where does 1/2 come from in capacitor energy calculation

[brownt]

Hi,

where does 1/2 come from with the 1/2CV² calculation for energy in a capacitor?

[Bassman59]

Hi,

where does 1/2 come from with the 1/2CV² calculation for energy in a capacitor?

integration!

[GregDunn]

When you integrate C*V wrt V you get 1/2 CV².  Simple as that.

[intabits]

(similar for kinetic energy: E=1/2MV^2, And I think I've seen another one for some other form of energy)

see here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html#c3
(You may need to click the "show more" hypertext)

[EEVblog]

The more interesting question is, where does half the energy go when charging a capacitor

[tszaboo]

And where does the energy go with the two capacitor paradox.
https://en.wikipedia.org/wiki/Two_capacitor_paradox

[Siwastaja]

You can and often will charge a capacitor with at least 95% efficiency, especially when it comes to large capacitors used for energy storage.

Half of the energy is not going anywhere else, nor is there any special magic with "half the energy". It's completely normal and typical to charge a capacitor, or transfer energy between two capacitors so that about 2% energy is lost, nothing special on the 50% mark.

Of course, doing anything with a voltage source and series resistance such that, on average, half the voltage is lost on that resistor, will effectively waste half of the energy on that resistor. This has nothing to do specifically with capacitors, but this is a great way to confuse people - especially since we are taught to work with models that lack necessary parasitic circuit elements that are always present, but we are not taught to work with the math accordingly (including how to deal with infinity properly).

[woodchips]

Umm, no.

You have two caps, each 1F 10V, one fully charged, one not. Energy 1/2CV^2 is 1/2x1x10^2 = 50J.

Connect in parallel, now have two caps total of 2F at 5V, Energy 1/2CV^2 is 1/2x2x5^2 = 25J.

To me, 25 is half of 50?

As far as I am aware, this halving always happens irrespective of how the cap is charged. Something to do with the fact that the cap charge is stored on the surface of the plates, not electrochemically as in a battery.

[StillTrying]

As a SMPS spends all of its time charging and discharging the output capacitor, how can its energy efficiency ever be more than 50%.

[NiHaoMike]

And where does the energy go with the two capacitor paradox.
https://en.wikipedia.org/wiki/Two_capacitor_paradox

Had the problem come up by accident in real life. I was working on a power inverter with two 330uF, 450V DC bus caps. Because the PCB services at Texas A&M were not very good, a via connecting the two had blown from previous testing so that when I powered it up the next time, the first cap charged up to the full voltage of the DC/DC converter (270V) and the via arced over, connecting the second cap in parallel with the first. The "missing" energy ended up making a small arc flash, scaring away the rest of my senior design team.

[tszaboo]

As a SMPS spends all of its time charging and discharging the output capacitor, how can its energy efficiency ever be more than 50%.

I give you one better: How a switching capacitor voltage converter can work with 50%+ efficiency, while it just charges capacitors? Sometimes even 3.

[Siwastaja]

Umm, no.

You have two caps, each 1F 10V, one fully charged, one not. Energy 1/2CV^2 is 1/2x1x10^2 = 50J.

Connect in parallel, now have two caps total of 2F at 5V, Energy 1/2CV^2 is 1/2x2x5^2 = 25J.

To me, 25 is half of 50?

As far as I am aware, this halving always happens irrespective of how the cap is charged. Something to do with the fact that the cap charge is stored on the surface of the plates, not electrochemically as in a battery.

It's unbelievable how this misconception sticks and lives its own life. It has been discussed to death.

Short answer: No. Nope. It's not like that. Capacitors are and can be very efficient, easily 99%. Charging circuits can be 98-99% efficient as well, easily 90%. This is why capacitors are and have been used in energy storage for example, regenerative braking in train systems.

Charging a cap from a voltage source through a resistor (implicit or parasitic), or using a simple linear supply, of course wastes at least 50% of the energy, since the average voltage during the charging over that resistor is half of the source voltage. This is true with charging a cap, or driving any other load using a linear or resistor-dropper supply to drop half of the voltage.

An SMPS can adjust its output voltage and keep high efficiency over the operating range.

Discretize the problem into ten timesteps and you'll figure it out easily. At the first step, the receiving capacitor is at 0V and the full 10V voltage is over the resistance. You'll see instantly where power is dissipated and by how much. Now replace the resistor circuit with an SMPS that can directly provide you with 1V, instead of 10V. Then set it for 2V for the next timestep, etc...

[intabits]

I was curious about this apparent paradox, and thought I had it worked out with this simulation, but now I'm a bit confused.
First, my explanation:- 

Using 15 ohms to exaggerate the parasitic resistances of the wires and capacitors.


S1 closes for 1 second between T=1 and T=2, and C1 charges via RC, and reaches the full 10v by T=2.
At T=3, S2 closes to transfer the charge between the caps. C1 discharges and C2 charges, reaching equilibrium at 5v.

The two spikes at T=1 and T=3 are the power dissipated in RC and RD, and the areas beneath them is the energy lost by these resistors.
It can be seen that the second spike's area, representing the missing energy,  is around half of the first. Any minor discrepancy would presumably be due to the internal resistance of the caps.

But now I realize that while the energy of the second spike can only have come from C1, I had assumed the area of the first spike represented the energy going into C1 to start with.
But now I can't see why the energy lost out of RC would at all have to be the same as the energy going into C1???.
(the large RX was needed to force C1 to initially be 0v. Not sure why, maybe some quirk of the voltage controlled switches?)

[EEVblog]

Therefore, the other half of energy went to resistor, no matter how high the resistance is or how low it is.

Well, actually, when the resistance approaches zero, the current approaches infinite, and the energy starts to get lost as electromagnetic radiation at that point.

[T3sl4co1l]

Well, actually, when the resistance approaches zero, the current approaches infinite, and the energy starts to get lost as electromagnetic radiation at that point.

Actually that can have better efficiency as stray inductance start to soft charge ("soft" comparing with resistance). The above analysis only assumes RC circuit without inductor.

1. There is ALWAYS loss.  Even a superconducting resonator (a more subtle kind of inductor-capacitor circuit) has a finite Q (around 10^8 -- better than most quartz crystals, giving a time constant of ~seconds at a resonant frequency of 100s MHz, not bad eh?).  Over a short time period, we can get reasonably lossless exchanges of energy, but over a long time scale, thermodynamics ALWAYS wins.

The equations for charge conservation, capacitor charging (RC or otherwise), and so on, are taken at \$ t \rightarrow \infty \$, so thermodynamics is the name of the game here.

2. There is no circuit without inductance, or capacitance, either.  Indeed, L and C are the electrical manifestations of physical causality: anywhere the physical length of some system is nonzero, and where the speed of light* is finite, there must be components of L and C!

*Generally, the velocity of propagation.  Equivalents can be derived in any wave system: e.g., in mechanics, the speed of sound causes finite and nonzero elastic modulus and mass.

So this teaches us a fallacy of our schematic notation: that all signals are instantaneous, or that the circuit has zero dimension, or the speed of light is infinite (all equivalent).  We must think carefully to ensure we aren't deceiving ourselves with an overly greedy simplification, and draw circuits that are physically realistic.

This is most applicable to SPICE modeling, where it is easy to create perfectly valid, but physically impossible, systems.

We're basically SPICEing this circuit by eye.

So there must, necessarily, be some L in the real circuit this represents.  And that L (and the C as well) will couple into ambient fields, however slight (there is no perfect shielding, either).  So that, even with a truly perfect resonator (better than any superconducting one we can actually make), the energy inside will inevitably leak into free space.  At infinite time, the final state will be the steady state given exactly by thermodynamics.

Tim

[hamster_nz]

And where does the energy go with the two capacitor paradox.
https://en.wikipedia.org/wiki/Two_capacitor_paradox

It is because energy isn't the conserved unit in electronics, current is (unless you are keeping track of lost energy, such as heat, light, EM radiation and stray electrons)

It is like momentum (mass * speed) is the conserved unit in collisions, not kinetic energy ( energy = 1/2 * mass & velocity ^2)

If you have a slow noise to tail, with a car of equal weight then your car will stop, and the other car will roll off with the approximately same speed - as long as the energy is transferred efficiently (much like Newton's Cradle https://en.wikipedia.org/wiki/Newton%27s_cradle). In this case momentum and energy are conserved - it has just been moved to the other car. This is like the energy transfer in a switch mode power supply - it transforms the voltage and current to ensure a good match between the source and load, allowing for efficient transfer (usually by moving energy in and out of an inductor).

However, if you have a more energetic crash nose to tail crash, one where the structure of the car cannot transfer the energy efficiently both cars will tend to stick together and move off at half the original speed. In this case only momentum is conserved, but half the energy is missing (as twice the mass is moving at half the speed) - much like the energy in the two capacitor 'paradox'.

So where does the missing energy go? Humm, I don't know, but just maybe that big bang and the crumpled metal had something to do with it...

[SparkyFX]

You have two caps, each 1F 10V, one fully charged, one not. Energy 1/2CV^2 is 1/2x1x10^2 = 50J.

Connect in parallel, now have two caps total of 2F at 5V, Energy 1/2CV^2 is 1/2x2x5^2 = 25J.

To me, 25 is half of 50?

Sorry to answer 2 years later, but
(1/2x2x5)^2 = 25J.
(1/2x1x10)^2 = 25J.

[Nominal Animal]

Sorry to answer 2 years later, but
(1/2x2x5)^2 = 25J.
(1/2x1x10)^2 = 25J.

True, but irrelevant in this context.  The energy E stored in a capacitor with capacitance C and voltage V across the capacitor is E=0.5CV², not (0.5CV)².

[SparkyFX]

Right, the problem is the assumption of half voltage, while the charge curve indicates an exponential function per energy stored.
The voltage in the new parallel system will be 7.07V, given the same energy is stored, then it will match the new capacity.

I started from the wrong end of the wrong statement in the wrong way, sorry