SWITCHED CAPACITOR

(Redirected from Switched-capacitor filter)
'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.
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 an 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, $C\_\{fb\}$, 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 $C\_f$:
:
And inserting 1):
:
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 $C\_s$ (thanks to the op-amp, sure). This is what you must store in your brain.
However, there is a more elegant way to formulate this fact if $T$ is very short. Let us introduce $dtleftarrow\; T$ and $dV\_\{fb\}leftarrow\; V\_\{fb\}(t)-V\_\{fb\}(t-dt)$ and rewrite the last equation divided by dt:
:
Therefore, the op-amp output voltage takes the form:
:
Note that this is an integrator with an "equivalent resistance" . 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).

Contents

References

References

★ Mingliang Liu, ''Demystifying Switched-Capacitor Circuits'', ISBN 0-7506-7907-7