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From: Winfield Hill on 7 Jul 2010 14:52 Jim Thompson wrote... > Winfield Hill wrote: >> Jim Thompson wrote... >>> John Larkin wrote: >>>> Adrian Jansen wrote: >>>>> Jim Thompson wrote: >>>[snip] >>>>>> >>>>>> Depends on the definition of "depends" :-) >>>>>> "Charge" IS conserved. So if you transfer Q from C1 to C2 >>> >> >>>>> 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 says, "...charge is not conserved." >>> Newbies are invited to Google on "conservation of charge". >>> (AND run the math problem I previously posted.) >>> John is so full of it I'd bet his eyes are brown ;-) >>> >>> Unfortunately, Adrian Jansen mis-states the results as well :-( >> >> I haven't been following this thread, but I have a comment. >> >> The operative phrase must be, "and do it efficiently." >> >> This is easy to do, with a dc-dc converter for example, or a >> mosfet switch and an inductor. In these cases it's easy to >> manipulate E1 and E2, C1*V1^2 = C2*V2^2. Forget about charge. > > "Forget about charge."??? > I guess it depends on your definition of "depends" :-( > We've gone from "dumping" to running a switcher. I imagine one can use the word dump. Consider this setup: A charged C1 with switch S1 to the top of grounded inductor L1, and from the inductor another switch S2 to a discharged cap C2. Close switch S1 and the inductor current builds, open S1 when C1 is discharged to 0V and the L1 current has reached a peak. Simultaneously close S2 and L1's current continues to flow, charging C2, until the inductor current stops, then open S2. (One can avoid time-precise switches by using series diodes, but they do have loss.) So C1's energy is dumped first into L1, and then into C2. If C2 is much smaller than C1, it'll end up with a much higher voltage, and the charge out of C1 will greatly exceed the charge into C2 (same current, shorter time), even though E1 = E2, exactly, at least in principle. It must meet John's definition of "do it efficiently." > Got one at 100%? Power-Out == Power-In ?? I made one of these where C1's starting voltage was 1.2kV, and the inductor's peak current was about 1.5kA. I used large IGBT switch modules with Vce(sat) of 3V, or about 0.3% loss. Pulse cap esr and a litz-wire inductor were responsible for most of the losses, about 5% or so, IIRC. -- Thanks, - Win
From: Jim Thompson on 7 Jul 2010 16:00 On 7 Jul 2010 11:52:06 -0700, Winfield Hill <Winfield_member(a)newsguy.com> wrote: >Jim Thompson wrote... >> Winfield Hill wrote: >>> Jim Thompson wrote... >>>> John Larkin wrote: >>>>> Adrian Jansen wrote: >>>>>> Jim Thompson wrote: >>>>[snip] >>>>>>> >>>>>>> Depends on the definition of "depends" :-) >>>>>>> "Charge" IS conserved. So if you transfer Q from C1 to C2 >>> >>> >>>>>> 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 says, "...charge is not conserved." >>>> Newbies are invited to Google on "conservation of charge". >>>> (AND run the math problem I previously posted.) >>>> John is so full of it I'd bet his eyes are brown ;-) >>>> >>>> Unfortunately, Adrian Jansen mis-states the results as well :-( >>> >>> I haven't been following this thread, but I have a comment. >>> >>> The operative phrase must be, "and do it efficiently." >>> >>> This is easy to do, with a dc-dc converter for example, or a >>> mosfet switch and an inductor. In these cases it's easy to >>> manipulate E1 and E2, C1*V1^2 = C2*V2^2. Forget about charge. >> >> "Forget about charge."??? >> I guess it depends on your definition of "depends" :-( >> We've gone from "dumping" to running a switcher. > > I imagine one can use the word dump. Consider this setup: > A charged C1 with switch S1 to the top of grounded inductor > L1, and from the inductor another switch S2 to a discharged > cap C2. Close switch S1 and the inductor current builds, > open S1 when C1 is discharged to 0V and the L1 current has > reached a peak. Simultaneously close S2 and L1's current > continues to flow, charging C2, until the inductor current > stops, then open S2. (One can avoid time-precise switches > by using series diodes, but they do have loss.) > > So C1's energy is dumped first into L1, and then into C2. > > If C2 is much smaller than C1, it'll end up with a much > higher voltage, and the charge out of C1 will greatly > exceed the charge into C2 (same current, shorter time), > even though E1 = E2, exactly, at least in principle. > It must meet John's definition of "do it efficiently." > >> Got one at 100%? Power-Out == Power-In ?? > > I made one of these where C1's starting voltage was 1.2kV, > and the inductor's peak current was about 1.5kA. I used > large IGBT switch modules with Vce(sat) of 3V, or about > 0.3% loss. Pulse cap esr and a litz-wire inductor were > responsible for most of the losses, about 5% or so, IIRC. Overall efficiency, Power-into-load/Power-supplied-to-system ?? Larkin vaguely started his thread with no mention of an inductor whatsoever, then added the inductor and claimed "sloshing" forever. What-a-pile of BS... use real switches and real inductors and real capacitors. Real switches, in Larkin's "episode", certainly have losses. I know that even ideal switches used to dump one capacitor into another exhibit "magical" losses (not magical to me ;-) I'll have to analyze such switching with an inductor... I'd guess there is indeed energy loss "magically" as well. ...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: A reincarnation of Nixon, narcissistically posing in politically-correct black-face, but with fewer scruples.
From: AM on 7 Jul 2010 16:24 On Wed, 07 Jul 2010 10:39:10 -0700, John Larkin <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >On the other hand, energy is always conserved. > >John If you are around to observe and tell about it.
From: AM on 7 Jul 2010 16:34 On Wed, 07 Jul 2010 13:24:08 -0700, AM <thisthatandtheother(a)beherenow.org> wrote: >On Wed, 07 Jul 2010 10:39:10 -0700, John Larkin ><jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: > >>On the other hand, energy is always conserved. >> >>John > > If you are around to observe and tell about it. B A N G !!! Energy finds its way out of the box(es)!
From: John Larkin on 7 Jul 2010 17:30
On 7 Jul 2010 11:52:06 -0700, Winfield Hill <Winfield_member(a)newsguy.com> wrote: >Jim Thompson wrote... >> Winfield Hill wrote: >>> Jim Thompson wrote... >>>> John Larkin wrote: >>>>> Adrian Jansen wrote: >>>>>> Jim Thompson wrote: >>>>[snip] >>>>>>> >>>>>>> Depends on the definition of "depends" :-) >>>>>>> "Charge" IS conserved. So if you transfer Q from C1 to C2 >>> >>> >>>>>> 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 says, "...charge is not conserved." >>>> Newbies are invited to Google on "conservation of charge". >>>> (AND run the math problem I previously posted.) >>>> John is so full of it I'd bet his eyes are brown ;-) >>>> >>>> Unfortunately, Adrian Jansen mis-states the results as well :-( >>> >>> I haven't been following this thread, but I have a comment. >>> >>> The operative phrase must be, "and do it efficiently." >>> >>> This is easy to do, with a dc-dc converter for example, or a >>> mosfet switch and an inductor. In these cases it's easy to >>> manipulate E1 and E2, C1*V1^2 = C2*V2^2. Forget about charge. >> >> "Forget about charge."??? >> I guess it depends on your definition of "depends" :-( >> We've gone from "dumping" to running a switcher. > > I imagine one can use the word dump. Consider this setup: > A charged C1 with switch S1 to the top of grounded inductor > L1, and from the inductor another switch S2 to a discharged > cap C2. Close switch S1 and the inductor current builds, > open S1 when C1 is discharged to 0V and the L1 current has > reached a peak. Simultaneously close S2 and L1's current > continues to flow, charging C2, until the inductor current > stops, then open S2. (One can avoid time-precise switches > by using series diodes, but they do have loss.) > > So C1's energy is dumped first into L1, and then into C2. > > If C2 is much smaller than C1, it'll end up with a much > higher voltage, and the charge out of C1 will greatly > exceed the charge into C2 (same current, shorter time), > even though E1 = E2, exactly, at least in principle. > It must meet John's definition of "do it efficiently." > >> Got one at 100%? Power-Out == Power-In ?? > > I made one of these where C1's starting voltage was 1.2kV, > and the inductor's peak current was about 1.5kA. I used > large IGBT switch modules with Vce(sat) of 3V, or about > 0.3% loss. Pulse cap esr and a litz-wire inductor were > responsible for most of the losses, about 5% or so, IIRC. I saw a datasheet recently, can't remember exactly where, for a chip that sure looked like a flying-inductor switcher. John |