Cut off adjustment for LNA
时间:04-05
整理:3721RD
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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 , width, , , etc.
So far, the cut-off frequency is calculated as
= ; [ =Gate-source cap]
Then, I proceeded to calculate as:
= ....Eq.(1)
Say, I got 4.25x10^11 for , and I have a of .167mA/V2.
Then, the = 0.17nH,
but due to technical limitation, I need to use = .21nH.
Now, I need to change the the capacitance by adding something to so that my
does not change.
Now, the problem is only equation that I find is Eq.(1) which relates to , but I need to counter the extra (it changed from .17nH to .21nH) in some way like a equation with .
The problem is, I cannot find an equation to tackle this.
Any help would be appreciated.
[Just FYI: it is not . In that case, the increase in "L' would mean a decrease in 'C'.
In my case, the increased is countered by adding some more capacitance to ]
Thank you for your time!
In the new (or the proper) approach,
I kept unchanged.
So, I went like this:
Then,I used
Thank you all!
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 , width, , , etc.
So far, the cut-off frequency is calculated as
= ; [ =Gate-source cap]
Then, I proceeded to calculate as:
= ....Eq.(1)
Say, I got 4.25x10^11 for , and I have a of .167mA/V2.
Then, the = 0.17nH,
but due to technical limitation, I need to use = .21nH.
Now, I need to change the the capacitance by adding something to so that my
does not change.
Now, the problem is only equation that I find is Eq.(1) which relates to , but I need to counter the extra (it changed from .17nH to .21nH) in some way like a equation with .
The problem is, I cannot find an equation to tackle this.
Any help would be appreciated.
[Just FYI: it is not . In that case, the increase in "L' would mean a decrease in 'C'.
In my case, the increased is countered by adding some more capacitance to ]
Thank you for your time!
Finally found a problem in my understanding, (and an equation).
I tried to keep
In the new (or the proper) approach,
I kept unchanged.
So, I went like this:
Then,I used
Thank you all!
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