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summi Community Member Offline Posts: 38	intuitive explanation of LNA input impedance Dec 26^th, 2011, 2:26pm Dear forum, can any one please explain how input impedance of an common source amplifer with inductive degeneration will pure resistive nature. I have derived and got answer but trying to understand through concepts. Br, Summi.
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aaron_do

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Re: intuitive explanation of LNA input impedance
Reply #1 - Dec 26^th, 2011, 5:30pm

Hi,

It is only purely resistive at one frequency. You are simply creating a resistive element to add to the input impedance and then resonating out the imaginary part.

Maybe you will find intuition if you derive it in a different way. The direct method is of course an easy way to derive it, but you can instead treat the amplifier as a feedback system and find the loop gain. Feedback changes the input impedance.

Not sure exactly what you're looking for here.

cheers,
Aaron

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raja.cedt

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Re: intuitive explanation of LNA input impedance
Reply #2 - Dec 27^th, 2011, 2:08am

hello may be like this,
let us say i have applied current with 0 phase shirt at the gate of the transistor, it flows through Cgs of the transistor and creates voltage of 90 degree lag, then 90 degree lag current will flow through transistor (because i=gm*vgs), this current will flow through inductor which creates 0 phase shift voltage, means some resistive impedance is appearing, but this will happen only at some frequency. Another way to look at this loop gain is a kind of high pass filer and now find the input impedance.

Thanks,
raj.

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RFICDUDE

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Re: intuitive explanation of LNA input impedance
Reply #3 - Dec 27^th, 2011, 6:04am

The fundamental concept is that "local or series feedback" provided by Ls modifies Vgs (voltage drop across Cgs) such that there is a real component to the input impedance that is equal to

Rs = gm*Ls/Cgs

This real resistance is somewhat broadband, but of course there is reactance that resonates only at one frequency (as has been mentioned). The series feedback that produced this result is, as mentioned in raja's post, that the voltage drop across Ls is dependent on Vgs, via Is=gm*Vgs, this results in a real impedance when measuring impedance into the gate.

You have to be a little careful with this simplified analysis because it ignores the fact that there are other capacitors besides Cgs.

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