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

[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...