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[SOLVED] is there any relation between IR drop and Noise Margin ?

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akhil_psm

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In ASIC design, is there any relation between IR drop and Noise Margin ? and If yes, then how?
 
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Re: ASIC Physical Design

Yes.

Your noise margin (for a fully settled output) is the
distance from rail voltage to threshold voltage. At rails
come together (I*R applies to both vdd and vss) the
distance must shrink. And that's just for the simple
static case.

Now consider a case where a gate at the worst "sagged"
point of the core, tries to drive another gate that's well
supplied (say, right up near the periphery bussing). The
core gate will have a vss potential higher and a vdd
potential lower than the peripheral gate, so the offset
subtracts from static noise margin and also gives sub-par gate
drive levels to the receiving gate. This can cause timing
to push out, possibly making logic fail for causes that
are not comprehended by static timing analysis.

Then there's that the I*R drop is not at all static, and things
like simultaneous switching can bounce the rails and signals
driven from them, big time - not just I*R, but L*dI/dt on
top of that. With about 1nH/mm, and chip sizes well above
10mm, and tens of pS risetimes, that's some ugly arithmetic.
Then think about hundreds or thousands of registers flipping
on "special occasions", cranking their output through high
drive (hence high shoot-through) buffers and millimeters
of wireload apiece....
 

Re: ASIC Physical Design

In ASIC design, is there any relation between IR drop and Noise Margin ? and If yes, then how?

Yes, there is. The explanation above is pretty thorough.

I would only add that in digital design we often don't think in terms of noise margin, we think in terms of library margins. If your library was characterized to work at nominal VDD-10%, that is where you draw the line.
 

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