Capacitor Bank Switching

[lauraj]

Can someone explain to me the theory behind why switching on a capacitor bank causes an oscillating transient. When I think of capacitors, I think of them as opposing a rapid change in voltage, whereas this does not seem to be the case with the transient.

 

[charlie b]

In simplest terms, the way in which it opposes a rapid change in voltage (e.g., at the moment of switching) is to insert a high amount of current. After some time, when things settle down, there will still be some current flowing in the capacitors. But it won't be anywhere near the initial amount of current. During the time it takes to "settle down," you will observe the transient oscillations that you mention.

A more detailed explanation might involve calculus. So I'd rather not go there just now. Did this at least give you an idea of what you were asking?

 

[lauraj]

I still don't quite understand. I would expect a sudden high current to draw the voltage source down, not up.

 

[charlie b]

The capacitor is not just one more load drawing current from the voltage source (which would, as you observed, increase the voltage drop internal to that source and thereby lower the voltage at its terminals). The capacitor acts as a current source in its own right, at least during the initial transient period.

Are you looking at a record of an actual transient event? Can you describe the behavior that you are trying to understand?

 

[lauraj]

I'm just looking at a textbook oscillatory transient, which they say is most often caused by capacitor bank switching. It just shows a transient that oscillates for about 1/4 to 1/2 cycle before dying out.

 

[charlie b]

Are you looking at a voltage transient? If so, voltage across what component?

Are you looking at a current transient? If so, current through what component?

 

[lauraj]

This is a voltage transient shown on just one phase. These are textbook examples, so it doesn't give much more info.

 

[Zifkwong]

The simplest way I can think of conceptualizing it is as follows:

Given that it is impossible for a voltage to instantaneously change across a capacitors terminals, when you initially energize a capacitor, if the voltage across the terminals was not zero just before the capacitor was energized, the capacitor will force the voltage to zero - by acting like a short. The capacitor will then begin to charge and increase its impedance. (Sorry for the run on)

Do you think that is a fair statement Charlie?

 

[steve66]

Usually oscillating transient waves are caused by a combination of capacitance and inductance. In this case, the inductance may be motor loads, or it may just be the 20 miles of wire back to the POCO.

The capacitor stores energy in an electric field, and the inductor stores energy in a magnetic field. When a sudden change occurs (like switching capacitors into a circuit), energy can get traded back an forth between the capacitors and inductors. This happens at the resonant frequency of the circuit. Eventually, the loads and losses in the circuit dampen the oscillations, and the energy trading becomes "steady state". Steady state is basically what the power triangle and Volts-Amperes-Reactive describe.

Steve

 

[lauraj]

Thanks Steve. From that definition, it seems to me like we would general see this type of event when switching on capacitors rather than switching them off. Is this correct?

 

[steve66]

Thanks Steve. From that definition, it seems to me like we would general see this type of event when switching on capacitors rather than switching them off. Is this correct?

That's a harder question It seems like we could just switch the capacitor off without much effect on the rest of the circuit, but its not quite that simple. In the steady state condition, the capacitor would always be charging and discharging. When we switch it off, we are switching off a current flow. If we assume there are inductors in the circuit, they are going to push that current somewhere else - somewhere like into stray capacitances. Since the stray capacitances are much smaller than the capacitor bank, the transient will probably die out much quicker than when you switch a capacitor on. But I am guessing that the voltage peaks could actually be higher when switching a capacitor off.

Of course, all this could vary a lot depending on the values of the circuit elements and the exact configuration.