switch capacitor multiplier: op-amp with non-ideal DC gain

continue the discussion in previous post: here we assume a non-ideal op-amp, i.e. op-amp gain is equal $A \neq \infty$. For simplification, lets assume the circuit is in phase 2 and $V_d=0$ (similar to case b). Again, in equilibrium, sampling charge on $C_1$ and $C_f$ is conserved by the feedback loop (as much as DC gain of the op-amp allows). We have:

$$\begin{eqnarray} q_f-q_1 &=& C_f\left(V_o-V_x  \right) - C_1 V_x  \nonumber  \\              &=& \left( C_1+C_f \right) V_i. \end{eqnarray}$$

If $V_o=-A \times V_x$ then

$$\begin{eqnarray} V_o C_f + \frac{V_o}{A} \left(C_1+C_f\right) &=& \left(C_1+C_f \right) V_i   \\       V_o &=& \frac{C_1+C_f}{C_f+\frac{C_1+C_f}{A}} V_i \\        &=& \frac{1}{\beta+\frac{1}{A}} V_i, \label{eq:3} \end{eqnarray}$$
where $\beta=\frac{C_f}{C_1+C_f}$. If $A \rightarrow \infty$ then $V_o^{*} = \frac{V_i}{\beta}$. Given Equation $\eqref{eq:3}$, The output voltage error, $V_{\epsilon}^{\text{output}}$ due to non-ideal gain of op-amp is given by

$$\begin{eqnarray} V_{\epsilon}^{\text{output}} &=& \left( \frac{1}{\beta} - \frac{1}{\beta+\frac{1}{\beta}} \right) V_i \\       &=& \frac{1}{\beta} \left(  \frac{1}{1+\frac{1}{\beta A}} -1  \right) V_i       &=& \left( \frac{\Delta}{1+\Delta}  \right) V_o^{*}, \end{eqnarray}$$
where $\Delta=\frac{1}{\beta A}$. For simplification, we assume that $\beta A \gg 1$; therefore we have:

$$\begin{eqnarray} V_{\epsilon}^{\text{output}} &\cong& \Delta \times V_o^{*} \end{eqnarray}$$