Win Hill: Inverse Marx Generator ??
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From: John Larkin on 16 Jul 2010 15:01 On Fri, 16 Jul 2010 13:45:59 -0500, "Tim Williams" <tmoranwms(a)charter.net> wrote: >"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 Since the electrons in the plate are moving around, the population on the surfaces may fluctuate, changing the perceived field at the other plate, so I'd go for a definite maybe. John
From: John Larkin on 16 Jul 2010 15:09 On Fri, 16 Jul 2010 13:45:59 -0500, "Tim Williams" <tmoranwms(a)charter.net> wrote: >"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 http://en.wikipedia.org/wiki/Johnson%E2%80%93Nyquist_noise#Thermal_noise_on_capacitors That works out to about 4e-8 VRMS on a 1 uF cap, if I got that right, 64 uV on 1 pF. That could matter. John
From: George Herold on 16 Jul 2010 15:12 On Jul 16, 2:45 pm, "Tim Williams" <tmoran...(a)charter.net> wrote: > "John Larkin" <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote in messagenews: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 Yeah, another step or two. The cap has one degree of freedom so it gets kT/2 (according to the equipartion theorem.) this energy is equal to CV^2/2 so the mean square voltage on the cap V^2 = kT/C. One fun thing to ask is, "The temperature of what?" It turns out there has to be some resistance between the ends of the cap. And it is the temperature of the resistor and not the temperature of the cap that matters. (I've done this experiment, which is kinda cool.) This is just a corallary(sp) to Johnson noise. George H.
From: Paul Hovnanian P.E. on 16 Jul 2010 16:04 John Larkin wrote: > > On Thu, 15 Jul 2010 13:36:44 -0700, "Paul Hovnanian P.E." > <paul(a)hovnanian.com> wrote: > > >Jim Thompson wrote: > > > >> On Fri, 09 Jul 2010 20:46:10 -0700, "Paul Hovnanian P.E." > >> <Paul(a)Hovnanian.com> wrote: > >> > >> [snip] > >>> > >>>Oooh. Its the old "where did the energy go" two cap puzzle. > >>> > >> [snip] > >> > >> Which is trivial to solve :-) > > > >I posted my solution on a.b.s.e. as Two Cap Puzzle. > > All that math is admirable, but there's a simpler way to calculate the > energy lost in the resistor. > > Connect two caps, charged to diffferent voltages, with a resistor. > Observe the voltage waveform across the resistor. It starts at some > initial voltage Ve = Vc1-Vc2. It decays with some time constant tau = > R * Ce, where Ce is the equivalent value of C1 in series with C2. > > As far as the resistor can tell, it has just been connected to a > capacitor Ce, charged to V, and it discharges that. The waveform > across the resistor is identical. So it burns energy equal to 1/2 * Ce > * Ve^2. > > That concept can be handy in other situations. > > John Right. If what you're after is the expression for i(t). But I figured I had to drag both of those caps through the equations since the original question was: Where did the original C1 energy go? Make too many simplifications and some folks will call 'Bullshit!'. -- Paul Hovnanian mailto:Paul(a)Hovnanian.com ------------------------------------------------------------------ Diplomacy is the art of saying "nice doggy" while looking for a rock.
From: Jim Thompson on 16 Jul 2010 16:15
On Fri, 16 Jul 2010 12:01:14 -0700, John Larkin <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >On Fri, 16 Jul 2010 13:45:59 -0500, "Tim Williams" ><tmoranwms(a)charter.net> wrote: > >>"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 > >Since the electrons in the plate are moving around, the population on >the surfaces may fluctuate, changing the perceived field at the other >plate, so I'd go for a definite maybe. > >John Real capacitors _are_ noisy. Some foundries now characterize the noise in the Spice subcircuit model. ...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 |