SaM - High Frequency (AC) Signals

Remeber:

If the signal measured by a capacitive sensor has DC components, to “catch them” we need an AC exitation, as we have said before. While if we have a pure AC signal (meaning with no DC components), so it has a spectrum like this:
⇒ Then DC exitation is possible.

With a simple setup we can obtain the value of our measured quantity $x$:

• $C_1=\frac{C_0}{1-x}$
• $C_2=\frac{C_0}{1+x}$
• The output will be, if we perform the calcualtions:$V_{out}=\frac{V_{DC}}{2}(1-x)$Where: the offset $\frac{V_{DC}}{2}$ can be corrected, and the final output $\left(-\frac{V_{DC}}{2}x\right)$ can be amplifed and inverted.

We have seen the example of a capacitive microphone:
The measured $C_S$ value is very small, and the nedded force $F$, to move the plate depends on: $F\propto \frac{A}{h^2}$Where $A$ and $h$ are the physical dimensions of the sensors ⇒ the minimum force which has to be overcome to move the movable plate is really small.NOT_SURE_ABOUT_THIS So the needed DC exitation to get a readable signal would be really high, $V_{DC}$ usually, it’s a large value ($\sim 100V$).IMPORTANTENOT_SURE_ABOUT_THIS

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Memory Card

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• When we deal with AC signals and usually high frequencies signals, which means that the spectrum of $x(t)$: $X(f)$, so if the sensor measure has no DC components.
  ⇒ Then DC exitation is possible.
• While in AM and FM we had only AC exitation.
• A simple capacitive microphone uses this formula:$C_{out}=\frac{C_0}{(1-x)}$
• With a simple voltage divider using two sensors with “opposed signs”, we have:$V_{out}=\frac{V_{DC}}{2}(1-x)$
• Also note that the offset of $\frac{V_{DC}}{2}$ can be removed, and the theoretical output value can become: $V_{out}=-\frac{V_{DC}}{2}x$Of course the output can be then amplified and inverted.
• $V_{DC}$ usually, it’s a large value ($\sim 100V$).IMPORTANTE

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Here we can see a typical solution in case of a capacitive microphone:

• Capacitive microphones have a small mechanical load represented by $R$.
• Since the $C_S$ value is very small, and the nedded force $F$, to move the plate depends on: $F\propto \frac{A}{h^2}$Where $A$ and $h$ are the physical dimensions of the sensors ⇒ the minimum force which has to be overcome to move the movable plate is really small.

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