Win Hill: Inverse Marx Generator ??
in [Design]
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From: ehsjr on 12 Jul 2010 21:32 John Fields wrote: <snip> >> >>We all understand that you were born that way, AlwaysWrong. > > > --- > He's not always wrong, and I don't understand why you guys don't cut > him some slack. > > Here lately he was talking about easing up on his rancor, and he even > asked a couple of questions neutrally, but all it seemed to gain him > was more hostility. > > It's not like he doesn't want to be here, so if he makes mistakes, > like we all do, why not just correct them, graciously, in order to > help him see where his errors lie instead of just beating him up for > making them. > Exactly right, on all points. Ed
From: Jim Thompson on 12 Jul 2010 21:34 On Mon, 12 Jul 2010 21:32:08 -0400, ehsjr <ehsjr(a)nospamverizon.net> wrote: >John Fields wrote: > ><snip> > >>> >>>We all understand that you were born that way, AlwaysWrong. >> >> >> --- >> He's not always wrong, and I don't understand why you guys don't cut >> him some slack. >> >> Here lately he was talking about easing up on his rancor, and he even >> asked a couple of questions neutrally, but all it seemed to gain him >> was more hostility. >> >> It's not like he doesn't want to be here, so if he makes mistakes, >> like we all do, why not just correct them, graciously, in order to >> help him see where his errors lie instead of just beating him up for >> making them. >> > >Exactly right, on all points. > >Ed Then there's Larkin, make grotesque errors, then stubbornly defend them ;-) ...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 isn't going to raise your taxes...it's Bush' fault: Not re- newing the Bush tax cuts will increase the bottom tier rate by 50%
From: George Herold on 12 Jul 2010 22:21 On Jul 12, 11:56 am, o pere o <m...(a)somewhere.net> wrote: > George Herold wrote: > > <snip> > > > > > You don't need Hobbs or Hill high power thought, this is simple > > freshman physics. Energy conservation and charge conservation are > > always true. In the above case the total charge in the system is near > > zero and doesn't change with time. There is charge separation in the > > cap and charge motion in the inductor, both store energy. > > > George H. > > Absolutely right. In the circuit involved charge conservation is only > used to write Kirchoff Current Laws. For any two terminal device, this > means than a charge q1 flowing into terminal 1 is equal to the same > amount of charge q1 flowing out of terminal 2. It may seem funny, but a > capacitor does not store (net) charge. > > As soon as both "charged" are connected in series, some positive charges > are annihilated by the same amount of negative charges, leaving the same > net balance, i.e. zero. > > Pere Yeah, and yet we talk about the charge on the capacitor. It's easy to be confused. I TA'ed freshman physics lab for a year (20 years ago). I remember how hard it was to get an isolated charge on a piece of metal. It kept leaking off.... George H.
From: JosephKK on 13 Jul 2010 00:31 On Sun, 11 Jul 2010 09:28:52 -0500, John Fields <jfields(a)austininstruments.com> wrote: >On Sun, 11 Jul 2010 07:20:02 -0500, "Tim Williams" ><tmoranwms(a)charter.net> wrote: > >>"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. > >--- >Indeed, since: > >q_tot = integral I*dt from 0 to infty > >should read: > >q_tot = integral I*dt from 0 to t, > >I believe. ;) > > >From: > >http://www.thefreedictionary.com/ampere-second > >"ampere-second - a unit of electrical charge equal to the amount of >charge transferred by a current of 1 ampere in 1 second." > >So, for one coulomb of charge to be transferred through a one ohm >resistor in one second, the voltage would have to remain at one volt >for one second. > >Such is not the case when a one farad capacitor is charged to one volt >and connected across a 1 ohm resistor for one second, since the >voltage will decay from 1V to 0.368V during that time and there'll be: > > Q = CV = 1F * 0.368V = 0.368 coulomb > >still left in the cap when it's disconnected. JF: Dude, an ampere*second (== 1 coulomb) is a product unit. It can be one kA for 1 ms or 1 uA for 1e6 seconds. Or any other combination, including non-constant current, whose time integral of current equals one ampere*second.
From: John Larkin on 13 Jul 2010 00:48
On Mon, 12 Jul 2010 21:31:39 -0700, "JosephKK"<quiettechblue(a)yahoo.com> wrote: >On Sun, 11 Jul 2010 09:28:52 -0500, John Fields ><jfields(a)austininstruments.com> wrote: > >>On Sun, 11 Jul 2010 07:20:02 -0500, "Tim Williams" >><tmoranwms(a)charter.net> wrote: >> >>>"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. >> >>--- >>Indeed, since: >> >>q_tot = integral I*dt from 0 to infty >> >>should read: >> >>q_tot = integral I*dt from 0 to t, >> >>I believe. ;) >> >> >>From: >> >>http://www.thefreedictionary.com/ampere-second >> >>"ampere-second - a unit of electrical charge equal to the amount of >>charge transferred by a current of 1 ampere in 1 second." >> >>So, for one coulomb of charge to be transferred through a one ohm >>resistor in one second, the voltage would have to remain at one volt >>for one second. >> >>Such is not the case when a one farad capacitor is charged to one volt >>and connected across a 1 ohm resistor for one second, since the >>voltage will decay from 1V to 0.368V during that time and there'll be: >> >> Q = CV = 1F * 0.368V = 0.368 coulomb >> >>still left in the cap when it's disconnected. > >JF: Dude, an ampere*second (== 1 coulomb) is a product unit. It can be >one kA for 1 ms or 1 uA for 1e6 seconds. Or any other combination, >including non-constant current, whose time integral of current equals one >ampere*second. The word "integral" scares a lot of people, maybe because so many math profs insist on keeping it abstract. All it means is "how much stuff has flowed." John |