Daniel R Plante wrote:

DAP wrote:

devotEE wrote:

In this device the charge is stored on the electrodes.

EEventually wrote:

…The field remains because of the charge on the plates. We call them insulators because they don't move charges.

I respectfully disagree. If Eestor wanted to make their PET/coated-CMBT composite as insulating as possible (therefore ensuring that charge remains on the aluminum plates), they would have never poled it.

See ** Analysis of electric conduction mechanisms in polyethylene terephthalate**. In particular, see Fig. 1 which shows that the repeated cycling of applied high voltage increases the current observed when applying that voltage in later cycles.

I agree more with the following sentiment: http://www.theeestory.com/posts/117061.

Dan, I read the paper and I don't see what you're referring to. The paper basically profiles a very good insulating material, made even better by poling that produces bi-polar space charges. This is basically an electret, and the reduction in leakage current due to the space charges in electrets (in plastics as well as ceramics like alumina) is well documented in the literature.

From figure 1, the leakage current at 1.5 V/um is extraordinarily low at aprox 1.5 pA/cm^{2} on their 6cm diameter sample (gee, I wonder how they measured 45pA ;)

Even at 20 V/um it's only about 1.5 nA/cm^{2}. The difference between the poled measure of 175 pA and the unpoled measure of 45pA might seem relatively large, but in absolute terms they are equally tiny.

I also don't see any attempt by them to find a space charge saturation point by leaving their moderate (not high) fields applied for a considerable time (minutes to days). If they did they might have noticed an eventual drop in leakage current, or a flattening of the curve with higher applied fields. That was not the focus of their experiment though.

DAP,

In the research paper there seems to be no indication of what units they are using in their graphs of current - is it amps for the whole specimen, amps per square metre, or amps per square cm?

In the absence of any indication perhaps they should be interpreted as current through the specimen.

Since the electrodes extend only over a 5cm diameter, then it is better to use this rather than a 6cm diameter as current cannot flow through an area of specimen without electrodes. So the area through which the current flows is pi x 5^{2} / 4 (to square the radius not the diamter) = 19.6 cm^{2}. Call it 20 cm^{2} to make all the maths easier. Since there are 10,000 cm^{2} in a square metre then we must multiply current figures by 500 to get A/m^{2}.

If you assume ohmic behaviour, then around 3 x 10^{-7} amps for the specimen at 180 volts for 10 um thickness requires we multiply by 500 (for area) and 20 (for voltage compared with 3,600V). The current would then become around 3 x 10^{-3} A/m^{2} = 3mA/m^{2}. Since the EESU has 60 coulombs per square metre of charge then at the full voltage 60 C would flow in 60/0.003 seconds = 20,000 seconds or 6 hours through the PET.

Then we have to take into account that the PET represents only say 4%, which increases the time up to 150 hours, or getting on for a week to mostly discharge.

Hopefully DW's PET is much much better than the stuff measured in this paper, or the current measurments are amps per m^{2} or something . Otherwise the EESU is dead in the water!

So it is easy to see why you might be concerned!!

If the currents are in A/m^{2} then we can kick off by multiplying by 500, so the discharge times then become around 500 weeks or approximately 10 years, which is more in line with the EESU patent, though not quite as good. That's not the whole story though, as Poole-Frenkel currents rise exponentially with the square root of the voltage rather than linearly.

I've emailed Eugen Neagu to ask for clarification on the scaling of the currents in the graphs.

Regards,

Peter

Last edited Wed, 04 Jan 2012, 10:30am
by Technopete

Assumptions: 1) E=^{1}/_{2}CV^{2}. (Only dummies assume this). (I am one of these dummies).