Capacitor Discharge Circuit with Energy Recuperation - Page 1
Forum section: electronics / projects

author: luky315
author_rank: Regular Contributor
timestamp: November 09, 2012, 09:29:22 am
content: I have another HV-Problem:A capacitor (47µF) is charged to 100V. I need a circuit to discharge it in less than 5ms. My problem is that the circuit must be small and I can't simply dissipate the Power (for thermal reasons). The discharge Current must also be limited to minimize EMV-Problems. There are a lot of circuits for this problem which reuse the energy to recharge the capacitor (widely used for actuator drivers) but I don't need to recharge the capacitor, therefore I need a solution to use the energy for something else (powering the circuit or "putting it back" to the DC Power supply).Any Ideas?

author: Psi
author_rank: Super Contributor
timestamp: November 09, 2012, 10:38:26 am
content: There aren't many options left if you eliminate heat.motors/solenoids, light, sound,Sound and motors/solenoids isn't going to be small and light isn't very efficient so you're back to heat.

author: luky315
author_rank: Regular Contributor
timestamp: November 09, 2012, 11:11:50 am
content: I can't dissipate the heat on the PCB. The idea is to give the energy back to the DC input which powers the whole system including the charging circuit.

author: jeroen74
author_rank: Frequent Contributor
timestamp: November 09, 2012, 11:32:08 am
content: A buck converter with a slightly higher output voltage then the input voltage of the system.

author: Simon
author_rank: Unknown
timestamp: November 09, 2012, 12:58:33 pm
content: I'm afraid we can't break physics for you, you can't "vaporize" power in 5mS and make no heat. What is your power limit ?, just shorting the cap with a mosfet or similar sounds like the best option to me. What are you actually doing ?

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: November 09, 2012, 01:41:25 pm
content: If you want to get exotic there are some phase-change materials that will absorb heat by turning from solid to liquid, and release it over a longer period of time.However a single discharge of 47uf 100V - about a quarter of a joule - doesn't seem like too much heat to deal with if it doesn't happen very frequently . Compared to the volume of the capacitor, you wouldn't need much metal to store that & let it out more slowly - maybe a to220 sized MOSFET. Just need to make sure your heat-absorbing device has sufficiently low transient thermal impedance.Also, think about whether you actually have to discharge it all the way to zero in the 5ms.

author: luky315
author_rank: Regular Contributor
timestamp: November 09, 2012, 01:54:21 pm
content: I have to discare it frequently, to 0V and I don't want to "vaporize" the power.Just give it back to the main power rail.

author: Simon
author_rank: Unknown
timestamp: November 09, 2012, 01:58:41 pm
content: what is the main power rail ?, by vaporize i mean dissapere without trace, which is what your almost asking, nothing can be destroyed, everything is transformed. Even if you can feed this "power" back to the input that won't discharge the capacitor to 0V

author: jeroen74
author_rank: Frequent Contributor
timestamp: November 09, 2012, 02:15:25 pm
content: without some more details its impossible to help you. What is frequently? One per second, once per hour? What is the main rail voltage? How much power does the device use? What exactly is this HV cap used for? Why charge it up, then discharge? That's like filling the bathtub with the tap running and filling a bucket at the same time from that same tap, and when the bucket is full dump it into the bathtubA similar application: old flash units that measure actual exposure use a very small flash tube to dump the excess energy in the main cap when enough light has been produced by main xenon tube. That will only reduce that cap voltage to 50V or so when the tube stops conducting, but at least you have lost 75% of the energy.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: November 09, 2012, 02:22:59 pm
content: Quote from: luky315 on November 09, 2012, 01:54:21 pmWill whatever is providing the main rail deal with it? What happens if it has a light loat - can it tolerate a high voltage spike?if so then the only way would be a boost converter (could be a very crude, loosely regulated one). You will need significant space for an inductor though. Minimum hardware would be a switching transistor, diode, inductor and source of frequency to drive the transistor. You'd probably want to sweep the duty cycle to even out the current over the 5mS. You'll probably also want something to stop it if the rail voltage gets too high to prevent killing the transistor.

author: G7PSK
author_rank: Super Contributor
timestamp: November 09, 2012, 07:28:49 pm
content: Sounds like a perpetual motion device.

author: luky315
author_rank: Regular Contributor
timestamp: November 09, 2012, 09:32:38 pm
content: The main power comes from a big 24VDC industrial Power supply with big output capacitors.The discharge Frequency is 100Hz (5ms to charge, 5ms to discharge). So yes, heat is a problem.And no, it's not a perpetual motion device but a test circuit for the charging unit. I tried with a simple circuit but it won't work for very long because of the thermal issues.

author: Simon
author_rank: Unknown
timestamp: November 09, 2012, 10:09:05 pm
content: if it's a test circuit why is the heat a problem. use suitable parts and you will be ok

author: luky315
author_rank: Regular Contributor
timestamp: November 09, 2012, 11:13:59 pm
content: I have a very limited space and almost no air (vacuum!). The main power comes from the outside. I can't dissipate the heat inside the system.I won't be asking for a circuit for energy recuperation if a simple Mosfet would solve the problem...

author: Simon
author_rank: Unknown
timestamp: November 10, 2012, 07:50:50 am
content: then take a pair of wires out of the box to discharge the cap from with a mosfet on the outside. remember what I said at the beginning ? you can't break physics. you are asking for the power to be made to vanish, can't be done. Do you need to discharge to 0V ?

author: luky315
author_rank: Regular Contributor
timestamp: November 10, 2012, 09:34:14 am
content: I don't want the power to vanish. I want to "give it back" to the main supply rail.I tried a lot of "simple" solutions but they don't work in my case.I need a circuit to recover the energy. Nothing else will work in this case.And I said already that I need to discharge it to 0V.

author: Simon
author_rank: Unknown
timestamp: November 10, 2012, 09:46:38 am
content: Quote from: luky315 on November 10, 2012, 09:34:14 amThe two above statements make your requirement impossible, you have set a specific discharge time and want it to fully discharge the capacitor but you also want to "take the time" to reprocess the power and if your power input is 24V you won't get the capacitor down to 0V as there will be a minimum voltage for the buck/boost circuit.I'm no expert of buck/boost circuits but i think you need to get it down to about 10-24V and then short the capacitor for the remaining voltage, you need to design a circuit to cut in the buck/boost converter at the right time and cut it off again. you also need to know that the power supply has enough load on it to use that power fast enough and your output voltage will need to be slightly higher than then main input voltage so that it takes over in order to discharge the capacitor fast enough.Good luck

author: jeroen74
author_rank: Frequent Contributor
timestamp: November 10, 2012, 10:57:20 am
content: What supplies the energy to the cap?

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: November 10, 2012, 10:58:33 am
content: Quote from: Simon on November 10, 2012, 09:46:38 amA simple boost circuit could achieve 0V without additional hardware as you have a DC path through the inductor and switch. Remember you still have the incoming supply available to power the converter.

author: Jovian
author_rank: Contributor
timestamp: November 10, 2012, 11:59:52 am
content: If you want to put the energy from the capacitor into the 24V rail then you will need a buck converter. Then once you have the capacitor down to say 30V you can dissipate the rest with a resistor. But 5mS is not long, and you are going to have to dissipate about 50W on the 24V rail. Will the power supply mind that it is supplying a load that effectively pulses at 100Hz?If you are testing a charger why use a capacitor as the load, unless you want to test high current impulses, which could easily be emulated with a power transistor (or multiple in parallel depending on the current). Could you use a resistor as the load, or put a resistor in series with the cap so that you have much less energy to deal with? (Remember the energy stored in a capacitor is proportional to the voltage squared.)Regards,Jovian.

author: Simon
author_rank: Unknown
timestamp: November 10, 2012, 12:10:36 pm
content: Quote from: mikeselectricstuff on November 10, 2012, 10:58:33 amok but surely at near 0V that will happen slowly ? can all this happen in 5mS ?

author: G7PSK
author_rank: Super Contributor
timestamp: November 10, 2012, 05:15:31 pm
content: Use the heatsink of the switching transistor as the discharge resistance, paint it black and it will radiate the heat in a vacuum. Just give up on trying to recover all the energy,surely it cant be that important as it is just for testing purposes.

author: Simon
author_rank: Unknown
timestamp: November 10, 2012, 05:39:22 pm
content: that is what I don't understand, if this is a test why all this fuss

author: free_electron
author_rank: Super Contributor
timestamp: November 10, 2012, 05:50:30 pm
content: pumping energy back into the rail isone thing. it needs to be consumed there as well. if there is no load on the rail that can handle this energy than nothing happens....if you are in vacuum your only option is to radiate it off as photons. dump the excess energy in a lightbulb... or a bunch of led's. the emitted light is effectively energy being wicked away...

author: KedasProbe
author_rank: Frequent Contributor
timestamp: November 11, 2012, 04:49:18 pm
content: Does it have to be recovering the energy in every cycle or would it be OK to load to somewhere else and then later with a much lower frequency give it back to the rail.Connecting a 10 times bigger capacitor parallel will drop the voltage to 1/11 without much loss of energy. The last bit of load on your cap (like 10%) you can decide to lose in heat to get to 0V.You can mux between a few large caps and feed them later together to the rail or battery. (maybe connect them in serie)basically a DC-DC converter.Theoretically a variable capacitor would allow you to play with the voltage without changing the load.Just an idea not really sure about the schematic.

author: Rerouter
author_rank: Super Contributor
timestamp: November 12, 2012, 01:58:53 am
content: you could possibly get away with a small dumbed down fly-back converter, short out the cap to ground through a small transformer via a transistor or fet, and on the fly-back kicks out a large enough voltage to feed most of it back into the supply rail. while at the same time pulling a negative on your cap which provided you limit it with a diode, should be able to discharge it in a sufficient time frame,

author: NiHaoMike
author_rank: Super Contributor
timestamp: November 12, 2012, 02:57:41 am
content: What about make the charging converter bidirectional? Just adding a MOSFET and drive circuit to the secondary side might be all that's needed.

author: Simon
author_rank: Unknown
timestamp: November 12, 2012, 06:56:44 am
content: How about using the power to run a 30-50V motor ?I mean 5mS charge and 5mS dicharge will be something like 30-50% duty

author: KedasProbe
author_rank: Frequent Contributor
timestamp: November 12, 2012, 10:21:36 am
content: Some other numbers to help ideas.The average current from the capacitor will be about 1A.100V*47µF/5ms = 940mA100²*0.000047 / 2= 0.235 Joules (1 charge)0.235J/0.005s = 47W (average inside the 5ms, apply 'duty cycle' to this)

author: mzzj
author_rank: Super Contributor
timestamp: November 25, 2012, 10:18:23 am
content: Quote from: Simon on November 12, 2012, 06:56:44 amEnergy does not disappear in the motor. Motor converts electrical energy to mechanical energy and small portion goes to losses mainly as thermal energy.Then you have to get rid of the mechanical energyInstall brake to motor and then you got unwanted thermal energyMore crazy ideas (instead of dumping energy back to charging converter or supply rail):Convert it to RF-energy and let it radiate to free space

author: G7PSK
author_rank: Super Contributor
timestamp: November 26, 2012, 03:53:34 pm
content: Use the motor to drive a dynamo and feed that power back int the system. But to my mind unless you are testing thousands of these capacitors simultaneously it would be more economical to just dump the waste power.

author: Hugoneus
author_rank: Frequent Contributor
timestamp: November 26, 2012, 11:45:30 pm
content: I skimmed through this, so I apologize if I am missing something.It seems that you are charging and discharging a cap at a rate of 100Hz. If you are charging a cap, by any traditional means, you will dissipate as much power as you put into the cap. E = (1/2)*C*V^2.Note how this equation does not have a resistance term in it! This means, even if you have two ideal capacitors (with ideal 0-ohm resistors) and one is fully charged and the other fully discharged, as soon as you connect them in parallel, some charge sharing will happen. After that, the total energy stored in both caps is half of what used to be stored in the first fully charged cap. The second half will disappear! In the case of ideal caps and wires, the energy will dissipate in the form of light, sound, and EM waves (from a spark). If there is a resistor in the middle, it will be dissipated mostly as heat.Considering that, I don't see how you can charge and discharge that cap at 100Hz without dissipating (at least) as much power as you put into the cap.

author: Simon
author_rank: Unknown
timestamp: November 27, 2012, 07:56:56 am
content: that is what we have all been trying to explain to him

author: luky315
author_rank: Regular Contributor
timestamp: January 06, 2013, 03:02:46 pm
content: The old story that you loose 1/2 of the energy if you connect 2 capacitors is only true if you connect them directly. With 2 Diodes and an inductor you can discarge one capacitor into another with >90% efficiency.

author: Kremmen
author_rank: Super Contributor
timestamp: January 06, 2013, 10:32:32 pm
content: WTF???What story is this? Do you seriously propose that energy _disappears_ somewhere? Certainly not - you must have heard about conservation of energy? (or mass-energy if you insist on picking nits).Assuming ideal caps and conductors and forgetting sparking, the energy goes nowhere else than into the caps.Further assuming you have equal caps, one discharged and one charged to voltage V, connecting them to parallel will result in both having half the original energy resulting in voltage V/sqrt(2) for both.

author: jeroen74
author_rank: Frequent Contributor
timestamp: January 06, 2013, 10:36:37 pm
content: No, half the voltage. There are quite a few discussions and papers on this on the net. It's an age old problem.

author: Kremmen
author_rank: Super Contributor
timestamp: January 07, 2013, 07:21:22 am
content: Do you have link(s)? That would be real interesting, because the energy has to go somewhere. Stating that half disappears just is not an answer.P.S. Ah, i see it now. You are right as far as the voltage goes. In my consternation over the energy loss i neglected to account for all physical effects actually occurring in the cap.Indeed, you only see half the voltage, but that definitely does not result from energy being lost somewhere. The "missing" energy is needed to maintain the charge concentration in the electrode plates of the cap. So no mystery after all.But please, do not say that some of the energy is lost because that is not the case. The total system of 2 ideal caps maintains its energy perfectly, some of the energy is just in a form that is not visible as voltage across the cap.P.P.S. Regarding charging and discharging the OP's cap is a different question and there of course you need to transfer the full cap energy back and forth to accomplish the goal. That energy must go to / come from somewhere at each cycle.

author: Hugoneus
author_rank: Frequent Contributor
timestamp: January 08, 2013, 01:44:26 am
content: Quote from: Kremmen on January 07, 2013, 07:21:22 amIf you connect a fully charges cap across a fully discharged cap the total energy that remains in both caps after everything is settled is half as much as you started. If the wires are ideal and there are no resistances, the energy dissipates into the atmosphere in the form of EM energy, light, heat and sound when a spark happens.

author: Hugoneus
author_rank: Frequent Contributor
timestamp: January 08, 2013, 03:43:37 am
content: Quote from: luky315 on January 06, 2013, 03:02:46 pmCan you provide a circuit for this?

author: Simon
author_rank: Unknown
timestamp: January 08, 2013, 06:53:10 am
content: Quote from: Hugoneus on January 08, 2013, 01:44:26 amyou sure? light?

author: Kremmen
author_rank: Super Contributor
timestamp: January 08, 2013, 10:37:57 am
content: Quote from: Hugoneus on January 08, 2013, 01:44:26 amThat cannot be right. Assuming the system is ideal there by definition will not be any losses. So no heat, no light no nothing related to any imperfection in the connection itself.If we are extremely strict, then yes all accelerated electrons do radiate EM quanta and you can't move the electrons from one cap to another without accelerating and decelerating them.Hmm, let's think about that for a moment... OK, heat and light are EM radiation so while there are no losses in an ideal system, the acceleration radiation technically qualifies because it is a property of quantum mechanics and EM in nature. Also now that i am thinking about it, there are 2 further possibilities; either we assume the connection to be totally lossless or then there are wires that have a distributed inductance and capacitance. The wire capacitance can be assumed to be not significant but the inductance will create a lossless resonant circuit with the original caps.So in the case of no inductance and no resistance we will have one hell of a current pulse when the charge carriers balance (no losses remember). I am too lazy to think how to calculate the current magnitude, but probably you need to do it the hard way from first principles. Electric field force accelerating the mass of an electron or whatever you want to use as the charge carrier. Nope, too lazy to start that. The peak current will anyway be _high_.And if we do have inductance, there will be a lossless resonant circuit. Only not totally lossless, because accelerating the charges back and forth will again cause EM radiation only on a vastly lower scale than in the previous case. So eventually this damped oscillation will stabilize (provided the universe lasts long enough) and the charges in the caps will be equal. Of course i just say so; a lossless oscillation never stops but at least you can calculate the limit value where the stable point would be.Without actually calculating it i am prepared to accept that the remaining charge in the caps is only half the original. The main thing was that the energy does not "disappear" into nevernever land.It is a bit surprising but on the other hand it shouldn't. I vaguely recall some theory related to power transmission, where you can only get maximum of half to the other end. Been too long, can't really recall any more but must be the same thing...

author: jeroen74
author_rank: Frequent Contributor
timestamp: January 08, 2013, 10:52:27 am
content: Infinitely short current pulse of infinite magitude (Dirac pulse).

author: Kremmen
author_rank: Super Contributor
timestamp: January 08, 2013, 11:11:01 am
content: In pure theory yes, but if you include the mechanism of current flow (i.e. mobile charge carriers) then no, the pulse will not be infinite even in the ideal case. Dirac delta function is an abstraction that cannot be realized outside pure math. Not even in a lossless ideal electric circuit, provided you account for the necessary physical phenomena, i.e. provide a mechanism for the events in the circuit but neglect losses. And you must provide the mechanism, otherwise the fundamental explanation of energy dissipation does not work.

author: luky315
author_rank: Regular Contributor
timestamp: January 15, 2013, 10:49:32 am
content: A circuit from another Forum:http://www.mikrocontroller.net/attachment/91031/umlad.png