This is from Slack, hope Texasbear reads it.

Copper wrote 4/4/17 8:36 a.m.

Matrixman, I applaud your efforts to understand the science behind this.  You're doing great!  Let me weigh in on what I think is going on.  I'm always a bit reluctant to do so on this forum because that kind of thing isn't always welcome but here goes.  Based on publicly-released 3rd part testing I believe what they have as their "cmbt" is a run-of-the-mill ferroelectric material.  You can get k = 20,000 without too much trouble that way and, when mixed as a powder with polymer, you would expect k in the tens. There is a pervasive error going around here that having k = 16,000 at high voltage means it's paraelectric.  It's not the voltage that counts, its the voltage gradient, the voltage divided by thickness.  So with a 4.66 mm thick sample (4,660 microns) 500 volts is only 0.11 V/micron which is next to nothing.  I think they'll find a very different result if they try to put this 500 V over 70 microns. If it's ferro they will see a vast drop in k at this voltage gradient (7.1 V/micron, which it should tolerate).  I note that they measured k at very low voltage gradient and are just assuming that this will hold at a much higher gradient and that's something that needs to be proven.  I bet it won't.

https://eestor.slack.com/archives/C14UJG00N/p1491395776843795

Picoman wrote 8:51 a.m.

@copper, 'run-of-the-mill ferroelectric'. Really? okay. You are not saying it is, just what you believe. Fair enough. And the recent puck doesn't disprove that, because of low v/um. The drop in k to 30 with 80% cmbt in their hv caps would be true due to bruggeman with para or ferro material. The Dr. Golla tests showing constant K around 22,000 over frequency and temperature does show paraelectric. The constant K (though only 30) at  >40V/um in the high voltage caps does show paraelectric. The xrd showed "The XRD pattern shows cubic BaTiO3 at room temperature. " But okay. believe what you want to.

Copper wrote 8:52 a.m.

They don't give the volume of the parts (that I saw) but you can get a clue to the ED.  For part 7 the capacitance is 4.04 nF so (1/2)(4.04e-9)(500^2) = 5.1e-4 J = 1.4e-7 Wh.  Without the part's volume we can't get straight to ED but, for comparison, a D cell battery has 9.56 Wh of energy so it would take 68 million of these parts to have the same energy as one D cell.  Check my math but I think that's right.