Switched capacitor

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Switched Capacitor is a circuit design technique for discrete time signal processing. It works by moving charges between different capacitors when switches are opened (off) and closed (on). Usually, non-overlapping signals are used to control the switches, so that not all switches are on simultaneously.

Voltage amplification can be achieved by moving a charge from a large capacitor to a small capacitor.^{[citation needed]}

Voltage amplification can be achieved by repeatedly switching capacitors from a parallel arrangement with regard to the supply to a series arrangement with regards to the load. This arrangement is called a charge pump.

The simplest Switched Capacitor (SC) circuit is made of one capacitor and two switches which connect the capacitor with a given frequency alternately to the input and output of the SC. This simulates the behaviour of a resistor, so SCs are used in integrated circuits instead of resistors. The resistance is set by the frequency.

  S1    S2
    /    /
o--/ ---/ --o
      |
in    |    out
     ===
      | Cs
      |
o-----------o

Often you will find this structure in place of the resistance of an integrator; see Operational amplifier applications. In turn, filters implemented with these integrators are termed Switched capacitor filters.

Let us analyze what happens in this case. Denote by $T = 1 / f$ the switching period. Recall that in capacitors charge = capacitance x voltage. Then, at the instant when S1 opens and S2 closes, we have the following:

1) Because $C s$ has just charged:

$Q_s(t) = C_s \cdot V_s(t)\,$

2) Because the feedback cap, $C f b$ , is suddenly charged with that much charge (by the opamp, which seeks a virtual shortcircuit between its inputs):

$Q_{fb}(t) = Q_s(t) + Q_{fb}(t-T)\,$

Now dividing 2) by $C f$ :

$V_{fb}(t) = \frac {Q_s(t)}{C_{fb}} + V_{fb}(t-T)\,$

And inserting 1):

$V_{fb}(t) = \frac {C_s}{C_{fb}} \cdot V_s(t) + V_{fb}(t-T)\,$

This last equation represents what is going on in $C f$ -- it increases (or decreases) its voltage each cycle according to the charge that is being "pumped" from $C s$ (due to the op-amp).

However, there is a more elegant way to formulate this fact if $T$ is very short. Let us introduce $dt\leftarrow T$ and $dV_{fb}\leftarrow V_{fb}(t)-V_{fb}(t-dt)$ and rewrite the last equation divided by dt:

$\frac {dV_{fb}(t)}{dt} = f \frac {C_s}{C_{fb}} \cdot V_s(t)\,$

Therefore, the op-amp output voltage takes the form:

$V_{OUT}(t) = -V_{fb}(t) = - \frac{1}{\frac{1}{fC_s}C_{fb}} \int V_s(t)dt \,$

Note that this is an integrator with an "equivalent resistance" $R_{eq} = \frac{1}{fC_s}$ . This allows its on-line or runtime adjustment (if we manage to make the switches oscillate according to some signal given by e.g. a microcontroller).