Switched capacitor

A switched capacitor is an electronic circuit element used for discrete time signal processing. It works by moving charges into and out of capacitors when switches are opened and closed. Usually, non-overlapping signals are used to control the switches, so that not all switches are closed simultaneously. Filters implemented with these elements are termed 'switched-capacitor filters'. Unlike analog filters, which must be constructed with resistors, capacitors (and sometimes inductors) whose values are accurately known, switched capacitor filters depend only on the ratios between capacitances. This makes them much more suitable for use within integrated circuits, where accurately specified resistors and capacitors are not economical to construct.

The switched capacitor resistor

Switched-capacitor resistor

The simplest switched capacitor (SC) circuit is the switched capacitor resistor, made of one capacitor C and two switches S₁ and S₂ which connect the capacitor with a given frequency alternately to the input and output of the SC. Each switching cycle transfers a charge $q$ from the input to the output at the switching frequency $f$. Recall that the charge q on a capacitor C with a voltage V between the plates is given by:

$q=CV\$

where V is the voltage across the capacitor. Therefore, when S₁ is closed while S₂ is open, the charge transferred from the source to C_S is:

$q_{IN}=C_{S}V_{IN}\$

and when S₂ is closed while S₁ is open, the charge transferred from C_S to the load is:

$q_{OUT}=C_{S}V_{OUT}\$

Thus, the charge transferred in each cycle is:

$q=q_{OUT}-q_{IN}=C_{S}(V_{OUT}-V_{IN})\$

Since a charge q is transferred at a rate f, the rate of transfer of charge per unit time is:

$I=qf\$

Note that we use I, the symbol for electric current, for this quantity. This is to demonstrate that a continuous transfer of charge from one node to another is equivalent to a current. Substituting for q in the above, we have:

$I=C_{S}(V_{OUT}-V_{IN})f\$

Let us define V, the voltage across the SC from input to output, thus:

$V=V_{OUT}-V_{IN}\$

We now have a relationship between I and V, which we can rearrange to give an equivalent resistance R:

$R={V \over I}={1 \over {C_{S}f}}\$

Thus, the SC behaves like a resistor whose value depends on C_S and f.

The SC resistor is used as a replacement for simple resistors in integrated circuits because it is easier to fabricate reliably with a wide range of values. It also has the benefit that its value can be adjusted by changing the switching frequency. See also: operational amplifier applications.

The Parasitic Sensitive integrator

File:Parasitic Sensitive Inverter.png

A Simple Switched Capacitor Parasitic-Sensitive Integrator

Often switched capacitor circuits are used to provide accurate voltage gain and integration by switching a sampled capacitor onto an op-amp with a capacitor Cfb in feedback. One of the earliest of these circuits is the parasitic-Sensitive integrator developed by the Czech engineer Bedrich Hosticka. 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 short circuit between its inputs):

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

Now dividing 2) by $C_{f}$:

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

And inserting 1):

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

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).

See also

• Switched-mode power supply
• Charge pump

References

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