Reversible and Irreversible Processes (redux)

[parbro]

Thanks. I see the flaw in my thinking. 

[Siwastaja]

Charge 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)

... 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.

[electr_peter]

Charge 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)

... 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...

I don't think it can be called ESR, because basic model assumes ideal capacitors 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.

[Siwastaja]

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.

[T3sl4co1l]

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

[suicidaleggroll]

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.

Exactly.  This whole discussion is like a dog chasing its tail.  You can't remove the only source of loss in 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 the only 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".

[parbro]

I always thought ESR was a measure of the imperfection of the dielectric ( for example leakage).

[T3sl4co1l]

Remembering it's equivalent series 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