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

[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?