wwm101 wrote on May 1st, 2008, 7:45am:why the follow circuit diagram which input impedance equal to 1/(gm1a+gm1b) in the radio frequency?
If it is not in the radio frequency, it is very simple.Transconductance of inverter is Gm=gm1a+gm1b.
So input admittance Yin seeing right at node X is like following.
Yin=GF*(YL+Gm)/(YL+GF)
where GF is feedback conductance for self bias, so GF=1/R, R is resistor between X and Y.
YL is load admittance at node Y.
If ZL=1/YL=infinity, Yin=Gm=gm1a+gm1b.
If ZL=1/YL=0, Yin=GF=1/R.
If GF=0, Yin=0.
So if impedance seeing right at node Y is enough high, ZL >> 1/Gm, 1/GF=R
an input impedance Zin seeing right at node X will be 1/Gm=1/(gm1a+gm1b).
For the radio frequency, this is not true.