Inverse Marx generator
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From: John Larkin on 6 Jul 2010 13:16 On Tue, 6 Jul 2010 11:33:11 -0500, "Tim Williams" <tmoranwms(a)charter.net> wrote: >"John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message news:jtl6361te3ku4ukh0jhp05tab4tvhjjqru(a)4ax.com... >> Connect an inductor across C1 until you've bled it down to half its >> charge. Now connect that inductor to C2 and charge it up to the same >> charge as C1 has. Now disconnect the inductor. If you keep the L >> shorted, you can save the residual energy for reuse later. > ^ ^ ^ ^ >Ha, so charge wasn't conserved after all. See? ;-) > >Tim All my cases involved controlling the charges on two capacitors. I never proposed violating conservation of energy. My point is that inductors can be useful in some situations. John
From: Jim Thompson on 6 Jul 2010 13:40 On Tue, 06 Jul 2010 10:16:23 -0700, John Larkin <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >On Tue, 6 Jul 2010 11:33:11 -0500, "Tim Williams" ><tmoranwms(a)charter.net> wrote: > >>"John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message news:jtl6361te3ku4ukh0jhp05tab4tvhjjqru(a)4ax.com... >>> Connect an inductor across C1 until you've bled it down to half its >>> charge. Now connect that inductor to C2 and charge it up to the same >>> charge as C1 has. Now disconnect the inductor. If you keep the L >>> shorted, you can save the residual energy for reuse later. >> ^ ^ ^ ^ >>Ha, so charge wasn't conserved after all. See? ;-) >> >>Tim > >All my cases involved controlling the charges on two capacitors. I >never proposed violating conservation of energy. My point is that >inductors can be useful in some situations. > >John Your "point" is, as usual, vagueness and subterfuge... and BS. For amusement, newbies are invited to MATHEMATICALLY analyze this simple case... Two equal value capacitors (C), one charged to Vo, the other uncharged (zero volts). Connect together with a switch, start with a finite resistance value, analyze; then reduce the resistance, re-analyze; continue this analysis, approaching zero in the limit. Then scratch your head in surprise... where did the energy go ?:-) ...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 | Obama: A reincarnation of Nixon, narcissistically posing in politically-correct black-face, but with fewer scruples.
From: John Larkin on 6 Jul 2010 15:43 On Sat, 3 Jul 2010 01:15:13 -0500, "Tim Williams" <tmoranwms(a)charter.net> wrote: >Let's say you had a really high voltage, low current source. Not very useful. Think nuclear battery, or lightning. > >So let's say you use it to charge a stack of caps. Then you rewire the caps in parallel. It's like an inverse Marx generator, or a synchronous C-W multiplier. How would you do it? > >Tim A *real* inverse Marx would charge a string of inductors in series, then connect them in parallel to get lots of output current. That actually wouldn't be a bad topology for certain switching regulators. That would work step-up, too, and avoid some capacitance problems. Ooh, use step-recovery diodes for the opening switches! John
From: whit3rd on 6 Jul 2010 15:46 On Jul 6, 6:53 am, John Larkin <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: > On Mon, 5 Jul 2010 22:28:44 -0700 (PDT), whit3rd <whit...(a)gmail.com> > wrote: > > >On Jul 5, 9:41 pm, John Larkin > ><jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: > > >> You can have two caps, C1 charged and C2 not, and transfer all the > >> charge from C1 to C2, without loss. In fact, you can slosh the charge > >> between them, back and forth, forever. Just don't use resistors. > > >It has to be identical size capacitors, otherwise 'all the charge' > >can't be transferred without adding/losing energy... > > Not so. Put a microcoulomb of charge on a 1 uF capacitor. Transfer it all to a 2 uF capacitor. The first state of the system holds twice the energy of the second. You can get that charge back onto the 1 uF capacitor, but it'll take work to do it.
From: John Larkin on 6 Jul 2010 15:59
On Tue, 6 Jul 2010 12:46:20 -0700 (PDT), whit3rd <whit3rd(a)gmail.com> wrote: >On Jul 6, 6:53�am, John Larkin ><jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: >> On Mon, 5 Jul 2010 22:28:44 -0700 (PDT), whit3rd <whit...(a)gmail.com> >> wrote: >> >> >On Jul 5, 9:41�pm, John Larkin >> ><jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: >> >> >> You can have two caps, C1 charged and C2 not, and transfer all the >> >> charge from C1 to C2, without loss. In fact, you can slosh the charge >> >> between them, back and forth, forever. Just don't use resistors. >> >> >It has to be identical size capacitors, otherwise 'all the charge' >> >can't be transferred without adding/losing energy... >> >> Not so. > >Put a microcoulomb of charge on a 1 uF capacitor. Transfer it all to >a 2 uF capacitor. The first state of the system holds twice the >energy of the second. Well, depends on words now. I can transfer "all the charge that's in C1 to C2" (ie, wind up with C1 at zero volts, and no energy lost) but the numerical amount of coulombs must change if the cap values are different, to conserve energy. I can move the charge back into C1, and return the system to its original state. My point was that you can move charge between caps, without losing energy, but not by using resistors. John |