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You should note that the intial charge in the circuit is due to Vin. Hence the initial charge is Vin(Cs+Cf), assuming no offset in the opamp. Anyways, in the sampling phase, both the inputs are grounded. In the amplification phase, the charge in the Cs is brought down to zero, while Cf will drive the output. Hence the effective charge to the ouput = Vo(Cf). Hence, using law of conservation of charge in both the phases, Vin(Cs+Cf) = Vo(Cf). Hence Vo/Vin should be equal to 1+Cs/Cf.
Note that there is no inversion of voltage because the sampling and amplification phases see the same capacitor terminal.
the feedback factor of the circuit is Cf/(Cf+Cs)
it express the output have what effect on the input.
the most intuitive thing is that,if the UGB of the amp is Wu
then the -3db frequency of the circuit is F*Wu
then F is the feedback factor
Only in the amplifer phase, the loop is close loop. then a feedback factor is meaning. other time , the feedback factor is no meaning due to open loop.
while the loop is close, open output node of feedback capacitor, then insert a voltage souce, then derive the loop gain G(s), after this, we know the forward gain of op is A(s) , then we can get the feedback factor is G(s)/A(s).
Based on the principle, the feedback factor is Cf/(Cf+Cs).
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