Win Hill: Inverse Marx Generator ??
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From: Tim Williams on 16 Jul 2010 07:41 "Artemus" <bogus(a)invalid.org> wrote in message news:i1of1p$g66$1(a)news.eternal-september.org... > Said another way - the external circuit uses energy to move charge > (electrons) from one plate of the capacitor to the other. No net electrons > are added to the capacitor, ergo it has no charge with respect to the rest > of the circuit. To be specific, that current is the displacement current. Curiously, that current even appears to flow in free space, in the absence of matter. I don't know if the quantum vacuum has anything to say about that. Maxwell's original derivation of displacement current assumed polarization charge seperation occured in a polarizable medium. That works for plastic, and it could even work for air, but it's a lot harder to explain for a vacuum. Tim -- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms
From: Jim Thompson on 16 Jul 2010 10:48 On Fri, 16 Jul 2010 08:23:36 +0100, John Devereux <john(a)devereux.me.uk> wrote: >Jim Thompson <To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> writes: > >> On Thu, 15 Jul 2010 18:34:23 -0700, "Artemus" <bogus(a)invalid.org> >> wrote: >> >>> >>>"Jim Thompson" <To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> wrote in message >>>news:1ibv36d2lmvvptmtq7e58cjvb8qqhq90ci(a)4ax.com... >>>> >>>> It's _really_ easy once you stare at the waveforms for awhile... I >>>> _knew_ charge has to be conserved, but I finally saw the answer while >>>> half dozing at a granddaughter's swim party :-) >>>> >>>> ...Jim Thompson >>> >>>A capacitor in an electronic circuit doesn't store charge, it stores energy. >>>The circuit passes a charge thru a capacitor. For every electron that enters >>>the + end another electron leaves the - end. So the capacitor doesn't store >>>charge with respect to the external circuit any more than a resistor does. >>>Art >>> >> >> Huh? > >That's pretty much my model too. So why not enlighten us? > >Of course there are usually several ways of describing any physical >system that end up amounting to saying the same thing in a different >way. I suspect that your epiphany may turn out to be something like >that. But we'll never know if you just sit there going "huh?". > >Put up or shut up Jim :) I'm waiting to see if there's someone out there who can solve it. I should have time to write it up this weekend. Of course both you and Artemus need to study up on the basics of capacitors. "That's pretty much my model..." is not the _accepted_ condition :-) ...Jim Thompson -- | James E.Thompson, CTO | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona 85048 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | Friday is Wine and Cheeseburger Day
From: John Larkin on 16 Jul 2010 12:56 On Fri, 16 Jul 2010 06:41:16 -0500, "Tim Williams" <tmoranwms(a)charter.net> wrote: >"Artemus" <bogus(a)invalid.org> wrote in message news:i1of1p$g66$1(a)news.eternal-september.org... >> Said another way - the external circuit uses energy to move charge >> (electrons) from one plate of the capacitor to the other. No net electrons >> are added to the capacitor, ergo it has no charge with respect to the rest >> of the circuit. > >To be specific, that current is the displacement current. > >Curiously, that current even appears to flow in free space, in the absence of matter. I don't know if the quantum vacuum has anything to say about that. > >Maxwell's original derivation of displacement current assumed polarization charge seperation occured in a polarizable medium. That works for plastic, and it could even work for air, but it's a lot harder to explain for a vacuum. > >Tim I like to think of vacuum having a dielectric constant of zero, not unity. Electrons added to one plate repel electrons on the opposite plate, which in turn prefer to exit to friendlier climes, hence the displacement current. Looking at things that way, the vacuum does nothing, stores no energy. Add a dielectric, and the dipoles within the dielectric act like little polarized springs that twist in the field and store energy and somewhat shield the plates from one another in the process. Of course, treating vacuum as having Er=1 is computationally handy. Since all those little dipoles in a dielectric are moving thermally, does a charged capacitor make noise? John
From: John Larkin on 16 Jul 2010 13:00 On Thu, 15 Jul 2010 18:34:23 -0700, "Artemus" <bogus(a)invalid.org> wrote: > >"Jim Thompson" <To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> wrote in message >news:1ibv36d2lmvvptmtq7e58cjvb8qqhq90ci(a)4ax.com... >> >> It's _really_ easy once you stare at the waveforms for awhile... I >> _knew_ charge has to be conserved, but I finally saw the answer while >> half dozing at a granddaughter's swim party :-) >> >> ...Jim Thompson > >A capacitor in an electronic circuit doesn't store charge, it stores energy. >The circuit passes a charge thru a capacitor. For every electron that enters >the + end another electron leaves the - end. So the capacitor doesn't store >charge with respect to the external circuit any more than a resistor does. >Art > That's silly. I can pump ampere-seconds into a resistor for years, and can recover none of them. If I pump amp-seconds into a cap, I can get them back. And how many go in versus how many come out is mathematically predictable, not to mention useful in real life. John
From: Tim Williams on 16 Jul 2010 14:45
"John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message news:p331461c7sl5547u0chhdnr9kjv46hbum8(a)4ax.com... > I like to think of vacuum having a dielectric constant of zero, not > unity. Electrons added to one plate repel electrons on the opposite > plate, which in turn prefer to exit to friendlier climes, hence the > displacement current. Looking at things that way, the vacuum does > nothing, stores no energy. Add a dielectric, and the dipoles within > the dielectric act like little polarized springs that twist in the > field and store energy and somewhat shield the plates from one another > in the process. > > Of course, treating vacuum as having Er=1 is computationally handy. > > Since all those little dipoles in a dielectric are moving thermally, > does a charged capacitor make noise? Sure. Voltage is one degree of freedom, so throw in a k_B*T/e. It's a capacitor, so it's a lowpass filter, so you'll see, well, not 1/f noise, but something along those lines. So is the current noise white? dV/dt = 0 on a shorted capacitor, so there's no more filtering going on. Question: if vacuum is truely empty, then does a vacuum capacitor have any noise? Tim -- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms |