HdrChopper
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Hi Joeb,
The formula you mentioned can be derived considering the clock phase is such that Cs is connected to the OTA´s input and assuming a linear small signal model for the OTA.
You can consider the feedback network is such that you are feeding back a voltage signal VI' = VI'+ - VI'-, which is the OTA´s differential input signal. If this is the case, then the feedback factor will be the transfer function given by VI'/VO (for VI=0). Considering this transfer function the feedback factor is readily deduced as the voltage divider between Cf and the parallel combination of Cs and Cp (or Cin in your figure):
VI'/VO (VI=0) = [1/s(Cs+Cp)] / [1/s(Cs+Cp)+1/sCf] = Cf/(Cf+Cs+Cp) (1)
Now, the feedback network can also be considered as connected in series with the input of your controlled source (OTAs linear model), or in other words, it feeds back a current. Since the output signal for your feedback network is now a current, the y parameters apply and you get -1/Zf, where Zf=1/sCf. Here the feedback factor is given by II'/VO @ VI=0, where II' is the differential input current coming out of the feedback network. However, the INPUT signal to your system now must be a current and not a voltage. Such input input current can be found as:
VI/ZI, where ZI= 1/s(Cs+Cp). Thus VI' = VI-I×ZI = VI+[(VO-VI')×sCf]×1/s(Cs+Cp).
If VI=0 you get VI'/VO=[1/s(Cs+Cp)] / [1/s(Cs+Cp)+1/sCf] = Cf/(Cf+Cs+Cp) , which is the same as (1)
So it is not that the y-parameters are wrong, it is just that you were considering different feedback networks when comparing the formulas. Therefore, the feedback factor will depend on what you consider is your feedback network and your input variables, but both must give you the same overall results.
Tosei
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