Capacitors Paradox: Can Energy Transfer Require Less?

[jim hardy]

Great post indeed !

I think the 'paradox' arises from the connection of two unequally charged ideal capacitors through ideal wires. That causes division by zero, I = ΔVolts/zeroOhms , so anything goes.

Adding an ideal inductor prevents division by zero. Further, an ideal inductor can dissipate no energy so the energy should all show up in the capacitors afterward.
Ahh if only you had an ideal diode...

It's not unlike the thermo question - a gas is confined to one half of a chamber, the other half is evacuated. What happens if the wall separating the chamber halves is suddenly removed?

 

[The Electrician]

With no diode the result is:

Note that the ground reference for the two voltage traces (yellow and green) is at the bottom 1 cm; you can see a ground symbol down there at the far left. The final result is 15 volts on both caps.

Here I have inserted a 10 ohm resistor in series. Now we're overdamped.

Still 15 volts on both caps at the end.

Here's a picture of the experimental setup:

 

[The Electrician]

@electrician...this is brilliant! I would love to check your numbers...what is the time scale on the plot?
What do you get without the diode? Do you get an oscillation.
I have no means of reproducing your measurements but I am impressed by this...one of the best things I have seen on PF and in the true spirit of physics...experimental evidence to discus.
Congratulations

You can see the horizontal sweep speed at the top of the image--500 μs/cm for the first image. Later images are 2 ms/cm. Channel 1 and 2 are 5 V/cm, and the current scale is 5 A/cm.

 

[jim hardy]

Those are very nice traces. What machine do you use to make them? I'm still on a analog HP180...

 

[The Electrician]

Those are very nice traces. What machine do you use to make them? I'm still on a analog HP180...

It's one of these:

http://www.home.agilent.com/en/pd-6...hannels?nid=-536902766.536908096&cc=US&lc=eng

For a while recently it looked like they were going to discontinue it, but apparently they decided to raise the price and continue making it!

 

[Dadface]

I like this. It's interesting to note that if the diode is not used both capacitors eventually reach 15V with each capacitor having the same charge and the energy loss being 50 percent.
If the circuit is switched off before charging/discharging is complete then the energy loss is reduced. The same applies when the capacitors are connected by wires. The energy loss is 50 percent only when charging/discharging is complete.

 

[jim hardy]

Interesting that the energy loss is same for the real case of resistance in wires and theoretical one of zero resistance in wires.

There must be a relation between capacitor ratio and energy lost.
That'd be an interesting piece of algebra, and might uncover one of those "little gems of wisdom" a fellow likes to keep in his bag of tricks.

I'd just embarrass myself with clumsy Latex if I tried...

old jim

 

[TurtleMeister]

Yes, this is a good learning exercise. If the circuit is allowed to oscillate (without the diode) then energy is lost mostly through the resistive component of the inductor. So in that sense it is similar to an RC circuit except that it oscillates.

Since no one has volunteered to answer my question in post #30, I will attempt to answer it myself. The question is: With the diode in the circuit will the polarity on the inductor reverse after the peak current is reached?

The answer is yes it will. This may seem non intuitive at first because the direction of current flow through the inductor does not change. During the first quarter cycle energy is being stored in the magnetic field of the inductor with C1 being the source. At the peak of current flow the voltage on C1 and C2 are equal (15 volts) and the voltage on the inductor is zero (voltage and current on the inductor is 90 degrees out of phase). During the second quarter cycle the magnetic field of the inductor is collapsing with the inductor being the source. The energy stored in the inductor reduces the voltage on C1 to it's final 5 volt value and increases the voltage on C2 to it's final 25 volt value. The diode prevents oscillation. The circuit cannot go for more than one half cycle, giving a maximum transfer efficiency. You can see the one half cycle of current in the oscilloscope trace. Note: because of losses, voltages may be slightly less than those stated.

 

[The Electrician]

Since no one has volunteered to answer my question in post #30, I will attempt to answer it myself. The question is: With the diode in the circuit will the polarity on the inductor reverse after the peak current is reached?

I thought your question was rhetorical. The polarity indication on the inductor is an artifact of the spice program I used to produce the schematic. Spice indicates polarity on inductors simply as a way of declaring the reference direction for inductor voltage when plotting waveforms. It was happenstance that the polarity indication matches the initial polarity in the behavior of the circuit.

Your detailed explanation is right on the money. The person who hasn't seen this type of circuit in action before could persuade his intuition about the inductor's voltage polarity in two ways. Remembering that for an inductor e=L*di/dt, we see that the slope of the inductor current versus time changes sign at the peak of the inductor current, hence the voltage polarity changes at the peak; after the first quarter cycle, as you point out.

Also, because the inductor is connected between the two capacitors, the voltage across the inductor is the difference of the capacitor voltages. The capacitor voltages cross over at 15 volts at the same time as the inductor current reaches its peak, and that's when the inductor voltage polarity changes.

This circuit is a simple demonstration of how "resonant switching" works in switching power supplies.

Thanks for your detailed further explanation of the circuit's operation.