Win Hill: Inverse Marx Generator ??
in [Design]
|
From: Artemus on 15 Jul 2010 22:10 "Jim Thompson" <To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> wrote in message news:t0ev36tbdm1gnr920kia6ab421085ksh6r(a)4ax.com... > 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? > > ...Jim Thompson > -- 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. Using Larkin's paradox as an example. Charge 2 parallel 2uF caps to 0.5V. Total charge moved from - to + is 2 coulombs, 1 coulomb per cap. Rewire the caps to be in series and discharge. Total charge leaving each cap is 1 coulomb. No electrons were created or destroyed in this process. Art
From: John Larkin on 15 Jul 2010 22:37 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
From: Jim Thompson on 16 Jul 2010 00:16 On Thu, 15 Jul 2010 19:10:18 -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:t0ev36tbdm1gnr920kia6ab421085ksh6r(a)4ax.com... >> 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? >> >> ...Jim Thompson >> -- > 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. > >Using Larkin's paradox as an example. >Charge 2 parallel 2uF caps to 0.5V. Total charge moved from - to + >is 2 coulombs, 1 coulomb per cap. Rewire the caps to be in series and >discharge. Total charge leaving each cap is 1 coulomb. No electrons >were created or destroyed in this process. >Art > Omigod! A convert! Better go back and study some basics :-( ...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: Jim Thompson on 16 Jul 2010 00:17 On Thu, 15 Jul 2010 19:37:54 -0700, John Larkin <jjlarkin(a)highNOTlandTHIStechnologyPART.com> 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 OK. No problem. Where'd the charge come from in your "example"? ...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 Devereux on 16 Jul 2010 03:23
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 :) -- John Devereux |