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Cpar_ are interconnect parasitics. Cparm is one such
(last character varied for uniqueness). The transistor-
as-capacitor elements are plainly shown as such.
I doubt that, as the schematic calls out VINT=2.0V,
you are looking at 0.14u technology. Unless the thick
oxide devices (?) are used.
You would have to descend into the inverter schematics
(or perhaps check properties, if the L is simply not
displayed but is passed down - this varies by library)
to see what the embedded L values are. Assuming is
bad. But you could assume that the WN, WP are used
literally since they end in "U".
Would be, yes. But I don't see any 0.14U spec. in your schematic, just 1.4U . This would agree with my assumption below.so 0.14U is same as 0.14um in the schematic?
Depends on your process size and its gate thickness tox. See below the cap calculation from process size and its toxHow much the capacitance each transistor as capacitor carry?
Cpar_ are interconnect parasitics. Cparm is one such
(last character varied for uniqueness). The transistor-
as-capacitor elements are plainly shown as such.
Of course they have relevance: they actually exist and are necessary for realistic simulation.are Cparm, Cpart only for spice simulation only? They have no relevance in layout and design?
These are intentionally instantiated to achieve the required capacitive load.the Cparm values are there for line parastic capacitance, then what are the M1-M4 capacitors there for?
Of course they have relevance: they actually exist and are necessary for realistic simulation.
.
They are a result of said fabrication, like it or not.
Model them during design, or be surprised at test.
approximation you may count with a total gate-oxide area of W*3L per transistor .
As an example, this would arise to a gate-cap of ≈6.3 fF from the M3/M4 PMOS transistors in your schematic.
why area is W*3L? shouldn't that be W*L?
Check the drain & source connection areas of these transistors: these also are active areas and count for capacitance. As an approximation, the lengths of these connection areas correspond to to the gate length between them (and their widths correspond to the gate width).
why not W*2L, W*4L? I still don't see how you derived the capacitance area of W*3L?
L[SUB]ges[/SUB] = L[SUB]d[/SUB]+L[SUB]g[/SUB]+L[SUB]s[/SUB] = 3L[SUB]g[/SUB]