
Thanks again for your time. I was hoping for a "true" derivation without the equation to start with. I mean, would you get the same result, if you didn't know the original equation?
– potato
Nov 22 '18 at 11:37
@potato: I'm not a telepath, so I can only show you an example of how such an approximation can be derived. If we assume dies can be at the very edge of the wafer, the estimate $maxleft(N_0(r/L), N_1(r/L)right)$ I derived here says we can fit many more dies on each wafer than the estimate you are asking about. [...]
– Nominal Animal
Nov 22 '18 at 16:47
[...] If, however, there was a reason to estimate that near the wafer perimeter, an area equal to an annulus with thickness (difference in radii) half the diagonal of the die is unusable, then you'd get the approximation/estimate you are talking about (ignoring the irrelevant small constant number of dies). Is there a reason for that? I don't know. You need to ask about that in the physics or possibly electrical engineering subforums, because it really depends on what kind of limitations the manufacturing method has.
– Nominal Animal
Nov 22 '18 at 16:52
add a comment |
$$text{Dies per wafer} = frac{ pi times (text{Wafer diameter} / 2)^2}{text{Die area}} - frac{pi times (text{Wafer diameter})}{sqrt{2 times text{Die area}}}$$
Because $text{Wafer diameter} / 2 = text{Wafer radius}$, $pi times text{Wafer diameter} = text{Wafer circumference}$, and $sqrt{text{Die area}} = text{Square die edge length}$, the above is equivalent to
$$text{Dies per wafer} = frac{text{Wafer area}}{text{Die area}} - sqrt{frac{1}{2}} frac{text{Wafer circumference}}{text{Square die edge length}}$$
The left side is the number of dies you could get, if the wafer had a shape that could be exactly packed with dies. The right side is the estimated number of dies "lost" because of the disc shape.
The way you can derive this estimate is simple. Let's assume $r$ is the wafer radius, and $L$ is the edge length for each square die, and that the area unusable for dies corresponds to the area within $L/sqrt{2}$ of the perimeter; within half the diagonal of each die. That gives us the estimate on the number of dies $N$ we can fit on the wafer as
$$begin{aligned}
N(r, L) &approx frac{pi (r - sqrt{1/2}L)^2}{L^2} \
; &= frac{pi r^2}{L^2} - frac{2 pi sqrt{1/2} r L}{L^2} + frac{pi L^2}{2 L^2} \
; &= pi left(frac{r}{L}right)^2 - frac{2 pi r}{sqrt{2} L} + frac{pi}{2} \
end{aligned}$$
The third term $pi/2 approx 1.57$, and is obviously insignificant for the estimate, so we can drop it. Our estimate then becomes
$$N(r, L) = frac{pi r^2}{L^2} - frac{2 pi r}{sqrt{2} L}$$
If we write $r$ as $text{Wafer diameter} / 2$, and $L^2$ as $text{Die area}$, it becomes
$$text{Dies per wafer} = frac{pi times (text{Wafer diameter} / 2)^2}{text{Die area}} - frac{pi times (text{Wafer diameter})}{sqrt{2 times (text{Die area})}}$$
which is exactly the estimate at hand.
Of course, we don't know if the original authors derived it this way, but this is definitely a sensible way one can derive the estimate.
Comparing to bounds estimates shown here, $lambda = text{Wafer radius} / text{Square die edge length}$. The above estimate of the number of dies then equates to
$$N(lambda) = pi lambda^2 - sqrt{frac{1}{2}} 2 pi lambda = pi lambda left ( lambda - sqrt{2} right )$$
which is smaller than the upper bounds derived in that other answer. In other words, the $text{Dies per wafer}$ estimate is quite conservative.
answered Nov 21 '18 at 14:02
Nominal Animal
6,7952517
$$text{Dies per wafer} = frac{ pi times (text{Wafer diameter} / 2)^2}{text{Die area}} - frac{pi times (text{Wafer diameter})}{sqrt{2 times text{Die area}}}$$
Because $text{Wafer diameter} / 2 = text{Wafer radius}$, $pi times text{Wafer diameter} = text{Wafer circumference}$, and $sqrt{text{Die area}} = text{Square die edge length}$, the above is equivalent to
$$text{Dies per wafer} = frac{text{Wafer area}}{text{Die area}} - sqrt{frac{1}{2}} frac{text{Wafer circumference}}{text{Square die edge length}}$$
The left side is the number of dies you could get, if the wafer had a shape that could be exactly packed with dies. The right side is the estimated number of dies "lost" because of the disc shape.
The way you can derive this estimate is simple. Let's assume $r$ is the wafer radius, and $L$ is the edge length for each square die, and that the area unusable for dies corresponds to the area within $L/sqrt{2}$ of the perimeter; within half the diagonal of each die. That gives us the estimate on the number of dies $N$ we can fit on the wafer as
$$begin{aligned}
N(r, L) &approx frac{pi (r - sqrt{1/2}L)^2}{L^2} \
; &= frac{pi r^2}{L^2} - frac{2 pi sqrt{1/2} r L}{L^2} + frac{pi L^2}{2 L^2} \
; &= pi left(frac{r}{L}right)^2 - frac{2 pi r}{sqrt{2} L} + frac{pi}{2} \
end{aligned}$$
The third term $pi/2 approx 1.57$, and is obviously insignificant for the estimate, so we can drop it. Our estimate then becomes
$$N(r, L) = frac{pi r^2}{L^2} - frac{2 pi r}{sqrt{2} L}$$
If we write $r$ as $text{Wafer diameter} / 2$, and $L^2$ as $text{Die area}$, it becomes
$$text{Dies per wafer} = frac{pi times (text{Wafer diameter} / 2)^2}{text{Die area}} - frac{pi times (text{Wafer diameter})}{sqrt{2 times (text{Die area})}}$$
which is exactly the estimate at hand.
Of course, we don't know if the original authors derived it this way, but this is definitely a sensible way one can derive the estimate.
Comparing to bounds estimates shown here, $lambda = text{Wafer radius} / text{Square die edge length}$. The above estimate of the number of dies then equates to
$$N(lambda) = pi lambda^2 - sqrt{frac{1}{2}} 2 pi lambda = pi lambda left ( lambda - sqrt{2} right )$$
which is smaller than the upper bounds derived in that other answer. In other words, the $text{Dies per wafer}$ estimate is quite conservative.
answered Nov 21 '18 at 14:02
Nominal Animal
6,7952517
answered Nov 21 '18 at 14:02
Nominal Animal
6,7952517
answered Nov 21 '18 at 14:02
Nominal Animal
6,7952517
answered Nov 21 '18 at 14:02
Nominal Animal
6,7952517
6,7952517
Thanks @Nominal Animal for taking your time. I have a subquestion. Can you further explain this sentance: "area unusable for dies corresponds to the area within L/2‾√ of the perimeter". How did you get that L/2‾√ relation?
– potato
Nov 21 '18 at 22:37
@potato: $L / sqrt{2}$ is half the length of the diagonal of a square with edges of length $L$. I actually derived the logic "backwards", writing it as $N(r, L) = frac{pi r^2}{L^2} - frac{2 pi r}{sqrt{2} L}$ first. The right side is the ratio of the wafer perimeter to the die edge length, divided by $sqrt{2}$. Complete to a square (with an additional small constant not depending on $r$ or $L$), and you get to my original description. (I did not know the definition before I got there, but I knew I could work it out when I saw the formula.)
– Nominal Animal
Nov 22 '18 at 1:39
Thanks again for your time. I was hoping for a "true" derivation without the equation to start with. I mean, would you get the same result, if you didn't know the original equation?
– potato
Nov 22 '18 at 11:37
@potato: I'm not a telepath, so I can only show you an example of how such an approximation can be derived. If we assume dies can be at the very edge of the wafer, the estimate $maxleft(N_0(r/L), N_1(r/L)right)$ I derived here says we can fit many more dies on each wafer than the estimate you are asking about. [...]
– Nominal Animal
Nov 22 '18 at 16:47
[...] If, however, there was a reason to estimate that near the wafer perimeter, an area equal to an annulus with thickness (difference in radii) half the diagonal of the die is unusable, then you'd get the approximation/estimate you are talking about (ignoring the irrelevant small constant number of dies). Is there a reason for that? I don't know. You need to ask about that in the physics or possibly electrical engineering subforums, because it really depends on what kind of limitations the manufacturing method has.
– Nominal Animal
Nov 22 '18 at 16:52
add a comment |
Thanks @Nominal Animal for taking your time. I have a subquestion. Can you further explain this sentance: "area unusable for dies corresponds to the area within L/2‾√ of the perimeter". How did you get that L/2‾√ relation?
– potato
Nov 21 '18 at 22:37
@potato: $L / sqrt{2}$ is half the length of the diagonal of a square with edges of length $L$. I actually derived the logic "backwards", writing it as $N(r, L) = frac{pi r^2}{L^2} - frac{2 pi r}{sqrt{2} L}$ first. The right side is the ratio of the wafer perimeter to the die edge length, divided by $sqrt{2}$. Complete to a square (with an additional small constant not depending on $r$ or $L$), and you get to my original description. (I did not know the definition before I got there, but I knew I could work it out when I saw the formula.)
– Nominal Animal
Nov 22 '18 at 1:39
Thanks again for your time. I was hoping for a "true" derivation without the equation to start with. I mean, would you get the same result, if you didn't know the original equation?
– potato
Nov 22 '18 at 11:37
@potato: I'm not a telepath, so I can only show you an example of how such an approximation can be derived. If we assume dies can be at the very edge of the wafer, the estimate $maxleft(N_0(r/L), N_1(r/L)right)$ I derived here says we can fit many more dies on each wafer than the estimate you are asking about. [...]
– Nominal Animal
Nov 22 '18 at 16:47
[...] If, however, there was a reason to estimate that near the wafer perimeter, an area equal to an annulus with thickness (difference in radii) half the diagonal of the die is unusable, then you'd get the approximation/estimate you are talking about (ignoring the irrelevant small constant number of dies). Is there a reason for that? I don't know. You need to ask about that in the physics or possibly electrical engineering subforums, because it really depends on what kind of limitations the manufacturing method has.
– Nominal Animal
Nov 22 '18 at 16:52
Thanks @Nominal Animal for taking your time. I have a subquestion. Can you further explain this sentance: "area unusable for dies corresponds to the area within L/2‾√ of the perimeter". How did you get that L/2‾√ relation?
– potato
Nov 21 '18 at 22:37
Thanks @Nominal Animal for taking your time. I have a subquestion. Can you further explain this sentance: "area unusable for dies corresponds to the area within L/2‾√ of the perimeter". How did you get that L/2‾√ relation?
– potato
Nov 21 '18 at 22:37
@potato: $L / sqrt{2}$ is half the length of the diagonal of a square with edges of length $L$. I actually derived the logic "backwards", writing it as $N(r, L) = frac{pi r^2}{L^2} - frac{2 pi r}{sqrt{2} L}$ first. The right side is the ratio of the wafer perimeter to the die edge length, divided by $sqrt{2}$. Complete to a square (with an additional small constant not depending on $r$ or $L$), and you get to my original description. (I did not know the definition before I got there, but I knew I could work it out when I saw the formula.)
– Nominal Animal
Nov 22 '18 at 1:39
@potato: $L / sqrt{2}$ is half the length of the diagonal of a square with edges of length $L$. I actually derived the logic "backwards", writing it as $N(r, L) = frac{pi r^2}{L^2} - frac{2 pi r}{sqrt{2} L}$ first. The right side is the ratio of the wafer perimeter to the die edge length, divided by $sqrt{2}$. Complete to a square (with an additional small constant not depending on $r$ or $L$), and you get to my original description. (I did not know the definition before I got there, but I knew I could work it out when I saw the formula.)
– Nominal Animal
Nov 22 '18 at 1:39
Thanks again for your time. I was hoping for a "true" derivation without the equation to start with. I mean, would you get the same result, if you didn't know the original equation?
– potato
Nov 22 '18 at 11:37
Thanks again for your time. I was hoping for a "true" derivation without the equation to start with. I mean, would you get the same result, if you didn't know the original equation?
– potato
Nov 22 '18 at 11:37
@potato: I'm not a telepath, so I can only show you an example of how such an approximation can be derived. If we assume dies can be at the very edge of the wafer, the estimate $maxleft(N_0(r/L), N_1(r/L)right)$ I derived here says we can fit many more dies on each wafer than the estimate you are asking about. [...]
– Nominal Animal
Nov 22 '18 at 16:47
@potato: I'm not a telepath, so I can only show you an example of how such an approximation can be derived. If we assume dies can be at the very edge of the wafer, the estimate $maxleft(N_0(r/L), N_1(r/L)right)$ I derived here says we can fit many more dies on each wafer than the estimate you are asking about. [...]
– Nominal Animal
Nov 22 '18 at 16:47
[...] If, however, there was a reason to estimate that near the wafer perimeter, an area equal to an annulus with thickness (difference in radii) half the diagonal of the die is unusable, then you'd get the approximation/estimate you are talking about (ignoring the irrelevant small constant number of dies). Is there a reason for that? I don't know. You need to ask about that in the physics or possibly electrical engineering subforums, because it really depends on what kind of limitations the manufacturing method has.
– Nominal Animal
Nov 22 '18 at 16:52
[...] If, however, there was a reason to estimate that near the wafer perimeter, an area equal to an annulus with thickness (difference in radii) half the diagonal of the die is unusable, then you'd get the approximation/estimate you are talking about (ignoring the irrelevant small constant number of dies). Is there a reason for that? I don't know. You need to ask about that in the physics or possibly electrical engineering subforums, because it really depends on what kind of limitations the manufacturing method has.
– Nominal Animal
Nov 22 '18 at 16:52
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