Do you know this? I didn't! ~~ to capacitors - Page 1
Forum section: electronics / beginners

author: tatus1969
author_rank: Super Contributor
timestamp: August 20, 2016, 06:41:07 am
content: I just stumbled across a thought experiment regarding capacitors that surprised me. I thought that i knew everything about them - I didn't.I would like to make this a contest. Maybe it is just me being this blind. Everyone of you who already knows the answer, please just do the poll, and don't disclose anything. If you didn't know the answer then post it here. The first one receives a thumbs up O0Here's the experiment:- take two capacitors of same capacity- charge up one of them to a certain voltage, leave the other one empty- now connect them together, allowing the charge to distribute between themWhat do we expect? If we charged the capacitor to 1.0 volts, then we expect both of them to have 0.5 volts after we have shorted them, right?You can check this by simulation, you can do this with real capacitors, the result will always be the same.The explanation of that is easy, because a capacitor collects charge. The formula for that is Q = C * U. Charge in a capacitor shows up as voltage. Double the charge, doubles the voltage. When we short the two, then the (at this time, constant) charge just evenly distributes between them. After the charge distribution, each capacitor now receives half the charge, which means that we measure only half the voltage across each.So far so good. Now lets add another formula: the stored energy is 0.5 * C * U^2.Let's assume the two capacitors have 1 Farad each, and that we charged up the first one to 1 volts.The energy that we needed to charge up the first capacitor was then: 0.5 * 1.0 * 1.0^2 = 0.5 joules.Now lets calculate the total energy after we shorted them. Remember, we just evenly distributed the charge in that step: 0.5 * 1.0 * 0.5^2 = 0.125 joules per capacitor. So 0.25 joules in total.Wait, we had 0.5 joules before. Where did that 0.25 joules go??

author: Simon
author_rank: Unknown
timestamp: August 20, 2016, 06:54:04 am
content: Look at the formula I highlighted in red. If you put your new 2F into that you now have 1 joules..... now that would get you the 0.25 each in your second formula.Quote from: tatus1969 on August 20, 2016, 06:41:07 am

author: tatus1969
author_rank: Super Contributor
timestamp: August 20, 2016, 07:03:02 am
content: Quote from: Simon on August 20, 2016, 06:54:04 amNo. 0.5 * 2.0 * (0.5^2) = 0.25. Remember the voltage across each of them is now only 0.5 volts.

author: helius
author_rank: Super Contributor
timestamp: August 20, 2016, 07:06:33 am
content: The 0.25 J was lost as heat, since this is a thermodynamically favored change (a chemist would call it "exergonic").Think about the reverse case: If you connect two 1 F capacitors each charged to 0.5 V, would that charge spontaneously move into just one capacitor, leaving the other fully discharged? No, because that is a thermodynamically unfavored change. The entropy of two capacitors in parallel, charged to voltages A and B, is at its maximum when A=B. So the system will move to this state and the energy difference is released as heat.

author: tatus1969
author_rank: Super Contributor
timestamp: August 20, 2016, 07:11:29 am
content: Quote from: helius on August 20, 2016, 07:06:33 amVery close. If you can narrow down where that heat is relased then you get that thumbs up. Remember that this thought experiment also works with ideal capacitors, which have no means to dissipate heat.

author: Simon
author_rank: Unknown
timestamp: August 20, 2016, 07:11:53 am
content: U is voltage right ? assuming you had charged the parallel pair together you would have 1J

author: tatus1969
author_rank: Super Contributor
timestamp: August 20, 2016, 07:13:49 am
content: Quote from: Simon on August 20, 2016, 07:11:53 amSorry for that U, I had mistakenly used the German letter for voltage. In the experiment, I only charge one of the capacitors, and leave the other empty in the first step.

author: SeanB
author_rank: Super Contributor
timestamp: August 20, 2016, 07:40:19 am
content: Most of the energy is released in the spark as you connect them. Some as light, some as heat and some is used to vapourise the ends of the contact wires. The remainder is lost in internal resistances of the foils, lead wires and the wires joining the capacitors. A very small amount is lost in internal chemical changes of the 2 capacitors most commonly all bundled together under dielectric absorption.

author: helius
author_rank: Super Contributor
timestamp: August 20, 2016, 07:41:53 am
content: Quote from: tatus1969 on August 20, 2016, 07:11:29 amEven if the parasitic resistances are all idealized to be zero, there is a sudden movement of the electric field, and therefore an electromagnetic wave is emitted. The spark gap demonstrated by Heinrich Hertz is just a capacitor discharging through the air, and it emits EM energy over a broad spectrum.

author: bobaruni
author_rank: Regular Contributor
timestamp: August 20, 2016, 08:11:44 am
content: Doubling the capacitance using half the voltage will not give you the same charge because it's a squared relationship.Where has the charge gone? nowhere until you use it but now you have twice the capacitance.

author: tatus1969
author_rank: Super Contributor
timestamp: August 20, 2016, 08:34:35 am
content: Quote from: blueskull on August 20, 2016, 07:43:05 amhelius and SeanB already gave part of the answer: it is lost as heat. But your formulas also prove that the amount is independent of the rate at which you do that balancing process. Either if you allow it to develop sparks smoking, or you connect meg ohms between the two, you always lose exactly half of the energy.

author: tatus1969
author_rank: Super Contributor
timestamp: August 20, 2016, 08:38:00 am
content: Quote from: bobaruni on August 20, 2016, 08:11:44 amIt is not about losing charge (that is kept constant and shared between the two) but about losing energy. Half of the energy that you need to charge up the first capacitor is lost in heat (and raditation, which will eventually turned later into heat as well) during the balancing process when shorting the capacitors.

author: Zero999
author_rank: Super Contributor
timestamp: August 20, 2016, 08:50:28 am
content: Yes, it's true. This has been discussed before but I can't remember the thread.Here's another question: A 10,000µF capacitor is connected directly to a 10V power supply. How much energy is taken from the power supply to charge the capacitor? The answer is 1J, not 0.5J but why when E = 0.5CV2? It's same as above.

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: August 20, 2016, 09:09:29 am
content: I think in that other thread, there was a good analogy.Something like 2 similar water tanks,1 full and one empty, then you let them drain to contain half the water each.It is easy to see you have lost energy.

author: Zero999
author_rank: Super Contributor
timestamp: August 20, 2016, 09:14:45 am
content: Quote from: blueskull on August 20, 2016, 09:01:48 amYes, you're right. Missed out the 0.5! Post edited, thanks.

author: tatus1969
author_rank: Super Contributor
timestamp: August 20, 2016, 10:49:41 am
content: Quote from: Hero999 on August 20, 2016, 08:50:28 amI think the question needs to be extended: a 10mF capacitor is connected to a 10V power supply through a resistor. In that case you need 1.0J from the DC source. 0.5J goes into the capacitor, 0.5J goes into the resistor.In contrast, if you have a decent switch mode lab supply that has no internal losses, then you only need 0.5J from it to charge the capacitor.Reflecting this back to my question, this means: if we have a lossless way to transfer energy from one to the other capacitor, for example a current limited switching regulator, then we preserve all (most) energy during the balancing process.

author: tatus1969
author_rank: Super Contributor
timestamp: August 20, 2016, 10:54:19 am
content: Quote from: HackedFridgeMagnet on August 20, 2016, 09:09:29 amI don't get that. The water tanks store potential energy, which is m*g*h. This is linear, so the two half full water tanks account for the same amount of energy as one full tank.

author: ruffy91
author_rank: Regular Contributor
timestamp: August 20, 2016, 11:36:38 am
content: Quote from: tatus1969 on August 20, 2016, 10:54:19 amBefore:1*m*g*h=1 energyAfter:2*0.5m*g*0.5h=0.5 energy (0.25 energy per tank)

author: Kevin.D
author_rank: Frequent Contributor
timestamp: August 20, 2016, 11:39:50 am
content: Quote from: tatus1969 on August 20, 2016, 07:11:29 amThe energy will always be lost in any esr no matter how small it is, e.g reducing esr by half results in double current so I*I*R (power dissipation in esr doubles) but time to charge halves, so power*time remains a constant. An ideal capacitor (zero esr which of course doesn't exist) would imply infinite initial current flow and infinite current through zero esr would dissipate infinite power or zero power depending on what you consider infinity*zero to be, but calculating with infinities as variables always gives odd solutions .

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 20, 2016, 12:07:10 pm
content: Quote from: tatus1969 on August 20, 2016, 07:13:49 amIf you do it with literally nothing but capacitors, in the abstract, everything ideal,You divide by zero.Because such a situation is meaningless and impossible.At best, you could have a superconducting wire, between a pair of superconducting capacitors, in vacuum.In that case, the wire has equivalent inductance, so the result is an oscillator.The energy difference is transformed into AC resonant energy.The meaning of "voltage", in this case, is the average voltage on each capacitor, which follows charge conservation, as it must. It is not the instantaneous voltage, because that requires a more particular analysis: namely that an inductor's voltage equals the inductance times the second time derivative of charge (d^2 Q / dt^2). Put that together with the capacitor equation (Q = V*C) and you get a second order differential equation, which has an oscillating solution.Again, in reality, even the superconducting apparatus will dissipate power; real superconducting resonators (of a waveguide type construction, in polished niobium at 4K) exhibit a Q factor of several million at ~500MHz, so the AC decays in a few milliseconds. Maybe a few seconds for our hypothetical lumped-capacitor-inductor system that we might have resonate at a few MHz instead. (The reason for any dissipation at all, isn't well understood. Suffice it to say: superconductors are only so at DC. Indeed, type II superconductors are so bad, you can see it macroscopically: the phenomenon known as flux pinning, which holds a superconductor levitating in relative position over a magnet, is due to hysteresis loss in the superconductor. Every time you change the superconductor's relative position, you dissipate energy in the superconductor!)Note that two capacitors in series is just another capacitor; we can cut the problem in half, down the symmetry line, and have a wire shorting one capacitor to ground. The result is the same: zero average voltage, implying zero average energy, but AC oscillation is present, until it dissipates.Tim

author: firewalker
author_rank: Super Contributor
timestamp: August 20, 2016, 12:57:31 pm
content: It was a question in a uni exam I had years ago.A detailed explanation:https://arxiv.org/pdf/1309.5034.pdfAlexander.

author: Cerebus
author_rank: Super Contributor
timestamp: August 20, 2016, 02:38:44 pm
content: Quote from: SeanB on August 20, 2016, 07:40:19 amA further large amount of energy is used swearing and cursing at the burnt fingers caused by the spark.

author: tatus1969
author_rank: Super Contributor
timestamp: August 20, 2016, 03:23:27 pm
content: Quote from: Kevin.D on August 20, 2016, 11:39:50 amAs this is a linear problem, you could set up and solve the limes and you would still get the same result as for finite ESR values.

author: tatus1969
author_rank: Super Contributor
timestamp: August 20, 2016, 03:27:51 pm
content: Quote from: ruffy91 on August 20, 2016, 11:36:38 amOkay, that way it makes sense. I was thinking of pumping water into elevated reservoirs. In that case you can treat h as a constant.In that similarity (and if we consider that the water has no inertia, the energy loss would take place when it flows through the balancing pipe, in form of turbulence and friction.

author: tatus1969
author_rank: Super Contributor
timestamp: August 20, 2016, 03:31:35 pm
content: Quote from: firewalker on August 20, 2016, 12:57:31 pmThanks, very nice!

author: IanB
author_rank: Super Contributor
timestamp: August 20, 2016, 03:35:57 pm
content: Also, here is the earlier thread with some discussion of this:https://www.eevblog.com/forum/projects/why-trying-to-store-energy-in-a-capacitor-can-be-less-efficient-than-you-think/

author: chris_leyson
author_rank: Super Contributor
timestamp: August 21, 2016, 11:58:02 am
content: Thanks Tim, I liked your explanation of other possible loss mechanisms. The problem becomes even more interesting if consider a single variable capacitor instead of two fixed value capacitors. Take a 350pF+350pF dual gang AM variable capacitor for example, connect both sections in parallel and fully mesh the plates so you've got about 700pF, charge to say 30V from a bench supply via a series blocking diode and unmesh the plates as fast as you can, now you can generate enough voltage across the capacitor to fire a 90V neon. I'm guessing you would have to put mechanical energy into the system to overcome the electrostatic force between the capacitor plates, now it gets tricky

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 21, 2016, 12:27:08 pm
content: Yup! Cranking a capacitor requires mechanical work, which puts energy into the system.If the capacitor were part of an oscillator, and you pumped it so it gets a little charge during the low part of the wave, then gets cranked and dumps it into the big part of the wave -- congratulations, you now understand what a parametric amplifier is!Now here's a riddle: semiconductor junctions do it as a function of voltage. So, as the voltage is going up, capacitance drops, which draws work (W = Q*V, in joules, coulombs and volts, respectively) from the electric field -- because, where else can it come from?So, what is the fastest rate at which capacitance can decline? That is, if C(V1) = Co, what's the lowest voltage V2 where C(V2) = Co/10, say?Tim

author: tatus1969
author_rank: Super Contributor
timestamp: August 21, 2016, 02:18:17 pm
content: Quote from: T3sl4co1l on August 21, 2016, 12:27:08 pmDidn't think about this yet. But my guess would be that, because the electron-free region of the junction grows with increasing voltage, the work in question would be, separating the charges against electrostatic forces.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 21, 2016, 02:52:31 pm
content: Yes, that's what the work goes into. And it comes from the terminals.Suppose C(V) was a step function, so below some voltage, it's a larger capacitance, and above, it's smaller. Does that make sense -- is it reasonable? Possible? Consider: as voltage just barely ticks past the threshold, capacitance drops, so voltage jumps up suddenly. Either those intermediate voltages can't be reached (the capacitor exhibits hysteresis!), or if voltage is forced to be constant, then the capacitor suddenly dumps a wad of charge into the supply -- releasing energy!This is actually a bad riddle, because nothing is still violated in the above case, as weird as it is. So let me modify it a little: why don't you get a big lump of chargeoutof a semiconductor junction as you charge it?

author: alanb
author_rank: Frequent Contributor
timestamp: August 21, 2016, 05:04:47 pm
content: Quote from: tatus1969 on August 20, 2016, 06:41:07 amI think that the assumption that they will each have 0.5V is wrong. I would expect them each to have 0.707 volts (i.e. 1 divided by root 2). This is of cause in an ideal loss free situation. The energy has been conserved but the capacitance has doubled.

author: dmills
author_rank: Super Contributor
timestamp: August 21, 2016, 05:32:00 pm
content: 0.707V violates conservation of charge, which requires 0.5V (And is in fact correct).Regards, Dan.

author: alanb
author_rank: Frequent Contributor
timestamp: August 21, 2016, 05:39:28 pm
content: The charge is being conserved but the capacitance has altered.

author: tatus1969
author_rank: Super Contributor
timestamp: August 21, 2016, 05:41:15 pm
content: Quote from: T3sl4co1l on August 21, 2016, 02:52:31 pmI'll give it a try. Charge is equivalent to the number of electrons that we have collected on the one plate of the capacitor, respectively the amount missing on the other. When the step capacitor changes its capacitance because of that one too many electron, then their count does not change. But we have to do something inside the "step" capacitor, either by replacing dielectrica, or moving the plates away from each other. Both requires mechanical work, and that is what would cause the capacitor release energy. This answers the situation above, but I still don't have a clue of your first riddle.Quote from: T3sl4co1l on August 21, 2016, 02:52:31 pmHmmm....

author: tatus1969
author_rank: Super Contributor
timestamp: August 21, 2016, 05:42:38 pm
content: Quote from: alanb on August 21, 2016, 05:39:28 pmcharge before: 1.0V x Ccharge after in your case: 2 x 0.707V x C

author: alanb
author_rank: Frequent Contributor
timestamp: August 21, 2016, 06:11:15 pm
content: Ok. I should have checked that my brain was in gear before posting!

author: Jeff1946
author_rank: Regular Contributor
timestamp: August 21, 2016, 07:00:55 pm
content: It is the same math as two identical cars colliding, one going x miles per hour and the other zero. If they collide and stick together they will be going x/2 mph. Momentum is conserved always but one half the energy lost.