can any one tell this...

[philba]

having worked for a semiconductor company for 7 years, I know that wafer allocation and defect models are a huge area of concern. every wafer has defects and it behoves the manufacturer to figure out how to minimize the number of die that fall on obvious defect areas. not simple at all. more viable die per wafer, means a greater yield. since the wafer is a fixed cost, greater yield means lower cost per die. which means more profit (or greater market share due to lower price).

 

[_nox_]

having worked for a semiconductor company for 7 years, I know that wafer allocation and defect models are a huge area of concern. every wafer has defects and it behoves the manufacturer to figure out how to minimize the number of die that fall on obvious defect areas. not simple at all. more viable die per wafer, means a greater yield. since the wafer is a fixed cost, greater yield means lower cost per die. which means more profit (or greater market share due to lower price).

What exactly causes such defects ?

 

[philba]

What exactly causes such defects ?

in 25 words or less?? I'm no expert but I believe the wafers themselves will have crystaline defects. then there is dirt - I don't recall the size they filter for but its a religion in the industry. vibrations. other things. some can be known in advance some are inherent in the process. ooops, too many words.

 

[itsallabtamrit]

mathematically..

As an intellectual exercise you might want to compute how many rectangles measuring L by W you can fit on a circular disc of diameter D. For extra credit you can compute the amount of waste as a percentage of the original area.

Guys who can do this are in high demand.

well i guess we can do it mathematically too..within 10 min..i've solved..tell me it's wrong or right..
according to question we get an equation..
consider the number of rectangles be 'n' and waste area be 'x'.. so the equation is.... nLW+x=area of circle..(i cant write pie as symbol..)
so 'x'=area-nLW
%waste=x/area of circle*100
i.e area-nLW/area*100
i.e [1-(nLW/area)]*100
that means nLW/area shud always be less that 1
calculating it we get a constant 4/pi=1.3
so the equation becomes [1-1.3(nLW/D^2)]*100
that means 1.3(nLW/D^2)<1
i.e nLW/D^2<1/1.3
i.e nLW/D^2<.76(approx)
so n<D^2*.76/LW
so the waste area% is <[ 1-(1.3*.76)]=.01(approx)
so the ans is <10%..now u tell me if i'm wrong or right..

 

[TKS]

uhh wasn't the position also needed to be usefull? because else the maximum wasted size is the size of the die

Tks

 

[_nox_]

because else the maximum wasted size is the size of the die

(wafer area) - (number of dies)(single die area) = (wasted wafer area)

@itsallabtamrit:
i assume it's partially correct, but that's not really useful, as TKS noticed.

 

[Roff]

It depends on the size of the die relative to the size of the wafer. The worst case (square) is a single square die inscribed in a circle, where the waste is approximately 36%. The other limit is where the die area approaches zero and the wafer area approaches infinity, in which case the waste is zero.

 

[_nox_]

It depends on the size of the die relative to the size of the wafer. The worst case (square) is a single square die inscribed in a circle, where the waste is approximately 36%. The other limit is where the die area approaches zero and the wafer area approaches infinity, in which case the waste is zero.

A pretty self-explanatory comment.

 

[TKS]

mhh mathematically thinking i also think that even if the wafer woud be a number then the more to 0 its is the smaller the waste (closer to 0 to)

What happend to your avatar Ron H

Tks

 

[Roff]

A pretty self-explanatory comment.

Not sure exactly what you mean by that, but I was aiming my comments at itsallabtamrit, whose equation had no provisions for relative die size, as far as I could tell. I got lost in his maths, partly because he was not rigorous with ths parentheses.

 

[Roff]

What happend to your avatar Ron H

Tks

I had a bad day.

Actually, I was bored, so I did a little cosmetic enhancement. I think I look a lot better now, don't you?

 

[_nox_]

Not sure exactly what you mean by that, but I was aiming my comments at itsallabtamrit, whose equation had no provisions for relative die size, as far as I could tell. I got lost in his maths, partly because he was not rigorous with ths parentheses.

ah forget it It wasn't thought for me...

 

[dknguyen]

I had a bad day.

Actually, I was bored, so I did a little cosmetic enhancement. I think I look a lot better now, don't you?

Your new eyes are freakishly high on your forhead. I would lower them a smidegen to make them look more like your eyes.

 

[Roff]

Your new eyes are freakishly high on your forhead. I would lower them a smidegen to make them look more like your eyes.

Maybe I just have freakishly high eyes. Or maybe a freakishly low forehead.

 

[dknguyen]

Maybe I just have freakishly high eyes. Or maybe a freakishly low forehead.

I think it's because your new eyes are bright white and looking straight out of the picture even though the picture is taken at an agle which kind of makes them jump out bobble eyes. Maybe if you changed them to red or something it would lessen the effect.

 

[Roff]

I think it's because your new eyes are bright white and looking straight out of the picture even though the picture is taken at an agle which kind of makes them jump out bobble eyes. Maybe if you changed them to red or something it would lessen the effect.

I think I'll just switch to Charles Manson. He's scarier-looking than anything I could draw.

 

[dknguyen]

I think I'll just switch to Charles Manson. He's scarier-looking than anything I could draw.

There, much more believable, dare I even say slightly more lifelike? lol. Those cold eyes have more life in them than the bobble eyes.

 

[itsallabtamrit]

(wafer area) - (number of dies)(single die area) = (wasted wafer area)

@itsallabtamrit:
i assume it's partially correct, but that's not really useful, as TKS noticed.

i must say sometimes intellects do some mistakes..the question never asked to calculated maximum area..otherwise the area of the die can be the answer..since it's possible that an rectangle of given dimension couln't fit into that die of diameter D..but if we assume that rectangles do fit then i don't think my ans is wrong..
and wat RON H said..exactly in worst case wen only one square is possible 36% is the answer..but we r here to assume the hardest case nt the simplest one..
isn't it..?

 

[HiTech]

That nut-case Manson is scheduled for a parole hearing next year. Then again, it's just a parole hearing... doesn't mean he'll be approved! Here's Mr. Benevolence himself looking the part of a cross between a Buddhust monk and Osama binLaden:

**broken link removed**

 

[_nox_]

Here's Mr. Benevolence himself looking the part of a cross between a Buddhist monk and Osama binLaden:

drop that! there's no comparison between pure wisdom and infinite badness.
This sentence is a defamation....