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}

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}

switch capacitor multiplier: theory of operation

This note is helping me to understand the theory of operation in MDAC (e.g. pipeline ADCs).

phase 1) sampling: when $S_{1t}$  is high, input voltage is sampled on $C_1$ and $C_f$. For simplicity, let's assume $C_1 = C_f = C$. Total charge saved on $C_1$ and $C_f$ is $q=2 \times C \times V_i$.

phase 2) amplification: when $S_2$ is high and $S_1$ is low, pre-charged $C_1$ and $C_f$ capacitors are placed in a feedback loop with an amplifier. let's review the following cases:

case (a) $C_1$ left plat is floating and $C_f$ is placed in the feedback loop.

case (a) $C_1$ left plate is floating.

in this case, given $C_1 = C_f = C$, $q=C V_i$. large gain of op-amp, would like to force $V_x$ voltage to $0$ volt through the feedback loop. If left plate of $C_1$ is floating, $C_1$'s left plate potential settles at $V_i$ and $C_1$ continues holding $q$ coulomb of charge. On the other hand, $V_o$ must settle to $V_i$ voltage to make sure that the rest of sampled charges are kept on $C_f$. The important point is that because node $V_x$ is a high impedance node and left plate of $C_1$ is also floating, op-amp output cannot inject any charge in the loop.

case (b) left plate of $C_1$ capacitor is connected to ground. In this case, op-amp feedback loop pulls node $V_x$ to zero volt.This means that $C_1$ holds zero coulomb of charge. In addition, node $V_x$ is a high impedance node; in equilibrium, the original $q$ charge that was saved on $C_1$ needs to be restored at node $V_x$. In other words, in equilibrium, $C_f$ needs to store $2\times q$ coulomb of charge, i.e. $V_o=2\times V_i$. This is possible because op-amp can inject charge into the feedback loop at its output (low impedance node).

case (b) $C_1$ left plate is ground.

case(c) left plate of $C_1$ is connected to a voltage source $V_d$. Following the logic of case (b), in equilibrium, $C_1$ stores $q_1 = \left(V_d-V_x \right)C_1 = V_d C_1$ coulomb of charge. originally, $C_1$ stored $q = C_1 V_i$ coulomb of charge. The difference, between $q$ and $q_1$ should be stored on $C_f$ (because node $V_x$ is high impedance).  Consequently, in equilibrium,

$V_o = \frac{q+q_2}{C_f} = \frac{2q-q_1}{C_f} = C_1 \frac{2V_i-V_d}{C_f} = 2V_i-V_d$.

case (c) $C_1$ left plate is connected to $V_d$.

case (c) is a representation of MDAC (in this example 1 bit) with 2x amplification gain. $V_o$ is the residual voltage that is transferred to the next pipeline stage.