the series R of varactor is different between m=2 and f=2 ?

Fahmy said:

If the varactor is based on Mosfet transistor then :
M=2 is equivalent to repeating the transistor two times , each instance have its own source/ drain . so we have two gates and four S/D regions.
F=2 is equivalent of having Transistor with two fingers . i.e we have one source and two drains.So we have only 3 S/D regions.

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walkingsun said:

from the simulation, the equivalent res of nmos varactor with M=2 F=1 is smaller than the one of with M=1 F=2.

and, if the number of M*F is large, the difference of equivalent res is very obvious.

so, could i say that the more fingers mean smaller area, and reasonable multiplier means suitable resistance?

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Fahmy said:

walkingsun said:

of course as the number of fingers increases, the series resistance decreases but also the effective capacitance per unit area decreases , so it is a tradoff between large capacitance and small series resistance given the same area. It actuallly depends on the application and the required specs.
By the way, don't put M with a value more than 1 except when you are doing the same thing in the layout . M means that you have duplicate instances of the device each separated from the other (they are not merged together). The LVS tools usually don't differentiate between (M=1,F=2) and (M=2,F=1) cases , although they give different results , so take care.

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but the interesting thing is, by the simulation, i find that the equivalent series res of the m=1 f=100 and the m=100 f=1 is very very diff.

is it right?

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walkingsun said:

is it right?

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