Win Hill: Inverse Marx Generator ??
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From: John Devereux on 11 Jul 2010 08:47 John Fields <jfields(a)austininstruments.com> writes: > 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. That is, 1 coulomb. > > 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. What on earth are you talking about? This is pretty much the *definition* of capacitance. I.e., from Q = CV = "Ampere Seconds". No wonder John's having trouble convincing you of anything... -- John Devereux
From: John Fields on 11 Jul 2010 09:14 On Sun, 11 Jul 2010 13:47:20 +0100, John Devereux <john(a)devereux.me.uk> wrote: >John Fields <jfields(a)austininstruments.com> writes: > >> 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. > >That is, 1 coulomb. > >> >> 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. > >What on earth are you talking about? This is pretty much the >*definition* of capacitance. I.e., from Q = CV = "Ampere Seconds". > >No wonder John's having trouble convincing you of anything... --- Not of anything, just of some things. About the ampere-seconds thing though: If you connect a 1VDC supply across a 1 ohm resistor for 1 second then the amount of charge tranferred will be 1 coulomb. Then, since it got transferred in one second, the rate at which it was tranferred was one coulomb per second, which is one ampere. Now, replace the DC power supply with a capacitor charged to one volt, connect it to the resistor, and then disconnect it after one second. Will one coulomb of charge have been transferred?
From: John Fields on 11 Jul 2010 10:28 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.
From: John Devereux on 11 Jul 2010 11:00 John Fields <jfields(a)austininstruments.com> writes: > On Sun, 11 Jul 2010 13:47:20 +0100, John Devereux > <john(a)devereux.me.uk> wrote: > >>John Fields <jfields(a)austininstruments.com> writes: >> >>> 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. >> >>That is, 1 coulomb. >> >>> >>> 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. >> >>What on earth are you talking about? This is pretty much the >>*definition* of capacitance. I.e., from Q = CV = "Ampere Seconds". >> >>No wonder John's having trouble convincing you of anything... > > --- > Not of anything, just of some things. > > About the ampere-seconds thing though: > > If you connect a 1VDC supply across a 1 ohm resistor for 1 second then > the amount of charge tranferred will be 1 coulomb. > > Then, since it got transferred in one second, the rate at which it was > tranferred was one coulomb per second, which is one ampere. Sure > > Now, replace the DC power supply with a capacitor charged to one volt, > connect it to the resistor, and then disconnect it after one second. > > Will one coulomb of charge have been transferred? Of course not, but Larkins original statement was "eventually", not "after one second". You are the only one to add this condition, obviously if you stop discharging the cap early you will not get all the charge out! >>>>Slap a 1-ohm resistor across the 1F/1v cap and discharge it. You'll >>>>get 1 ampere-second out of it eventually. One ampere-second does not have to mean that "one amp flowed for one second", It means an amount of charge *equal to* that transferred by one amp flowing for one second. It could be 2 amps for half a second, 10 amps for 0.1s, 1000 amps for 1 millisecond. Or an exponential decay. -- John Devereux
From: Tim Williams on 11 Jul 2010 11:01
"John Fields" <jfields(a)austininstruments.com> wrote in message news:osjj36lr6hhtvasmda6r795t8kq5tf3a8s(a)4ax.com... > Indeed, since: > > q_tot = integral I*dt from 0 to infty > > should read: > > q_tot = integral I*dt from 0 to t, > > I believe. ;) Well, no. The man said: "John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message news:06ah36h7br6k0i7n10isg2heuvtjiravh4(a)4ax.com... > OK, enlighten me. > > Slap a 1-ohm resistor across the 1F/1v cap and discharge it. You'll > get 1 ampere-second out of it eventually. ^ ^ ^ ^ ^ ^ The mathematical definition of "eventually" is, "as t approaches infinity". Larkin never stated that the capacitor was to discharge for only one second. I don't know where you got that from. Tim -- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms |