Capacitor charging loss (Not the two capacitor issue.)

In summary: C (V_1\,-\,V_0)^2[/itex]would only apply when the power supply is supplying a constant voltage (in other words, not changing). In summary, the third reference that spaceball3000 mentioned seems to back up his argument that charging a capacitor through a resistor is very inefficient unless the applied voltage stays close to the voltage across the capacitor. However, charging a capacitor through an inductor is not inefficient at all.
  • #1
spaceball3000
41
0
I got some guy who says when a capacitor is fully charged, that half of the input energy from the power source gets lost as heat. Now I do understand the two cap problem (links below), but I think he is wrong because one can use dc switching and inductors to improve the 50% loss of energy. (but he says 1/2 loss is still happening.)

I would find it hard to believe that for DL ultracaps/supercapacitors used on EV's/hybrid's for quick storage would be loosing 1/2 of the power to charge them as heat.

He's asking me to prove him wrong (he has same links I've listed below), is there any other references that anyone has that I use to back up my argument?

http://www.hep.princeton.edu/~mcdonald/examples/twocaps.pdf
http://puhep1.princeton.edu/~mcdonald/examples/EM/powell_ajp_47_460_79.pdf
http://puhep1.princeton.edu/~mcdonald/examples/EM/mita_ajp_67_737_99.pdf

Thanks!
 
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  • #2
Hi spaceball3000! :smile:

Up to 50% is lost even with one capacitor , unless it's done through an inductor …

see this from the PF Library (unfortunately, no details for the inductor part) …

Energy loss:

Energy lost (to heat in the resistor):

[tex]\int\,I^2(t)\,R\,dt\ =\ \frac{1}{2}\,C (V_1\,-\,V_0)^2[/itex]

Efficiency (energy lost per total energy):

[tex]\frac{V_1^2\,-\,V_0^2}{V_1^2\,-\,V_0^2\,+\,(V_1\,-\,V_0)^2}\ =\ \frac{1}{2}\,\left(1\,+\,\frac{V_0}{V_1}\right)[/tex]

Accordingly, charging a capacitor through a resistor is very inefficient unless the applied voltage stays close to the voltage across the capacitor.

But there is no energy loss on charging a capacitor through an inductor, basically because the applied voltage then appears across the inductor instead of across the capacitor.​
 
  • #3
Thanks Tiny-tim,

I hope that will be enough for him to see the light.
 
  • #4
Well that didn't convince the guy at all.

I even talked about LC Tank circuits and how every LC Oscillation periods C is being charge and discharged. And that there doesn't seem to be a rule where there has to be an minimum of 50% energy loss with every LC oscillation period in a LC Circuit. Where the charge has is moved once from the inductor to capacitor (or the opposite), the total energy in the capacitor by his logic it would have to be reduced by a minimum of 50% each time period (when the capacitor charged.)

Can't someone provide some proof so that he doesn't confuse more people that how capacitors work.
 
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  • #5
The third reference you gave:
http://puhep1.princeton.edu/~mcdonald/examples/EM/mita_ajp_67_737_99.pdf

explains things. On the second page, near equation (11), the author discusses the situation where an inductor in introduced into the current path. He mentions in footnote 7 that he considers the case where only a single cycle of the oscillation that occurs in this situation is allowed to happen.

In the case where the circuit includes a small parasitic resistance as well as the inductance, and where a voltage source is suddenly connected to a capacitor (plus L and parasitic R in series), the current will oscillate until the oscillations finally die out. In that case, I think the parasitic resistance will in fact dissipate half the energy delivered by the power supply.

But, if the oscillations are stopped after one cycle by the simple expedient of placing a diode in the current path, then the capacitor will contain essentially all of the energy delivered by the power supply.

Furthermore, if a variable power supply is used as the source, starting with the supply set to zero volts out and connecting a capacitor to the supply, if the voltage is turned up gradually then the capacitor will contain essentially all of the energy delivered by the supply.
 
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  • #6
Hi 'The Electrician',

I 100% agree on your thoughts on this subject.

He basically ignored that third reference saying it's "contradicts itself", and saying

. said:
"Despite all their moving around with complicate mathematic tools (Heaviside, Dirac etc.) these Ladies have just calculated the energy temporarily in the R or L. This energy is the ½ QV but together with ½ QV on the capacitor we have 1/1 for the battery. They claim others are using idealized circuits while their assumptions are just appalling ideal (without a R).

These ladies make then assumptions in equation (13) for a step of the voltage, but in next equation (14) they just assume the usual voltage ramp as to V= Q/C and this over the battery. ...

In the second last paragraph they realize that the step function become ½ for continuous operation, which would let fall out the ½ for the energy calculation, but they call this a 'pure mathematical curiosity' not important here."

Then later on he points me to this link --> http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html#c1 saying

"the formula from the University of Georgia is a physical principle, it is not limited to how the energy is fed, you assumed a constant voltage, but could also be a constant current, or via any other element etc. This is physics: it is like that because it is like that!"He does seems smart, but I think he is missing something, the equation in the link he gave, seems to only apply to scenarios when a capacitor is being charging through a resistor. Thus the equation he gave (I think) would need to be changed to take into account when the inductor is added.

Now adding the inductor, doesn't prevent the oscillations (which would generate RF/EM, i.e. losses) and you are correct a diode would prevent that.

So in a nutshell, can someone find proof that his equation (that he linked) Does Not apply to the scenario where an capacitor is being charged via inductor?
 
  • #7
As near as I can tell, the link http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html#c1

merely gives the classical formula for the energy stored in a capacitor, c*v^2/2, or q*v/2 which is the same thing. This formula is correct no matter how the capacitor gets charged.

However, it isn't correct to say that whenever a capacitor is charged from some kind of energy source, without exception, only half the energy provided by the source ends up in the capacitor.

Sometimes it is true, as in the case where a source of constant voltage is suddenly connected to a series combination of a resistor and capacitor.

If the circuit consists of an inductor in series with a capacitor, with negligible parasitic resistance, then suddenly connecting this circuit to a constant voltage source will transfer almost all of the energy provided by the source to the capacitor, IF the voltage source is disconnected at the right time, namely when the current goes to zero at the end of one half cycle of oscillation. A simple diode can provide this disconnect.

Another way to ensure that all the energy provided by the source ends up on the capacitor is to connect the capacitor to a variable voltage source initially set to zero, and then gradually turn up the voltage to some final value.

Even if there is a resistor of non-negligible size in series, if the voltage is turned up slowly so that the charging current remains small at all times, most of the energy ends up in the capacitor, with very little dissipated in the resistor.

This is what is described in your original third reference, in the vicinity of equations 13 and 14. By making the voltage steps smaller and more numerous, the effect is the same as turning up a variable voltage source gradually.
 
  • #8
Ah, that makes sense to me.

Me just repeating this to him will likely not convince him at all, I'm trying to find a logical error in his thinking.

So just to confirm, the classical formula for the energy stored in a capacitor never changes, but to calculate the energy lost he likely using this equation from --> Energy lost (to heat in the resistor): (look under "Energy lost (to heat in the resistor):"

But since there is no energy loss on charging a capacitor through an inductor, so does this same energy loss (resistor) equation still apply (newly added inductor)?
 
  • #9
If there's no resistance (even the smallest parasitic) in the circuit, with an inductor and capacitor, the oscillations will go on forever, and in that case the resistor-capacitor equation doesn't apply. But then how can we say that the capacitor reaches a final state of charge?

I've attached an analysis of the problem for the R-C and the R-L-C cases.

But, the fact that the energy that ends up in the capacitor in these two cases is only half the energy provided by the voltage source doesn't mean that every method of charging the capacitor loses half the energy from the source. I described two methods that don't suffer from that loss in a previous post.
 

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  • #10
Wonderful! Think this will help a lot, Thanks!
 
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  • #11
You only need one contrary example to disprove an assertion. But it's not very convincing to say that no energy is lost while the current is still sloshing back and forth between capacitor and inductor. Will adding a switch change this?
 
  • #12
Phrak said:
You only need one contrary example to disprove an assertion. But it's not very convincing to say that no energy is lost while the current is still sloshing back and forth between capacitor and inductor. Will adding a switch change this?

Ok dropped the word sloshing.

In an LC Tank circuit when the energy is being transferred from one to another, disconnecting L when it's empty (i.e. switch) will prevent any more oscillations. At that point most all of the energy that was in L should be transferred to C.
 
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  • #13
Yes if you are allowed to use an inductor and a switch it's easy to do.

Just get a voltage_source, an inductor, a switch and the capacitor all in series. Let the initial voltage on the capacitor be zero and the value of the voltage source be one half of the final voltage that you wish to charge the capacitor to.

Close the switch and then re-open it on the next zero crossing of the current waveform (at [itex]t = \pi \sqrt{LC}[/itex]). The capacitor will then be charged to twice the voltage source value, ideally without loss. This is the basis of how a resonant mode DC-DC converter operates.
 
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  • #14
It's a tank, it's sposta slosh. :smile:

As uart was saying, the voltage across the capacitor will be double the supply voltatge. Connect a switch, an inductor and a capacitor across 10V.

Initially, the current in the inductor is zero. The voltage across the capacitor is zero. Close the switch at time t=0.

The voltage across the capacitor will be
[tex] V_C = 10 - 10 cos ( 2 \pi f t ) [/tex]

The current through the inductor will be
[tex] I_L = I_{peak} sin ( 2 \pi f t ) [/tex]

The peak current can be determined from conservation of energy.
[itex] L \cdot I_{peak}{}^2 + C \cdot 10^2 = C \cdot 20^2 [/itex] Joules.

Open the switch after half a cycle, as uart was saying, at time t = 1/(2f).

The energy stored in the capacitor will be
[tex](1/2)C \cdot 20^2[/tex]

The energy delivered by the 10 volt source will be
[tex]\int_{0}^{1/(2F)}10I_L dt[/tex]
 
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  • #15
Again, you guys have been great!

Honestly this has helped me more than the guy who I was trying to convince (I learned a few new things), he will not admit he is wrong in lieu of the facts. Only reason I can think he's ignoring the facts, is that he trying to convince people that all capacitors have 50% losses when charging (all methods including an inductor.) So that gives his new battery/capacitor invention an edge, i.e. he says his capacitor doesn't have this 50% problem.

If your curious more on this discussion, just Google with these keywords "quantum battery rolf" , and look at the various forums.
 
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  • #16
Low loss translation of current and voltage using inductors and capacitors and switches (transistors and diodes), done in a dozen different ways, is the basis of switched mode power supplies.
 
  • #17
he says I'm wrong again, anyone interested to point out his math errors?
See attached file.
 

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  • #18
spaceball3000 said:
he says I'm wrong again, anyone interested to point out his math errors?
See attached file.

Yes he's totally wrong on the 8th line :

Code:
Work Performed = Uc Qc

This should read :

Power supplied by Battery = Ub Qc = 2 Q Ub, exactly the same as the energy delived to the capacitor.
 
  • #19
Well I mentioned what you said, and also saying 4 Q Ub is wrong and that 2 Q Ub is the Work performed. He replied (below) and also added and additional equation.

Response said:
you can also put your figure into the equation and with my second post you find out that the integration of WC gives only half of it. You can put in for WB = 4QU or 2QU or 1QU the "1/2" remains.
..
The are no 4Q in the first cycle only 2 Q but with 2UB so the energy gets 2*2*Q*U.

I'm going to assume he messed up like in his first example. Taking a look I see the equation where Ipeak has 4Ub, I'm pretty sure it should be 2Ub, am I correct?
 

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  • #20
spaceball3000 said:
Well I mentioned what you said, and also saying 4 Q Ub is wrong and that 2 Q Ub is the Work performed. He replied (below) and also added and additional equation.
I'm going to assume he messed up like in his first example. Taking a look I see the equation where Ipeak has 4Ub, I'm pretty sure it should be 2Ub, am I correct?

Yeah there are quite a few errors here as well. He seems to like just throwing in factors of 2 and 4 at will without any explanation to make the results fit his theory.

The first error is that Ipeak is only one times Ub w c, not 4 times as claimed (line 4). Interestingly however he then makes a mistake in the integration giving a factor of 1/2 (integral over half period of w sin(wt) dt ) whereas the correct value for this integral is 2 (line 7). So it turns out that these two mistakes cancel out and that the expression he obtains, Wc = 2 Q Ub is actually correct (where Q = C Ub).

Note however that Wb = 2 Q Ub (not 4 Q Ub as claimed by this guy) so once again we have Wb = Wc.

So (apart from some new mistakes that fortunately canceled out) the common mistake from both posts is his belief that Wb = 4 C Ub^2 whereas the correct value is only 2 C Ub^2. This is pretty trivial, the charge drawn by the capacitor (up until t=T/2) is C Uc(T/2) = 2C Ub. So the energy drawn from the DC supply is just Ub times the charge drawn which gives the value of 2 C Ub^2. I have no idea where he gets the 4 from here, and he gives no explanation, so it's all very dubious.
 
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  • #21
I've worked out the full-bore mathematical analysis of an underdamped RLC circuit, with a suddenly applied voltage step.

I've plotted the energy loss in the resistor vs. time. It can be plainly seen that if the oscillations are allowed to continue until they damp out, the loss in the resistor reaches a final value of c*v^2/2. But, it can also be seen that in the first half cycle, the loss is nowhere near this much, so that if a switch is opened to stop the oscillations (such as a diode) after one half cycle, the efficiency of charging the capacitor can be much higher than 50%.
 

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  • #22
I tried to post an image that is too big, and the forum shrunk it.

I'll try posting in two parts.
 

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  • #23
Thanks for the RLC info I'll pass that on.

He says he fixed his mistake (see attached file), and says that your guys math is wrong (see his response below.) I'm sure he trying to hide/confuse the math even more now, I'm having hard time following his new cryptic changes though.

Response said:
Again my strong believe in physics says that a capacitor of U has a charge of Q = CU and to get to this state an energy of QU had to be performed such that the battery pumps QU into the system but eventually only half of it is on the C. If you increase the voltage the energy to doing it is proportional to the square of it. So doubling to 2U needs an energy of 4UQ.

I guess that the confusion that most people have is because they take as prove by integrating Ub*I*sin(wt)dt over battery and Ub(1-cos(wt))*Isin(wt)dt over the C and get equal results = no loss. But sin(wt)dt and sin(wt)*(1-cos(wt)dt over pi give both 2. Must be same.

To find the real input energy one has to include the L as an additional serial voltage source of full Ub at the end of the cycle. This is indicated with the result in my calculation with the term 2*Ub.
 

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  • #24
I added a plot of the energy in the capacitor vs. time to the plot.

You can see that after many oscillations, the energy lost in the resistor equals the energy stored in the capacitor, but if the oscillations are stopped after one half cycle, the energy lost in the resistor is a small fraction of the energy in the cap.
 

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  • #25
Hi Spaceball. Once again he gets the current wrong. Once again he arbitrarily includes a factor of two without explanation (and where it doesn't belong).

Specifically he claims that :

[tex] I_{peak} = 2 w C U_B[/tex]

btw previously he claimed it was 4 w C Ub and that was wrong too.

The correct value is :

[tex] I_{peak} = w C U_B[/tex]

BTW, this is a 100% certainly, no if's no but's. There is absolutely no controversy about this result, it could be found in any textbook and could be derived by just about anyone who's completed even half of an EE degree.

Specifically the DE's are (where "i" is the inductor current and "v" is the capacitor voltage)

[tex]di/dt = (U_B - v)/L[/tex]

[tex]dv/dt = i/C[/tex]

with initial conditions : [itex]i(0) = 0[/itex] and [itex]v(0)=0[/itex], hence [itex]di/dt (0) = U_B/L[/itex]

Combining the two DE's gives :

[tex] d^2 i / d t^2 = - i /(LC)[/tex]

Which has the solution :

[tex]i = A \cos(w t) + B \sin(wt) [/tex]

where "A" and "B" are constants and [itex] w = 1 / \sqrt{LC}[/itex]

Applying the initial condition on the current we get [itex]A=0[/itex] and from the initial condition on the current derivative we get [itex]B w = U_B/L[/itex].

Rearranging the last equation gives : [itex]B = U_B\,\sqrt{C/L} = w C \,U_B[/itex].

That is :

[tex]i = w C U_B \sin(wt)[/tex].
 
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  • #26
Again my strong believe in physics says that a capacitor of U has a charge of Q = CU and to get to this state an energy of QU had to be performed such that the battery pumps QU into the system but eventually only half of it is on the C. If you increase the voltage the energy to doing it is proportional to the square of it. So doubling to 2U needs an energy of 4UQ.

OK let me reply directly to this. Firstly, let's be specific about which voltages we're talking about here.

- "Again my strong believe in physics says that a capacitor of [itex]U_C[/itex] has a charge of [itex]Q = CU_C[/itex]"

This is correct.- "to get to this state an energy of [itex]QU_C[/itex] had to be performed such that the battery pumps [itex]QU_C[/itex] into the system"

This is not correct. Assuming this "U" is the final capactor voltage (2 U_B) then no. At the capacitor terminals the charge is entering while the terminal voltage is less then this (final value) during most of the charging period, which is why the stored energy on the capacitor is only half of that value. Also, at the battery terminal the voltage is fixed at [itex]U_B[/itex], so the energy taken from the supply is only [itex]U_B Q_C[/itex] and not [itex]U_C Q_C[/itex] as claimed.

In summary [itex]W_C = 1/2 \, U_C Q_C[/itex] and [itex]W_B = U_B Q_C[/itex], which are of course the same since [itex]U_C = 2 U_B[/itex]
 
  • #27
In his corrected paper, why does he refer to "half cycle" when he apparently should be using "quarter cycle"?

For example, just to the left of the red and blue plots, he has "First half cycle = second half cycle", when he clearly should say "First quarter cycle = second quarter cycle". The second half cycle is shown as a dashed line, and apparently never considered.

His paper is hard to follow, and I have to go to bed.
 
  • #28
Thank Uart & Electrician for the followups, I'll passed your info, almost seems like he want's to give out misinformation.
 
  • #29
uart said:
That is :

[tex]i = w C U_B \sin(wt)[/tex].

Notice uart's expression for i is the same as my expression for i(t) shown in the first image of post #22, except I have taken into account resistance in the circuit. I think it is without question that the factor of 2 is wrong.
 
  • #30
spaceball3000 said:
Thank Uart & Electrician for the followups, I'll passed your info, almost seems like he want's to give out misinformation.

You should be sure to show him the image I included in post #24. It really tells the whole story.
 
  • #31
I see his latest response posted an hour or so ago.

He says: "Because it is irrespective of the element value it cannot be calculated, it is just there. So also with a simulation program it cannot be calculated, when the loss is independent of the element value!"

You probably aren't going to get very far with someone who believes this effect is so magical that it can't be calculated, and it can't be simulated.

I wonder what he would say if offered experimental evidence? It wouldn't be hard to solder up an RLC circuit and capture some current and voltage waveforms.
 
  • #32
The Electrician said:
I see his latest response posted an hour or so ago.

He says: "Because it is irrespective of the element value it cannot be calculated, it is just there. So also with a simulation program it cannot be calculated, when the loss is independent of the element value!"

You probably aren't going to get very far with someone who believes this effect is so magical that it can't be calculated, and it can't be simulated.

I wonder what he would say if offered experimental evidence? It wouldn't be hard to solder up an RLC circuit and capture some current and voltage waveforms.

Sigh, your likely right.
I thought of doing that, but my thought he would just say my testing equipment is inaccurate when testing my RLC circuit.
 
  • #33
My father used to say:

"There are none so blind as those who will not see."
 
  • #34
It occurred to me to wonder why, if he really believes that "Because it is irrespective of the element value it cannot be calculated, it is just there. So also with a simulation program it cannot be calculated, when the loss is independent of the element value!", why would he provide calculations attempting to prove his own point of view is correct?

Wouldn't his own calculations be incorrect, because "...it cannot be calculated..."?
 
  • #35
The Electrician said:
It occurred to me to wonder why, if he really believes that "Because it is irrespective of the element value it cannot be calculated, it is just there. So also with a simulation program it cannot be calculated, when the loss is independent of the element value!", why would he provide calculations attempting to prove his own point of view is correct?

Wouldn't his own calculations be incorrect, because "...it cannot be calculated..."?

Thats a good point.
 

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