# Help with Capacitor Theory

#### itsme83

Joined Jul 11, 2011
6
I hope some of you can help me with some capacitor theory, because I'm getting a bit stuck on this one...

Take a look at these two circuit diagrams:

This is the equivalent circuit for a piezo actuator - essentially two capacitors of a similar value, driven by two high voltage sine waves that are the inverse of each other. So at any given time there will be a voltage on the top of C1 and a symmetric negative voltage on the bottom of C2. It is the electric field present on each of these 'capacitors' which makes the actuator move, so by applying inverse sine waves to each side, the actuator basically 'wiggles' back and forth by a distance proportional to the amplitude of the waves.

As you can see in (1) the path between the capacitors is grounded, which is how these actuators are typically wired up. However I'm trying to work out the theoretical implications of removing this ground connection, as shown in (2).

One thing I know for sure is that the actuator still 'works' like this, but something in my head is nagging about the build up of charge, leakage etc. and whether there may be a long term problem in driving the circuit this way due to the voltages and capacitances being slightly asymmetrical.

Would anyone care to comment on the difference of (2) from (1)?

Thank you.

#### t_n_k

Joined Mar 6, 2009
5,455
Presumably each of the supplies V1 and V2 have ground as reference. In the case with the ground connected, each device voltage is equal only to the applied voltage at the individual device's non-grounded (active) terminal. In the second case any difference in capacitance values would lead to (small) differences in device applied voltage as the total voltage is shared in relation to the individual capacitance values. Exactly equal capacitance values should give equal voltage sharing of the total voltage V1-V2. How important is the matching of mechanical vibration between the actuators themselves?

If the magnitude of displacement in an actuator is exactly proportional to the applied voltage only - i.e. not related to actual capacitance - then a small C difference in the ungrounded case would be observed as a small difference in mechanical response when compared with the grounded case where the mechanical responses should be identical (albeit in anti-phase). This also depends on the degree of matching between the V1 & V2 magnitudes.

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#### itsme83

Joined Jul 11, 2011
6
Thanks t_n_k, that makes sense.

But I also wonder whether there is the possibility of charge building up between the two capacitors over time due to these differences in C and V, and then taking a while to leak away when the signals are shut off? I'm concerned that if this could happen then it might not only degrade performance but also be a safety issue due to charge being stored between actuators even when turned off.

Does that make sense? This is where my theory is falling down really - will operating the ungrounded circuit (2) with slightly different values for C1/C2 and V1/V2 result in an imbalance which worsens over time, or is that total hooey?!

#### t_n_k

Joined Mar 6, 2009
5,455
The total series mean voltage drop across the capacitors can only ever equal the difference in DC mean value between the sources V1 & V2. If you were worried about any potential charge buildup on the caps after removal of the source voltages, you could place a reasonably high value resistance of suitable wattage across (i.e. in parallel with) the two capacitors to discharge any residual voltages.

#### itsme83

Joined Jul 11, 2011
6
I see, thanks for clearing that up.

#### Lundwall_Paul

Joined Oct 18, 2011
229
Use the circuit with the ground. If you leave off this ground, and V1 is noisy you will just pass on the noise to V2. The ground should help eliminate noise on both sides.