[SOLVED] Cut off adjustment for LNA

Hello,
I am designing an LNA (in Cadence for class project). I am having a simple problem, (sorry for my representation here). I am going Cascode with this one.

I have calculatad values for \[{g }_{m}\], width, \[{w }_{T }\] , \[{F }_{min,p}\], etc.

So far, the cut-off frequency is calculated as
\[{w }_{T }\]= \[\frac{{g }_{m}}{{C}_{gs}}\] ; [\[{C}_{gs}\] =Gate-source cap]

Then, I proceeded to calculate \[{L}_{s}\] as:
\[{L}_{s}\]=\[\frac{{R}_{s }}{{w}_{T}}\] ....Eq.(1)

Say, I got 4.25x10^11 for \[{ w}_{T }\], and I have a \[{g }_{m}\] of .167mA/V2.
Then, the \[{L}_{s}\] = 0.17nH,
but due to technical limitation, I need to use \[{L}_{s}\] = .21nH.


Now, I need to change the the capacitance by adding something to \[{C}_{gs}\] so that my
\[{w }_{T }\] does not change.

Now, the problem is only equation that I find is Eq.(1) which relates to \[{w }_{T }\], but I need to counter the extra \[{L}_{s}\] (it changed from .17nH to .21nH) in some way like a equation with \[{C}_{gs}+{C}_{gs}\].

The problem is, I cannot find an equation to tackle this.
Any help would be appreciated.

[Just FYI: it is not \[{w }_{0 } = \frac{1 }{\sqrt{L C } \]. In that case, the increase in 'L' would mean a decrease in 'C'.
In my case, the increased \[{L}_{s}\] is countered by adding some more capacitance to \[{C}_{gs}\] ]

Thank you for your time! :-D

Finally found a problem in my understanding, (and an equation).

I tried to keep \[{ w}_{T }\] unchanged. However, it was to change.
In the new (or the proper) approach,
I kept \[{ g}_{m }\] unchanged.

So, I went like this:
\[{ w}_{T-modified}=\frac {Impedance}{{L}_{s -modified} } \]

Then,I used
\[{ w}_{T-modified}=\frac {{g}_{m}}{{C}_{gs}+{C}_{extra} }\]

Thank you all!