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Why is the minimum feature size of filler ( 1u ) s...
# ihp-sg13g2
t
Why is the minimum feature size of filler ( 1u ) so large compared to the minimum feature of the equivalent metal layer ( 0.2u ) ? It does make it kid of hard to stick filler inside densish-routing ?
k
Since the purpose of fillers ensure the surface uniformity I guess that it makes sense, If you have dense routing on a layer you just do not need filler there.
t
Heh, that's just not the case in practice.
If every single routing track was used, the density would be 41.6%.
But if you look at something like :

https://i.imgur.com/byS3GqS.png▾

It's sort of dense ... and it's hard to find holes of 1.84u ( absolute min space you'd need to put filler ). But it's not enough to meet a 25% density target.
k
I guess that's why you have global and local densities. I guess that then this filler shapes contribute to the density and compensate the situation you have on the image.
t
But the issue is that if you have large digital blocks, it's hard to hit the windowed density when one of the window is pretty much only that digital block ...
( That 25% min is the 800u windowed target. )
And it would be much easier to hit if I could put 0.5u squares of filler ... ( in addition to the larger ones )
k
If this is the case maybe it is a high time to lower routing congestion. Not very productive result but still.
I can also ask some more experienced designer how do they compensate these low density areas.
t
But I can draw 0.5u squares on the 
drawing
layer, so why not on the 
filler
layer, that's was the original question intent. They both end up on the same mask in the end right ?
k
you are right, the drawing and the filler converge to a physical layer
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