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

[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.