Reversible and Irreversible Processes (redux) - Page 1
Forum section: electronics / projects

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 30, 2015, 08:23:07 pm
content: I would like to continue this thread,https://www.eevblog.com/forum/projects/why-trying-to-store-energy-in-a-capacitor-can-be-less-efficient-than-you-think/but as it's already ballooned to six pages in the time I've seen it, I don't want my reply to get buried within. Hope that's not a problem.In knowledge, there is a constant back-and-forth between "aha, it's this !" "Well yes it's that, but it's also [so-and-so]." "You're both wrong, it's , which is far richer than either of you imagined, and includes both of your cases, when taken to very specific limits." (Case in point; BJTs current vs. voltage. I won't bore you with the thread...)In this situation, the usual background goes thus:Perhaps in high school science or intro university, you learned that charge is always exactly conserved between capacitors. If you take two equal value capacitors, one charged and one discharged, and connect them together, the final voltage will be exactly half: because the sum capacitance is double, and Q=VC, and charge conservation, blah blah blah, it works. With the curious note that "itdoesn't matterhow we got there, how fast or how slow; it simply happens, perfect every time."As with many things you were taught, perhaps you simply left it at that, stashed it in the corner of your mind for the test, then forgot about it. Because who measures charge, anyway? Multimeters don't, so unless you're doing fancy experiments, who cares. (And honestly, I wouldn't say this is an unfair conclusion, because current is all-around easier to deal with, in a great many cases.)And as with so many things, the assumptions and exceptions are easily more complex, important, and numerous, than the simplified core of the statement itself! And so often forgotten or omitted outright!So let us begin.(Note: I have left the algebra in, in the hopes that the intrepid student will follow along. Yes I know it's a strain on the eyes, and ASCII makes for poorly formatted math besides; but if you force yourself to read it, it's really quite easy, and very illuminating to get a feel for how the quantities are connected by the process.)I haven't scanned the above thread for important data, so this is mainly a reply or addendum to the first post.Let me be clear:everything that IanB has written is true!What may not be clear, is the extent of that truth.Gotcha #1. "50% of the energy is always lost."My emphasis: of theenergy difference, not the absolute energy.If you take two 1uF capacitors, one charged to 1uC and the other to 2uC (thus, 1 and 2V respectively), and connect them together, you end up with 3uC in 2uF, or 1.5V.The energy states changed thus:E1 = 0.5 * (1uF) * (1V)^2 = 0.5uJE2 = 0.5 * (1uF) * (2V)^2 = 2uJsum = 2.5uJEfinal = 0.5 * (2uF) * (1.5V)^2 = 2.25uJIn fact, only 10% of the total energy (0.25uJ out of 2.5uJ) has been lost! This must be some serious bullshit going on here. Apparently both of us are lying: it's not 50%, nor is it 50%of the difference(which should be (2.5 - 0.5) / 2 = 1uJ lost, for a final energy of 1.5uJ).But, the latter is a somewhat closer approximation (1.5 is closer to 2.25 than 1.25 is), and we can refine its overly simple language to give a better fit (in this case, exact). Bear with me.If we crank the numbers, we can find the "difference" that we should be referring to.Problem Statement:Suppose we have two capacitors, C1 and C2, initially at voltages V1 and V2 respectively (thus we know their charges, but again, charge is more useful as concept than a quantity, so I'll be eliminating those momentarily), then connected together until they reach equilibrium. What is the energy loss? Show your work.The final voltage will be,Vf = Qtot / Ctot= (Q1 + Q2) / (C1 + C2)= (V1 C1 + V2 C2) / (C1 + C2)and energy,Efinal = Ctot Vf^2 / 2= (C1 + C2) [(C1 V1 + C2 V2) / (C1 + C2)]^2 / 2the (C1 + C2) cancels, leaving one on the bottom,= (C1 V1 + C2 V2)^2 / (2 (C1 + C2))The loss is then,Eloss = E1 + E2 - Efinal= C1 V1^2 / 2 + C2 V2^2 / 2 - (C1 V1 + C2 V2)^2 / [2 (C1 + C2)]Uhm let's see, kind of a mess... we can multiply through by 2 to leave that pesky half alone...We probably also want to distribute the squared term, and multiply the left terms by (C1 + C2) so we can get the right hand side over a common denominator (or throw it on the left hand side just the same).2 Eloss (C1 + C2) = C1 V1^2 (C1 + C2) + C2 V2^2 (C1 + C2) - C1^2 V1^2 - 2 C1 C2 V1 V2 - C2^2 V2^2See anything dropping out? Ehh... maybe... let's distribute some more.= C1^2 V1^2 + C1 C2 V1^2 + C1 C2 V2^2 + C2^2 V2^2 - C1^2 V1^2 - 2 C1 C2 V1 V2 - C2^2 V2^2the doubly-squared terms drop out, leaving= C1 C2 V1^2 + C1 C2 V2^2 - 2 C1 C2 V1 V2= C1 C2 (V1^2 + V2^2 - 2 V1 V2)= C1 C2 (V1 - V2)^2Going back to Eloss alone, we findEloss = C1 C2 (V1 - V2)^2 / [2 (C1 + C2)]Thus, the difference does indeed show up: the simple difference between voltages. Energy always goes as V^2, so we have to square the difference. Handy to note: squaring always eliminates the sign, so we don't care which voltage is higher or lower than the other, or whether either is positive or negative: just as it should be -- we're only asking about the absolute energy difference, and this has given it.Capacitor energy always goes as C/2, but what is the capacitance this time? It's the everything-else:Ceff = C1 C2 / (C1 + C2)But what is this? This is the formula for capacitors in series! So although we might claim we've connected capacitors in parallel; really, with only two nodes, two components and one loop, it's both series and parallel, so this is a perfectly reasonable outcome, logical even!Thus, the energy lost due to connecting capacitors together is given by the series equivalent capacitance, and the difference in voltage.A handy result of this is, being able to calculate the efficiency of a switched-capacitor regulator. (Equivalent terms: "flying capacitor" switcher, charge pump.) Yes, such a device exists, and yes, they aren't terrible! Since they use resistive switches, you should intuitively expect they can do at least as well as linear regulators (which only drop voltage, at complete loss), while also implementing useful  ratios (like 1:2 or 1:-1). What may not be intuitive is, how they can ever possibly do anybetterthan a linear regulator. But that's the trick. By designing for multiple ratios, and selecting between them like a car's manual transmission, the amount of voltage required to drop (as losses) can be optimized down. It can't be eliminated, and the losses taken on thatdifferencewill always be total, in the sense that I've covered above: but since overall performance is key, it's not nearly as bad as it sounds, especially if many ratios (thus giving closely spaced options) are provided.Gotcha #2. Irreversible systems give total losses.Problem: they don't need to.Besides resistors and capacitors, there are also problems from thermodynamics and other parts of physics, which have exactly the same concerns.And as you might be guessing, following the earlier section... well damn, can we just keep on reducing losses arbitrarily?Maybe.The trick is, most irreversible processes (such as charging capacitors, or compressing a gas) are locally reversible -- that is: over a small enough trajectory, you lose much less energy than you've exchanged in that small step. What you have to remember is, when you implement such a piecewise path... you have to implement every single step along the path.Offhand... I can't quite convince myself that itcanbe done with capacitors, say by "bucket brigade"-ing one capacitor into another, using a smaller "bucket" capacitor and many steps, or by using many buckets in a chain. I've written a fairly convincing spreadsheet which demonstrates this for 10 buckets and 100 steps (giving the expected "lossy" result with no path dependence -- that is, the small steps still lose a proportional amount of energy).One example where it does work: counter-flow heat exchangers. The temperature of one fluid can be fully swapped with another (or in whatever ratio their heat capacities suggest), to the level of 90% efficiency or better.Gotcha #3. Equilibrium....So, what about inductors and switches?Aha. The truth hidden in the background of this one is, we're only talking resistors and capacitors. (Actually, whether we have inductors or not, doesn't matter, as long as there's some equivalent resistance somewhere.) We're also talking thermodynamics: the exchange ofenergy, not ofpower. What happens at equilibrium, not in the process. You can't do thermo by looking at two arbitrary points in time, no; that requires kinetics and flow and, oh god, all sorts of horrible things! Thermo demands the result at t --> infty.In the thread, Jay Diddy has presented a couple simple cases where energy can be transferred without the apparent "50% loss".These examples are actuallymore wrongthan the simple "50% total losses" statement is....BUT... this is only true when taken within the same (or similar) problem statement: when taken to infinity time.What's more, they're far more useful, to us beings of finite life span, than near-philosophical matters of zero and infinity.In those circuits, a diode, active switch or something even more complex has been used to achieve single-event energy efficiency. But these are inherently non-equilibrium cases, because the diodes or switches do not have zero leakage current. Eventually, when left alone, they will come back to equilibrium, and (assuming that leakage was between the capacitors, which are otherwise ideal capacitors, and not to ground), the eventual final result will be identical to the thermodynamic case.To correct the apparent contradiction of wrongness, we must apply a different energy rule, using quasi-static averages, rather than thermodynamic eventualities. This is the domain of switching power supplies, where one can evaluate events in a stepwise manner, without individually needing to know how they proceeded during that interval (no need to solve, for example, voltage as a function of time -- which might involve a lengthy and perhaps unsolvable differential equation!), as long as they proceeded according to the assumed rules (such as the conservation of energy between reactive elements, without dissipating too much of it in losses).An illustrative example is any rough switching supply calculator tool: this one's my favorite.http://schmidt-walter.eit.h-da.de/smps_e/smps_e.htmlGive it a romp: select a topology, put in some goofy numbers (or let it give you the defaults), and hit calculate. You will see straight line, ideal waveforms, which arise from calculating only the averages, assuming all ideal components. What's actually calculated, is not the exact waveforms in any meaningful sense, but only the corners where they change direction -- what happens between those points doesn't matter, and indeed might proceed quite differently (most noticeably around switching edges, which can be very complicated waveforms indeed), but so long as the waveforms transition between those corners in a lossless (path-independent / reversible) fashion, the result will, in fact, remain perfectly valid! Your challenge, as the designer and engineer utilizing these tools, is to ensure, as well as possible, that this condition is met!Connecting with not only thermodynamics, but some famous physics as well, Richard Feynman was quite fond of methods like these. In physics, you can find the Lagrangian of the system, which describes how the energy levels evolve over time. You can apply exactly the same reasoning to these systems, in a stepwise or assumed-reversible fashion, to get a very good idea of what the system is doing, without having to go to the trouble (or often, outright impossibility!) of finding the exact functions which describe all quantities in the system (positions, voltages, quantum probabilities, etc.).Tim

author: IanB
author_rank: Super Contributor
timestamp: January 30, 2015, 10:12:04 pm
content: Thanks for the detailed treatise. It is too much to absorb in detail until I get home from work later, but here is a follow up to one item:Quote from: T3sl4co1l on January 30, 2015, 08:23:07 pmI came across this reference, which describes a demonstration experiment intended to show how the irreversible process of charging a capacitor can be brought asymptotically closer to reversibility by breaking it down into smaller incremental steps:http://www2.fisica.unlp.edu.ar/materias/Experi_2008/Proc_irrev.pdfIt looks like the apparatus required could be assembled by someone with moderate technical abilities and a reasonably well equipped lab.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 31, 2015, 09:19:37 pm
content: Yes, doing it from voltage sources in steps should give the "reversible-ized" result, as is consistent with the first part of my post. In that case, the only "50% loss" step is the first one, but as it delivers a voltage of 1/N times the final voltage, it's a small part of the total; and each subsequent step loses less and less.Tim

author: Zero999
author_rank: Super Contributor
timestamp: January 31, 2015, 11:08:30 pm
content: Thanks for posting this.Yes, you're right. The correct statement is energy transfer between a voltage source and capacitor is 50%. None of the energy already stored in the capacitor is dissipated (except for the leakage current). Half of the energy transferred from the voltage source to the capacitor is wasted as heat.

author: IanB
author_rank: Super Contributor
timestamp: February 01, 2015, 12:32:29 am
content: Quote from: T3sl4co1l on January 30, 2015, 08:23:07 pmI found this analysis fascinating, so I took the liberty of working it through myself and setting it out in printed form for reference. See below.Quote from: T3sl4co1l on January 30, 2015, 08:23:07 pmYes, that is really rather nice.

author: IanB
author_rank: Super Contributor
timestamp: February 01, 2015, 07:11:19 pm
content: I realized there is a simpler way to approach this problem. If you analyze it visually, by rearranging components and reference points, the result can be determined without the messy algebra. Two capacitors equalizing is seen to be equivalent to discharging a capacitor, with the expected outcome of the stored energy being lost:

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 01, 2015, 11:22:02 pm
content: Another way to view it: the initial voltage on a capacitor can be expressed as a voltage source in series with an uncharged capacitor. So, you add up all the initial voltages, then do the charge equalization thing.When you run a SPICE simulation with source ramping enabled (RAMPTIME > 0), this is exactly what's done: all capacitors and inductors are set to zero, then ramped up to their nominal initial conditions (.IC) as the simulation proceeds.Tim

author: Galaxyrise
author_rank: Frequent Contributor
timestamp: February 02, 2015, 03:03:44 am
content: Quote from: T3sl4co1l on January 31, 2015, 09:19:37 pmI think you've got a constant current source as N approaches infinity, so that's another way it's consistent.Stepping capacitance looks to do the same thing, a thought experiment I undertook after AG6QR pointed out where the loss was when using a solar cell to charge capacitors. But varying capacitance is much less straight-forward (at least to me.) I did a cheesy little "merge sort" switching approach on 8 capacitors, which reduced energy waste in LTSpice by about 70%.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: February 02, 2015, 05:27:31 pm
content: Quote from: Galaxyrise on February 02, 2015, 03:03:44 amExactly. The most efficient way to charge the capacitor is for the source voltage to always be just a LITTLE higher than the capacitor voltage, with the desired charge current being equal to (Vs-Vc)/ESR. This, of course, is nothing more than a constant current charger.

author: parbro
author_rank: Contributor
timestamp: February 02, 2015, 06:12:16 pm
content: Thank you Tim and IanB for the detailed and thoughtful analysis. I am still digesting.I have a question about modeling this in LTSpice. I would like to get the accurate resistor dissipation involving complex impedance when simulating a capacitor discharging into a series RC circuit through a switch. The series RC works fine with an AC source but I want to use a charged cap and switch instead and get accurate dissipation. Any suggestions?

author: IanB
author_rank: Super Contributor
timestamp: February 02, 2015, 06:47:02 pm
content: The complex impedance approach is only applicable if we make the assumption of steady processes with time invariant periodic wave forms. It doesn't really apply to transient situations with non-periodic waves. (The complex impedance is a fiction introduced to make the calculations for steady AC systems look elegant. But nature does not actually have complex numbers and complex voltages or currents do not actually exist. Every voltage and current in system at any instant is a real number.)When discharging a capacitor through a switch you need a time domain transient simulation, starting from a particular initial condition and integrating through to the final steady state. To find the resistor dissipation you then need to integrate the instantaneous power dissipation over time to get the total energy dissipated. I think that LTspice can do this for you, but I am not expert enough in the tool to tell you how to do it.

author: Galaxyrise
author_rank: Frequent Contributor
timestamp: February 02, 2015, 06:48:10 pm
content: Quote from: parbro on February 02, 2015, 06:12:16 pmSomething like Jay_Diddy did inthe capacitor energy thread?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 02, 2015, 07:16:09 pm
content: Quote from: IanB on February 02, 2015, 06:47:02 pmFYI, not strictly true. To be perfectly fair; there is no "real" signal, but for reasons other than "imaginaries".Complex numbers arise as a description of phase shift; the shift is always complementary against inversion -- i.e., real signals are Hermitian. Real signals are always positive AND negative frequencies. The analytical signal -- a fiction used to facilitate analysis -- is complex, and therefore "unreal".Depending on your philosophical stand on causality and eternity, AC steady state signals are inapplicable because sines and cosines are only strictly true if they exist for all time. In practice, we can locally approximate a function we don't know much about (i.e., the entire time-history, from before the universe to after its probable heat death) quite effectively by approximating parts of the frequency domain function. Namely, that harmonics of a "sine" wave are not perfectly well defined -- besides issues of phase noise and such, the simple fact that a signal hasn't existed forever puts a hard limit on line width, in the nHz range. Most signals are already much wider bandwidth than this (whether intentionally or not), so we rightfully scoff at such pedantism as purely academic.Tim

author: IanB
author_rank: Super Contributor
timestamp: February 02, 2015, 07:29:32 pm
content: Quote from: T3sl4co1l on February 02, 2015, 07:16:09 pmOK, I modified my statement slightly: we assume steady state signals for the analysis, although recognizing that steady state is an abstraction never perfectly fulfilled in reality. Our calculated results may therefore contain a certain degree of error based on the validity of our assumptions, and we need to estimate the magnitude of this error to ensure it is small enough before relying on the results.Quote from: IanB on February 02, 2015, 06:47:02 pmMy intent here is to say that if you measured the instantaneous value of the voltage or current at any microsecond (or nanosecond, or picosecond...) then that would be a value that you could state as a real number. (But by the uncertainty principle, the more precisely you try to measure the value, the less precisely you get to measure its rate of change.)

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 02, 2015, 10:39:51 pm
content: Yeah. And actually, in practice, assuming "steady state" works out surprisingly well, even over a fairly short term. I once did a sort of VNA calculation with live ADC data on a frequency control process; the FPGA was doing a single-bin, variable-length DFT per cycle. Even though the frequency was running in a control loop (varying relatively slowly with operating condition, and relatively quickly by control loop dither), the results (real + imaginary, voltage and current) were quite reasonable.Tim

author: Dr. Frank
author_rank: Super Contributor
timestamp: February 03, 2015, 10:36:06 am
content: The question is, where the energy is lost, or in other words, into which state the difference is going to.As far as I remember my electrodynamics lectures @ university, (decades ago) , that's simply the change (setup and decrease) of the electrical field between the capacitor plates..So if you change that field, electro-magnetic energy is emitted. The capacitor plates can be understood as an ordinary antenna in this case.So if you apply Maxwells equations to this problem, you'll easily find the missing energy budget.(I'll search for that problem in Jackson, Classic Electrodynamics)Frank

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 03, 2015, 11:18:07 am
content: What if the (implied) inductor and capacitor are inside a sealed metal box?What if everything is superconducting?*(*Even the best superconducting resonators lose some power at AC, so this is only a distraction.)Tim

author: parbro
author_rank: Contributor
timestamp: February 03, 2015, 04:39:12 pm
content: Quote from: Galaxyrise on February 02, 2015, 06:48:10 pmI tried that but LTSpice assumes that the power dissipated in the load (resistor) is just a function of the charging current. I am trying to model the real world power dissipation of a load between a capacitor discharging into another capacitor. Otherwise known as the famous Dave Wing experiment.

author: peter.mcnair
author_rank: Contributor
timestamp: February 03, 2015, 04:45:42 pm
content: Quote from: Dr. Frank on February 03, 2015, 10:36:06 amPerhaps the electrons would need tochange direction(more than once!) for that to happen...the frequency of radiated 'light' being the frequency with which the electrons change direction (i.e. accelerate)...this would normally be accomplished using anACsupply.

author: IanB
author_rank: Super Contributor
timestamp: February 03, 2015, 04:54:11 pm
content: Quote from: parbro on February 03, 2015, 04:39:12 pmCan you say what you think is wrong with that assumption?

author: electr_peter
author_rank: Unknown
timestamp: February 03, 2015, 05:05:48 pm
content: This paper has a very clear explanation of what happens with two caps:The Paradox of Two Charged CapacitorsCharge carrying particles are accelerated and slammed into charging capacitor. Energy goes into acceleration which results in heat of charged capacitor (this can cover 100% of lost energy), sound/light/EM radiation (which can not cover 100% of lost energy according to the paper). Oscillations may occur if inductance is included.Almost identical analogy to two capacitor case - two water columns connected via pipe (one full to 2*H, other empty). Average water level goes from 2*H to H (potential energy decreases), but during transition kinetic energy increases (which results in splashing, waves and heating of water).In all cases part of energy is lost. To increase efficiency, either use that "wasted" energy for a useful work (add heater resistor or water wheel), or use different efficient method of charging capacitor.

author: parbro
author_rank: Contributor
timestamp: February 03, 2015, 06:22:30 pm
content: Quote from: IanB on February 03, 2015, 04:54:11 pmI worded that poorly. What I meant is LTSpice shows power dissipation that is independent of load resistance when the source is a capacitor.

author: IanB
author_rank: Super Contributor
timestamp: February 03, 2015, 06:39:08 pm
content: Quote from: parbro on February 03, 2015, 06:22:30 pmAnd...?Do you think LTspice is giving the wrong answer?

author: parbro
author_rank: Contributor
timestamp: February 03, 2015, 06:48:42 pm
content: I guess I am trying to reconcile the fact that with an AC source you have to use the angle of the dangle thing to get the dissipation of the load and voltage across the cap. I know with a charged capacitor as the source the complex impedance wouldn't apply because we don't have a periodic signal. But wouldn't something like the impulse response of the RC combination be needed to determine how much the load used and what the resulting voltage on the capacitor would be.

author: IanB
author_rank: Super Contributor
timestamp: February 03, 2015, 07:18:35 pm
content: Quote from: parbro on February 03, 2015, 06:48:42 pmYes, you effectively need to integrate the step response from time zero out to infinity. Did you see my development of this at the beginning of this other thread? -->https://www.eevblog.com/forum/projects/why-trying-to-store-energy-in-a-capacitor-can-be-less-efficient-than-you-think/Notice the conclusion at the end: "The amount of lost energy is independent of the resistance."When you simulate this in LTspice it is running a numerical simulation of the same situation.Quote from: parbro on February 03, 2015, 06:22:30 pmThis is good, then. It shows that LTspice works.

author: parbro
author_rank: Contributor
timestamp: February 03, 2015, 07:26:34 pm
content: Thanks. I see the flaw in my thinking.

author: Siwastaja
author_rank: Super Contributor
timestamp: February 04, 2015, 04:43:14 am
content: Quote from: electr_peter on February 03, 2015, 05:05:48 pm... which is exactly normal behavior for capacitors, normally modeled as "ESR", causing all of the losses, which typically vary between 0.1% and 50% depending on how the caps are used... And yes, surprise surprise, it's practically only heat being generated, which is kinda boring.Of course, the resistance of the metal foils inside a capacitor is also there in "ESR" which is just the sum of all those forms of resistances. And in the end, it doesn't matter, but yeah, resistance is not a description of any single concrete physical phenomenon, but a roof term for something that can be caused by several different mechanisms, but cause the same result - a voltage drop dependent on current. Usually, but not necessarily, with energy lost as heat.This is even more obvious when batteries are considered, because they have A LOT* of "kinetic energy" losses because ions swim slowly. And people on forums are REALLY confused as they think this is something "different" from just having resistance, but it isn't. Voltage drop is a voltage drop and the batteries heat up just like the drop predicts, losses are ~I^2, efficiency depends linearly on charge/discharge times, and nothing fancy happens here, because a slowly swimming ion loses its energy sweating just like an electron in a wire does, and in the end, an electric motor loses its efficiency when overworked exactly in the same way a battery does.You can't say whether it is a resistance caused by electrons frictioning when trying to run in a metal lattice (resistance of a copper wire, for example), or ions running in an "sticky" electrolyte, or electrons banging on the foils in a capacitor. This is just ESR and we have to face it everyday, in all cases.*) compared to capacitors, at least; in a battery, the metal foils do not contribute to the resistance that much, but with a capacitor, I wouldn't necessarily say that.

author: electr_peter
author_rank: Unknown
timestamp: February 04, 2015, 09:25:06 am
content: Quote from: Siwastaja on February 04, 2015, 04:43:14 amI don't think it can be called ESR, because basic model assumesidealcapacitors with no series resistance. If cap is modelled with ESR, cap behaves like a charging/discharging battery (with heat and voltage drop with both charging and discharging - in a case model heat comes from charging only).Heating effect in a model comes from the fact that acceleration energy has to be dumped somewhere (and electrons prefer transferring energy to heat capacitor), not that capacitor is by default with an ESR.

author: Siwastaja
author_rank: Super Contributor
timestamp: February 05, 2015, 03:26:15 am
content: Well, what you are saying is simply that you want to have an imaginary capacitor without mechanisms to have losses, but still you want them to have losses because physics say so. This just doen't make sense and is utter waste of time.Capacitors do have losses. Even imaginary capacitors must have losses, or otherwise your mental model of a capacitor is broken so badly it's useless in what you are trying to make it explain for you.You can't have or even really model an ideal (as in: non-lossy) capacitor. If you want to ignore this fact, then your result for the 50% loss problem will be just wrong.People just seem to think that "ESR" means only wire / metal resistance inside the capacitor and could be ignored because of that, but this is not the case at all. ESR models exactly all the phenomena that have been discussed here, and by very definition, the losses you will see are ESR losses. ESR is just the term for the numerical presentation for the losses due to voltage drop (and charging a cap 0V-->3V from a 3V source will inevitably have an average voltage drop of 1.5 volts, no matter what hat tricks you try to do with the model!). That's why there's "equivalent" there; it's not physically like a resistor, but has an equivalent ohmic value.I think the problem in this capacitor discussion is that people seem to want to ignore the answer, by denying ("modeling as zero") the very exact mechanisms of energy loss they are frantically looking for.Science doesn't work like this. You don't take an existing phenomenon and then decide to remove it from your model (i.e., model it as a zero) and expect to get correct explanations or results after that!tldr; ESR cannot be set to zero for modeling the 50% loss problem.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 05, 2015, 01:47:51 pm
content: The case for zero losses, by the way, is very simple to solve analytically: the differential equation becomes omega^2 V - d^2 V / dt^2 = 0, with the solution A sin(omega t) + B cos(omega t) for arbitrary coefficients depending upon the initial conditions. Note that sin/cos goes on forever, so the lossless solution is indeed perfectly oscillatory, and the energy is not lost.It's noteworthy that, if you look at the DC level of such a system, you will still see the correct result. Indeed, the energy that is "lost" is "lost" to an oscillating mode of the system, and by definition does not contribute to the DC level. If you change your definition to this (that you are measuring the DC level, not the final voltage at t --> infty), then the conclusion is instantaneous, and completely independent of loss, including zero or negative values.Tim

author: suicidaleggroll
author_rank: Super Contributor
timestamp: February 05, 2015, 04:26:40 pm
content: Quote from: Siwastaja on February 05, 2015, 03:26:15 amExactly. This whole discussion is like a dog chasing its tail. You can't remove theonly source of lossin the system, and then ask what's causing the loss. It makes no sense. You can only use the model of an ideal capacitor when its ESR is several orders of magnitude smaller than some other source of loss in the system. If the capacitor's ESR is theonly source of loss, you can't just ignore it, it's a pointless exercise.It's just like the scenario I laid out earlier. Take an ideal capacitor with no ESR/ESL, charge it up to some known voltage, and then short out the terminals using a wire with no impedance (resistance, capacitance, or inductance). Where does the power go? The answer is "that's an unrealistic scenario and a waste of time".

author: parbro
author_rank: Contributor
timestamp: February 05, 2015, 11:16:31 pm
content: I always thought ESR was a measure of the imperfection of the dielectric ( for example leakage).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 06, 2015, 01:50:01 am
content: Remembering it'sequivalentseries resistance, it's an expression of losses in the capacitor, whether series, parallel or even more complicated than that (e.g., ionic diffusion in electrolyte, skin effect in conductors).ESR is more popular than EPR because it's more accurate over a wider range, for most capacitor types. Conversely, EPR tends to be more representative for inductors.Tim