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

[IanB]

It was suggested I start a new thread with this rather than having it lost in the noise of the free energy thread. It shows us that we can't "just store energy in a capacitor". There are important factors to consider if attempting to use a capacitor to capture energy is not to result in huge system losses.

We see that naively attempting to charge a capacitor from a voltage source is going to result in the loss of half the energy supplied (a 50% system efficiency):

[Zucca]

Interestig!

BUT slow down a little bit....

I am still thinking the above is applicable only if R is different from 0 or infinite.
In theory with R=0 there is no lost energy, neither with R=infinite (here there will be no energy transfer at all).

Now let´s put an Rp in parellel to the C and call the R above Rs.
The lost energy will be:

1/2 C [V*Rp/(Rp+Rs)]²

{I did no calculations for what I wrote; just used the Thevenin/Norton theorem}

Now you can`t eliminate the Rs anymore.

So.... I would say since in reality there will be always an Rp in parallel to the C to optimize the energy transfer you want Rp as big as possible and Rs as small as possible. fuuu... the world makes sense again?

BTW the 1/2 max energy efficiency in an real power source (i.e. with Rs in serie in a V) is a common result. I think even with inductors or other stuff it would be the same. Maybe just generalize with an impedance and you got the above results too...

It could be I am wrong. In this case sorry in advance.

[Galaxyrise]

When you posted that before, you added a comment about not charging from a constant voltage source if you want efficiency.  That got me curious what the efficiency is from a constant current supply.  Let's see if I can avoid careless mistakes...

Charging capacitance C to voltage V with current I will take CV/I seconds.  The power dissipated by the resistance is I²R, so energy is I²RCV/I = IRCV joules. 

So trickle charging a capacitor from a photo cell might be decently efficient?

[Zucca]

So trickle charging a capacitor from a photo cell might be decently efficient?

It depends on how you put on paper reality. Surely charging a C with a photo cell is not a stupid idea, because a photo cell can be modelized very well with an ideal I source...

[Galaxyrise]

I am still thinking the above is applicable only if R is different from 0 or infinite.
In theory with R=0 there is no lost energy, neither with R=infinite (here there will be no energy transfer at all).

Yeah, it's got a R/R in it, so it's not defined with R = 0 or R = infinity.  I stopped and thought about R=0 for awhile, too.  It doesn't ever apply to reality, though: you can't have infinite current even with superconductors.

If you're going to add in the parasitics like parallel resistance, you should consider inductance and rise time as well; but they all only serve to make the loss larger; the moral of "less efficient than you think" still holds!

[AG6QR]

So trickle charging a capacitor from a photo cell might be decently efficient?

It depends on how you put on paper reality. Surely charging a C with a photo cell is not a stupid idea, because a photo cell can be modelized very well with an ideal I source...

Assume the solar cell is an ideal one, a constant current source, at any voltage from 0 up to Vmax.  If you connect that to a capacitor and charge the capacitor, then yes, there will be no energy loss in the electrical circuitry between the solar cell and the capacitor.  All the electrical energy produced by the cell will go into the capacitor.  No loss.

Or is there loss?

The solar cell will have been producing half of its rated power over the charge cycle.  The power is V*I, so when V is low, most of the solar radiation falling on the cell is not being converted to electrical energy.  In order for the cell to produce maximum rated power, it must output its constant current at Vmax.

You have still lost half of the available solar energy.  It's just that it wasn't wasted in resistance of the circuit; it was wasted by having your constant current source operating at, on average, half of its rated voltage.

[dannyf]

While simplifies to:

No, it doesn't.

Therefore the energy loss from power dissipated in the system resistance is equal to the power stored in the capacitor, and is independent of the resistance

All you need to do is to step back and read that sentence of yours and think, big picture, what it means. It means that regardless of your component parameters, such a system is always 50% efficient.

Just think about that for sanity's sake and you will realize that something is terribly wrong here.

[IanB]

All you need to do is to step back and read that sentence of yours and think, big picture, what it means. It means that regardless of your component parameters, such a system is always 50% efficient.

Just think about that for sanity's sake and you will realize that something is terribly wrong here.

Perhaps you can point to the mistake?

[Zucca]

No, it doesn't.

Please provide the correct result...

[suicidaleggroll]

It's just an oversimplification of the problem.  It is only true if you have an ideal voltage source capable of infinite current with zero voltage drop.  In reality, as you start drawing current from the source, the voltage will dip and it will look more and more like a current source (at least until it blows), which breaks the assumptions made in the first post.

I agree though, that for an ideal voltage source, the system efficiency will be 50%.  A similar argument could be made for batteries as well though.  If you want to charge a battery efficiently, you don't drive it with a fixed voltage source through a resistor, you use constant current up to a set point and then CV afterward, same with a capacitor, if energy storage and efficiency is your goal at least.

[IanB]

It's just an oversimplification of the problem.  It is only true if you have an ideal voltage source capable of infinite current with zero voltage drop.  In reality, as you start drawing current from the source, the voltage will dip and it will look more and more like a current source (at least until it blows), which breaks the assumptions made in the first post.

Note that the dotted line box represents a "real" voltage source, where R represents the internal resistance of the source. Current flowing through R (non-zero) will cause the observed voltage to dip at the output terminals as expected. Since the resistance of the system is non-zero, the current will never be infinite.

I agree though, that for an ideal voltage source, the system efficiency will be 50%.  A similar argument could be made for batteries as well though.  If you want to charge a battery efficiently, you don't drive it with a fixed voltage source through a resistor, you use constant current up to a set point and then CV afterward, same with a capacitor, if energy storage and efficiency is your goal at least.

As shown in the analysis, the efficiency will be 50% even with real voltage sources.

Yes, a constant current source will improve matters considerably, but it must be an active source designed to minimize internal losses.

[Hugoneus]

This is a very classic example of charging a capacitor. The system will dissipate as much energy as it will store. Nothing wrong with the calculations.

[suicidaleggroll]

Note that the dotted line box represents a "real" voltage source, where R represents the internal resistance of the source. Current flowing through R (non-zero) will cause the observed voltage to dip at the output terminals as expected. Since the resistance of the system is non-zero, the current will never be infinite.

Sorry, hadn't had lunch yet, my brain was off in left field

Yes, a constant current source will improve matters considerably, but it must be an active source designed to minimize internal losses.

But this is no different than any other storage device.  You don't just dump a voltage source through a resistor into your storage system if you care about efficiency.  The efficiency for a capacitor will be 50% if you do this, but it's not going to be much better for a battery either.  The difference is a capacitor's voltage increases linearly from 0 with a CC input, while a battery's voltage increases rapidly from 0 to the nominal voltage, where it stays through the majority of the charge cycle, before increasing rapidly at the end.  So the average voltage for the capacitor during the charge cycle is half of the nominal voltage (meaning the other half is burned off in the resistor), while for a battery it might be more like 80% (meaning the other 20% is burned off in the resistor).  Either way it's crap if you care about efficiency.

[Hugoneus]

But this is no different than any other storage device.  You don't just dump a voltage source through a resistor into your storage system if you care about efficiency.  The efficiency for a capacitor will be 50% if you do this, but it's not going to be much better for a battery either.

Yes, I think the reason this example is so popular is that even if you let R approach zero, the system would still lose 50% of its delivered energy (in that case mostly in the form of broadband EM radiation and sound).

[Zucca]

Or is there loss?

Yeah, it is basically right what you say. BTW I meant a solar cell can be modelized with a Is source and a Rp in parallel. The Vmax you mention is just Vmax=Is*Rp, in other words the open circuit voltage.

OFF TOPIC: This web site is fantastic to understand solar cells: http://pveducation.org/pvcdrom

[IanB]

Yes, I think the reason this example is so popular is that even if you let R approach zero, the system would still lose 50% of its delivered energy (in that case mostly in the form of broadband EM radiation and sound).

I think this example is interesting because it may run counter to intuition and can present a trap for the unwary. For example, what will it do to your system if you try to put a capacitor across the output of a PWM to smooth the voltage...?

[Hugoneus]

Yes, I think the reason this example is so popular is that even if you let R approach zero, the system would still lose 50% of its delivered energy (in that case mostly in the form of broadband EM radiation and sound).

I think this example is interesting because it may run counter to intuition and can present a trap for the unwary. For example, what will it do to your system if you try to put a capacitor across the output of a PWM to smooth the voltage...?

It will smooth the PWM alright!   

[dannyf]

Two points for you to think about:

The capacitor and constant voltage 50% power loss is very classic in power electronics.

As the power loss is independent of the resistance in the circuit, what happens to the other 50% of energy when you charge up an ideal capacitor with an ideal power source?

The best way to charge capacitors efficiencly is constant current charging.

When you increase R, a voltage source approaches a current source. So if you believe that the 50% efficiency is independent of the resistance, charging a battery with a current source is every bit as inefficient as charging a battery with a voltage source.

Enjoy.

[AG6QR]

Two points for you to think about:

The capacitor and constant voltage 50% power loss is very classic in power electronics.

As the power loss is independent of the resistance in the circuit, what happens to the other 50% of energy when you charge up an ideal capacitor with an ideal power source?

It's dissipated in the resistor.

Changing the value of resistance changes the rate at which the charging happens, as well as changing the rate at which energy is dissipated.  It changes both rates in equal proportions.  Since power is rate of energy (more precisely, it's the derivative of energy with respect to time), changing the resistance changes the power dissipated in the resistor, as well as changing the power delivered to the capacitor.  But changing the resistance does not change the ratio of those two powers, as integrated over the charge cycle.

[Hugoneus]

Two points for you to think about:

The capacitor and constant voltage 50% power loss is very classic in power electronics.

As the power loss is independent of the resistance in the circuit, what happens to the other 50% of energy when you charge up an ideal capacitor with an ideal power source?

Mathematically this would mean an infinitely high pulse of current which has an infinitely short duration. In real life the resistance can approach zero, but never reach zero. In the case of very small resistance, not all of the power loss would be thermal. Most of it would be in the form of broadband EM emission, light and sound.

The best way to charge capacitors efficiencly is constant current charging.

When you increase R, a voltage source approaches a current source. So if you believe that the 50% efficiency is independent of the resistance, charging a battery with a current source is every bit as inefficient as charging a battery with a voltage source.
Enjoy.

This is incorrect. A voltage source with a large series resistor is not a current source.

[c4757p]

Would it kill you to quote with the author's name so it's a bit easier to piece together the conversation without re-reading the page every time somebody adds a comment?

[tggzzz]

I find it confusing that you intermix power and energy willy-nilly.

Why do you start with an equation of how much power is dissipated in a resistor? Starting with how much energy is dissipated is far more appropriate.

[IanB]

The capacitor and constant voltage 50% power loss is very classic in power electronics.

I first became aware of this phenomenon yesterday, so I'm gratified to learn that it is "common knowledge" among those whose job it is to know such things.

It puts me in mind of this particular comic:

[IanB]

I find it confusing that you intermix power and energy willy-nilly.

Why do you start with an equation of how much power is dissipated in a resistor? Starting with how much energy is dissipated is far more appropriate.

It is a document editing error. I will fix it up later. I was trying to say that total power loss is power integrated is energy, but the wording came out badly.

[Galaxyrise]

When you increase R, a voltage source approaches a current source. So if you believe that the 50% efficiency is independent of the resistance, charging a battery with a current source is every bit as inefficient as charging a battery with a voltage source.

You've made an inefficient current source that way, but hey, at least there's almost no energy loss outside your current source

[Hugoneus]

The capacitor and constant voltage 50% power loss is very classic in power electronics.

I first became aware of this phenomenon yesterday, so I'm gratified to learn that it is "common knowledge" among those whose job it is to know such things.

When it comes to issues of humility, I think Christopher Hitchens said it very nicely:

"And though I have met many people much wiser and more clever than myself, I know of nobody who could be wise or intelligent enough to say differently."

[dannyf]

You've made a very inefficient current source that way

No - since the efficiency is always 50%, regardless of the source resistance.

You cannot have it both ways, as the two statements are mutually exclusive.

[Hugoneus]

You've made a very inefficient current source that way

No - since the efficiency is always 50%, regardless of the source resistance.
You cannot have it both ways, as the two statements are mutually exclusive.

Again, a voltage source with a large series resistor is NOT a current source. 

[c4757p]

Again, a voltage source with a large series resistor is NOT a current source. 

If you keep doing for long enough, it might start to look like one...

hmm...

[paulie]

Would it kill you to quote with the author's name so it's a bit easier to piece together the conversation without re-reading the page every time somebody adds a comment?

LOL. My buddy westfw does that too. In dannyf's case it appears to be the only way he can get away with altering the quotes. He's done that twice with me.

I still love the guy though. Specially that last ghetto/el-cheapo opto-isolator trick.

[electr_peter]

The capacitor and constant voltage 50% power loss is very classic in power electronics.

I first became aware of this phenomenon yesterday, so I'm gratified to learn that it is "common knowledge" among those whose job it is to know such things.

I was aware of this for a long time. However, at first it struck me that there can be 50% losses by default. This issue is similar (in principle) to resistance/impedance matching to have 1:1 source/load proportion.

dannyf, stop trolling and quote other users properly

[Kevin.D]

The storage efficiency  will depend on the voltage difference between the Cap and the supply .
The smaller the difference between V source and cap Voltage the greater the ratio of energy stored to that supplied

We have IT=Q  . where T= time I= Current  Q =charge ,
and also  Q=CV   where  C= capacitance ,V=voltage
 
now it takes a fixed amount of charge Q(Coulombs in C) to raise a fixed cap voltage by 1V  .

Example - Cap is 1 Farad and V supply = 10V

So take a 1 F cap and charge it to 1V . this takes 1 C of charge.(use Q=CV)
 to supply 1 C then our battery must supply 1A for 1 sec (using Q=IT)so  (using P=V*I) battery delivers 10 Watt seconds (10 Joule's) of energy to deliver 1 C of charge .

So our cap stored only (use 1/2 C V squared)  0.5 Joules of our 10 J supplied . (only 5% was stored)

but if we start the Cap at 9V then charge to 10V we now stored 50(final energy of cap)-40.5(starting energy) = 9.5 J of our 10 joules supplied by battery ( 95% stored) .
 

[Galenbo]

In theory with R=0 there is no lost energy,

But chances are that you have a lost capacitor.

[Zucca]

to supply 1 C then our battery must supply 1A for 1 sec (using Q=IT)

who is telling you the current is constant at 1A in that second? In other words, can you show us the circuit you are talking about?

[Zucca]

But chances are that you have a lost capacitor.

...and my mind too  (just a joke...). That C on paper can not burn, it has no Vmax, no ESR, no polarity and can store all the Q he wants up to infinite.

[Kevin.D]

The Current alone is irrelevant because it will be I * T  that will be a constant in order to supply a given charge  .Since Vsupply is also constant then V*I*T = energy is also constant for a given charge . I  chose 1A for 1 sec for these two values since that  made it easier to work out power.

[electr_peter]

If we make a little step into the real world, calculations should take into account transmission line effects of wires and capacitor.
Thus model could be voltage source, transmission line (charged or uncharged) and an ideal capacitor. Voltage pulse starts, encounters transmission line impedance, reaches capacitor and start bouncing back and forth depending on impedance mismatch.
By the way, capacitor charge curve is only approximated by exponential curve, in reality it is made of little bumps/steps (and steps can be quite big).

Capacitor charging efficiency would be essentially the same, but limit case of [charge time->0] would have different interpretation

[IanB]

One can work directly with charge and have no need to bring current or time into the analysis. The result depends only on the initial and final states of the system and is independent of the intermediate path followed.

Take my example from the top of the thread. Let's say that when the capacitor reaches full charge it holds a charge Q.

Then the energy introduced into the system by the voltage source is:

    E_S = QV_S

And the energy stored by the capacitor is:

    E_C = ½QV_S

[Zucca]

Then the energy introduced into the system by the voltage source is:

    E_S = QV_S

And the energy stored by the capacitor is:

    E_C = ½QV_S

+1 and thank you to save me the time to post that. Of course you didn't mention the delta voltage (final-initial) stuff, but the principle is more than correct... the rest is just fine tuning.

[electr_peter]

One can work directly with charge and have no need to bring current or time into the analysis. The result depends only on the initial and final states of the system and is independent of the intermediate path followed.

I agree that final result will be the same (with simple and basic calculations or with more detailed model) - 50% efficiency. It is just interesting where, how and when 50% of energy manages to dissipate as it is not immediately intuitive.

[tggzzz]

Let's cut to a traditional conundrum think about where the energy has gone.

Consider a circuit consisting of an two equal identical ideal capacitors and some ideal wire. No resistance, no inductance, no dielectric absorbtion etc.

Start with capacitor C1 charged to V volts, and capacitor C2 discharged.
The total charge in the circuit is therefore Q=VC coulombs, and the total energy is ¹/₂CV²

Now connect the two capacitors. Since charge is conserved (unless you believe electrons evaporate!) and the capacitors are equal, the charge is equally distributed across the two capacitors and the voltage across each capacitor is therefore 0.5V (i.e. since CV = 2*C *V/2). Simple so far.

But the total energy in the circuit is then 2*¹/₂C(0.5V)² = ¹/₄CV² i.e. 50% has disappeared.

Where has the energy gone?

[The Electrician]

This problem has been around a long time; see:

http://physics.princeton.edu/~mcdonald/examples/seriesrlc.pdf

and the included references.

Also: http://physics.princeton.edu/~mcdonald/examples/twocaps.pdf

This was the subject of a number of threads on other sites also:

https://www.physicsforums.com/threads/capacitor-charging-loss-not-the-two-capacitor-issue.292838/

Besides using an inductor and diode, the capacitor can be charged from a variable voltage source if the votage is turned up slowly.

[georges80]

...

But the total energy in the circuit is then 2*¹/₂C(0.5V)² = ¹/₄CV² i.e. 50% has disappeared.

Where has the energy gone?

Quite obviously it has gone into the quantum vacuum to later be extracted by circuits designed specifically to harness that energy. Haven't you learned anything from certain youtube videos recently??




cheers,
george.

[IanB]

Now connect the two capacitors.

Depending on the size of the capacitors, there will tend to be a large bang and a flash when you make this connection, indicative of a short circuit being closed. Short circuits across power sources are not good...

In thermodynamics we would call this an irreversible process. Irreversible processes are associated with an increase in entropy of the system, and increases in entropy are associated with the conversion of other forms of energy into heat.

[The Electrician]

In thermodynamics we would call this an irreversible process. Irreversible processes are associated with an increase in entropy of the system, and increases in entropy are associated with the conversion of other forms of energy into heat.

Have a look at the mechanical analogy given starting at this post in the thread I referenced:

https://www.physicsforums.com/threads/capacitor-charging-loss-not-the-two-capacitor-issue.292838/page-3#post-2107781

[mrflibble]

Let's cut to a traditional conundrum think about where the energy has gone.

Consider a circuit consisting of an two equal identical ideal capacitors and some ideal wire. No resistance, no inductance, no dielectric absorbtion etc.

Start with capacitor C1 charged to V volts, and capacitor C2 discharged.
The total charge in the circuit is therefore Q=VC coulombs, and the total energy is ¹/₂CV²

Now connect the two capacitors. Since charge is conserved (unless you believe electrons evaporate!) and the capacitors are equal, the charge is equally distributed across the two capacitors and the voltage across each capacitor is therefore 0.5V (i.e. since CV = 2*C *V/2). Simple so far.

But the total energy in the circuit is then 2*¹/₂C(0.5V)² = ¹/₄CV² i.e. 50% has disappeared.

Where has the energy gone?

Last time I checked our universe was rather partial to conservation of energy and indeed conservation of charge. I didn't get the memo on conservation of field potential (aka voltage here).

[Someone]

Besides using an inductor and diode, the capacitor can be charged from a variable voltage source if the votage is turned up slowly.

The obvious example being a large capacitor connected across the mains to provide power factor correction, marginal losses there.

[parbro]

But the total energy in the circuit is then 2*¹/₂C(0.5V)² = ¹/₄CV² i.e. 50% has disappeared.

Where has the energy gone?

I think the 50% energy can be thought of as the energy required to establish the electric field within the dielectric.

[IanB]

Last time I checked our universe was rather partial to conservation of energy and indeed conservation of charge. I didn't get the memo on conservation of field potential (aka voltage here).

Ah, but it rather seems that voltage is conserved here.

Initially:

    V₁ = V_S
    V₂ = 0
    V₁ + V₂ = V_S

Finally:

    V₁ = V_S/2
    V₂ = V_S/2
    V₁ + V₂ = V_S

Conservation of voltage: verified   

[IanB]

The obvious example being a large capacitor connected across the mains to provide power factor correction, marginal losses there.

I dunno. If you connected a 3000 µF capacitor across the mains, I think you might encounter the loss of your mains supply...   

[IanB]

Have a look at the mechanical analogy given starting at this post in the thread I referenced:

There's an even better mechanical analogy. Imagine a 1 kg sticky ball moving at 10 m/s colliding with a second, stationary, sticky ball.

The first sticky ball has a momentum of 1 x 10 = 10 kg m/s and a kinetic energy of 0.5 x 1 x 10² = 50 J.

Upon collision the sticky balls remain stuck together and continue as one.

After the collision momentum is conserved, so the velocity of the combined balls is 10 / 2 = 5 m/s. Therefore the kinetic energy of the combined balls is 0.5 x 2 x 5² = 25 J.

It appears that 25 J of energy has been lost. Where did it go?

[parbro]

Here is an interesting article I found about whether constant current charging is more efficient. The comments kind of mirror our discussion.
http://www.olino.org/us/articles/2006/11/22/charge-efficiency-capacitor

[Marco]

Where has the energy gone?

If ESR is 0 the current is inifinite ... the power lost in the ESR is 0*0*infinite = 1/2 CV^2.

[Galaxyrise]

You've made a very inefficient current source that way

No - since the efficiency is always 50%, regardless of the source resistance.

You cannot have it both ways, as the two statements are mutually exclusive.

I'm suggesting abandoning the constant supply voltage.  Even just charging in two voltage steps improves efficiency quite a bit.  If you make a current source which perfectly varies the supply voltage (not by resistively dropping the voltage!) and keep R constant, then you end up with the "RICE" loss given in the formulas posted earlier. 

[owiecc]

If ESR is 0 the current is inifinite ... the power lost in the ESR is 0*0*infinite = 1/2 CV^2.

Why not 0*0*infinite = 10*CV^2 ? Power lost in resistance equal to zero will be zero.
There is another physical phenomenon connected with moving electrons in a wire loop. The remaining energy will be radiated away.

[tggzzz]

Now connect the two capacitors.

Depending on the size of the capacitors, there will tend to be a large bang and a flash when you make this connection, indicative of a short circuit being closed. Short circuits across power sources are not good...

Irrelevant. These are ideal components. Theoretical gedankenexperimenten are designed to cut through lots of obfuscating details and get to the heart of the problem.

Of course if you tried to practically verify the gedankenexperiment thenyou would have to assess the magnitiude of the effect, minimise it, and account for its affect in the numbers

In thermodynamics we would call this an irreversible process. Irreversible processes are associated with an increase in entropy of the system, and increases in entropy are associated with the conversion of other forms of energy into heat.

Indeed, but that sheds no light on where the energy goes.

[tggzzz]

Let's cut to a traditional conundrum think about where the energy has gone.

Consider a circuit consisting of an two equal identical ideal capacitors and some ideal wire. No resistance, no inductance, no dielectric absorbtion etc.

Start with capacitor C1 charged to V volts, and capacitor C2 discharged.
The total charge in the circuit is therefore Q=VC coulombs, and the total energy is ¹/₂CV²

Now connect the two capacitors. Since charge is conserved (unless you believe electrons evaporate!) and the capacitors are equal, the charge is equally distributed across the two capacitors and the voltage across each capacitor is therefore 0.5V (i.e. since CV = 2*C *V/2). Simple so far.

But the total energy in the circuit is then 2*¹/₂C(0.5V)² = ¹/₄CV² i.e. 50% has disappeared.

Where has the energy gone?

Last time I checked our universe was rather partial to conservation of energy and indeed conservation of charge. I didn't get the memo on conservation of field potential (aka voltage here).

Just so. What conservation of voltage are you thinking of? The voltage wasn't conserved.

OTOH charge was conserved, and "invisible missing energy" is the whole point of the conundrum.

[tggzzz]

Where has the energy gone?

If ESR is 0 the current is inifinite ... the power lost in the ESR is 0*0*infinite = 1/2 CV^2.

That is both mathematically and physically invalid, of course.

[tggzzz]

But the total energy in the circuit is then 2*¹/₂C(0.5V)² = ¹/₄CV² i.e. 50% has disappeared.

Where has the energy gone?

I think the 50% energy can be thought of as the energy required to establish the electric field within the dielectric.

Er, that's the mechanism of capacitance! Which is already included in the equation.

[tggzzz]

...

But the total energy in the circuit is then 2*¹/₂C(0.5V)² = ¹/₄CV² i.e. 50% has disappeared.

Where has the energy gone?

Quite obviously it has gone into the quantum vacuum to later be extracted by circuits designed specifically to harness that energy. Haven't you learned anything from certain youtube videos recently??

Since that effect has been shown to be equivalent to harmonic morphic resonance, I have to accept that as the second best answer

The best answers are referred to by The Electrician in his post https://www.eevblog.com/forum/projects/why-trying-to-store-energy-in-a-capacitor-can-be-less-efficient-than-you-think/msg597970/#msg597970

[opty]

But the total energy in the circuit is then 2*¹/₂C(0.5V)² = ¹/₄CV² i.e. 50% has disappeared.

Where has the energy gone?

I think the 50% energy can be thought of as the energy required to establish the electric field within the dielectric.

Er, that's the mechanism of capacitance! Which is already included in the equation.

Ehm, but then plain (ideal) LC circuit will oscillate forever, won't it? So 'energy required to establish the electric field within the dielectric' must be recoverable and is seems to be missing from above equation?

[Kalvin]

Does this suggest that the capacitor-based charge pumps may actually be only 50% efficient at maximum, although the datasheets claim over 90% efficiency?

[Jay_Diddy_B]

Hi,
There are ways of efficiently charging and discharging capacitors. Here is one way using a resonant circuit and a diode:



Here are the results:



You will see that at some point there is 1/2 the voltage on each capacitor,  2x 1/2C (V/2)²  = 1/2 of 1/2 CV², but energy is also stored in the inductor 1/2LI²

And:



I have attached the LTspice model

Note: 1 is one Henry and 1 is 1 Farad. If you write 1F you get 1 femto Farad which is really small.

Regards,

Jay_Diddy_B

[max_torque]

To quote our illustrious leader, this thread is "really rather fascinating!"


And i thought i knew how capacitors worked!   These paradoxes of science, where the maths/formulas show an effect, that can be measured in practical circuits to be happening, but that are somewhat illogical or unexpected when considered directly, just goes to show the elegance of the maths and how we can use it to define and explain our world!

[ruffy91]

While charging the capacitor it has an impedance other than 0, even in an ideal circuit. Half the energy is lost in this impedance as heat.

[Alex30]

So we can make a system more efficient by only using 25% of the capacity of the capacitor? Ie oversize the cap so that it charges to 75% and then we only use the rest of the capacity of the capacitor in our circuit?

[Marco]

Why not 0*0*infinite = 10*CV^2 ? Power lost in resistance equal to zero will be zero.

No, power in a resistance of 0 with infinite current is undefined. I was being facetious, you have to take the asymptotic limit as ESR goes to 0 ...

[ruffy91]

If you use a series resistor of infinite you get a current of 0A and the efficiency will go to 100% (no energy lost if you don't transfer the energy), inbetween 0 and infinite it is 50% and with 0 it is actually not defined (division by zero).

[IanB]

In thermodynamics we would call this an irreversible process. Irreversible processes are associated with an increase in entropy of the system, and increases in entropy are associated with the conversion of other forms of energy into heat.

Indeed, but that sheds no light on where the energy goes.

I would say it does. If we have an irreversible process then the system is going to dissipate some of its energy as heat to the surroundings. These two things go hand in hand. Since connecting a charged capacitor to an uncharged capacitor is clearly an irreversible process we can know with certainty that the system will dissipate some of its energy.

[ruffy91]

Typically the energy will dissipate in the spectrum somewhere between DC and light (up to gamma for very high energies). The energy dissipates as electromagnetic waves, as you have changing electric and magnetic field.

[tggzzz]

In thermodynamics we would call this an irreversible process. Irreversible processes are associated with an increase in entropy of the system, and increases in entropy are associated with the conversion of other forms of energy into heat.

Indeed, but that sheds no light on where the energy goes.

I would say it does. If we have an irreversible process then the system is going to dissipate some of its energy as heat to the surroundings. These two things go hand in hand. Since connecting a charged capacitor to an uncharged capacitor is clearly an irreversible process we can know with certainty that the system will dissipate some of its energy.

Yeah, well, everything ends up as heat, so that's a boring trivial statement - especially since it doesn't indicate where the heat appears. In this ideal case, with only lossless components, what is the route by which it ends up as heat and where is the heat finally found? To concentrate the mind and without loss of generality, assume the components are in a vacuum.

The useful interesting correct answer for the ideal case (i.e. this classic conundrum) has been pointed to The Electrician, as I previously indicated.

[tggzzz]

Typically the energy will dissipate in the spectrum somewhere between DC and light (up to gamma for very high energies). The energy dissipates as electromagnetic waves, as you have changing electric and magnetic field.

Yes, except that the EM waves don't dissipate the energy, they transmit it to some body in which it is dissipated as heat.

[Kalvin]

Taking two 100uF capacitors. The other is completely discharged and the other contains a charge giving terminal voltage of 3V. Connecting those two capacitors in parallel doesn't end up with the terminal voltage of 1.5V?

[Jay_Diddy_B]

Hi,
If you connect two capacitors in parallel one charged, the other discharged, half the energy is lost if the mechanism is resistive.

The energy can be preserved if an ideal diode and inductor is used. It can also be preserved if a switching power supply is used, In this model I have a SEPIC supply with an input capacitor charged to 19.2 Volts. I have no load on the output. I have arranged the feedback divider to stop charging the output capacitor when the voltage reaches 13.2, with this circuit the voltages on the input capacitor and capacitor are equal. 92.4% of the energy was preserved. Power was lost in the IC, the diode and the MOSFET.




Regards,

Jay_Diddy_B

[IanB]

I would say it does. If we have an irreversible process then the system is going to dissipate some of its energy as heat to the surroundings. These two things go hand in hand. Since connecting a charged capacitor to an uncharged capacitor is clearly an irreversible process we can know with certainty that the system will dissipate some of its energy.

Yeah, well, everything ends up as heat, so that's a boring trivial statement - especially since it doesn't indicate where the heat appears. In this ideal case, with only lossless components, what is the route by which it ends up as heat and where is the heat finally found? To concentrate the mind and without loss of generality, assume the components are in a vacuum.

The useful interesting correct answer for the ideal case (i.e. this classic conundrum) has been pointed to The Electrician, as I previously indicated.

What is interesting for one person may be different from what is interesting to another.

I find it interesting that we don't need to know¹ where the energy goes or how it gets there. Without knowing, we can still be certain from analysis that if an irreversible process is allowed to reach a settled, final state then some useful energy has been lost from the system, and what is more we can calculate how much energy that is.

It is supremely interesting that thermodynamics can tell us this, without us needing to know how or where that energy disappeared.

The calculations at the top of the thread do not need to be performed. The result is guaranteed without having to go through that algebra. It is fascinating to me from a physical sciences perspective that the web is full similar derivations, yet rarely is it pointed out that it is unnecessary to go through the rigmarole.

¹ Obviously as system designers and engineers we sometimes do need to know, to allow for the effect this may have on the working of our system.

[IanB]

The energy can be preserved if an ideal diode and inductor is used. It can also be preserved if a switching power supply is used, In this model I have a SEPIC supply with an input capacitor charged to 19.2 Volts. I have no load on the output. I have arranged the feedback divider to stop charging the output capacitor when the voltage reaches 13.2, with this circuit the voltages on the input capacitor and capacitor are equal. 92.4% of the energy was preserved. Power was lost in the IC, the diode and the MOSFET.

What happened with regard to conservation of charge in this system? Is charge conserved?

[MrAl]

Hello,

The basic question here is an old one that comes up now and then on forums like this one.

The original assertion is true, that the energy lost in the resistor is the same as that which gets transferred into the capacitor.  This is part of what limits the top end frequency of computer CPU's due to the constant switching.

Also true is that for any capacitor starting voltage other than zero, then additional energy transferred to the cap is the same as that lost in the resistor, and it does not matter how close the initial capacitor voltage is to the final voltage.  To total energy in the cap is not lost here, it's just the additional energy transferred.  So we always lose as much energy as that which gets transferred to the cap.

This is of course for a voltage source, resistor, and capacitor, with no other parts, and ignoring any kind of radiation or other loss.

This question is similar to the one about using pure PWM with an LED in a flashlight.  The initial thought is that PWM by itself somehow raises the efficiency of the flashlight.  That's not true however because just like the cap, there is always some series resistance.  The only way to boost the efficiency is with an inductor.

[albert22]

As EEs we apply Kirchhoff's laws to solve circuits. Which do not apply in every case. Really what works in every case are Maxwell equations which are a nightmare to solve even in simple circuits. KVL and KCL are simplifications of MEs when certain conditions are met: total electric charge is constant, no linkage of magnetic flux, low frequency compared to wire length (no radiated energy). We must comply with the lumped element model.
The "ideal" components that we use in our circuit analysis are defined in the LEM.
When we try to solve circuits with zero resistance, ideal capacitors and ideal voltage sources. Strange things occurs. The capacitor gets charged

by infinite current in zero time. Yet the product of an infinite current by a zero time gives a finite amount (the charge). When two capacitors are connected together, a finite charge is transfered in zero time.
When we introduce finite resistance things get more manageable, equations dont get weird and we are much more near to what we see in practice.
My point is that we need to be very carefull when we use a model and push it to mathematical limits of zero or infinity. Otherwise we may violate the conditions that made valid that model.
In the context of Kirchoff laws and ideal circuit elements (RLC), the only way that energy is lost is by heat in a resistor. There is no EM waves, sparks or any other means. To take into account this effects we need to apply Maxwell equations or add parasitic components to our analysis.
For example, the analysis of an antenna or a transmision line.

In the case of the two capacitors I like more the following explanations which do not involve radiation. (The 2nd link explains why not)
http://arxiv.org/pdf/0910.5279.pdf
http://arxiv.org/pdf/1309.5034.pdf

This thread is very interesting. Hope that I reached two cents

[Kalvin]

Taking two 100uF capacitors. The other is completely discharged and the other contains a charge giving terminal voltage of 3V. Connecting those two capacitors in parallel doesn't end up with the terminal voltage of 1.5V?

Confirmed. My capacitors don't lose half of the energy due to resistance. Maybe these chinese capacitors are better than those used in the calculations?

[MrAl]

Hi,

Dont forget that 1/2 the voltage across a capacitor means 1/4 of the energy.
Thus two caps with 1/2 voltage means they now have 1/4 of the total energy they had before.
(1/2)*C*V^2
Two caps with 1/4 the energy sums to 1/2 the energy.

[Kalvin]

That is correct. However, the energy loss is not due to any resistance what so ever.

[tggzzz]

This is part of what limits the top end frequency of computer CPU's due to the constant switching.

For today's very fine geometry CPUs, leakage current is highly significant, IIRC 1/3 or 2/3 (which?) of the total dissipation.

[tggzzz]

I would say it does. If we have an irreversible process then the system is going to dissipate some of its energy as heat to the surroundings. These two things go hand in hand. Since connecting a charged capacitor to an uncharged capacitor is clearly an irreversible process we can know with certainty that the system will dissipate some of its energy.

Yeah, well, everything ends up as heat, so that's a boring trivial statement - especially since it doesn't indicate where the heat appears. In this ideal case, with only lossless components, what is the route by which it ends up as heat and where is the heat finally found? To concentrate the mind and without loss of generality, assume the components are in a vacuum.

The useful interesting correct answer for the ideal case (i.e. this classic conundrum) has been pointed to The Electrician, as I previously indicated.

What is interesting for one person may be different from what is interesting to another.

I find it interesting that we don't need to know¹ where the energy goes or how it gets there. Without knowing, we can still be certain from analysis that if an irreversible process is allowed to reach a settled, final state then some useful energy has been lost from the system, and what is more we can calculate how much energy that is.

It is supremely interesting that thermodynamics can tell us this, without us needing to know how or where that energy disappeared.

The calculations at the top of the thread do not need to be performed. The result is guaranteed without having to go through that algebra. It is fascinating to me from a physical sciences perspective that the web is full similar derivations, yet rarely is it pointed out that it is unnecessary to go through the rigmarole.

¹ Obviously as system designers and engineers we sometimes do need to know, to allow for the effect this may have on the working of our system.

That's too close to "Daddy, why does X happen?". "It just does, OK".

[suicidaleggroll]

Taking two 100uF capacitors. The other is completely discharged and the other contains a charge giving terminal voltage of 3V. Connecting those two capacitors in parallel doesn't end up with the terminal voltage of 1.5V?

Confirmed. My capacitors don't lose half of the energy due to resistance. Maybe these chinese capacitors are better than those used in the calculations?

That is correct. However, the energy loss is not due to any resistance what so ever.

So what was the initial voltage and final voltage of both capacitors, and where do you think the energy went?

[Kalvin]

Taking two 100uF capacitors. The other is completely discharged and the other contains a charge giving terminal voltage of 3V. Connecting those two capacitors in parallel doesn't end up with the terminal voltage of 1.5V?

Confirmed. My capacitors don't lose half of the energy due to resistance. Maybe these chinese capacitors are better than those used in the calculations?

That is correct. However, the energy loss is not due to any resistance what so ever.

So what was the initial voltage and final voltage of both capacitors, and where do you think the energy went?

The capacitors C1 and C2 are both 100uF. Initially, the voltage of the C1 is 0V the voltage of the C2 is 3V. Then we connect the capacitors in parallel, and the resulting voltage will be V1 = V2 = 1.5V. The energy of the C2 is only 1/4 compared to its energy in the beginning, and the combined energy of the capacitors is only half compared to the beginning. Where the energy went? That is a good question. Did it go into heating the resistor? Or was the energy required to move the charge from one capacitor to another until their charges were equal and their voltages were equal, too.

[nctnico]

All you need to do is to step back and read that sentence of yours and think, big picture, what it means. It means that regardless of your component parameters, such a system is always 50% efficient.

Just think about that for sanity's sake and you will realize that something is terribly wrong here.

Perhaps you can point to the mistake?

Energy cannot be lost. Work out the energy balance of the system (IOW: point out where the loss goes).

[suicidaleggroll]

Taking two 100uF capacitors. The other is completely discharged and the other contains a charge giving terminal voltage of 3V. Connecting those two capacitors in parallel doesn't end up with the terminal voltage of 1.5V?

Confirmed. My capacitors don't lose half of the energy due to resistance. Maybe these chinese capacitors are better than those used in the calculations?

That is correct. However, the energy loss is not due to any resistance what so ever.

So what was the initial voltage and final voltage of both capacitors, and where do you think the energy went?

The capacitors C1 and C2 are both 100uF. Initially, the voltage of the C1 is 0V the voltage of the C2 is 3V. Then we connect the capacitors in parallel, and the resulting voltage will be V1 = V2 = 1.5V. Where the energy went? That is a good question. Did it go into heating the resistor? Or was the energy required to move the charge from one capacitor to another until their charges were equal and their voltages were equal, too.

It went into heating both capacitors via ESR, as well as a small amount of heat in the wire connecting them (depending on the ratio of the resistance of the wire and the capacitors' ESR.

[IanB]

All you need to do is to step back and read that sentence of yours and think, big picture, what it means. It means that regardless of your component parameters, such a system is always 50% efficient.

Just think about that for sanity's sake and you will realize that something is terribly wrong here.

Perhaps you can point to the mistake?

Energy cannot be lost. Work out the energy balance of the system (IOW: point out where the loss goes).

Who are you addressing this to?

[nctnico]

Have a look at the mechanical analogy given starting at this post in the thread I referenced:

There's an even better mechanical analogy. Imagine a 1 kg sticky ball moving at 10 m/s colliding with a second, stationary, sticky ball.

The first sticky ball has a momentum of 1 x 10 = 10 kg m/s and a kinetic energy of 0.5 x 1 x 10² = 50 J.

Upon collision the sticky balls remain stuck together and continue as one.

After the collision momentum is conserved, so the velocity of the combined balls is 10 / 2 = 5 m/s. Therefore the kinetic energy of the combined balls is 0.5 x 2 x 5² = 25 J.

It appears that 25 J of energy has been lost. Where did it go?

It can't be gone and it doesn't:
http://en.wikipedia.org/wiki/Elastic_collision

[IanB]

Have a look at the mechanical analogy given starting at this post in the thread I referenced:

There's an even better mechanical analogy. Imagine a 1 kg sticky ball moving at 10 m/s colliding with a second, stationary, sticky ball.

The first sticky ball has a momentum of 1 x 10 = 10 kg m/s and a kinetic energy of 0.5 x 1 x 10² = 50 J.

Upon collision the sticky balls remain stuck together and continue as one.

After the collision momentum is conserved, so the velocity of the combined balls is 10 / 2 = 5 m/s. Therefore the kinetic energy of the combined balls is 0.5 x 2 x 5² = 25 J.

It appears that 25 J of energy has been lost. Where did it go?

It can't be gone and it doesn't:
http://en.wikipedia.org/wiki/Elastic_collision

Why do you refer to an elastic collision when I described an inelastic collision in the example?

It may help you to read the thread from the beginning. You are going back over a lot of old ground here.

[suicidaleggroll]

Have a look at the mechanical analogy given starting at this post in the thread I referenced:

There's an even better mechanical analogy. Imagine a 1 kg sticky ball moving at 10 m/s colliding with a second, stationary, sticky ball.

The first sticky ball has a momentum of 1 x 10 = 10 kg m/s and a kinetic energy of 0.5 x 1 x 10² = 50 J.

Upon collision the sticky balls remain stuck together and continue as one.

After the collision momentum is conserved, so the velocity of the combined balls is 10 / 2 = 5 m/s. Therefore the kinetic energy of the combined balls is 0.5 x 2 x 5² = 25 J.

It appears that 25 J of energy has been lost. Where did it go?

It went into the plastic deformation of the balls that enabled an inelastic collision to take place, which ultimately means heat.  Same with a car crash...a car driving at 30 mph hits a stopped car of the same mass in a perfectly inelastic collision.  The pair continue at 15 mph due to conservation of momentum, and the extra energy goes into the deformation of the body/frame of the cars.  If no deformation took place, it would be an elastic collision, and the first ball/car would stop while the second ball/car continued at the same velocity as the first (think billiard balls).

[nctnico]

Have a look at the mechanical analogy given starting at this post in the thread I referenced:

There's an even better mechanical analogy. Imagine a 1 kg sticky ball moving at 10 m/s colliding with a second, stationary, sticky ball.

The first sticky ball has a momentum of 1 x 10 = 10 kg m/s and a kinetic energy of 0.5 x 1 x 10² = 50 J.

Upon collision the sticky balls remain stuck together and continue as one.

After the collision momentum is conserved, so the velocity of the combined balls is 10 / 2 = 5 m/s. Therefore the kinetic energy of the combined balls is 0.5 x 2 x 5² = 25 J.

It appears that 25 J of energy has been lost. Where did it go?

It can't be gone and it doesn't:
http://en.wikipedia.org/wiki/Elastic_collision

Why do you refer to an elastic collision when I described an inelastic collision in the example?
It may help you to read the thread from the beginning. You are going back over a lot of old ground here.

In either case it wouldn't be a surprise where the energy went. Like in my post before: it would be nice to point out where the energy actually goes instead of stating 'it's just gone'.

[Kalvin]

Taking two 100uF capacitors. The other is completely discharged and the other contains a charge giving terminal voltage of 3V. Connecting those two capacitors in parallel doesn't end up with the terminal voltage of 1.5V?

Confirmed. My capacitors don't lose half of the energy due to resistance. Maybe these chinese capacitors are better than those used in the calculations?

This is embarrassing, I made a mistake  The capacitors do lose energy, although the total charge is preserved. Sorry about that. However, the energy will be lost even if the capacitors are ideal and no resistance is in the circuitry. The energy was needed to move the charge to build up the electrical field until the voltages of the capacitors were equal.

[parbro]

Some thoughts....

Within a capacitor are 2 forces. An attractive force between the plates. And a repulsive force within each plate. If you connect a load to a charged capacitor the energy expended is due to the repulsive forces within the plates. The electric field between the plates is responsible for the 1/2 power term and represents unrecoverable energy.
Say you have a capacitor charged to some voltage V and connect it to a voltage source. You must expend a sufficient amount of energy W just to maintain voltage V on the capacitor. If you want to increase capacitor charge from V to 2*V you must expend 2*W.
I think converter efficiency is a measure of the parasitic losses and doesn't refer to the underlying physics involved.

[suicidaleggroll]

It's just because of the ESR (or the resistor you purposefully stick there, if you use one)...there's no reason or need to make it more complicated than that.  You have the exact same effect if you try to charge a battery with a CV source through a series resistor too!  Use a CC charger and the problem all but disappears (at that point the efficiency is just a function of the cap's ESR and the charge rate, same as a battery).

Things only get "complicated" when you start introducing theoretical components with zero ESR, supplies with infinite current capability and zero ESR, connected by wires with zero resistance, etc., which has no basis in reality and would cause just as much "0/0*infinity" confusion as trying to simulate the impulse response of any reactive system with the same unrealistic conditions imposed on it.

[Jay_Diddy_B]

Hi group,
I am going to immortalize IanB's stick balls with a LTspice model. I suppose these are 'Spicy Sticky Balls' 



I have modelled a 1 Farad capacitor charged to 1V, charged with 0.5 Joule.

I close the switch, I used a 10u Ohm switch. I did this with two values of limiting resistors 1m and 10m Ohms.



The results show the predicted results 50% of the energy is left in the combined capacitors and 50% is dissipated as heat in the resistor. No energy is lost or unaccounted for.

Regards,

Jay_Diddy_B

[MrAl]

That is correct. However, the energy loss is not due to any resistance what so ever.

Hello there,

But what you are suggesting is not real because there is ALWAYS some resistance no matter how small, and  considering that i had qualified my previous statement about not having any other losses.
We must first allow other losses to exist before we can think otherwise.  For example, if we allow radiation then we can say that it becomes an antenna.

[MrAl]

This is part of what limits the top end frequency of computer CPU's due to the constant switching.

For today's very fine geometry CPUs, leakage current is highly significant, IIRC 1/3 or 2/3 (which?) of the total dissipation.

Hello,

Well leakage current will add to the dissipation at any frequency.  Capacitance causes an increase in dissipation as frequency increases.  Since this thread is about charging capacitances, i use that as an example of how this concept affects us in everyday life.

[tggzzz]

This is part of what limits the top end frequency of computer CPU's due to the constant switching.

For today's very fine geometry CPUs, leakage current is highly significant, IIRC 1/3 or 2/3 (which?) of the total dissipation.

Well leakage current will add to the dissipation at any frequency.  Capacitance causes an increase in dissipation as frequency increases.  Since this thread is about charging capacitances, i use that as an example of how this concept affects us in everyday life.

Of course, as implied by my statement.

But it is the proportion attributed to leakage current that is relevant. As far as I can tell with a quick google, leakage is about 1/3 the total dissipation, rising with increasing temperature.

[MrAl]

This is part of what limits the top end frequency of computer CPU's due to the constant switching.

For today's very fine geometry CPUs, leakage current is highly significant, IIRC 1/3 or 2/3 (which?) of the total dissipation.

Well leakage current will add to the dissipation at any frequency.  Capacitance causes an increase in dissipation as frequency increases.  Since this thread is about charging capacitances, i use that as an example of how this concept affects us in everyday life.

Of course, as implied by my statement.

But it is the proportion attributed to leakage current that is relevant. As far as I can tell with a quick google, leakage is about 1/3 the total dissipation, rising with increasing temperature.

Hi,

Well i am not sure what you are trying to say here.  We are talking about capacitance and loss DUE TO THAT capacitance, so i mentioned in passing:
"This is part of what limits the top end frequency of computer CPU's due to the constant switching"

So who cares about leakage current when we are studying the effects of power loss due to the charge/discharge mechanism of capacitance in switching circuits?

[dannyf]

We are talking about capacitance and loss DUE TO THAT capacitance

You cannot lose energy into an ideal capacitor (or an ideal inductor).

That's fundamentally the issue some folks are having a very difficult time comprehending and explaining when R -> 0, or R -> infinity.

[Siwastaja]

This phenomenon is actually very intuitive, and most people would understand and agree on this, it's just about how to express things without overcomplicating them. Don't go too theoretical unless you really can do that math thing...

A fixed voltage source cannot do voltage conversion at high efficiency, we all know this. An empty capacitor starts at 0 volts. If you connect a 3 volt voltage supply (say, two AA batteries, or a LM317 regulator) to an empty capacitor, you will have a rising voltage curve. And as we know, if you short your voltage supply to near 0 volts, it has horrible efficiency; there is no real power conversion done, the excess voltage is only simply dissipated and energy is lost. If you charge a capacitor this way, the average voltage of the charging step will be 1.5 volts, which means the average voltage drop will also be 1.5 volts, and in a linear regulator, this is all losses. Two AA batteries in series behaves the same; if real output voltage is lower than the open-circuit voltage, there can only be losses. The fix is to use a switch mode power supply. We all know this: the idea of a switch mode PSU is to convert energy to any voltage, whereas a linear supply or battery converts energy to one fixed voltage and dissipates any excess in circuit resistances. In the battery case, it's the chemical energy being converted to 1.5 volts.

Of course the resistance doesn't matter - it only defines how quickly the capacitor is charged, but the average voltage drop, defining the energy loss, will still be 1.5 volts. This only gets difficult when you start assuming unreal, "ideal" components with zero resistance but don't have guts or mathematical skills to go all the way there.

In real life, you encounter exactly this when calculating a snubber; although the snubber loss happens in the resistor, only the capacitor value defines the amount of loss - the average power - (as long as the resistor is small enough so that the capacitor gets fully charged/discharged on every cycle). The resistor only defines how quickly the energy is dissipated.

Battery charging has the same problem, but fortunately, compared to the capacitor, the voltage curve is more like a horizontal line, so the amount of loss is much less. For example, if you charge a lead-acid battery from 11V (empty) to 14V (full), using a fixed-voltage 14V supply and a current-limiting resistor, your average voltage drop is only 1.5V and you have relatively manageable efficiency, which is why to this day simple car battery chargers use transformer, bridge rectifier and a limiting resistor (can be constructed in the transformer, using thinner wire), as it is a simple construction. Using this structure for charging capacitors would always result in efficiency below 50% upper limit.

[dannyf]

A fixed voltage source cannot do voltage conversion at high efficiency,

What is "voltage conversion"?

which is why to this day simple car battery chargers use transformer, bridge rectifier and a limiting resistor (can be constructed in the transformer, using thinner wire), as it is a simple construction.

Depending on how you define "simple": most car battery chargers today are smps.

Using this structure for charging capacitors would always result in efficiency below 50% upper limit.

That 50% efficiency number is the upper limit, lower limit, and middle limit as well.

[Siwastaja]

You cannot lose energy into an ideal capacitor (or an ideal inductor).

That's fundamentally the issue some folks are having a very difficult time comprehending and explaining when R -> 0, or R -> infinity.

You need to think about limit (the math thing). The energy lost is constant; the power goes up when R -> 0, but the time it takes goes ->0. So, at R = 0, you have infinite power loss for zero time. What is the energy loss then? I'm not a math guru, but I'd say that is something called "undefined". Sorry. So, sometimes math gives you an answer: "Can't calculate". The "limit" thing is there for you. But for understanding the physics (and to be able to do simple numerical calculation, which is what SPICE simulation is all about, for example), it is easiest to assume some resistance, and you will find it doesn't matter whether it's 0.00000001 ohms or 1000 ohms.

[dannyf]

What is the energy loss then?

the energy loss is 50%, regardless of the resistance in the circuit.

So as R approaches 0 from the right, or infinity from the left, the energy loss is always 50%.

It creates two issues:

1) as R approaches 0, the circuit approaches an ideal capacitor, which cannot lose energy. So where did that other 50% energy go?

2) as R approaches infinity, the circuit approaches a constant current charger, which everyone has asserted to be infinitely inefficient. But it should have a 50% efficiency figure. So how do you reconcile the two?

[Siwastaja]

What is "voltage conversion"?

I mean converting energy from one voltage level to another voltage level. You can do it by dissipating excess voltage (using a resistor, a diode drop, linear regulator, etc.), or using a switch mode converter. Both can be very good when the load wants that fixed voltage, but, unfortunately, the charging capacitor as a load "wants" all voltage levels starting from 0. That's the whole point people seem to easily miss.

Depending on how you define "simple": most car battery chargers today are smps.

Indeed; this has happened during the last 10-20 years. They figured out that iron and copper is expensive and wanted to get rid off that 50 or 60 Hz transformer. By today's standards, SMPS is simple as hell. But maybe saying "to this day" was a bit off, of course they are all SMPS now and has been for some time. (I'm starting to feel a bit old.)

[suicidaleggroll]

1) as R approaches 0, the circuit approaches an ideal capacitor, which cannot lose energy. So where did that other 50% energy go?

A pointless question.  Here's one for you.  You take an ideal 1uF capacitor with 0 ESR, charge it up to 3v, and then short the two leads to each other using a wire with zero impedance.  Where does the power go?

2) as R approaches infinity, the circuit approaches a constant current charger, which everyone has asserted to be infinitely inefficient. But it should have a 50% efficiency figure. So how do you reconcile the two?

A CV source with infinite output resistance is not a CC source...

[dannyf]

the charging capacitor as a load "wants" all voltage levels starting from 0.

I don't quite understand that one. A capacitor isn't a being so I don't where its "desires" or "wants" come from.

[Kalvin]

That is correct. However, the energy loss is not due to any resistance what so ever.

Hello there,

But what you are suggesting is not real because there is ALWAYS some resistance no matter how small, and  considering that i had qualified my previous statement about not having any other losses.
We must first allow other losses to exist before we can think otherwise.  For example, if we allow radiation then we can say that it becomes an antenna.

I do agree and understand what you and others are saying.

What really got me interested in to this problem was the 50% energy loss was independent of the resistor. So, if the 50% energy is lost no matter size resistor we put there, maybe there is something else behind this phenomenon.

The circuit analysis operate on voltages and currents, and we need to model this energy transfer problem in circuit analysis domain. Of course, using the circuit analysis, you will get absurdly high current etc. but the analysis will produce expected answer ie. it seems that 50% of energy is lost into resistor independently of the resistor value.

Ok, I went into physics 101 dealing with capacitors. The capacitance, charge and the voltage are related as follows:

Eq 1:


It is also said that work is needed to move charge from one capacitor plate to another, which will result higher electrical potential between the plates.

Eq 2:


The energy stored in the capacitor is the same as the work required to charge the capacitor:

Eq3:


Now, lets take two identical 100uF capacitors C1 and C2. The C1 and C2 has following initial terminal voltages:

V1 = 3V
V2 = 0V.

Thus, we can calculate the initial charge of those two capacitors Q=V*C:

Q1 = 300uC
Q2 = 0C.

We can also calculate the energies stored in those capacitors W=Q*V/2:

W1 = 450uJ
W2 = 0J.

Total charge in the system is Q = Q1 + Q2 = 300uC. Total system energy is W = W1 + W2 = 450uJ.

Now, let's start moving some charge from the C1 to C2 until the capacitors contain equal electrical potential ie. V1 = V2. In this case, the resulting voltage will be 1.5V:

V1 = 1.5V
V2 = 1.5V

The resulting charges will be equal as the both capacitors are 100uF:

Q1 = 150uC
Q2 = 150uC

Thus, the resulting energies will also be equal:

W1 = 112.5uJ
W2 = 112.5uJ

And the total energy in the system is W1 + W2 = 225uJ.

Yes, 50% of the system's energy has been lost while moving some charge from one capacitor to another. And look, no resistors here.

Of course, one might argue that in practice wires are required to move the charge from C1 to C2, and the wires will definitely contain resistance, thus the missing energy must have been lost in this resistance. This has also been confirmed with the circuit analysis and circuit simulator in this thread. However, we needed to introduce this [artificial] resistance in order to be able to apply the kirchhoff's equations for the circuit analysis, yet we only wanted to move some charge from one capacitor to another.

[IanB]

@Kalvin:

Your analysis is sound, and has touched on thermodynamics, which gets to the heart of the problem.

The key to it is when you used the symbol W and described it as the work required to charge the capacitor.

In thermodynamics, energy flows can be divided into two categories: work (usually given the symbol W), and heat, (usually given the symbol Q -- not to be confused with electric charge). To grossly oversimplify a complex subject, work is 100% useful and heat is only partially useful. Any time heat flows, entropy increases and some waste of energy occurs.

If you do an analysis and the work you can account for is less than the total energy involved, then the system has an efficiency less than 100%, some wasted energy occurs, heat is produced, and the temperature goes up somewhere.

In the capacitor circuit you do not actually need the resistance to turn some energy into heat. Even without the resistance some other mechanism will occur that produces the same result. Nature will always find a way. For example, the system may act like an antenna and radiate the energy away as EM radiation.

[albert22]

...  You take an ideal 1uF capacitor with 0 ESR, charge it up to 3v, and then short the two leads to each other using a wire with zero impedance.  Where does the power go?

Good question. What is the explanation for that?
Neither the capacitor or the wire can dissipate the energy.
It seems that the model of the ideal components cannot explain this.

[IanB]

If you do an analysis and the work you can account for is less than the total energy involved, then the system has an efficiency less than 100%, some wasted energy occurs, heat is produced, and the temperature goes up somewhere.

For example, consider a switched mode converter like a boost converter or a buck converter.

The power source does work on the input terminals of the converter. The output terminals of the converter do work on the load. If the work done by the output terminals is less than the work done on the input terminals, then the efficiency is less than 100% and the difference between input work and output work is the energy lost, dissipated as heat. The converter will get warm and suitable cooling arrangements will need to be provided.

[Marco]

Good question. What is the explanation for that?

I know it pisses off mathematicians ... but I think it really helps if you just tell people that 0*infinity can be a number.

[albert22]

Already told that on my previous post.

I still don find the answer using the ideal capacitor and the ideal conductor.
My point is that we are using models of ideal components and Circuit laws that represent just a tiny part of what happens in reality. And if we need more understanding we need to apply more precise models that include, radiation, conductivity,.... and perhaps for certain problems quantum mechanics.

[Galaxyrise]

2) as R approaches infinity, the circuit approaches a constant current charger, which everyone has asserted to be infinitely inefficient. But it should have a 50% efficiency figure. So how do you reconcile the two?

Trying again...

The assertion is that charging a capacitor from a constant current loses little energy outside of the current source.  You must then compute the energy losses inside the current source to get system efficiency.

While we can use a constant voltage and a varying resistance to control current, and such a thing is often called a constant current source, for the purposes of this discussion it's still a constant voltage source and a resistance.  The losses will still add up to >=50%.  This is the kind of current source your proposal "approaches", so it reconciles just fine.

There are other ways to produce a constant current, and they must be considered on their own merits.

[dannyf]

0*infinity can be a number.

It can be a number (inclusive of 0) and infinity, all depending on how big that "infinity" is.

Infinity divided by infinity can be a number (inclusive of 0) and infinity, all depending on which of the infinity is bigger.

[Zero999]

It's true: charge a capacitor from a constant voltage source and only half of the energy is transferred to the capacitor, the rest is dissipated in the resistance of the wire and the internal resistance of the capacitor. No energy is dissipated by the capacitance itself. It's the resistive element (however small or large it may be) which dissipates half of the the energy.

50% is the maximum efficiency when charging a capacitor from a voltage source. A little more energy will be lost due the leakage current inside the capacitor.

The points made about zero resistance and an ideal capacitor are irrelevant to the discussion because we live in the real world where there will be some resistance in the circuit.

This is very intuitive and doesn't require any knowledge of calculus: I figured this out before I even went to college.

It's true a switched mode constant current source can charge a capacitor with a much higher efficiency than 50%.

[tggzzz]

The points made about zero resistance and an ideal capacitor are irrelevant to the discussion because we live in the real world where there will be some resistance in the circuit.

You have a limited concept of the real world...
All the wires are superconducting, and therfore really do have zero resistance.
The capacitor is simply two parallel supercondicting plates in a vacuum.

[Jay_Diddy_B]

Hi,

Assume we *DO* have zero resistance, but we have some inductance, the circuit will resonate. The current will continue to circulate forever:







The frequency will be determined by the capacitor in series and the inductance:

Freq = 1/2 x pi SQRT (L x C)  where C is the capacitors in series.

and the peak current

= Vinitial / SQRT(L/C)

The Q is infinite because there are no losses so the sinewave does not decay.

Now if L goes to zero the frequency goes to infinity and peak current goes to infinity.


Regards,

Jay_Diddy_B

[Marco]

You have a limited concept of the real world...
All the wires are superconducting, and therfore really do have zero resistance.
The capacitor is simply two parallel supercondicting plates in a vacuum.

The current will exceed the threshold at which they can maintain superconductivity, unless the inductance is high enough ... in which case the final energy will be 0 (it will oscillate and radiate until that point).

[Zero999]

The points made about zero resistance and an ideal capacitor are irrelevant to the discussion because we live in the real world where there will be some resistance in the circuit.

You have a limited concept of the real world...
All the wires are superconducting, and therfore really do have zero resistance.
The capacitor is simply two parallel supercondicting plates in a vacuum.

Except for the fact that the metals will stop being superconductors once the critical current density has been exceeded, which will happen at some point.

Hi,

Assume we *DO* have zero resistance, but we have some inductance, the circuit will resonate. The current will continue to circulate forever

Except the current will not continue to resonate forever, even if the LC circuit is superconducting, there will be electromagnetic radiation radiated at the resonant frequency.

The thread is about RC circuits though so the inductance is negligible i.e. it's over-damped.

[T3sl4co1l]

Relevant info, posted separately for reference; https://www.eevblog.com/forum/projects/reversible-and-irreversible-processes-(redux)/

Regarding superconductors, the best resonators (which can be represented as LC lumped circuits) have Qs on the order of 10^8.  It's high, but still dies out in seconds, let alone thermodynamic infinity.

Tim

[ludzinc]

Hi,

Assume we *DO* have zero resistance, but we have some inductance, the circuit will resonate. The current will continue to circulate forever:







The frequency will be determined by the capacitor in series and the inductance:

Freq = 1/2 x pi SQRT (L x C)  where C is the capacitors in series.

and the peak current

= Vinitial / SQRT(L/C)

The Q is infinite because there are no losses so the sinewave does not decay.

Now if L goes to zero the frequency goes to infinity and peak current goes to infinity.


Regards,

Jay_Diddy_B

Turns out, the lost energy is radiated, even in the case or zero series resistance. 

http://www.physics.princeton.edu/~mcdonald/examples/twocaps.pdf

[parbro]

1/2*Q*V is a statement about capacitance. To me it doesn't make sense to ascribe the 1/2 power term to ESR. In the two capacitor example the fact that they reach an equilibrium is a demonstration that one energy holds back the other. In order to add energy you have to overcome the existing energy. If ESR was involved the capacitor with the lowest ESR would suck all the energy. To me capacitors are not electrical sponges. To charge a capacitor you have to push the charge carriers kicking and screaming so to speak.

[Kalvin]

What will happen if we connect the two capacitors in parallel using a lossless transmission line? Let's say the transmission line has a characteristic impedance of 50 ohms. As the transmission line is lossless, there is no resistance and the transmission line will not radiate any energy either. However, as we connect the capacitor to the transmission line, the current will be limited by the 50 ohm impedance, and we do not have a problem of zero resistor anymore.

[Zero999]

1/2*Q*V is a statement about capacitance. To me it doesn't make sense to ascribe the 1/2 power term to ESR. In the two capacitor example the fact that they reach an equilibrium is a demonstration that one energy holds back the other. In order to add energy you have to overcome the existing energy. If ESR was involved the capacitor with the lowest ESR would suck all the energy. To me capacitors are not electrical sponges. To charge a capacitor you have to push the charge carriers kicking and screaming so to speak.

If doesn't matter which capacitor has the higher ESR, when the steady state condition is reached, the voltage across the capacitors will be the same and half of the energy transferred will be lost in the wiring resistance and the ESR of the capacitors.

[IanB]

What will happen if we connect the two capacitors in parallel using a lossless transmission line? Let's say the transmission line has a characteristic impedance of 50 ohms. As the transmission line is lossless, there is no resistance and the transmission line will not radiate any energy either. However, as we connect the capacitor to the transmission line, the current will be limited by the 50 ohm impedance, and we do not have a problem of zero resistor anymore.

See this thread: https://www.eevblog.com/forum/projects/reversible-and-irreversible-processes-(redux)/

T3sl4co1l has done that analysis there with a fascinating result.

[Jay_Diddy_B]

Hi,

If the Transmission line, line of impedance Z, is long compared to the time constant C x Z

That the propagation time is Tp >> C x Z

The capacitor will discharge with an exponential waveform with peak current = Vcharge / Z

The pulse will travel down the line charge the capacitor at the other end and then reverse.



Jay_Diddy_B

[Siwastaja]

the charging capacitor as a load "wants" all voltage levels starting from 0.

I don't quite understand that one. A capacitor isn't a being so I don't where its "desires" or "wants" come from.

A capacitor "wants" exactly what voltage its terminals are currently at, and you have to go through all voltages from 0 to your max. This is an important distinction compared to an ideal battery, or even a real battery, which is a voltage source/sink with constant voltage itself. This difference explains the whole 50% loss thing which happens when charging a capacitor with a CV source, but not when charging a battery with a CV source.

[GeorgeOfTheJungle]

What will happen if we connect the two capacitors in parallel using a lossless transmission line? Let's say the transmission line has a characteristic impedance of 50 ohms. As the transmission line is lossless, there is no resistance and the transmission line will not radiate any energy either. However, as we connect the capacitor to the transmission line, the current will be limited by the 50 ohm impedance, and we do not have a problem of zero resistor anymore.

See this thread: https://www.eevblog.com/forum/projects/reversible-and-irreversible-processes-(redux)/

T3sl4co1l has done that analysis there with a fascinating result.

From an energy point of view, this behaves very much like communicating vessels.

Take two vessels one full and one empty, open the tap, let the water flow and you'll end up with 1/2 the initial (potential) energy in the system: 2 * (1/2 the mass at 1/2 the height).

Where did the other half go?

To move water from vessel a to b requires work. There's a mass of water that has aquired kinetic energy. It was still but now it's moving. There are losses due to friction while flowing, and the water arriving at b keeps revolving for a while in the vessel until any remainig kinectic energy is totally gone due to friction then stops. All is gone as heat.

s/water/electrons/g, and s/vessel/capacitor/g, and there you have it?

Electrons have mass, have kinetic energy, and there are energy losses when they flow through a wire.

[IanB]

From an energy point of view, this behaves very much like communicating vessels.

For a minute there I thought you meant ships at sea sending messages to each other   

But this overall situation is going to occur over and over in physics, in any situation where work is expressed as a potential difference times a flow, or a force times a velocity. It becomes an inevitable consequence of applied mathematics, and hence becomes a unifying element across many different phenomena.

[GeorgeOfTheJungle]

From an energy point of view, this behaves very much like communicating vessels.

For a minute there I thought you meant ships at sea sending messages to each other   

Google translate "vasos comunicantes"

But this overall situation is going to occur over and over in physics, in any situation where work is expressed as a potential difference times a flow, or a force times a velocity. It becomes an inevitable consequence of applied mathematics, and hence becomes a unifying element across many different phenomena.

Other way of thinking about this is: how much energy does it take to fill vessel b (the capacitor)?

If vessel a had initially 2 kg of water, and 2 meters of height, to fill vessel b we have to lift 1kg of water 1 meter, right?

From an energy point of view that's an increment of ~10 joules (mgh, g~=10) of (potential) energy.

But how much energy do you really need to lift 1kg 1 meter? No, it's not 10 joules, it's always *more* than that, no matter what!

Say you try to lift with 1 kg of force (kp). What happens? Nothing... because that's it's weight, it stays where it is! (at "potential 0").

So you've got to apply more than a kg of force (kp). Say 2 kp during 1 meter to lift it rapidly. How much work is that? 2kp * 1m ~= 20 julios (1kp ~= 10N).

That's *twice* the potential energy the water has gained. Where's the rest?

The water is now moving upwards, it has acquired kinetic energy (?Ek). F= 2kp(lift force)-1kp(weight) = 1kp = m*a => a ~= 10m/s ; v2= 2a*(x-x0)+0 => ?Ek= 1/2*m*v2= 1/2*1*2a*1+0= 10 joules.

So the faster you pull to lift it to the desired potential, the more energy wasted.

[GeorgeOfTheJungle]

So the faster you pull to lift it to the desired potential, the more energy wasted.

It also makes sense because energy losses in a wire are proportional to I squared: transfer a Coulomb in a second and the losses will be 4 times more than in 2 seconds. So be green and to save the planet charge your caps gently

[The Electrician]

There are errors in your analysis of the case of lifting water in a gravitational field.  You can stop the lifting force before the water reaches its final destination, let gravity decelerate it; the kinetic energy is converted to potential energy.  It doesn't matter how fast you move something from point A to point B in a gravitational field; as long as there are no frictional losses the change in potential energy only depends where point A and point B are.

See: http://en.wikipedia.org/wiki/Conservative_force

"Gravity is an example of a conservative force, while friction is an example of a non-conservative force."

The I^2*R losses are frictional losses--not conservative.

So the faster you try to charge a capacitor through a resistance the greater the losses.  Not so for moving mass from one point to another in a gravitational field (without friction).

[T3sl4co1l]

The gravitational potential is the same as the potential of the capacitor.  It takes energy to increase the potential of a capacitor incrementally; indeed, the amount required is exactly proportional to voltage (dE/dV = CV).

Note that dE/dV is... simply the charge Q.

Another way to think of it, you're winding up a spring.  Every increment of displacement upon the spring requires increasingly higher force, and therefore the energy required for a given small displacement rises proportional to the total displacement.

They are all conservative forces; the main difference with gravity is, it's a doubly small change (displacement << radius of Earth), so the force doesn't change noticeably with displacement, and we write U = m g h for the local gravity field, rather than U = 1/2 k x^2 for the spring.

Tim

[MrAl]

That is correct. However, the energy loss is not due to any resistance what so ever.

Hello there,

But what you are suggesting is not real because there is ALWAYS some resistance no matter how small, and  considering that i had qualified my previous statement about not having any other losses.
We must first allow other losses to exist before we can think otherwise.  For example, if we allow radiation then we can say that it becomes an antenna.

I do agree and understand what you and others are saying.

What really got me interested in to this problem was the 50% energy loss was independent of the resistor. So, if the 50% energy is lost no matter size resistor we put there, maybe there is something else behind this phenomenon.

The circuit analysis operate on voltages and currents, and we need to model this energy transfer problem in circuit analysis domain. Of course, using the circuit analysis, you will get absurdly high current etc. but the analysis will produce expected answer ie. it seems that 50% of energy is lost into resistor independently of the resistor value.

Ok, I went into physics 101 dealing with capacitors. The capacitance, charge and the voltage are related as follows:

Eq 1:


It is also said that work is needed to move charge from one capacitor plate to another, which will result higher electrical potential between the plates.

Eq 2:


The energy stored in the capacitor is the same as the work required to charge the capacitor:

Eq3:


Now, lets take two identical 100uF capacitors C1 and C2. The C1 and C2 has following initial terminal voltages:

V1 = 3V
V2 = 0V.

Thus, we can calculate the initial charge of those two capacitors Q=V*C:

Q1 = 300uC
Q2 = 0C.

We can also calculate the energies stored in those capacitors W=Q*V/2:

W1 = 450uJ
W2 = 0J.

Total charge in the system is Q = Q1 + Q2 = 300uC. Total system energy is W = W1 + W2 = 450uJ.

Now, let's start moving some charge from the C1 to C2 until the capacitors contain equal electrical potential ie. V1 = V2. In this case, the resulting voltage will be 1.5V:

V1 = 1.5V
V2 = 1.5V

The resulting charges will be equal as the both capacitors are 100uF:

Q1 = 150uC
Q2 = 150uC

Thus, the resulting energies will also be equal:

W1 = 112.5uJ
W2 = 112.5uJ

And the total energy in the system is W1 + W2 = 225uJ.

Yes, 50% of the system's energy has been lost while moving some charge from one capacitor to another. And look, no resistors here.

Of course, one might argue that in practice wires are required to move the charge from C1 to C2, and the wires will definitely contain resistance, thus the missing energy must have been lost in this resistance. This has also been confirmed with the circuit analysis and circuit simulator in this thread. However, we needed to introduce this [artificial] resistance in order to be able to apply the kirchhoff's equations for the circuit analysis, yet we only wanted to move some charge from one capacitor to another.

Hello again,

I think i understand what you are looking for here, but one thing we have to make clear first.  That is, theory is a measuring tool which which we can use to measuring real life systems.  Some things in theory can not be done in the real world, because they would violate basic physical principles.  The best we can do sometimes is create an experiment that gets close to what we are looking for.  In addition to that, we have to first make sure we are using the right theory.

For the example of two caps, we can not assume zero resistance because not only will that not be possible, there will also be some inductance in series.  Also, if the resistance was low enough, even a superconductor would not hold up because superconductors have limits just like regular conductors although they may be different.

If we had two capacitors of any physical size, there has to be some resistance, but if we let that resistance be small, then we are going to see some decent radiation as well.  If we let that resistance approach zero then we would see an increase in radiation.  The radiation would have to be an impulse, at least in theory.  But again, i dont think there is any way to get zero resistance because there is nothing that can conduct with zero resistance when the current is infinitely high.
Also, because we live in a physical space, there is some distance between the caps, so there would be some inductance, and that inductance would act as a delay in the rise of the current wavefront.  That would mean that we might see the effects we see with an inductor that was placed in the circuit on purpose, which could mean a more effective transfer of energy.

So rather than model it as a resistor and capacitor and ask the question about zero resistance, i think we have to model it as an inductor with the resistor and capacitor, because there will be some inductance there too.  If we let the distance approach zero, then the two capacitors would have to become one and the same capacitor, with twice the plate areas, where the charge still has to move from one plate area to the other.  Thus, the very act of moving them together would change the total energy in the system.  We'd have to get down to the theory of how they would be moved together to figure out how the energy changed.  They may move together under the force of the charges, or they may move apart, but i think they would actually rotate into a minimum energy position.  When the charge starts to move however, energy would radiate, so we'd loose energy as they moved closer to each other.

But you can start to see, and i think you already understood this, that when we ask questions that are more complicated we have to resort to a more physical model.  We can not use models that are based on simpler assumptions intended to quickly answer simpler questions to answer much more complex questions.  We have to choose a model that fits the question.  We try to use the simplest model that will be possible for that particular problem.  But we have to make sure we use a model that includes all the effects we want to look at. For a super simple example, we dont have to use a resistor with a small cap in parallel to answer a simple question about how a resistor voltage divider works, even though there is always some small capacitance present, but when we get into higher frequency circuits we can not leave out the small parallel capacitance or we will never know how the circuit really works.

[IanB]

Tanks filling with water can be analyzed like capacitors filling with charge.

The work to move water is given by volume moved times pressure difference. We can say:

    W = V * DP

(In SI units we have: {J} = {m³} * {N/m²} )

Suppose we want to fill our tank to a level h from a fill pipe at the base of the tank.

When the tank is empty there is no level and no pressure (head) resisting the flow of water. So the first litre of water can be put in the tank for free. But as the tank fills up the existing level of water imposes a back pressure on the fill pipe and we have to overcome this pressure to get more water in the tank. When the tank is nearly full the whole height of water is pushing back and opposing the last litre of water we want to push in there.

If we integrate this system, we find in an analogous way to capacitors that the energy stored in a tank of water is given by the formula:

    E = ½VP²

where V is the volume of water and P is the pressure at the bottom of the tank.

The pressure at the bottom of the tank is a function of the level and is given by:

    P = (rho) gh

where the Greek letter rho is customarily used to denote density except this forum software doesn't support Greek letters 

[GeorgeOfTheJungle]

There are errors in your analysis of the case of lifting water in a gravitational field.  You can stop the lifting force before the water reaches its final destination, let gravity decelerate it; the kinetic energy is converted to potential energy.  It doesn't matter how fast you move something from point A to point B in a gravitational field; as long as there are no frictional losses the change in potential energy only depends where point A and point B are.

See: http://en.wikipedia.org/wiki/Conservative_force

"Gravity is an example of a conservative force, while friction is an example of a non-conservative force."

The I^2*R losses are frictional losses--not conservative.

So the faster you try to charge a capacitor through a resistance the greater the losses.  Not so for moving mass from one point to another in a gravitational field (without friction).

Well yes that's true, absolutely.

But if you push the water trough the pipe too much (high pressure) / for too long (as in the example) it will acquire more kinetic energy than needed (you just need 10j+friction losses to fill it) and the losses will then be even higher (and dissipated as heat through friction). That's what I meant to say but screwed it up a little bit

[GeorgeOfTheJungle]

?
where the Greek letter rho is customarily used to denote density except this forum software doesn't support Greek letters 

Nor html entities: &#x3C1; <https://mothereff.in/html-entities>

[GeorgeOfTheJungle]

Tanks filling with water can be analyzed like capacitors filling with charge.

And it's the best way to visualize it...

[CatalinaWOW]

Water in tanks is a good analogy for electrical circuits.  I use it a lot myself.  But as Mr. Al says, you have to make sure your model is appropriate to the problem.  Electrons in movement create fields that are modeled by Maxwell's equations.  Water in tanks does not create fields on the same scale (there are probably gravity waves generated but the forces, unlike electric and magnetic fields, are incredibly weak).  For this reason there is no meaningful mechanism in the water analogy for energy to leave the system.  But we know that in the electrical system, energy really does leave the local system.  We all listen to radio and watch TV.

[suicidaleggroll]

For this reason there is no meaningful mechanism in the water analogy for energy to leave the system.

Friction in the pipe connecting the tanks generates heat just like resistance in the wire connecting the two capacitors.

[CatalinaWOW]

I agree, friction is equivalent to electrical resistance.  The missing thing in the water system is an equivalent to electromagnetic field radiation.

Any horse can be ridden too hard.  Some fairly good analogies for magnetic circuits can be generated with either water and tank systems, or electrical circuits.  But these analogies break down with low permeability materials, and become very labored with non-linear magnetic materials. 

Again, this is not to say that these analogies are not useful, just that you need to be very aware of their limitations.

[BravoV]

I agree, friction is equivalent to electrical resistance.  The missing thing in the water system is an equivalent to electromagnetic field radiation.

How about the sound generated by the water friction with the pipe ?

Of course the sound/noise here doesn't mean its always can be heard by human's ears. 

[CatalinaWOW]

If you can come up with a consistent and simple model to transfer the energy to sound, go for it.  Remember the purpose of all of this is to improve understanding and provide an easy to use numerical model for predicting results.  These analogies are tools, and tools are only useful when they make the job easier.

[Dr. Frank]

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.

I forgot to mention that the leads of the capacitor, they form the inductance.. the charging current creates a magnetic field, and a changing magnetic field also emits an electro-magnetic wave.

Frank

PS: Sorry to be a bore..

[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 capacitors 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.

Frank

Oh, don't be so boringly accurate.

Discussing incorrect (and therefore irrelevant) analogies, is so much more enlightening. Not.

[MrAl]

I agree, friction is equivalent to electrical resistance.  The missing thing in the water system is an equivalent to electromagnetic field radiation.

Any horse can be ridden too hard.  Some fairly good analogies for magnetic circuits can be generated with either water and tank systems, or electrical circuits.  But these analogies break down with low permeability materials, and become very labored with non-linear magnetic materials. 

Again, this is not to say that these analogies are not useful, just that you need to be very aware of their limitations.

Hi,

Very eloquently put.  I usually dont appreciate technicality laced with poetry, but i cant help but admire your short yet adequate comparison.

[Siwastaja]

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.

[GeorgeOfTheJungle]

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.

[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

[MrAl]

Hi again and thanks for the reply,

When you say "circuit" do you mean the circuit with RC or the circuit with RLC or the circuit with zero resistance with or without inductance?  We've been talking about some four or so different circuits.

Also, when you say "change" do you mean increase or decrease the 50 percent, or both increase and decrease?

[IanB]

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?

OK, let's take this step by step with a two capacitor example.

Problem statement: We have two identical capacitors, one charged to voltage V₁, one discharged. We connect them together and wait for the system to stop changing. What is the final result?

To solve such a problem we need to state our assumptions.

Assumption #1: Charge is conserved in the system (no charge is added or removed from outside)
Assumption #2: The system eventually settles out and the voltages stop changing (the system has at least some source of dissipative loss so that it doesn't oscillate forever)
Assumption #3: The capacitance of each capacitor is constant and does not change
Assumption #4: The capacitors are ideal and have no charge leakage between their plates

Given these assumptions we can proceed.

Firstly, determine the initial charge in the system:

    Q = CV₁

This is merely the charge stored in the first capacitor. The second capacitor has no charge.

Secondly, determine the final voltage V₂ when the two capacitors have equalized. By assumption #1 we must have:

    Q = 2CV₂

Therefore:

    V₂ = ½V₁

Thirdly, determine change in energy between the initial and final states. We have:

    E₁ = ½CV₁²
    E₂ = 2(½CV₂²) = CV₂²

But since V₂ = ½V₁, we find that:

    E₂ = CV₂² = ¼CV₁² = ½E₁

As you can see from the working, the change in stored energy is exactly 50%. Neither 49.999% nor 50.001%, but 50% exactly. The only way to change this result is to change the assumptions.

Does that help?

[dannyf]

E2 = CV2² = ¼CV1² = ½E1

So every time you split your capacitor into halves, you would double your energy.

As the capacitors get smaller and smaller, your total energy gets doubled, and redoubled and so on and so forth.

Don't you love math when you don't understand it?

[T3sl4co1l]

Hi again and thanks for the reply,

When you say "circuit" do you mean the circuit with RC or the circuit with RLC or the circuit with zero resistance with or without inductance?  We've been talking about some four or so different circuits.

Also, when you say "change" do you mean increase or decrease the 50 percent, or both increase and decrease?

Anything topologically equivalent; so, a series circuit between a discharged cap, a charged cap, and some (non capacitive) impedance, or a discharged cap, a voltage source, and some (non capacitive) impedance.

Obvious "no good" cases include anything where the capacitor is shunted by a resistor, thus its charge bleeds away at t --> infty.

Speaking of, note that switching converter circuits break this rule, because they exhibit switch leakage to ground or whatever.  The leakage is small enough to be of great value in the short term (microseconds to years), or under continuous operation (power transfer rather than energy transfer).  The latter of course is beyond the scope of thermodynamics, which deals with the balance of energy, not the flow of energy (power).

Tim

[CatalinaWOW]

E2 = CV2² = ¼CV1² = ½E1

So every time you split your capacitor into halves, you would double your energy.

As the capacitors get smaller and smaller, your total energy gets doubled, and redoubled and so on and so forth.

Don't you love math when you don't understand it?

I love it when a comment actually answers itself

[Galenbo]

...
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.

We handled this in university with a mechanical example: Making a shaft turn, by coupling a clutch, has an efficiency of 50%.
Independent of the time that is taken to do it, independent of rpm, Nm and so on.