Win Hill: Inverse Marx Generator ??
in [Design]
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From: Jim Thompson on 10 Jul 2010 13:55 On Sat, 10 Jul 2010 10:13:27 -0700, John Larkin <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >On Sat, 10 Jul 2010 09:51:13 -0700, "Paul Hovnanian P.E." ><paul(a)hovnanian.com> wrote: > >>John Larkin wrote: >> >>> On Sat, 10 Jul 2010 11:35:35 -0400, "tm" <noone(a)msc.com> wrote: >>> >>>> >>>>"John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in >>>>message news:o61h36lt8fvhsc00mrc9824ju0jd4hml8s(a)4ax.com... >>>>> >>>>> To celebrate the 21st century, I have composed a new riddle: >>>>> >>>>> Start with a 4 farad cap charged to 0.5 volts. Q = 2 coulombs. >>>>> >>>>> Carefully saw it in half, without discharging it, such as to have two >>>>> caps, each 2 farads, each charged to 0.5 volts. The total charge of >>>>> the two caps remains 2 coulombs, whether you connect them in parallel >>>>> or consider them separately. >>>>> >>>>> Now stack them in series. The result is a 1F cap charged to 1 volt. >>>>> That has a charge of 1 coulomb. Where did the other coulomb go? >>>>> >>>>> I think this is a better riddle. >>>>> >>>>> John >>>>> >>>>> >>>> >>>>One should not confuse charge with energy. >>>> >>>> >>> >>> Exactly the point I've been making. Some EEs seem to think that charge >>> is always conserved. >> >>It is. >> >>> Some physicists seem to think that energy is >>> always conserved. They can't both be right. >> >>It is. You're just not accounting for where part of it went. > >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. We started with 2 coulombs. >Where did the other coulomb go? > >John > A 1F/1V cap has ONE COULOMB you dummy ;-) Why won't you defend your "not conserved" with your inductor scheme? ...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: kevin93 on 10 Jul 2010 14:24 On Jul 10, 8:16 am, John Larkin <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: > On Sat, 10 Jul 2010 08:32:31 -0500, "George Jefferson" > > > > > > <phreon...(a)gmail.com> wrote: > >>>>>>If you conserve energy, then you must have > > >>>>>>C1*V1^2 = C2*V2^2 > > >>>>>Right. If you dump all the energy from one charged cap into another, > >>>>>discharged, cap of a different value, and do it efficiently, charge is > >>>>>not conserved. > > >>>>>John > > >>>>Would you care to prove that for us John? Mathematically, that is. No > >>>>hand-waving. After all you do claim trivial EE101 :-) > > >Larkin fails to realize that V1 = V2. > > >V1*Q1 = V2*Q2 > > >=> Q2 = V1/V2*Q1, V1 = V2 so Q1 = Q2. > > >You would think with all his vast knowledge that would understand basic 101 > >electronics. > > >+----+ > >| | > >C1 C2 > >| | > >+----+ > > >If you discharge C1 into C2 then the voltages across them will be equal > >after a given time. The total charge will still not have changed. > > >The total charge Q = Q1 + Q2. As electrons move from the cap of higher > >voltage to lower voltage we end up with Q = (Q1 -+ de) + (Q2 +- de) = Q. > > That simple riddle is ancient, possibly even older than JT. > > Obviously charge is conserved in this circuit, independent of what > impedance is used to bridge the caps or of when you observe the > system. I certainly wouldn't dispute that. > > What I *did* say is conveniently right at the top of your post. > > To celebrate the 21st century, I have composed a new riddle: > > Start with a 4 farad cap charged to 0.5 volts. Q = 2 coulombs. > > Carefully saw it in half, without discharging it, such as to have two > caps, each 2 farads, each charged to 0.5 volts. The total charge of > the two caps remains 2 coulombs, whether you connect them in parallel > or consider them separately. > > Now stack them in series. The result is a 1F cap charged to 1 volt. > That has a charge of 1 coulomb. Where did the other coulomb go? > > I think this is a better riddle. > > John Each of the two 2F caps each with 1 coulomb of charge will be discharged simultaneously. So even though just 1 Coulomb will flow through the external circuit it is used twice. Maybe you should not think about charge but about the voltage on a capacitor being the integral of the current over time. Then you would see that on all of the circuits you have described that the "Charge is conserved" mantra is essentially true In the case of the two or three component systems you have been describing (two caps or two caps with another component between them) all of the current from one of the capacitors flows into the other capacitor with none going anywhere else. In that scenario the integral of the current out of one is equal to the integral of the current flowing into the other. So whatever charge one loses, the other gains. kevin
From: John Larkin on 10 Jul 2010 14:37 On Sat, 10 Jul 2010 12:45:09 -0500, "George Jefferson" <phreon111(a)gmail.com> wrote: > > >"John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message >news:5d6h365rnvea7rv1cnttr5uq71sr7of954(a)4ax.com... >> On Sat, 10 Jul 2010 11:35:35 -0400, "tm" <noone(a)msc.com> wrote: >> >>> >>>"John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in >>>message >>>news:o61h36lt8fvhsc00mrc9824ju0jd4hml8s(a)4ax.com... >>>> >>>> To celebrate the 21st century, I have composed a new riddle: >>>> >>>> Start with a 4 farad cap charged to 0.5 volts. Q = 2 coulombs. >>>> >>>> Carefully saw it in half, without discharging it, such as to have two >>>> caps, each 2 farads, each charged to 0.5 volts. The total charge of >>>> the two caps remains 2 coulombs, whether you connect them in parallel >>>> or consider them separately. >>>> >>>> Now stack them in series. The result is a 1F cap charged to 1 volt. >>>> That has a charge of 1 coulomb. Where did the other coulomb go? >>>> >>>> I think this is a better riddle. >>>> >>>> John >>>> >>>> >>> >>>One should not confuse charge with energy. >>> >>> >> >> 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. >> > >No, your not making sense. When you use the term "conserve" you are implying >a net conservation. It makes no sense to say something isn't conserved when >you are not talking about the net effect because it is obvious. > >Obivous charge in the sense you are talking about is not conserved. Take a >battery. Electrons flow out of it... hence there is no "conservation". > >If you charge a cap up with a battery(analogous to your cap to cap) there is >no "conservation" in your sense because the cap "stole" electrons from the >battery. > >Yet the net CHARGE is CONSERVED and always will be(except possibly at scales >near the planck time). > >Your confusing conservation with distribution. If you take any distribution >of charge you can easily say that any part of it will not be conserved. To >prove this you can just move a charge out of that part under consideration. > >Again, if you want to use such an obtuse definition for conservation then >you are right. Generally when we talk about conservation we are saying so in >terms of a closed system else it is generally meaningless/useless. > >You can then say nothing is conserved. Heat, charge, momentum, etc... > >http://en.wikipedia.org/wiki/Conservation_law > >Read the first sentence: > >"In physics, a conservation law states that a particular measurable property >of an ****isolated**** physical system does not change as the system >evolves." > >Of course it requires a bit of intelligence to know what an isolated system >is and how to use it in practice to get any meaningful result. > >Now I suppose your argument will be that "conserved" has no relation to a >conservation law? > > > I'm an engineer. I design circuits. Philosophy is useless to me unless it allows me to quantify and measure things and predict what the numbers will mean. Take two caps. C1 is 1 farad, charged to 1 volt. C2 is 2 farads, zero volts. C1 stores 1 coulomb, C2 has zero. Connect a 1 henry inductor across C1 until its voltage is zero. All the energy has been transferred to the L. Now use the energy in L to charge C2, and disconnect it when the current decays to zero. All the energy that used to be in C1 is now in C2. C2 is now at 0.707 volts, and it holds 1.414 coulombs of charge. Even simpler: calculate the coulombs stored in (eg, ampere-seconds removable from) four identical charged caps in parallel. Now rearrange them in series to form a 2-terminal capacitor. Recalculate the coulombs available for extraction. Or: start with identical charged caps in parallel. Compute Q of the parallel combo. Disconnect them and reconnect with opposed polarity. Recompute Q. That's what "conservation of charge" means in my world: things I can calculate, measure, and use. Sometimes I can count on conservation of charge, sometimes I can't. I can always count on conservation of energy. John
From: John Larkin on 10 Jul 2010 14:49 On Sat, 10 Jul 2010 11:24:16 -0700 (PDT), kevin93 <kevin(a)whitedigs.com> wrote: >On Jul 10, 8:16�am, John Larkin ><jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: >> On Sat, 10 Jul 2010 08:32:31 -0500, "George Jefferson" >> >> >> >> >> >> <phreon...(a)gmail.com> wrote: >> >>>>>>If you conserve energy, then you must have >> >> >>>>>>C1*V1^2 = C2*V2^2 >> >> >>>>>Right. If you dump all the energy from one charged cap into another, >> >>>>>discharged, cap of a different value, and do it efficiently, charge is >> >>>>>not conserved. >> >> >>>>>John >> >> >>>>Would you care to prove that for us John? �Mathematically, that is. No >> >>>>hand-waving. �After all you do claim trivial EE101 :-) >> >> >Larkin fails to realize that V1 = V2. >> >> >V1*Q1 = V2*Q2 >> >> >=> Q2 = V1/V2*Q1, V1 = V2 so Q1 = Q2. >> >> >You would think with all his vast knowledge that would understand basic 101 >> >electronics. >> >> >+----+ >> >| � �| >> >C1 � C2 >> >| � �| >> >+----+ >> >> >If you discharge C1 into C2 then the voltages across them will be equal >> >after a given time. The total charge will still not have changed. >> >> >The total charge Q = Q1 + Q2. As electrons move from the cap of higher >> >voltage to lower voltage we end up with Q = (Q1 -+ de) + (Q2 +- de) = Q. >> >> That simple riddle is ancient, possibly even older than JT. >> >> Obviously charge is conserved in this circuit, independent of what >> impedance is used to bridge the caps or of when you observe the >> system. I certainly wouldn't dispute that. >> >> What I *did* say is conveniently right at the top of your post. >> >> To celebrate the 21st century, I have composed a new riddle: >> >> Start with a 4 farad cap charged to 0.5 volts. �Q = 2 coulombs. >> >> Carefully saw it in half, without discharging it, such as to have two >> caps, each 2 farads, each charged to 0.5 volts. The total charge of >> the two caps remains 2 coulombs, whether you connect them in parallel >> or consider them separately. >> >> Now stack them in series. The result is a 1F cap charged to 1 volt. >> That has a charge of 1 coulomb. Where did the other coulomb go? >> >> I think this is a better riddle. >> >> John > >Each of the two 2F caps each with 1 coulomb of charge will be >discharged simultaneously. So even though just 1 Coulomb will flow >through the external circuit it is used twice. Used twice? What does that mean? What meter would measure the "used twice" current waveform? > >Maybe you should not think about charge but about the voltage on a >capacitor being the integral of the current over time. Then you would >see that on all of the circuits you have described that the "Charge is >conserved" mantra is essentially true > >In the case of the two or three component systems you have been >describing (two caps or two caps with another component between them) >all of the current from one of the capacitors flows into the other >capacitor with none going anywhere else. In that scenario the >integral of the current out of one is equal to the integral of the >current flowing into the other. So whatever charge one loses, the >other gains. Obviously true, if you're only allowed to connect them with a series impedance. But that's a special case, not the only one I have suggested. John
From: Vladimir Vassilevsky on 10 Jul 2010 14:59
John Larkin wrote: > I'm an engineer. I design circuits. Philosophy is useless to me unless > it allows me to quantify and measure things and predict what the > numbers will mean. Yea, this is what good soldier Schweik used to say: "When a car runs out of gas, it stops. Even after been faced with this obvious fact, they dare to talk about momentum". Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com |