MDAC operation

according to [Pipeline ADC enhancement techniques; page 35], MDAC output voltage is given by:

\begin{eqnarray} V_{out} &=& \frac{\Sigma_{1}^{2^n} c_i}{c_f} V_{in} - \left[ \frac{\Sigma_{1}^{k} c_i}{\Sigma_{1}^{2^n} c_i} V_{ref} -\frac{\Sigma_{k+1}^{2^n} c_i}{\Sigma_{1}^{2^n} c_i} V_{ref} \right] \end{eqnarray}

question: is this equation accurate?

1- at sampling phase charge $q=V_{in} \Sigma_{1}^{2^n} c_i$ is stored on sampling capacitors $c_i$;

2- at amplification phase we have the superposition of the following cases:

• case A) $c_i$s' left plate is floating and op-amp settles to an equilibrium in which negative pin of op-amp is a virtual ground (all the charges that were stored are pulled to the left plate of $c_f$ by and electric field that is created by op-amp output voltage of the right plate of $c_f$. In this case, to achieve equilibrium, $V_out$ should be 

\begin{eqnarray} V_{out} &=&\frac{\Sigma_{1}^{2^n} c_i}{c_f} V_{in} \end{eqnarray}

• case B) left plate of $c_i$s are connected to either $V_{ref}$ or $-V_{ref}$ depending on sub-ADC output codes. Let's assume $k$ of sub-ADC codes are $1$ and the rest are $0$. In this condition, we have average charge of $\Sigma_{1}^{k} c_i V_{ref} -\Sigma_{k+1}^{2^n} c_i V_{ref}$ that must be absorbed to to right plate of $c_i$s in reciprocity with the electric field that is introduced by $\pm V_{ref}$ combinations. In addition, op-amp tends to holds the negative pin at virtual ground. Consequently, op-amp must create an opposite electric field to cancel the impact of input codes; The magnitude of this electric field (i.e. voltage at the right plate of $c_f$) should be enough to keep the balance of the charge at virtual ground, i.e. 

\begin{eqnarray}
V_{out} &=&- 1 \times \frac{1}{c_f} \left[ \Sigma_{1}^{k} c_i -\Sigma_{k+1}^{2^n} c_i \right] V_{ref} \end{eqnarray}

3- supposition of case A and B results in:

\begin{eqnarray} V_{out} &=& \frac{\Sigma_{1}^{2^n} c_i}{c_f} V_{in} - \left[ \frac{\Sigma_{1}^{k} c_i - \Sigma_{k+1}^{2^n} c_i}{c_f} V_{ref }\right] \\ &=& \frac{\Sigma_{1}^{2^n} c_i}{c_f} \left[ V_{in} - \frac{\Sigma_{1}^{k} c_i - \Sigma_{k+1}^{2^n} c_i}{\Sigma_{1}^{2^n} c_i} V_{ref } \right] \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.