op-amp with limited bandwidth

op-amp open loop transfer function: \begin{eqnarray} B \left( s \right) &=& \frac{V_o \left( s \right)}{V_x \left( s \right)} \\ &=& \frac{-A}{ 1+ \frac{s}{w_p} }, \label{eq:2} \end{eqnarray} where $w_p$ is the first pole of op-amp and $A$ is it's dc gain. Total charges at node $x$ in sampling phase: $-q_x = \left(C_1 + C_f \right) \times V_i \left( s \right) $. Ideally, charges at node $x$ cannot escape (no low impedance path exist); therefore, op-amp settles with respect to charge equilibrium at node $x$: \begin{eqnarray} -q_f + q_1 &=& -q_x \\ C_f \left( V_o \left( s \right) - V_x \left( s \right) \right) - C_1 V_x \left( s \right) &=& \left(C_1 + C_f \right) \times V_i \left( s \right) \\ \end{eqnarray} if $\beta = \frac{C_f}{C_1 + C_f}$, and given op-amp open loop transfer function (Eq. \eqref{eq:2}): \begin{eqnarray} V_o \left( s \right) &=& \frac{-B \left( s \right) }{1- \beta B \left( s \right)} V_i \left( s \right)\\ H \left( s \right) &=& \frac{B \left( s \right) }{\beta B \left( s \right) -1} \\ &=& \frac{A}{1+\frac{s}{w_p}+\beta A} \\ &=& \frac{1}{\beta + \frac{1}{A}} \frac{1}{1+\frac{s}{w_p \left( 1+\beta A \right)}} \end{eqnarray} where $H\left( s \right)$ is the closed loop transfer function of the circuit. step response of $H\left(s \right)$, $Y\left(s\right)$ is given by: \begin{eqnarray} Y\left(s\right) &=& \frac{1}{\beta + \frac{1}{A}} \left( \frac{1}{s} -\frac{1}{s+w_p\left(1+\beta A\right)} \right) \\ y\left( t \right) &=& \frac{1}{\beta + \frac{1}{A}} \left(1-e^{-w_p\left(1+\beta A\right) t} \right) \end{eqnarray} at the end of amplification period ($t=\frac{T_s}{2}$), gain error is equal to: \begin{eqnarray} G_{\text{err}} &=& \frac{1}{\beta} - y\left( t=\frac{T_s}{2} \right) \\ &=& \frac{1}{\beta} - \frac{1}{\beta+\frac{1}{A}} + \frac{1}{\beta+\frac{1}{A}} e^{\frac{-w_p\left( 1+\beta A \right)}{2 f_s}} \end{eqnarray} assumption 1- unity gain bandwidth of open loop op-am, $f_u$, given $A \gg 1$: \begin{eqnarray} w_u &=& w_p \sqrt{A^2-1} \\ &\approx& A w_p \\ f_u &=& \frac{A w_p}{2 \pi} \label{eq:15} \end{eqnarray} assumption 2- unity gain bandwidth of closed loop op-amp, $f_u^{*}$ given $\beta + \frac{1}{A} \approx \beta$: \begin{eqnarray} H\left( s \right) &=& 1 \\ \beta^2 \left( \frac{1}{\beta} + A \right)^2 + \frac{w_u^{*}}{w_p^2} &=& A^2 \\ w_u^{*} &=& A w_p \sqrt{1-\beta^2} \\ f_u^{*} &=& \frac{A w_p \sqrt{1-\beta^2}}{2 \pi} \end{eqnarray} for simplification, let's assume dc gain is relatively large; therefore, $\beta + \frac{1}{A} \approx \beta$: \begin{eqnarray} G_{\text{err}} &=& \frac{1}{\beta} e^{\frac{-A \beta w_p}{2 f_s}} \end{eqnarray} for an $N$-bit pipeline, input refered gain error, $G_{\text{err}}^{\text{input}} = \frac{G_{\text{err}}}{\frac{1}{\beta}}$, should be better than quantization error: \begin{eqnarray} G_{\text{err}}^{\text{input}} &<& 2^{-N} \\ \frac{A \beta w_p}{2 f_s} &>& N \ln \left(2\right) \\ f_u &>& \frac{N \ln \left(2\right) }{\pi \beta} f_s \end{eqnarray}