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ft simulation in Cadence Spectre

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ft simulation gm cgs spectre

Hi diemilio,
Hi sutapanaki !

Yesterday I was sure not to mention this MOSFET current gain stuff again.
However, in the mean time I have searched a bit around - and I must confess, that indeed in some books and articles a term "current gain" is used sometimes for MOSFET´s.
In this context, presently my knowledge and my "feeling" is as follows:
It was the aim of some researchers to define something like "transit frequency" not only for the BJT (where it is a quite logical parameter) but also for the MOSFET.
That means, a parameter had to be found which approaches "1" at a higher frequency and which - at the same time - is a measure for the amplifying properties of the device. Therefore, of course, the drain current had to be the key element for this definition - and another current had to be defined which could be used in order to reach the ratio "1" at a certain frequency.
Now, in spite of the fact, that the MOSFET is a voltage driven unit, the parasitic input current through the GS-capacity was used as the second parameter.
And as the result we get: 2*Pi*Ft=gm/Cgs.
This definition makes sense because - at the same time - it is something like a quality figure of the MOSFET taking into account the transconductance and the limiting G-S capacity at one specific bias point.
In summary, in principle you are right that some authors use the term "current gain" also for MOSFETs (and I have learned something).
Kind regards
 

ft simulation mos

Hi LvW,

Thanks for your reply and for sharing with us your findings on how the concept of ft arose for MOS devices.

Regards,

diemilio
 

cadence ft

hi dfrndez

Use a very high value (approx 10e+10)H in series with the Voltage source.. this will help in shorting the DC voltage to gate of Mosfet and the Ac current source will drive only the mosfet ... this will surely help ..as i always do this for finding ft by simulation
 

Hello everybody

I want as well to determine yhe ft of a mos transistor, but I had a question ,
ft = gm/cgs
which means that the ft depends on the bias condition , is there some standard definition of ft that states that "ok, the bias vltage of the gate should be half the supply, or quarter, or tenth , or .. I don't know"'

thank you in advance
 

yes. you are correct the ft depends on dc bias condition. the method i use which has worked for me is to have a diode-connected mosfet whose gate is exited by a dc source. i then sweep the dc voltage and plot gm/2*pi*cgs to get ft across vds, for diff L
this will be enough for u to tell the ft for a vds u operate in and for the L of the device u choose.
 
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