Analyzing large circuits

[duke37]

You say,
I think I have an fair idea of what's going on. A question for my own sanity though, doesn't the main current (the one not generated by the inductor) charge the capacitor? So if the inductor was to release the energy, wouldn't the capacitor be at it's limit? Or does the capacitor release the energy while the inductor is transferring its energy to the capacitor?

The capacitor will be charged by the supply in one direction and then will be charged in the reverse direction when the switch is turned off. This means that an electrolytic capacitor cannot be used. The capacitor may be at or past its limit if you get your calculations wrong.

Duke

 

[(*steve*)]

Ah, now I see what you're referring to -- a capacitor placed across a relay coil in place of a diode.

Let's go back to why we want to do this...

When the coil is carrying current it stores energy in the magnetic field that it generates. If current is suddenly stopped the collapsing magnetic field induces a current in the coil.

With DC coils, a diode reverse biased across the coil will effectively provide a low resistance path for that energy (remembering that it's energy stored).

With an AC coil a diode across the coil would effectively short out the power each alternate half-cycle, so clearly it's a bad thing, and we need something else.

Let's go back to that stored energy. The energy wants to go somewhere. if there is no path, then the voltage will rise until something gives. This may be a spark as you open a switch, or in the case of a semiconductor it may be the failure of the transistor.

If we place a capacitor across the coil, the energy from the collapsing magnetic field charges the capacitor. The voltage rises high enough so that the energy stored in the capacitor equals the energy dumped from the coil.

then, when the energy in the coil is exhausted, the capacitor discharges into the coil, which then charges the capacitor in the opposite direction (yes, this means that electrolytic capacitors are a poor choice), and then it keeps going back and forth until losses reduce the amplitude to something no longer measurable.

Another option, and one that will work for both AC and DC is to place back to back zener diodes across the relay. This will result in the voltage rising just above the zener's voltage (which mist be higher than the peak supply voltage)

This last technique will cause the energy in the coil to be dissipated rapidly, works with AC, and doesn't ring.

Capacitors may be cheaper though.

edit: to more directly answer Duke37's question. Yes, the main current holding the relay in also charges the capacitor. When that voltage is removed, the capacitor discharges through the relay, the inductive effect meaning that the charge goes through zero and up to a reverse value before falling back through zero to a positive value, and so on. It's a lot like a pendulum where the eight of the pendulum (potential energy) is analagous to the charge on the capacitor and the kinetic energy of the pendulum is the energy stored in the inductor. Just like a pendulum, the 2 forms of energy are swapped back and forth until the losses dissipate the energy.

 

[boogyman19946]

Ahh, that makes sense now Thanks guys!