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From: Jim Thompson on 6 Jul 2010 10:41 On Mon, 05 Jul 2010 21:41:36 -0700, John Larkin <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >On Mon, 5 Jul 2010 17:48:35 -0500, "Tim Williams" ><tmoranwms(a)charter.net> wrote: > >>"Fred Abse" <excretatauris(a)invalid.invalid> wrote in message news:pan.2010.07.05.19.11.23.376684(a)invalid.invalid... >>> It still implies that reducing V by increasing C involves a loss of >>> energy. >> >>So a flying capacitor converter is always 50% efficient? :-) >> >>Cap charging (in terms of conserved charge) is an irreversible process. But just like irreversible thermodynamic processes, if you make the steps small enough, it starts looking reversible. > > >You can have two caps, C1 charged and C2 not, and transfer all the >charge from C1 to C2, without loss. In fact, you can slosh the charge >between them, back and forth, forever. Just don't use resistors. > >John > John Larkin, Please explain how you do that? Magic switch? Or magic perfect inductor ?:-) ...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: Tim Williams on 6 Jul 2010 12:05 "John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message news:obd6369v1tabmic9tsmqpo4ms91qclt2aa(a)4ax.com... >>> You can have two caps, C1 charged and C2 not, and transfer all the >>> charge from C1 to C2, without loss. In fact, you can slosh the charge >>> between them, back and forth, forever. Just don't use resistors. >> >>What if you want equal charges on both? > > Easy. Without resistors? Prove it ;-) Note: no dangling currents. Inductors carry charge, too, so that wouldn't conserve it very well. Tim -- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms
From: Tim Williams on 6 Jul 2010 12:22 "Fred Abse" <excretatauris(a)invalid.invalid> wrote in message news:pan.2010.07.06.08.41.06.201326(a)invalid.invalid... > Switching capacitors from series to parallel does not change each individual C, > hence, neglecting switching losses, both charge and energy remain the > same. I assume that is what you mean by a flying capacitor converter. What I meant was charging a cap with another cap. Without tricky quasi-resonant or inverter circuits, you inevitably have to do this (under the presumption that a constant DC output is desired), and this inevitably leads to loss. But the 50% loss only occurs to the *change* in energy, so if you make this change an arbitrarily small fraction of the supply voltage, efficiency can be quite high. Hence why things like MAX232 can be ~95% efficient despite pumping caps into caps. When you mentioned a mechanical method, I envisioned a stack of capacitors charged by the high-voltage source, then a rotor to ferry a bit of charge at a time from the stack to a reservoir. The rotor would be equipped with capacitors, and contacts would be present on each side, so that the rotor is charged by the HV stack on one side, then connected in parallel on the other side to deliver its charge. The rotor has to be big enough, and insulating, so that arc-over doesn't occur at either end of the rotor. The capacitors on the rotor have to be big enough to deliver a useful amount of charge, enough times per second, to meet the design current and efficiency specs. Of course, no standing capacitor chain need be provided; the rotor can simply mesh with series-connected contacts, providing all the capacitance itself. Likewise, two or more rotors could be used, in make-before-break mode, to eliminate the DC link capacitor. YMMV; a standing cap bank would be wise for lightning collection, but unnecessary for experiments (in either direction, step-up or step-down). Three rotors in make-before-break would be quite suitable for supplying a conventional (inductor based) converter, transforming, say, 1-10kV into 1.5V or 12V or 160V, etc. OTOH, when a capacitance is changed, work is performed. An electrophorus works by applying force to seperate charges, increasing the voltage. If capacitance falls linearly, voltage rises linearly, but energy rises as voltage squared, so the energy rises linearly. The difference comes from the work input, which by hand, feels negligible against a 100g electrophorus. This might be confusing to perpetual motion types, who are fond of electric or magnetic devices with forces so weak, they seem to move of their own accord. Tim -- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms
From: John Larkin on 6 Jul 2010 12:25 On Tue, 6 Jul 2010 11:05:23 -0500, "Tim Williams" <tmoranwms(a)charter.net> wrote: >"John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message news:obd6369v1tabmic9tsmqpo4ms91qclt2aa(a)4ax.com... >>>> You can have two caps, C1 charged and C2 not, and transfer all the >>>> charge from C1 to C2, without loss. In fact, you can slosh the charge >>>> between them, back and forth, forever. Just don't use resistors. >>> >>>What if you want equal charges on both? >> >> Easy. > >Without resistors? Prove it ;-) > >Note: no dangling currents. Note: I make the rules. >Inductors carry charge, too, so that wouldn't conserve it very well. > >Tim Connect an inductor across C1 until you've bled it down to half its charge. Now connect that inductor to C2 and charge it up to the same charge as C1 has. Now disconnect the inductor. If you keep the L shorted, you can save the residual energy for reuse later. John
From: Tim Williams on 6 Jul 2010 12:33
"John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message news:jtl6361te3ku4ukh0jhp05tab4tvhjjqru(a)4ax.com... > Connect an inductor across C1 until you've bled it down to half its > charge. Now connect that inductor to C2 and charge it up to the same > charge as C1 has. Now disconnect the inductor. If you keep the L > shorted, you can save the residual energy for reuse later. ^ ^ ^ ^ Ha, so charge wasn't conserved after all. See? ;-) Tim -- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms |