Hybrid Dielectric in Capacitors

The air gap will light up. It takes a few usec to build a plasma - 5us RF pulses would get to a higher power level. If the solid dielectric is sapphire or alumina (er ~ 9) the air gap will break down at a lower voltage than it would with quartz (er ~ 4.5).

I believe that is the case, yes.

Only if the air gap is small enough that the capacitance of the "air dielectric" part is significantly larger than the capacitance of the "high dielectric" -- this follows from the "two caps in series" bit.

I'm running out of practical expertise here, but I suspect that if the voltage of the virtual air dielectric cap remains under it's breakdown voltage that you are OK.

If the arrangement is such that the voltage on the virtual air cap is higher than it's breakdown voltage -- I dunno. I suspect that you'll get conduction in some form to the surface of the dielectric.

I also suspect that there are so many variables at work here that if you want to do this for real you should do some experimenting.

Chief among the variables is how much the high dielectric strength material extends past the plates, and how well smoothed-off the edges of the plates are. But, this is definitely something that cries out for someone who works at a company that makes high-voltage RF caps to weigh in, or for someone to make up some parts and start playing around.

It only matters when the applied FIELD is greater than the breakdown field in the air gap; not all the voltage is dropped in air, it'll be partly in the higher dielectric. So, it matters what the dielectric constant is (at the frequency of interest), and how thick the two layers are.

If the air breaks down, your next concern will be the reactive ions that impinge on the 'high dielectric constant' material.

call this eps1 in a thickness d1

and this eps0 in a thickness d2

That's correct. On the boundary between the two dielectric layers the potential is uniform, so there could as well have been a metal layer. Or two, with a connector between them, separating the capacitor of 2 layers into two capacitors with one layer each.

Back to the single double-layer capacitor, the normal component of D doesn't change when passing from one layer to the other, because

div D = 0 (no charge 'sources').

The electric field (strength) however does change, according to

E = D/eps.

So when you pass from the high eps into the low eps area, your E suddenly increases. If its value exceeds the breakthrough field strength, then you get partial and/or complete discharge.

The voltage distribution follows from the voltage U across the capacitor, the values of eps0 and eps1, and the thickness of the dielectric layers:

U = U1 + U2 = E1.d1 + E2.d2 = D.d1/eps1 + D.d2/eps0 = (1) = D.(d1/eps1 + d2/eps0)

From this we get D = U/(d1/eps1 + d2/eps0) and the electric field

in layer 1 is E1 = D/eps1 and in layer 2 is E2 = D/eps2. (2)

In another way of writing, (1) into (2):

E1 = U.eps0/(d1.eps0+d2.eps1) (3) E2 = U.eps1/(d1.eps0+d2.eps1)

(Don't use these results without checking the math :)

If you look a bit at (3) you will see the problem: The field strength E2 in the air gap is proportional to the dielectric constant eps1 of the dielectric layer. So the 'better' your dielectric, the higher the strength in any air bubble or air gap.

That's also why you normally don't want air inside your dielectric, or between the layers of insulation tape, or etc..

Yes

Only if the thickness is still much less than that of the dielectric.

Well, if the air gap is large enough, the field can be small enough to prevent break down. I don't completely understand though what you mean so it's possible this comment made no sense to you.

You really need to translate this to voltage/distance, or electric field values. Or do you actually mean that the applied voltage is such that the electric field strength E2 in the air gap would exceed its breakdown value, but the E1 in the dielectric would not?

My guess is that you get a corona-like discharge through the air gap, possibly damaging the dielectric and the plate material. You will also get an E-field that is not a nice function any more of the distance where it occurs. And more important, I think, is that due to the (local) breakdown you get (locally) the full voltage applied over the dielectric only, possibly causing the E in the dielectric to exceed its breakdown value and to damage your nice (?) dielectric layer.

Then it will probably occur more rapidly.

joe

The problem is "corona discharge" through tiny air bubbles which only vacuum impregnation manufacturing techniques eliminate or winding in vacuo.

... Phil

The average electrical characteristic will be: a terminal capacitance somewhere between the two cases, and loss corresponding to the amount of power lost in the discharge.

You can draw the equivalent circuit: a large capacitor in series with a smaller one, with a lossy, variable capacitor in parallel with the smaller one. The ratios of these parts will determine the overall impedance, Q, whatever.

As for how lossy and variable the discharge actually is... who knows? It's tempting to assign a Q of 1 or something like that, but really, the voltage drop through the myriad discharge streamers will actually be pretty low, which implies the change from low to high capacitance will occur with fairly low losses. On the other hand, the discharges are literally discharging the capacitance of that gap, presumably consuming the energy stored there (which implies a low Q, i.e., it fully charges and self-discharges every cycle).

Tim

-- Seven Transistor Labs, LLC Electrical Engineering Consultation and Contract Design Website:

If a capacitor has two layers of dielectric, one with a high dielectric strength and high dielectric constant, the other air with a much lower dielectric strength and dielectric constant of 1. My understanding is that the single capacitor would be the equivalent of two capacitors in series as if there were a plate between the two dielectrics... in some respects.

If the air gap is small enough the capacitance of that virtual capacitor is large enough the total capacitance is that of the higher dielectric. If the air gap is larger the capacitance of the air gap is a lot less than the capacitance of the higher dielectric and so dominates the capacitance. This larger air gap capacitor will have nearly the full applied voltage to it. The air gap is not so large that the breakdown voltage is larger than the high dielectric.

So when the applied voltage is greater than the breakdown voltage of the air gap, but smaller than the breakdown voltage of the high dielectric, what will happen? The applied voltage is RF.

--

Rick

By the way, is this a re-trial of your inquiry in Message-ID: ? :)

joe