Smudge,
I having trouble understanding what you are proposing here!
I start with the fact that in non-electrolytic capacitors there is never any flow of charge into or out of the dielectric. When we charge the capacitor by applying current the electrodes gain or lose electron charge and thus create an electric field in the dielectric which becomes polarised; at the surfaces in contact with the electrodes that polarisation appears as though these surfaces have charge on them. Those induced surface charges create electric fields that oppose the flow of current onto the electrodes, so our charging system uses energy to supply the current. The current source is seeing a back emf. This is a slightly different viewpoint that to charge a capacitor you apply voltage and the capacitor demands current. Note that the charge separation in the dielectric is driven by the charge separation on the electrodes,
but there are two different charge separations; they normally have the same value. To differentiate between these, I used the term dielectric displacement for the charge separation in the dielectric (maybe I shouldn’t since displacement
D is a recognised EM vector alongside
E). The electrode charge separation is given by the integral of the current with respect to time. Within our circuit, when we do the math, that integral relates to voltage and we think of the capacitor being charged to that voltage.
Turning to the uncharged disc capacitor filling the hole of a ring core having no electrical connection to electrodes. The dielectric displacement will be driven by the
E field due to the changing flux in the core, not by an
E field from charge on the electrodes since there is no means for charge to get there, and that is a different ball game. The randomly orientated electric dipoles in the dielectric are aligned by that ring core
E field, but we cannot consider it to be a charged capacitor yet (there has been no conduction current so there is no electrode charge). If the
E field can be removed faster than the dielectric can relax and we put R across the electrodes the dielectric will quickly drive current to get the electrodes charged, losing a small amount of stored energy in doing so. That current spike ceases when the magnitude of charge on each electrode equals the opposite polarity bound surface charge on the dielectric. Thereafter we do have a conventional charged capacitor discharging exponentially through R with a reversed current. The loss of dielectric energy to get charge onto the electrodes is small so the voltage on this capacitor after that current spike is close to the earlier
E field potential difference across it.
Turning to the situation where we previously charge the capacitor (to 4V) and your specific queries.
What means are we using to stress this dielectric? Is this a poled dielectric that we then assemble with the top and bottom plates and insert into the core? Let's assume yes at this point.
I thought this was quite clear. There is no primary current and no flux in the core. We simply charge the capacitor conventionally (to 4V). This can be done outside the core and the charged capacitor then placed into the core, or it can be done with the capacitor in place. The dielectric is now stressed, has charge displacement, has internal charge separation held there by the charge on the electrodes.
We then energise the core to create an E field that doubles the stress in the dielectric. It has not doubled the charge on the electrodes, that charge Q=CV remains at the Q=4C value. The magnetic vector potential
A field penetrates the dielectric to add its
E=-d
A/dt field to the field already there from the +- Q on the electrodes.
So now we have "the dielectric has been stressed to be equivalent to what it would be under normal charging to 8V" as you stated above. (I don't understand why we wouldn't start with 4v!)
We did start with 4V charge on the capacitor.
Now we apply 4v/turn to the primary. In your case as stated, we would expect the voltage stress across the dielectric to now be 12v
No. We have already done that, we don't do it twice. The dielectric was stressed to 4V (and the electrodes carried the Q=4C value), now it is stressed to 8V (while the electrodes still carry Q=4C).
In all my experimentation with this type of charge separation, at no time have I ever seen energy taken from the primary for charge separation in any open circuited object.
With respect I am not aware that you used a disc capacitor as shown. In all your experiments I see a small purchased capacitor of unknown internal structure.
We can now apply the R load to said capacitor during the time 4v/turn is applied to the primary and we will see a drop in the cap voltage of 12v commensurate with the value of R.
No. We quickly turn the E field off so the 4V/turn on the primary is turned off. With resonance the primary current could be passing through zero at that voltage so this is not difficult, we hold the current at zero. Then we apply the R load.
I have already run tests on your basic scheme and the results were COP<1. However, I must say that I used vertically oriented plates for the capacitor and you state that these types would not work!
What capacitor did you use? Did it nearly fill the core hole with dielectric?
Smudge
Author
Topic: Transformer Induction (Read 26544 times)
