Picoahole's firemen buddies must be happy they invested in EEScam based on Picoahole's suggestion, hahaha!!! 

Hope you suckers also didn't invest in the PLI SCAM, hehehe. 

From the good old days @ Slack EEScam La La Land:

4/4/17 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.