Mark,

Some good effort put in here! You document your ideas exceedingly well.

I do have some comments after reading your report and trying your spreadsheet. I hope you will accept my comments in the spirit of wanting to be constructive and engaging in a free discourse where we all learn from each other. Initially I was surprised to see an argument about DC bias voltages levels using only capacitive analysis, but I wanted to see if you had a new angle. I followed the logic in your report until I came to the top of the last page where you said: “Each capacitor stores 6.092 uC, so…”. That’s where I hit a snag.

Let’s digress and try a thought experiment. I have two perfect capacitors, one is 24.4nF, and the other is 2.2nF, just to match your example. It doesn’t matter what the dielectrics are for this experiment. Assume there are no leakage paths. I momentarily connect each cap to an adjustable DC power supply to apply charges to each cap separately. I charge the 2.2nF cap to 2900 V DC and the 24.4nF cap to 100V, just arbitrarily. I know that I just induced a charge of 6.38uC on the smaller cap, and 2.44uC in the larger cap. I now connect the positive end of one cap to the negative end of the other, while leaving the outer leads of the caps open, unconnected to anything. I measure the voltage across these outer leads and see, as expected, 3000V. While the caps remain connected one-lead-to-one-lead, I measure the voltage across each cap and I still see 2900 V on the 2.2nF cap and 100 V across the 24.4nF cap. I must see these same voltages because no current was ever allowed to flow into or out of either cap after giving them the initial charges. No current could have flowed since the outer leads of each cap were always kept open (floating). I know that I now have a 2nF equivalent cap from the two series caps. With 3000 volts across this equivalent cap I calculate that I have a charge available to me from the outer leads of 6uC. I remember that I imparted 6.38uC and 2.44uC on the caps individually. But wait…

Just for grins, let’s now apply a new DC voltage to each of these same two caps. This time we’ll pick 1500 V for each one. After charging, I again place the caps in series (positive to negative) leaving the outer leads open. Again I measure 3000V across the pair since I allowed no current to flow through either cap – it couldn’t have flowed with the open outer leads. I calculate the charge on the 24.4nF cap at 36.6uC and the charge on the smaller cap is now 3.3uC. But since I still have a 2nF equivalent capacitance with 3000V across the outer terminals, I must still have a charge available to me on the outer leads of 6uC, same as in the prior case. Hmm…

Finally, I separate the two caps but I don’t discharge them. There is still 1500V measured across each one, as before. I now reconnect one lead from each cap together, but this time I connect the two positive leads together, leaving the negative leads open. I measure the voltage across the outer leads and I read ZERO volts. Just to double-check, I measure each cap again and each reads 1500V, as expected with no current being allowed to flow. As before the larger cap must be storing 36.6nC of charge and the smaller must have a charge of 3.3uC. If I mounted these charged caps in this series connection in a black (insulated) box with only the outer open pair of leads coming out, and asked an innocent observer to tell me how much charge is on the cap inside the black box, he’d measure the voltage at ZERO and correctly conclude that there is ZERO charge – AVAILABLE TO HIM OUTSIDE THE BOX. He would never know that 1500V lurked within and that there were 36.6uC and 3.3uC stored within separate caps. He could happily use this black-box 2nF cap (the series equivalent value) for years in various applications and he’d never realize the inner voltages and charges which are hidden from him (remember we said ideal caps with no leakage paths forever).

So now let’s discharge the same two weary caps to zero voltage each. Let’s then connect them in series again. No voltage across the pair, no voltage across either individual cap. A fresh start. Now let’s apply 3000V DC from a power supply across the outer leads. Remove the supply and connect a voltmeter. We read, as expected, 3000V across the outer leads. But what voltages would we predict to be across each one of caps individually? Answer: We can’t predict; it’s INDETERMINATE.

Before we jump off a cliff in despair, let’s return to the real world. In the real world, there is no such thing as an ideal cap. Each cap will have some leakage resistance, even if that resistance is enormously high. It is those resistances that determine the DC voltage ratios across series caps.

I will give one more analogy. If you are familiar with the principle of duality, we can use an example of inductors. We can draw direct analogies between the case of series capacitors with voltages, and the case of paralleled inductors with currents. That’s duality. If I place two ideal inductors (DCR=zero) in parallel and I then force a 1A DC current through this paralleled pair (rather than a voltage as with the cap examples), how much current will flow in each individual inductor? Answer: We can’t know; it’s indeterminate. We need real-world DCRs to decide that argument for us, just like we need real world leakage resistances to settle things for series caps.

So we return to what intuition should have told us in the very beginning: in deciding DC voltage ratios (including the electric fields for biasing electrostatic speakers) we must ignore capacitance altogether (inductance too). For DC calculations, we have to use only resistance calculations.

We should return to Janszen’s and Strickland’s work (see image below if I was able to post it).

There is always going to be some leakage through an air gap. Air starts to ionize significantly above 1000V per millimeter, but there can be minute ion currents below this level too. This effect sets an equivalent resistance for air (Strickland labels it Ria). The insulation will have a resistance of between about 10^14 ohms-cm for PVC to above 10^17 ohms-cm for Teflon and PE (Strickland labels this Rd). Humidity matters in complex ways to both air ionization and hygroscopic effects in the plastic material, by the way. And never mind air pollution or salt content on the coast.

In any case, you have a simple two-resistor voltage divider between diaphragm and stator. It is very possible, under real world humidity levels and high bias conditions to have more resistance in the stator insulator than in the air gap itself. The voltage divider calculation predicts that the bias voltage, intended for the air gap, is instead stolen by the insulator’s larger resistance. With a 3000 V bias supply, you could end up with 300 volts across the air gap and 2700 volts “wasted” across the dielectric, in an extreme example. That makes for a very quiet ESL!

The dielectric constant is not directly relevant. If it were, using Mark’s spreadsheet and default example values, we’d predict only a 2.7% difference in bias voltages in the air gap between Teflon and PVC.

However, from a resistance perspective, PVC is roughly 1000 times leakier than Teflon. That’s why such a big difference in performance between these two insulators has been observed, in favor of the lower resistance PVC.

Again Mark, I appreciate the effort you must have put into your analysis, but I believe one wrong assumption about charges and their effect on DC voltages threw you off course. It happens to all of us. You've made so many other great contributions in this forum. If I’ve made a mistake here, I hope someone sets me straight. I’m here to learn and to share.