Win Hill: Inverse Marx Generator ??
in [Design]
|
From: John Devereux on 11 Jul 2010 04:05 Jim Thompson <To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> writes: > In the next few days, when I have time, I will issue a mathematical > proof that Larkin is totally wrong. Watch for it ;-) > > Why haven't Win Hill and Phil Hobbs come to Larkin's defense? > > Bwahahahaha! I'm no Phil Hobbs, but isn't all this argument because we are conflating two different usages of "charge"? The "charge" on a capacitor, as somone pointed out already, is really charge *separation* (dilectric polarization). The Q=CV refers to a *separation* of charge, not an absolute quantity. The "absolute" charge - the total number of electrons minus the number of protons - is normally low or zero. Unless your whole circuit picks up an electrostatic charge from somewhere else. It is this "absolute" charge which is conserved, the "Q=CV" "charge" of normal electronics is not. Take a solar cell charging a battery for one obvious example. As Larkin would say, where did the charge come from? Photons don't carry charge! -- John Devereux
From: John Fields on 11 Jul 2010 06:27 On Sat, 10 Jul 2010 18:13:29 -0700, John Larkin <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >On Sat, 10 Jul 2010 19:13:31 -0500, "Andrew" <anbyvbel(a)yahoo.com> >wrote: > >>"John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message >>news:c19h36hekre5kldo38cmdt465f5consr42(a)4ax.com... >>> On Sat, 10 Jul 2010 11:31:15 -0500, Vladimir Vassilevsky >>> <nospam(a)nowhere.com> wrote: >>> >>>>John Larkin wrote: >>>> >>>>> Exactly the point I've been making. Some EEs seem to think that charge >>>>> is always conserved. Some physicists seem to think that energy is >>>>> always conserved. They can't both be right. >>>>> >>>>> I'll side with the physicists on this one. >>>> >>>> >>>>There is no physical laws of "conservation of ...". >>>>There are, however, artificially designed parameters such as "energy", >>>>"charge", "momentum", etc. Those parameters are *defined* in such way >>>>that their value is preserved through certain transformations of a >>>>physical system. The only purpose of this is simplification of math; so >>>>it is possible to balance the states of a system instead of solving >>>>differential equations. >>>> >>> >>> But it's convenient to balance the books by calculating the total >>> energy in a system and assuming it's constant. That can short-cut all >>> sorts of circuit and signal processing problems, avoiding the calculus >>> you suggest. I know of no cases where the energy balance thing has >>> been violated. It would make the front page of the New York Times if >>> it ever were. >> >>Every time it found to be violated new item was added to the definiton of >>"energy" to make it constant. >> >>Last time it was mc^2 if I remember correctly. > >Well, that was over 100 years ago. And even that addition is >irrelevant to electronic design. --- Tell that to Tektronix.
From: John Fields on 11 Jul 2010 07:11 On Sat, 10 Jul 2010 10:13:27 -0700, John Larkin <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >OK, enlighten me. --- OK. --- >Slap a 1-ohm resistor across the 1F/1v cap and discharge it. You'll >get 1 ampere-second out of it eventually. --- Sorry, Charlie, but no. An ampere-second is the amount of charge transferred by a current of 1 ampere in one second. In your example the current will be one ampere when the resistor is first connected, but will have decayed to about 368 mA after one second has passed, so there's no way you'll get one ampere-second out of it. JF
From: John Fields on 11 Jul 2010 07:28 On Sat, 10 Jul 2010 21:24:16 -0700, John Larkin <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >As an EE, I think that a 1F cap charged to 1 volt stores 1 coulomb of >charge, namely because I can observe 1 ampere-second of integrated >current if I connect its plates through a resistive conductor. --- Nope; an ampere-second is one ampere for one second. Plus, (just as an aside) to completely discharge the cap would take forever and, at the end of that time, what you'd get out of the cap would be: QV 1C * 1V W = ---- = --------- = 0.5 joule 2 2 which is exactly what you'd get out of your 4 farads charged to 0.5V. example.
From: Tim Williams on 11 Jul 2010 08:20
"John Fields" <jfields(a)austininstruments.com> wrote in message news:3o7j36d5jvgeg5276nkr2t1fuibdmd6fij(a)4ax.com... > In your example the current will be one ampere when the resistor is > first connected, but will have decayed to about 368 mA after one > second has passed, so there's no way you'll get one ampere-second out > of it. Instead of clucking around, you could actually do some math. Definition: q_tot = integral I*dt from 0 to infty Equation: I(t) = (V/R) * exp(-t/RC) So: q = V/R * integral exp(-t/RC) dt from 0 to infty = [-RC * V/R * exp(-t/RC)] from 0 to infty = -VC * [exp(-infty/RC) - exp(0/RC)] = -VC * [0 - 1] = VC V = 1V and C = 1F so q = 1C. QED. This is only highschool calculus, how embarrassing. Tim -- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms |