SaM - Switched Capacitors

Remeber:

A switched capacitor can be used for increasing the capacitance value, with respect to the sensor’s capacitance, without changing the actual capacitance.
This is like amplifying the capacitors of the sensor:

• So the starting phase, $S_3$ : closed, so that we start from $0V$.
• Then there is a sequence of repeated steps, which are this ones:
  1. $S_1$: ON, $S_2$: OFF, $S_3$: OFF.
     So in this phase, we have a charge, which is stored in the capacitance $C_S$, equal to:
     So we have this charge $Q_1$ stored in $C_S$ at this point.
  2. $S_1$: OFF, $S_2$: ON, $S_3$: OFF.
     At the beginning of this phase, we had the initial charge in $C_S$.
     While at the end the total charge reamins the same, it is now shared between $C_S$ and $C_E$, so $C_S$ has given some charge to $C_E$.
     So the value $V_{out}$ increases.
  3. So then we go back to the first phase (1.), so: $S_1$: ON, $S_2$: OFF, $S_3$: OFF.

Most important formula: $V_{out}^{(n)}=\frac{C_S{V}_R+C_R{V}_{out}^{(n-1)}}{C_S+C_R}$After $n$ steps we have that:$V_{out}^{(n)}=n\cdot C_S\cdot \Delta Q_n$Where: $\Delta Q_n=\frac{V_R}{C_R}$So to obtain the sensor’s capacitance we just use this formula:$C_S=\frac{V_{out}^{(n)}}{n\Delta Q_n}$

If I perform all the calculations, I put the steps in a graph, I will have this:

• $\frac{V_r}{C_R}$ is a constant term, which I call $\Delta Q_n$.
• So we have a linear behavior at the beginning, and then saturation, because we can go beyond this value, we will have a saturation at a certain point, the steps will become smaller and smaller until I reach the full charge of the capacitance.
  ==That’s why I can perform this “cheat” only untill the voltage $V_R\gg V_{out}$==.
  ⇒ If I limit the measurement ratio to $n_{MAX}$, opprotunaly choosen, I will remain in the linear range of the capacitance, and I have this very simple way of amplifying the signal due to the sensor.

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Memory Card

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• So the starting phase, $S_3$ : closed, so that we start from $0V$.
• Then there is a sequence of repeated steps, which are this ones:
  1. $S_1$: ON, $S_2$: OFF, $S_3$: OFF.
     So in this phase, we have a charge, which is stored in the capacitance $C_S$, equal to:
     So we have this charge $Q_1$ stored in $C_S$ at this point.
  2. $S_1$: OFF, $S_2$: ON, $S_3$: OFF.
     At the beginning of this phase, we had the initial charge in $C_S$, which was:
     And the charge at the end is the same obviously the same:
     But the voltage, since these two are put in parallel, at the end of this first phase, will be:
     So we have a first step of $V_{out}$, the output voltage, which is given by this value here.
  3. So then we go back to the first phase (1.), so: $S_1$: ON, $S_2$: OFF, $S_3$: OFF.

After $n$ repetitions, we will have:

This is like having a larger capacitance, with respect to the sensor’s capacitance.
This is like amplifying the capacitors of the sensor, and actually, if I perform all the calculations here, I put the steps, I will have this:

• $\frac{V_r}{C_R}$ is a constant term, which I call $\Delta Q_n$.
• So we have a linear behavior at the beginning, and then saturation, because we can go beyond this value, we will have a saturation at a certain point, the steps will become smaller and smaller until I reach the full charge of the capacitance.
  That’s why I can perform this “cheat” only untill the voltage $V_R\gg V_{out}$.
• If I limit the measurement ratio to $n_{MAX}$, opprotunaly choosen, I will remain in the linear range of the capacitance, and I have this very simple way of amplifying the signal due to the sensor.
• Moreover, the readout circuit, when it reads the voltage output, sees these capacitors that can be very large.
  ⇒ I don’t put the wiring in parallel to a small capacitance (less restrictions).
• Obviously, here I didn’t take into account the parasitic resistances, and I also considered all the switches ideal, and so this is something which is really simplified, but it any case the principle is this.
• Thing to remember:
  • Most important formula: $V_{out}^{(n)}=\frac{C_S{V}_R+C_R{V}_{out}^{(n-1)}}{C_S+C_R}$
  • After $n$ steps we have that:$V_{out}^{(n)}=n{C}_S\Delta Q_n$Where: $\Delta Q_n=\frac{V_R}{C_R}$
• So to obtain the sensor’s capacity we just use this formula:$C_S=\frac{V_{out}^{(n)}}{n\Delta Q_n}$