Why trying to store energy in a capacitor can be less efficient than you think

[tggzzz]

You're searching for a mechanism, where the charging energy is lost?

As far as I remember my Electrodynamics lectures, decades ago, the answer is not obvious, but simple in the end.

The capacitor builds an electrostatic field between his plates.

Every time, you CHANGE this field, an electro-magnetic wave is emitted, as the capacitor acts as an antenna; therefore energy from the charging / discharging process, which always changes the electrostatic field, is 'lost' due to the electro magnetic wave. Therefore, to solve this problem, you have to apply Maxwells equations, also.

So what is the amount of energy lost via this phenomenon you are discussing, and is it a constant amount, or does it vary (with charging time, for example)? Because this is something you cannot see in practical electronics (at least I cannot) - you can get something like 99.9%  efficiency of the capacitor with proper current-mode DC/DC charging and discharging, and that 0.1% is clearly coming from capacitor's ESR, which has a simple physical explanation (no antennas). Capacitors can be used, for example, in storing and reusing braking energy in urban transit systems, and these give great efficiency; the AC motor VFDs give the current control ("DC/DC") charging and discharging automatically.

The 50% loss described by the OP is very simple and has nothing to do with your explanation, but maybe your point works out to be important with extremely high (theoretical) currents resulting in from approaching zero ohms? So, please elaborate.

But the magnetic field created by a current doesn't necessarily generate losses unless there's another circuit magnetically coupled in which to induce a current. For example a transformer's primary with the secondary opened draws ~nil (just the losses due to alternating magnetization of the core) but with the secondary shorted the primary looks like a short too.

It is an electromagnetic field.

If you think that cannot carry energy away from the source, then I suggest you go outside and look up, anytime day or night, preferably when it is clear.

[GeorgeOfTheJungle]

The 50% loss described by the OP is very simple and has nothing to do with your explanation, but maybe your point works out to be important with extremely high (theoretical) currents resulting in from approaching zero ohms? So, please elaborate.

But the magnetic field created by a current doesn't necessarily generate losses unless there's another circuit magnetically coupled in which to induce a current. For example a transformer's primary with the secondary opened draws ~nil (just the losses due to alternating magnetization of the core) but with the secondary shorted the primary looks like a short too.

It is an electromagnetic field.

If you think that cannot carry energy away from the source, then I suggest you go outside and look up, anytime day or night, preferably when it is clear.

No. I haven't said that. I see the sun :-)

But a coil generates a magnetic field too. Won't that energy return to the circuit when the field collapses? Unless a secondary robs it?

In the capacitor, how much of that 50% would be lost as EM radiation (discounting heat due to resistances)? In which band of the spectrum would that be?

Please elaborate.

[IanB]

But a coil generates a magnetic field too. Won't that energy return to the circuit when the field collapses? Unless a secondary robs it?

The magnetic field around a coil holds stored energy. If the coil acts as an antenna and radiates energy away then there is less energy to store, ergo less of a magnetic field, ergo less energy to be returned to the circuit.

[T3sl4co1l]

You do still have analogies of electric and magnetic fields; the difference is, acoustic waves admit both transverse AND longitudinal modes, which has certain implications for things that I don't remember.   Whereas E&M is strictly transverse.

Tim

[Siwastaja]

Oh yeah, every wire creates a magnetic field that will couple to nearby or even non-nearby metallic objects and cause eddy losses and and this may cause a 0.00000001% efficiency drop or thereabouts. And the same goes for the electric field and capacitive coupling, sound being emitted from slightly vibrating wires etc. But, this has absolutely nothing to do with the 50% loss described by the OP, however, because we have a simple, measurable, real-world explanation: voltage mismatch resulting in average voltage of the charge/discharge process being only half of the final voltage, causing average dissipation of 50% (starting at 100%, ending at 0%) in all equivalent series resistances in the system, which always exist because we live in a real world, and even if we did model them as zero, the math would still work out (with infinite power loss for zero time) if you really understand math (I don't, so I skip it, but I understand the whole idea of the math thing enough so that I can say a generic answer will be possible with the correct math, even if not trivial).

And maybe with infinite or near-infinite power pulses, you will get all kind of UFO things, too. Sparks are quite nice and complex already if you think about it, but they either won't provide the answer to "where does the 50% go". Sorry, we have the answer already, even though it may be boring with nothing special but heat emitted.

[MrAl]

Hello again,

The energy in the magnetic field may be stored energy, but the energy that leaves the system is radiant energy that never returns unless it can be reflected back somehow, which usually isnt the case.  With a capacitor we probably have something like a dipole antenna where each plate acts as one side of the dipole.
Using a formula for the E and B fields we would come up with a value for E and B vectors, and then the power radiated per unit area would be the cross product of E and B times a universal constant.  This would never return to the source but would someday get absorbed by a surface that absorbs it, or partially absorbed and partially reflected.  The levels of the E and B field are probably due to the physical dimensions of the capacitor in the same way that a dipole radiator is dependent on dimensions relative to the frequency.  If we knew the impulse response of the dipole we might be able to calculate the fields and thus the power emitted.

What i think we are surmising in this thread is that if there is no resistance then it has to radiate, but how much we havent determined yet.  We cant use general circuit analysis because it's not really applicable to a physical system unless we include parameters in that analysis that includes the dimensions.  In general circuit analysis we usually assume a lumped condition where we only have to worry about current and voltage and time and not anything like length or width of the physical circuit elements.

[Dr. Frank]

Hello again,

The energy in the magnetic field may be stored energy, but the energy that leaves the system is radiant energy that never returns unless it can be reflected back somehow, which usually isnt the case.  With a capacitor we probably have something like a dipole antenna where each plate acts as one side of the dipole.
Using a formula for the E and B fields we would come up with a value for E and B vectors, and then the power radiated per unit area would be the cross product of E and B times a universal constant.  This would never return to the source but would someday get absorbed by a surface that absorbs it, or partially absorbed and partially reflected.  The levels of the E and B field are probably due to the physical dimensions of the capacitor in the same way that a dipole radiator is dependent on dimensions relative to the frequency.  If we knew the impulse response of the dipole we might be able to calculate the fields and thus the power emitted.

What i think we are surmising in this thread is that if there is no resistance then it has to radiate, but how much we havent determined yet.  We cant use general circuit analysis because it's not really applicable to a physical system unless we include parameters in that analysis that includes the dimensions.  In general circuit analysis we usually assume a lumped condition where we only have to worry about current and voltage and time and not anything like length or width of the physical circuit elements.

Well, that's roughly what has to be calculated.. If you start that process, this will easily lead to a very elaborate calculation, as all electrodynamics problems tend to be.. You would have to determine all the tensor algebra for the capacitor, inductance, reactance, and all the fields, assuming a special geometry.

I've scanned through several theoretical physics books ( Jackson, Landau Lifshitz, and others), but all did not deal with that special paradoxon in their example calculations.. and myself, I'm a little rusty in theoretical electrodynamics..


What's indeed striking, is the independence of this phenomenon from the resistance .. that's a very strong indication, that there is a deeper physical effect buried in the grains, something like a "universal" physical effect.

In the publication mentioned here, I've seen that there are several physicians who tried to explain this effect, including E.M. radiation, but obviously there is no commonly agreed conclusion yet..

And the publication itself is contradictory in itself.

That's very interesting, and I'll try further to find a smoother explanation.
Frank

[IanB]

What's indeed striking, is the independence of this phenomenon from the resistance .. that's a very strong indication, that there is a deeper physical effect buried in the grains, something like a "universal" physical effect.

There is. The deeper physical principle is thermodynamics: the difference between heat and work, and the consideration of entropy. When you equalize two capacitors, or when you charge a capacitor from a voltage source, entropy increases.

[Dr. Frank]

What's indeed striking, is the independence of this phenomenon from the resistance .. that's a very strong indication, that there is a deeper physical effect buried in the grains, something like a "universal" physical effect.

There is. The deeper physical principle is thermodynamics: the difference between heat and work, and the consideration of entropy. When you equalize two capacitors, or when you charge a capacitor from a voltage source, entropy increases.

Yes, that's the 2nd theorem of Thermodynamics.
Helmholtz Free Energy equals Inner Energy minus Temperature times Entropy: F = U - T*S
Fits here, because this problem is equivalent to every other irreversible process, comparable to dilution, for example.

But it should be possible to derive that strictly from the electrodynamics theory, so that you can exactly assign these thermodynamic entities to their electrodynamic counterparts or -effects.

That means, the entropy is increased in this system, but where does this Entropy go to?

In the resistive case, it's directly the Ohm power dissipation, but with zero resistance, it is not evident.
If the electrons are simply accelerated, in the end they must come to a rest, but by definition w/o direct  thermal dissipation (otherwise, it would not be zero ohm or superconducting). So the way, how the entropy increase in the system is happening, has to be explained. And only E.M. radiation, would unload energy, otherwise.

Frank

[IanB]

But it should be possible to derive that strictly from the electrodynamics theory, so that you can exactly assign these thermodynamic entities to their electrodynamic counterparts or -effects.

That means, the entropy is increased in this system, but where does this Entropy go to?

There's a paper here that looks at this: http://arxiv.org/pdf/1201.3890.pdf

Entropic considerations on the Two-Capacitor Problem
V.O.M. Lara, A. P. Lima, and A. Costa
Instituto de F´?sica - Universidade Federal Fluminense
Av. Litorˆanea s/n
24210-340 Niter´oi - RJ Brazil
(Dated: June 12, 2012)
In the present work we study the well-known Two Capacitor Problem from a new perspective.
Although this problem has been thoroughly investigated, as far as we know there are
no studies of the thermodynamic aspects of the discharge process. We use the Free Electron
Gas Model to describe the electrons’ energy levels in both capacitors in the low temperature
regime. We assume that the capacitors and the resistor can exchange energy freely with a
heat reservoir. We assume that the resistance is large enough to consider an isothermal heat
exchange between the resistor and the heat reservoir. Thereby we obtain a positive entropy
variation due to the discharge process, corroborating its irreversibility.

[AG6QR]

Yes, that's the 2nd theorem of Thermodynamics.
Helmholtz Free Energy equals Inner Energy minus Temperature times Entropy: F = U - T*S
Fits here, because this problem is equivalent to every other irreversible process, comparable to dilution, for example.

But it should be possible to derive that strictly from the electrodynamics theory, so that you can exactly assign these thermodynamic entities to their electrodynamic counterparts or -effects.

That means, the entropy is increased in this system, but where does this Entropy go to?

In the resistive case, it's directly the Ohm power dissipation, but with zero resistance, it is not evident.

Well, zero resistance isn't likely to happen in practice, but assuming you built the entire apparatus out of superconductors (let's pretend they're special superconductors that have no current limit), what would happen?

When the switch is closed, very high current would flow from the charged to the empty capacitor.  At first analysis, because there is zero resistance, it would seem to be infinite current, until you consider the inductance of the wire connecting them.  That inductance will impede the flow of current.  As current begins to flow, energy will be stored in the inductor's magnetic field.  Due to conservation of energy, the energy stored in the first capacitor would not disappear, but would be partly transferred to the second capacitor, and partly transferred to the magnetic field surrounding the wire.  When the voltages between the capacitors equalize, there is still a lot of energy (half the total original energy) stored in that inductor.  It will continue pushing charge from one capacitor to the other until all the charge goes into the second (originally uncharged) capacitor.  Then the process will reverse.

This is just like the "two buckets of water" problem, where a pipe connects the bottoms of the buckets.  In the total absence of friction (which ain't gonna happen in real life), water would slosh between the buckets forever.

But I've left out one other important point so far:  The inductors and capacitors are storing energy in electrical and magnetic fields, and these fields are constantly changing.  Maxwell tells us that changing electrical and magnetic fields will radiate energy away, as electromagnetic radiation.  This loss of energy will eventually damp the oscillations, causing the apparatus to reach the state of each capacitor having equal charge, just as though we had placed a resistor in the wire.  The energy left the system via EM radiation.

[GeorgeOfTheJungle]

This is just like the "two buckets of water" problem, where a pipe connects the bottoms of the buckets.  In the total absence of friction (which ain't gonna happen in real life), water would slosh between the buckets forever.

But I've left out one other important point so far:  The inductors and capacitors are storing energy in electrical and magnetic fields, and these fields are constantly changing.  Maxwell tells us that changing electrical and magnetic fields will radiate energy away, as electromagnetic radiation.  This loss of energy will eventually damp the oscillations, causing the apparatus to reach the state of each capacitor having equal charge, just as though we had placed a resistor in the wire.  The energy left the system via EM radiation.

Amazing. Even beautiful. I like it. I almost can see it!

[T3sl4co1l]

Even if you build the capacitor out of superconducting plates in vacuum, and the inductor out of superconductor and vacuum, and house everything within a superconducting shield chamber -- or better still, construct all three together as a resonator -- the energy decays, because superconductors aren't ideal at AC.

An ideal ideal superconductor would conserve energy for all time, but there is no physical realization of it.

Practical superconducting resonators (used for linear particle accelerators) have Q factors in the 3 x 10^8 range, which is better than a quartz crystal, but still results in a time constant of seconds at best.

The conclusion is, even if radiation (to infinity) is blocked by as ideal a shield as possible, there's still loss in the system somewhere, dissipating that energy as heat, so that in the end, thermodynamics wins.

Tim

[CatalinaWOW]

Even if you build the capacitor out of superconducting plates in vacuum, and the inductor out of superconductor and vacuum, and house everything within a superconducting shield chamber -- or better still, construct all three together as a resonator -- the energy decays, because superconductors aren't ideal at AC.

An ideal ideal superconductor would conserve energy for all time, but there is no physical realization of it.

Practical superconducting resonators (used for linear particle accelerators) have Q factors in the 3 x 10^8 range, which is better than a quartz crystal, but still results in a time constant of seconds at best.

The conclusion is, even if radiation (to infinity) is blocked by as ideal a shield as possible, there's still loss in the system somewhere, dissipating that energy as heat, so that in the end, thermodynamics wins.

Tim

The devil is always in the details.  Whether it is heat dissipated in a superconductor, or just a tiny incremental contribution to the cosmic background radiation, thermodynamics wins. 

The problem here, as in all engineering applications (theoretical physicists get to play their own games), is in finding an appropriate model for the system in question, that provides a "good enough" answer for the problem at hand.  Capacitors in the day of Faraday and Franklin could be modeled very simply.  Worked pretty well up through the 1920s or 1930s when radio frequencies got high enough that the inductance of the capacitor started to be a problem.  A few more years down the line and people started doing switched mode power supplies.  Lots of current flowing in and out of capacitors all the time and ESR started being important.  Think of varactors for another example of needing to tune a model for the application.

Thinking will never be outdated, but models often are.

[Dr. Frank]

Even if you build the capacitor out of superconducting plates in vacuum, and the inductor out of superconductor and vacuum, and house everything within a superconducting shield chamber -- or better still, construct all three together as a resonator -- the energy decays, because superconductors aren't ideal at AC.

An ideal ideal superconductor would conserve energy for all time, but there is no physical realization of it.

Practical superconducting resonators (used for linear particle accelerators) have Q factors in the 3 x 10^8 range, which is better than a quartz crystal, but still results in a time constant of seconds at best.

The conclusion is, even if radiation (to infinity) is blocked by as ideal a shield as possible, there's still loss in the system somewhere, dissipating that energy as heat, so that in the end, thermodynamics wins.

Tim

That's not correct.

Superconducting coils (like in the CERN accelerator ring, or in any MRT) can store magnetic energy indefinitely, as long the He4 cools the S.C.

These coil wires also can definitely withstand dI/dt, without becoming normal conductors.

Whether they dissipate some energy depends on the type of the S.C., i.e. either type I or type II.

Type II S.C. pin the magnetic flux and this pinning contains dissipative energy during changes.

But type I S.C., w/o pinning may bear relatively high currents, but without dissipation.

As the inductance of the wire also limits the current, it is really possible that a superconducting system, consisting of say 2 s.c. capacitors and 2 superconducting wires, really does not dissipate any heat energy.

You may as well do a gedankenexperiment  where you model this s.c. system w/o heat dissipation, for example additionally with infinitely small changes. This will lead to a lossless oscillator.

Every LC oscillator emits E.M. radiation; the amount of emitted energy strongly depends on the geometry of the capacitor and the wire (Hertz' dipole). This oscillation exists, until equilibrium is reached, i.e. the static case.

Therefore, nobody has to  search compulsorily for heat losses, if that energy dissipation path by e.m. wave explains the problem sufficiently well.

This is also done in the publication   "The two-capacitor problem with radiation" , Boykin, Hite, Singh, Am. J. Phys 70,415 (2002).

Frank

[MrAl]

Hi,

I could not help but add one more tiny post perhaps on a much lighter note than previous.

As we can see now this is a physics problem, and that means we've got to define every little smidget of everything everywhere in order to know for sure exactly what is happening and where it is happening and how much of everything changes, and we have to know where everything goes.
We can build it in a box where nothing can escape so we can keep track of everything, or can we?  What we end up with is Schrodinger's Oscillator as we dont know if it is oscillating or not :-)

On the more serious side, we have found that there will always be at least some radiation so the actual RC circuit will dissipate more than 50 percent of the total energy, even if it is only a little more sometimes.

[T3sl4co1l]

Oh my, this is quite a bit more advanced and quite a bit more fascinating than I was looking for, but anyway, it does cover the subject of kinetic inductance in superconductors:
http://web.physics.ucsb.edu/~bmazin/Papers/2008/Gao/Caltech%20Thesis%202008%20Gao.pdf

The reason I specified a resonator (an enclosed, self shielded kind, if I didn't make that clear) is because the external field, and therefore radiation, is absolutely zero.

The specific example I was thinking of is LINAC beam line resonators, which have a Q in the hundreds of millions.  Think about that number for a moment.  Consider that the equivalent inductance and capacitance of a resonator at 100MHz might be, oh, let's say it's out of 200 ohms, so, 0.318uH and 7.95pF.  Not that you can exactly measure those directly, as with wires, but they're there, more or less.  A Q factor that high means either the ESR is single microohms, or the EPR is tens of gigaohms.  It would be quite misleading to imagine such a system as being lossy from cycle to cycle!  But yet, over millions of cycles, it nonetheless decays, and therefore thermodynamics wins again.

It's also interesting to note that superconductors cease somewhere in the THz to far IR range -- consider that a hunk of niobium doesn't suddenly become shinier (at visible wavelengths) when cooled very low.  This frequency range corresponds to the energy required to disrupt Cooper pairs, which it would seem is the mechanism used in the above paper.  Neato.

Tim

[The Electrician]

The three laws of thermodynamics:

1. You can't win.

2. You can't even break even.

3. You can't get out of the game.

[VintageTekFan]

The three laws of thermodynamics:

1. You can't win.

2. You can't even break even.

3. You can't get out of the game.

I think I've found my sig.

[GeorgeOfTheJungle]

This thread has made me discover "The Electrician" http://en.wikipedia.org/wiki/The_Electrician... and lose countless hours reading the .pdfs of the few numbers that are floating in the internet. And I'm addicted now, I need to know and see how it all developed back in the 1800s in "The Electrician" pages. Does anybody know where can I get them all?

A pearl from back then:

"In his experiments Lodge had simulated lightning by discharging Leyden jars. Preece and others questioned whether this was a valid analogy, and it was indeed the weakest part of Lodge's argument. The more substantial issue, however, was the alleged effect of self-induction in radically increasing the impedance of a conductor to a rapidly alternating current such as resulted from a Leyden jar discharge or, according to Lodge, a lightning bolt. According to Maxwellian theory, as elaborated by Heaviside, Poynting, and Lord Rayleigh (J. W. Strutt), this was essentially a field effect, and it involved nothing less than the nature of an electric current.
To most "practical men" like Preece, a current in a wire was much like the flow of water in a pipe. There might be some modifications picturing the pipe as elastic, to simulate capacitance, for instance, or filled with baffles to simulate resistance but it was basically a simple and intuitive picture, and a remarkably useful one. The essential insight of Maxwell's theory, however, was to focus on the field, not the wire; as FitzGerald put it, "According to Maxwell's view, there is a great deal more going on outside the conductor than inside it." The most striking implication of this view was one Poynting (and independently Heaviside) had derived from Maxwell's theory in 1884.

14 Preece in Electrician, 1888, 21:646, 662; Heaviside in ibid., p. 772 (rpt. in Heaviside, Electrical Papers, Vol. II, p. 448).
15 G. J. Symons (ed.), Lightning Rod Conference Report (London: Spon, 1882). 16 Electrician, 1888, 21:676, 679-680; cf. Lodge, Advancing Science, pp. 96-97. 17 See, e.g., S. A. Varley, Electrical Review, 1888, 23:224."

BRUCE J. HUNT, "Practice vs. Theory" The British Electrical Debate, 1888-1891.
https://google.com/search?q="Practice+vs.+Theory,+The+British+Electrical+Debate,+1888-1891"

What still isn't clear to me is whether the radiation losses are in addition to the 50% or included in the 50%, nor the % in which energy is lost as heat in resistances vs the losses due to EM radiation, nor if a 50% loss is inevitable. Some have suggested in this thread that slow charging trough an inductor can reduce the losses below that 50%, is it true?

[dannyf]

-Some have suggested in this thread that slow charging trough an inductor can reduce the losses below that 50%, is it true?-

Those are the same people who have absolutely no clue what they are advocating.

The central point here is that the resistive loss in any r/c charging circuit is only independent of the resistance in the circuit. Actually, it is only dependent of the supply voltage and capacitance.

That means the circuit is as efficient the the resistance goes to zero (then where is the energy lost?) or infinity (the capacitor is charged by a constant current source).

[MrAl]

-Some have suggested in this thread that slow charging trough an inductor can reduce the losses below that 50%, is it true?-

Those are the same people who have absolutely no clue what they are advocating.

The central point here is that the resistive loss in any r/c charging circuit is only independent of the resistance in the circuit. Actually, it is only dependent of the supply voltage and capacitance.

That means the circuit is as efficient the the resistance goes to zero (then where is the energy lost?) or infinity (the capacitor is charged by a constant current source).

Hi,

Actually in an RC circuit it is true that the loss is 50 percent which is independent of the R value if we totally ignore radiation (more correctly irradiation).  But when we also include an inductor in series, if we break the circuit at a certain point we get more energy into the capacitor and thus we dont loose 50 percent.  In fact we can get the loss down much lower than that.
To understand how this works, all you have to do is do a time domain analysis on the RLC circuit, or you can do it with two capacitors (CRLC circuit).
This really does work, no kidding.  But the circuit does have to be broken at a certain time to see the increase.  If it is left on it's own forever it will degrade.

We also think that when charging an RC circuit that there will always be some energy lost to radiation, so that reduces the transfer of energy to below 50 percent, even if it is just a little less.  Depending on the value of the resistance, it could actually radiate more for the faster charge time, and if the physical dimensions are adjusted properly to match RC we might see a huge percentage of the energy lost as radiation.  This turns out to be much harder to calculate, but the RLC circuit is quite easy to calculate and we can do that here if you like.

[T3sl4co1l]

The "50% loss" condition, again, is a thermodynamic eventuality.  It does not matter how it is achieved.  Radiation as an equivalent loss resistance (or noise voltage/current source) is a useful tool, and behaves as any other resistance, hence its usefulness.

For detail about what happens at different initial voltage levels, https://www.eevblog.com/forum/projects/reversible-and-irreversible-processes-(redux)/

Tim

[MrAl]

Hi,

Not sure what you are trying to say here.  Isnt it good to know the various mechanisms instead of lumping them all into one category?  It's up to you though of course how you want to view this.

[T3sl4co1l]

The important nugget is that, no matter how the circuit distributes the losses, you can never change the 50% total.

Tim