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

[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.