----- Original Message ----- From: "Horace Heffner" <[EMAIL PROTECTED]> To: <[email protected]> Sent: Monday, June 25, 2007 3:47 AM Subject: Re: [Vo]:Air threads
> > On Jun 24, 2007, at 3:50 PM, Michel Jullian wrote: > >> I had some trouble coming to terms with this observation myself >> when I first experimented with corona discharges, but emitter >> current is equal to plate current at all times, not just on >> average, even though the ions take ages (milliseconds) to cross the >> gap. Try it if you don't believe me. > > But we may not have corona discharges here. Agreed, but corona generated ions are, just like droplets, slow air flying charge carriers, i.e. the conduction mode in which you seem to expect emitter and plate currents not to be equal, except in steady state. I maintain that they are equal at all times, including when the very first carrier starts crossing the gap. Of course if you add a grid or a ring, or any interfering grounded object in addition to the plate, as in your new circuit diagram below, things are different. Maybe I should have made it clear that what I said only applies when, apart from the plate, there are no other grounded objects around (I thought that was implicit since none was shown on the original circuit diagram). >> A more experienced friend gave me the answer to this apparent >> paradox: the plate charge is equal and opposite to the sum of the >> charges on the emitter and in the air. > > False. This is true only on average for drops. No, it is true at all times. The discharge device as a whole remains neutral at all times, like a capacitor or any component, so current in = current out. > This is assuming > steady state current flow, which drops do not provide. They impart > an AC signal and that signal couples in part directly to ground, and > thus puts the emitter AC signal through R4 in that circuit loop. > When the drops arrive, more of their charge is induced on the plates > (vs direct to ground), and that puts an AC signal on the plate and > thus through the R1 or R3 loops. If the drops have exploded in the > process then the AC signal on the plates would be diminished. > > >> >> In terms of circuit theory, as far as the ground connection is >> concerned, the PS + T1 + discharge device is a dead end (open >> circuit), so no current flows through it (the whole thing could be >> disconnected from ground and would still operate identically), it >> just sets a voltage reference. So if you apply KCL http:// >> en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws the current in R2 >> is equal to the current in R1 (at all times). > > > OK, so it appears you want to utterly dismiss the possibility of > drops or filaments, and the effect of the coupling of an AC signal to > other paths through ground, or through a conducting filament. You > seem to want to consider only DC steady state. No, what I said applies to instantaneous currents, whatever the waveform, whether steady state or not. > Within those confines > a ring-pass-through set-up should still be interesting. Bill did say > he had successfully done that. > --------------- > | | > Emitter V | > . | > | > . | > | > Ring O . O | > | o-----C1----o AC Signal (Optional) > . | | > | | > Plates ___ ___ | T1=== AC > | | | | > | | R4 | > | V3 o | P > | | | | > | R3 | | > | | | | > -o-R1-o--G---R2-o- > | | | > o o o > V1 G V2 > > Fig. 2 - Circuit diagram for ring pass-through drop/thread > detection > > > Fig. 2 is again a diagram of the ring pass through concept. It shows > that there is at least one alternate current path through the ring > and thus R4 to ground. This takes fixed font Courier to view. > > In the case of ions, at some ratio of distances, but still > maintaining the stream through the ring to the plates as Bill did, I > would expect a significant amount of the ions go to the ring and not > to the plates (but not so for a true conductive filament). From a > corona R4 would take a significant proportion of the current R1+R3+R4 > = R2. Not if the low current corona emitted a linear string of ions, which is my favorite theory currently. > Further, of those charges that make it to the plates, I would > not expect a sharp change in current from R3 to R1 as the emitter and > ring are passed from left to right over the plates, but would for a > filament. Ah, this I agree with, the landing area of the flying charges would necessarily be a few mm wide so when crossing the border we could have current on both plates at the same time, which could not happen with a molecular sized filament. Cheers, Michel
