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I recently bought two 3300uf 100v capacitors and have connected them in parallel. I charge them up to 100v and discharge them. I then hook up a multimeter and notice the voltage going up very slowly, about .01 volts every 20-40 seconds. So I discharge the capacitors and the voltage goes back to zero. When I woke up this morning, I checked the capacitors and it had gone up to 5 volts! And I am able to power an LED with them. What is going on here?

Edit:

Thanks to Robert's comment in one of the answers, I think he's right. This is probably dielectric Absorption.

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    \$\begingroup\$ WOW! Really good observation \$\endgroup\$ – slebetman Jan 29 '17 at 2:00
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What you've observed is called "dielectric absorbtion" or "recovery voltage phenomenon".

It's cause by kind of interia of the dipoles (ions) in the electrolyte while charging and discharging.

From wikipedia:

Dielectric absorption is the name given to the effect by which a capacitor, that has been charged for a long time, discharges only incompletely when briefly discharged. Although an ideal capacitor would remain at zero volts after being discharged, real capacitors will develop a small voltage from time-delayed dipole discharging, a phenomenon that is also called dielectric relaxation, "soakage", or "battery action". For some dielectrics, such as many polymer films, the resulting voltage may be less than 1–2% of the original voltage, but it can be as much as 15% for electrolytic capacitors.

Further:

When the capacitor is discharging, the strength of the electric field is decreasing and the common orientation of the molecular dipoles is returning to an undirected state in a process of relaxation. Due to the hysteresis, at the zero point of the electric field, a material-dependent number of molecular dipoles are still polarized along the field direction without a measurable voltage appearing at the terminals of the capacitor. This is like an electrical remanence.

From a Mouser note

7 Recovery Voltage

Where a capacitor is once charged and discharged with both of the terminals short-circuiting and then left the terminals open for a while, a voltage across the capacitor spontaneously increases again. This is called “recovery voltage phenomenon”. The mechanism for this phenomenon can be interpreted as follows:

When charged with a voltage, the dielectric produces some electrical changes within, and then the inside of the dielectric is electrified with the opposite polarities (dielectric polarization). The dielectric polarization occurs in both ways of proceeding rapidly and slowly. When a charged capacitor was discharged until the voltage across the capacitor disappears, and then being left the terminals open, the slow polarization will discharge within the capacitor and appear as recovery voltage. (Fig. 28).

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    \$\begingroup\$ SAFETY ISSUE: High-voltage capacitors, when in storage, should have a shorting wire installed. Otherwise they may rise to dangerous voltage over hours/days, from recovery of charge that had been injected into the dielectric. (Dielectrics aren't perfect insulators, they're just large resistors.) This effect is larger whenever the capacitor had been charged to high voltage for long periods. HV capacitors in circuit should have HV "bleeder resistors" in parallel, to eliminate the problem. \$\endgroup\$ – wbeaty Jan 29 '17 at 0:29
  • \$\begingroup\$ Where does the energy come from? Surely we cannot build a perpetual motion machine with disconnected capacitors. \$\endgroup\$ – dotancohen Jan 29 '17 at 13:42
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    \$\begingroup\$ Note: I'm not a physicist. In a capacitor energy is stored by orienting dipoles along an electrical field. Energy in = alignment. Energy out = restore disorder. Releasing that energy is archived by relaxing the dipoles. D/A appears when rapidly discharging a capacitor in short time. To my understanding it comes from some dipoles relaxing slower than others. So they provide energy with some time delay. You may ask another question focusing on the cause of D/A - either here or at the physics stack. There's is already a question with answer there \$\endgroup\$ – try-catch-finally Jan 29 '17 at 21:04
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    \$\begingroup\$ @dotancohen - Capacitors are imperfect, that's how. You put in 1J, in a normal discharge you get 0.9J back, with this you may get 0.05J more. (Numbers are just a guess). \$\endgroup\$ – TLW Jan 30 '17 at 5:32
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Actually the dielectric absorbtion story is ok, but complex To put it short:The electric field deforms the molecule structure just like some thick and soft cloth gets gets a dimple when you press it by your hand. By the time that dimple vanishes and the cloth rises a bit upper again against the gravity.

The molecular deformations in the electrolyte and insulation layer reverse gradually and the ionic particles return to their original places. That means a new electricity, because the charge distribution is changed.

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  • \$\begingroup\$ NO. DA is caused by the actual movement of electrons in the dielectric (sometimes called space charge). The electric force is very strong and the number of electrons involved is small. This has been studied to death. iequalscdvdt.com/miscellaneous.html \$\endgroup\$ – Robert Endl Feb 15 '17 at 18:29
  • \$\begingroup\$ @RobertEndl The molecular structure IS due the electrons and their possible states. Only electrons keep the material together. Plenty of them have loose orbits, but definitely some limits exist because this is not a conductor. There's no contradiction. The space charge = those electrons in the molecular structure that have plenty of variation possiblities in how they take their places in all possible states. \$\endgroup\$ – user287001 Feb 15 '17 at 19:20
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Dielectric absorption Equivalent Circuit

  • 100V to 5V C1V1=C2V2 before = after discharge after long time
  • Main Cap C1= 3300uF at V1=100V and V2=5V
  • therefore C2 = C1 * V1/V2= 66 mF equiv dielectric absorption Capacitance
  • ".01 volts every 20-40 seconds" or 10mV/20s=dV/dt thus Voltage rise on C2 at 100V and C1 at 0V
  • The discharge on C2 at V1=100V due to series ESR2 on absorption cap, C2
    • Ignoring leakage R for now,
    • V2/ESR2 = Ic2 = Ic1 = C1 * dV1/dt or
    • ESR2 = V1/C1 * dt/dV1 = 100V/66mF * 20s/10mV = 3MΩ

schematic

simulate this circuit – Schematic created using CircuitLab

Note for old E-Caps each component value in the equivalent circuit can be estimate by various tests. Your test estimates the ESR2 * C2 = T2 = 180ks C2/C1=20 with as the absorption/cap ratio.

Bed Side notes

  • if dV/dt was 10mV/30s , can we estimate the minimum amount of sleep you had?
    • if 60% charge time of 5V is reached at a 10mV/30s rate this would take 5V/10mV *30s = 15ks = 4.17 h without knowing the dV/dt in the morning rate, we can only assume it was much lower such as 2T or 3T meaning 8 h or 12 h sleep or the ESR2 reduced overnight.

Parallel Leakage R values are known to reduce with aging and conditioning old large E caps using a large Series R will raise the leakage R values towards the original values in many cases. This is a safe practice when dealing with big-old low ESR caps to prevent interlayer short circuits.

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Different possible explanations:

  • Static electricity: it's perfectly normal that there's a potential difference between ground and e.g. clouds. That can amount to quite a lot of V/m in "free air", but with very little charge behind it, ie. trying to measure that is practically impossible. However, if you have two electrodes stretched across that gradient, you'll likely be able to charge your caps.
  • RF harvesting: Everything acts as an antenna. Now, that doesn't matter because RF induction is by definition AC and will cancel itself out, but if you happen to have some rusty/salty/... conductor interface, that can act as an accidental diode. See: early crystal radios.
  • Residual charge: I don't know your caps well, but maybe it's chemically sound to assume that shorting them doesn't eliminate all energy stored within, but only what is accessible quickly. In that case, if you'd let caps shorted for longer, this effect wouldn't take place.
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your multimeter is charging up the capacitors.

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    \$\begingroup\$ I wouldn't assume that's the case for voltage measurements. For resistance measurement, yes. \$\endgroup\$ – Marcus Müller Jan 28 '17 at 18:07
  • \$\begingroup\$ I doubt it's the multimeter because I left them sitting overnight with the multimeter unconnected \$\endgroup\$ – Erik Jan 28 '17 at 18:08
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    \$\begingroup\$ I think you just discovered dielectric absorption. Look it up. \$\endgroup\$ – Robert Endl Jan 28 '17 at 18:12
  • \$\begingroup\$ @RobertEndl Reading about it, you're probably right. Thanks! \$\endgroup\$ – Erik Jan 28 '17 at 18:18

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