Curious case of the charging capacitor - Page 1
Forum section: electronics / beginners

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: February 20, 2019, 05:25:56 am
content: I'm going to revisit the classic capacitor charging case, which I just came across1. Is it true that half the energy fed by the power supply in charging the capacitor is lost, I mean if 100uJ is supplied by the power supply of 10VDC, finally the energy stored in the capacitor @ 10VDC would only be 50uJ.If this is true can charging a capacitor be considered inefficient?Following on are there efficient ways charging a capacitor?2. This case is of charging a capacitor with another capacitor. Eg. If I have a 1uF capacitor charged to 10VDC and another 1uF capacitor uncharged. If I connect both of these in parallel the final derived energy is again half the energy that we started off with. So where did the other half go?And if this is true then how come charge pumps achieve high efficiencies?On subsequent searches I also found several posts indicating that this is likely due to energy loss in the resistancewires.I would like to hear the views of the learned audience at eevblog on this subject.TIAPS: Would the above cases still be true if the capacitor was charged with constant current?

author: Doc Daneeka
author_rank: Contributor
timestamp: February 20, 2019, 06:10:39 am
content: The problem is the 'abstract model' of the circuit is not valid - the problem talks about the voltage across the capacitor as if it were directly connected across an idea voltage source, or as if there were a step change in voltage across the ideal capacitor - this is not a mathematically valid set of conditions - its not even that 'the energy is lost in the resistance' there is no resistance in the model. Its just that a step change in voltage gives undefined power into or out of the capacitor.Yes in a real circuit the energy goes into the resistance. As soon as you add a resistor to the model there is no problem and the model works perfectlyIf it is charged with a current source everything works out fine.Its just the old 1/0 trick in disguise - effectively the problem gives an initial and final condition for the capacitors, but no way for it to go from one to the other.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: February 20, 2019, 06:13:55 am
content: Quote from: Doc Daneeka on February 20, 2019, 06:10:39 amOk, so lets consider a one ohm resistor into the circuit. What I would like to know is If I put in say "x" amount of Joules in charging the capacitor would the final energy on the capacitor be less that 50% of what was fed in?

author: Doc Daneeka
author_rank: Contributor
timestamp: February 20, 2019, 06:18:27 am
content: I dont know, do the calculation or try in LT spice! The efficiency does depend on the rate of charging in a slightly surprising way...

author: Jwillis
author_rank: Super Contributor
timestamp: February 20, 2019, 09:30:50 am
content: Your greatest loss occurs just at the beginning of charge when the capacitor acts like a dead short.Then when it begins to charge the amount of current absorbing lowers exponentially as the voltage increases exponentially until the capacitors resistance reaches infinity,Then its considered fully charged.When you disconnect the battery and attach another fully discharged capacitor parallel to charged capacitor the same thing occurs until both capacitors reach a balanced state then both will discharge because neither have and infinite resistance.

author: Zero999
author_rank: Super Contributor
timestamp: February 20, 2019, 10:21:49 am
content: No, not if the power supply is switched mode and current limiting is done by the controller, rather than a simple series resistance. Energy efficiencies of 80% are reasonable.Yes, just connecting one charged capacitor to another, will result in half of the energy being lost in the series resistance.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 20, 2019, 01:21:23 pm
content: If you charge a capacitor from zero, through an ideal voltage source, ideal diode and inductor, it charges to double the input voltage at 100% efficiency.Tim

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: February 20, 2019, 01:31:30 pm
content: Quote from: T3sl4co1l on February 20, 2019, 01:21:23 pmWhat???!How does that work? And how does one get double the voltage??

author: Zero999
author_rank: Super Contributor
timestamp: February 20, 2019, 01:43:38 pm
content: Quote from: ZeroResistance on February 20, 2019, 01:31:30 pmTry simulating it in LTSpice, see attached.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: February 20, 2019, 02:24:09 pm
content: Quote from: Zero999 on February 20, 2019, 01:43:38 pmThanks for the file, did you whip that up in seconds?! That's awesome!!So after simulating, It looks like it works with some kind of resonance?The L and C, resonant frequency is around 50Hz or 20ms. So I would have expected the peak to reach in 1/4th of that time or 5ms, but the simulation returns 10ms as time to peak.So does it work on resonance?!

author: rstofer
author_rank: Super Contributor
timestamp: February 20, 2019, 04:28:33 pm
content: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html#c4

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 20, 2019, 07:23:21 pm
content: The lesson is this: a real physical circuit has R, L and C elements. The voltage source* and capacitor circuit is a fiction, adequate to explore conservation of charge, but completely inadequate for any other aspect -- such as time or frequency behavior, or energy or power balance.*Or capacitors of different charge, or any combination. Two capacitors in series, with some initial voltage on one, is equivalent to one capacitor of series equivalent value, plus one voltage source equal to the difference in initial voltages.The case with just L and C (no D) is actually identical to the C and C case, if you separate the solution properly:With no D, of course, the resonant wave continues on forever -- the diode simply stops it at a convenient point, turning the resonant energy into potential energy (capacitor final voltage and charge). If we leave it resonate instead, then we have a mean value on the capacitor, equal to the voltage source, and therefore the mean charge as well. And if we look at the energy in the capacitor, it's also oscillating up and down, with the average equal to the half charge value. Presto -- we've separated the energy components into a oscillating component, and a static component! The oscillating component, obviously, has equal energy -- it's oscillating up and down with amplitude equal to the average. If we calculate the energy of that sine wave by itself (Parseval's theorem), we find the same answer again.Incidentally, if you try simulating this, you may find the sine wave grows or shrinks over time, even with all components carefully set to zero ohms resistance. This is a consequence of numerical stability: it's very hard to calculate something with quite that much accuracy, when you're doing it incrementally, one tiny step at a time! This isn't a bad illustration of it actually, and getting a feel for how it varies with the various TOL parameters, and integration method (simulation settings).Tim

author: IanB
author_rank: Super Contributor
timestamp: February 20, 2019, 07:35:10 pm
content: Quote from: ZeroResistance on February 20, 2019, 05:25:56 amI posted this a while back. It was something curious to me at the time:https://www.eevblog.com/forum/projects/why-trying-to-store-energy-in-a-capacitor-can-be-less-efficient-than-you-think/About resonance, if the series resistance is large enough the system will be over damped and oscillations will be quenched. Then the simple analysis becomes (more) valid.

author: Jwillis
author_rank: Super Contributor
timestamp: February 20, 2019, 09:41:20 pm
content: Voltage doubler/multiplier work by taking the peek to peek AC current and charging two capacitors one positive and one negative .This is done with diodes one diode directs the first 0 to 180 degree half cycle of the AC positive to one capacitor and the second diode directs 180 to 360 degree half cycle of The AC to the second capacitor. For example you have 5 volts AC supply , one capacitor is charged to +5 volts from the first half cycle .the other capacitor is charged - 5 volts from the second half of AC cycle . As one is being charged the other is being discharged. The quantitative addition of positive 5 volts and negative 5 volts is 10 volts . Add another diode and capacitor (Tripler) adds another 5 volts for 15 volts total so on and so on. But as your voltage increases at the out the available current at output halves in a doubler and is 1/3 rd for a tripler. So if your available current is 1 amp at the AC source only 0.5 amps is available after the doubler and 0.3 amps available after a tripler . But that's in a perfect world condition.Actual available current and voltage will be much different than expected due to combined losses of capacitor reactance and voltage drops across the diodes. Also Reactance is directly proportional to frequency but for the sake of simplicity I didn't include that relationship.Models tend to fail because they assume a zero resistance between the voltage source and in the capacitor.But the reality is resistance is there however small it may be .Look at the the time constant T=R X C .If the resistance (including the resistance of the capacitor) always remains zero ,charge would also remain zero.This can be verified with Kirchhoff’s voltage law Vs - R x I(t) - Vc .Capacitor would remain a dead short.

author: not1xor1
author_rank: Frequent Contributor
timestamp: February 21, 2019, 05:51:17 am
content: Quote from: Doc Daneeka on February 20, 2019, 06:10:39 amin the real world part of the energy (how much depends on the features of the circuit) is also radiated as EM waves

author: Zero999
author_rank: Super Contributor
timestamp: February 21, 2019, 08:59:00 am
content: Quote from: ZeroResistance on February 20, 2019, 02:24:09 pmTry using a parallel circuit and a constant current source, rather than voltage. Perhaps that might be more intuitive.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: February 27, 2019, 07:54:39 am
content: QuoteThe capacitor and inductor seem to start charging at the same time and then the diode turns on after the capacitor has discharged.it seems the following may be occuring1. Current source charges the C and L2. Then C discharges into L3. Then L discharges thru the diodeI don't understand why the current source only charges the C up to only 300V and why it does not infinitely charge the C?

author: unitedatoms
author_rank: Frequent Contributor
timestamp: February 27, 2019, 11:46:37 am
content: Quote from: ZeroResistance on February 20, 2019, 05:25:56 amVoltage is not related to energy linearly. Its is squared in capacitor energy formula.