Win Hill: Inverse Marx Generator ??
in [Design]
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From: Jim Thompson on 11 Jul 2010 13:01 On Sat, 10 Jul 2010 21:24:16 -0700, John Larkin <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: [snip] > >I wonder if Jim will agree with you that a charged capacitor and a >discharged capacitor both retain the same amount of charge. Nowhere did I say that charge cannot be removed (via the resistor, or whatever) from a capacitor... you MORONIC OBFUSCATOR! > >Do you actually design electronics? Certainly! Do you? ...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: John Devereux on 11 Jul 2010 13:26 Jim Thompson <To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> writes: > On Sun, 11 Jul 2010 09:05:34 +0100, John Devereux > <john(a)devereux.me.uk> wrote: > >>Jim Thompson <To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> writes: >> >>> In the next few days, when I have time, I will issue a mathematical >>> proof that Larkin is totally wrong. Watch for it ;-) >>> >>> Why haven't Win Hill and Phil Hobbs come to Larkin's defense? >>> >>> Bwahahahaha! >> >>I'm no Phil Hobbs, but isn't all this argument because we are conflating >>two different usages of "charge"? >> >>The "charge" on a capacitor, as somone pointed out already, is really >>charge *separation* (dilectric polarization). The Q=CV refers to a >>*separation* of charge, not an absolute quantity. The "absolute" charge >>- the total number of electrons minus the number of protons - is >>normally low or zero. Unless your whole circuit picks up an >>electrostatic charge from somewhere else. It is this "absolute" charge >>which is conserved, the "Q=CV" "charge" of normal electronics is >>not. Take a solar cell charging a battery for one obvious example. As >>Larkin would say, where did the charge come from? Photons don't carry >>charge! > > The photons entered the game from "outside the box" as someone opined. But they don't carry charge. They do carry energy - so are you saying you can convert energy into "charge"? I agree - this is true for "Q=CV" "charge-as-used-in-electronics", but false for "number of electrons - number of protons" physicists charge. > As for your "explanation" above... :-( I have yet to see yours... > If I'm so wrong and Larkin is so right, WHY don't Hill and Hobbs jump > to his defense? Something about wrestling with pigs? :) -- John Devereux
From: Jim Thompson on 11 Jul 2010 13:41 On Sun, 11 Jul 2010 18:26:35 +0100, John Devereux <john(a)devereux.me.uk> wrote: >Jim Thompson <To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> writes: > >> On Sun, 11 Jul 2010 09:05:34 +0100, John Devereux >> <john(a)devereux.me.uk> wrote: >> >>>Jim Thompson <To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> writes: >>> >>>> In the next few days, when I have time, I will issue a mathematical >>>> proof that Larkin is totally wrong. Watch for it ;-) >>>> >>>> Why haven't Win Hill and Phil Hobbs come to Larkin's defense? >>>> >>>> Bwahahahaha! >>> >>>I'm no Phil Hobbs, but isn't all this argument because we are conflating >>>two different usages of "charge"? >>> >>>The "charge" on a capacitor, as somone pointed out already, is really >>>charge *separation* (dilectric polarization). The Q=CV refers to a >>>*separation* of charge, not an absolute quantity. The "absolute" charge >>>- the total number of electrons minus the number of protons - is >>>normally low or zero. Unless your whole circuit picks up an >>>electrostatic charge from somewhere else. It is this "absolute" charge >>>which is conserved, the "Q=CV" "charge" of normal electronics is >>>not. Take a solar cell charging a battery for one obvious example. As >>>Larkin would say, where did the charge come from? Photons don't carry >>>charge! >> >> The photons entered the game from "outside the box" as someone opined. > >But they don't carry charge. They do carry energy - so are you saying >you can convert energy into "charge"? I agree - this is true for "Q=CV" >"charge-as-used-in-electronics", but false for "number of electrons - >number of protons" physicists charge. Yes, they carry energy and "knocked loose" electrons. But the total charge contained "inside the box" stays _constant_, does it not? No charge was created, was it? Or destroyed? > >> As for your "explanation" above... :-( > >I have yet to see yours... > >> If I'm so wrong and Larkin is so right, WHY don't Hill and Hobbs jump >> to his defense? > >Something about wrestling with pigs? :) I agree, Larkin _is_ a pig... Thank You !-) ...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: John Larkin on 11 Jul 2010 14:27 On Sun, 11 Jul 2010 11:20:41 -0500, John Fields <jfields(a)austininstruments.com> wrote: >On Sun, 11 Jul 2010 08:36:11 -0700, John Larkin ><jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: > >>On Sun, 11 Jul 2010 08:14:39 -0500, John Fields >><jfields(a)austininstruments.com> wrote: >> >>>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? >> >> >>Quoting myself, >> >>"Slap a 1-ohm resistor across the 1F/1v cap and discharge it. You'll >>get 1 ampere-second out of it eventually." >> >>How did you miss the words "discharge" and "eventually"? I worded the >>situation as carefully as I could, figuring some whiney dork or >>another would get pretend-lawyer pickey. Sigh. > >--- >I didn't miss them, I just thought someone as vague as you are coudn't >possibly have meant "coulomb" since it's a much less confusing term. > >So you prefer ampere-second to coulomb? Same thing, different word. It's more common to actually measure in ampere-seconds using resistors, voltmeters, scopes, ammeters, stuff we generally have handy. I do actually own one meter that literally reads coulombs, a Keithley 610C ftp://jjlarkin.lmi.net/Keithley_1gig.JPG which is a wonderful old beast to have around. It's shown measuring a 1 Gohm resistor, a feat that has been declared impossible in this very newsgroup. Some scopes will calculate the area under a curve, allowing one to measure charge from a current waveform. Usually you can just eyeball it close enough for lots of apps, or calculate charge indirectly. >--- > >>How many ampere-seconds would you get if it was discharged by a 10 ohm >>resistor? > >--- >Depends on how long you left it on there. Good grief, you *don't* understand this stuff. You, like AlwaysWrong, are certainly smart enough to learn the basics of electrical circuit math, but for some emotional reason you have chosen not to. I see that a lot in techs. They compensate by attacking people who can do the arithmetic, calling them eggheads or "inexperienced" or argue over definitions and third-order effects to obscure the fact that there *are* calculable answers. John
From: John Larkin on 11 Jul 2010 14:32
On Sun, 11 Jul 2010 09:34:23 -0700 (PDT), Richard Henry <pomerado(a)hotmail.com> wrote: >On Jul 11, 8:46�am, Jim Thompson <To-Email-Use-The-Envelope-I...(a)On-My- >Web-Site.com> wrote: >> On Sat, 10 Jul 2010 09:07:22 -0700, John Larkin >> >> >> >> <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: >> >On Sat, 10 Jul 2010 11:35:35 -0400, "tm" <no...(a)msc.com> wrote: >> >> >>"John Larkin" <jjlar...(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. >> >> >John >> >> "Side" with whomever you like. �But both "laws" apply simultaneously. >> >Energy is always conserved. > >Charge gets moved around. Whether you find it to be "conserved" or >not depends on where you look for it. Right. "Conservation of charge" in an electronic circuit has to be cited under exactly specified conditions. In some circuits, it makes no sense at all, so it should only be used carefully. I can extract all the energy from one cap and dump it into another one such that the summed charge of both caps is not conserved. John |