9
\$\begingroup\$

Capacitors store energy then give it back once required. A perfect capacitor is nearly lossless on DC power because you only fill it once then it keeps energy in it until you discharge it so no power loss to mention, but on AC the capacitors will be charged then discharged all the time which in my theory seems to waste power, is that true? If it is true then how can I measure that loss for a 250V 8000µF capacitor connected to 170V AC power line?

\$\endgroup\$
  • 2
    \$\begingroup\$ Yes. That is why AC not so good for undersea power cables. It is also why processors have dynamic power consumption (charging and discharging the transistor parasitic capacitance on every switch.. \$\endgroup\$ – DKNguyen Feb 19 at 15:31
  • 2
    \$\begingroup\$ You have to consider the ripple of current, then you will be able to calculate the RMS current, then ESR*I^2 will give you the losses. Datasheet of capacitors gives you the max ripple current admissible, if the ripple is too high your capacitor will get too hot and the lifetime will be shortened. \$\endgroup\$ – Delphesk Feb 19 at 15:56
  • 5
    \$\begingroup\$ A perfect capacitor wastes no energy at all when hooked up to a AC load. Power losses happen in real capacitors because they are imperfect. \$\endgroup\$ – user1850479 Feb 19 at 17:00
  • 8
    \$\begingroup\$ Perfect capacitors don't consume power. Real capacitors do. It may help you to google "capacitor ESR" and "capacitor loss tangent". Note that the ESR and loss tangent vary with frequency (in some cases it is a huge difference). So try to use the loss tangent at 50-120 Hz, not, say, 1 MHz. \$\endgroup\$ – mkeith Feb 19 at 19:23
  • 2
    \$\begingroup\$ @mkeith From first principles it follows that even a perfect capacitor (superconducting, whatever) consumes power because it must involve accelerating charges which always radiate photons. \$\endgroup\$ – Peter - Reinstate Monica Feb 20 at 16:39
16
\$\begingroup\$

There would be power loss because with real life capacitors, there are parasitic losses, this means that the capacitor can not act as a pure capacitive load in real life, this is mainly because of the building process/materials/sizes.

In real life capacitors have an ESL (Equivalent Series Inductance), an ESR (Equivalent Series Resistance), and a Leakage Resistance in parallel with the capacitor which is commonly notated as Rleak.

enter image description here

You would need to know the ESR to calculate power loss. But keep in mind that this is a parameter that degrades with component use.

As per the industry standards EIA-463-A, MIL-C-62F, under nominal operating conditions, capacitors are considered completely degraded and not usable in the circuit when it's ESR value reaches 2.8 times of the initial ESR value and capacitance decrease in excess of 20% of the initial value.

Link to original text.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ Why is the acronym for Equivalent Series Inductance ESL? \$\endgroup\$ – reggaeguitar Feb 20 at 21:36
  • \$\begingroup\$ @reggaeguitar acronyms.thefreedictionary.com/ESL says: "Equivalent Series Inductance (L is inductance)" \$\endgroup\$ – Crowley Feb 20 at 21:55
  • 1
    \$\begingroup\$ @reggaeguitar L is the standard symbol for inductance. Capital i had already been taken by current before inductance was discovered, unfortunately. \$\endgroup\$ – Mark Snyder Feb 20 at 22:06
9
\$\begingroup\$

There are three loss mechanisms within the capacitor, all of which are fairly minor, and one that it causes to the power supply, which depending on how you're billed for your electricity, may or may not worry you.

Within the capacitor, the electrodes have resistance, which causes \$I_{terminal}^2R_{electrode}\$ losses. The dielectric has a conductivity which is usually very very small indeed in plastic dielectrics, more significant in electrolytics, also causing \$I_{leakage}^2R_{dielectric}\$ losses. The dielectric also has hysteresis loss, also negligible in most plastics, more significant in electrolytics.

The charging/discharging current flows in your supply lines, but there's no nett power transferred to you by this, so most domestic meters will not charge you. This is usually called reactive power, or VAr. Commercial customers may be charged for VAr, as it causes the electricity infrastructure supplier to have invest capital in thicker supply cables than would otherwise be needed. However, as most commercial users tend to have inductive loads like motors, capacitors across the line will tend to reduce their VAr. This is why there is a market in large power factor correction capacitors to be used in shunt with large factories.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ "Leakage" seems to me a clearer description than "I^2R" of conductivity through the dielectric. It's a very different I than in the other formula, so I'd suggest not saying "also causing"; that wording seems to wrongly imply there's more similarity between those losses than both being resistive. Also, "net" has one "t", unless that's a technical term I'm unfamiliar with. \$\endgroup\$ – Peter Cordes Feb 20 at 5:37
  • \$\begingroup\$ @PeterCordes Leakage is the flow of current across the cap. However the OP asked about power, so I expressed that as power. Current flowing through a high value resistor, even if its resistance is Gohms or Tohms, still dissipates power. Of course the leakage I is a different measurement to the charging I, which loses power in the electrode resistance. I am aware of the correct spelling of net, it appeared in an anagram puzzle with my wife only last week. However, it looks wrong to me, and I reserve my right to use personal aesthetics rather than accuracy when not on the English Spelling stack. \$\endgroup\$ – Neil_UK Feb 20 at 6:03
  • \$\begingroup\$ Ok, fair points, thanks for explaining your reasoning. Glad I only commented instead of suggesting an edit. :) \$\endgroup\$ – Peter Cordes Feb 20 at 6:06
  • \$\begingroup\$ @PeterCordes I've taken your technical point, and clarified my answer. The spelling stays. \$\endgroup\$ – Neil_UK Feb 20 at 6:09
  • 1
    \$\begingroup\$ British variant spelling according to Collins. 'Dated' according to Wiktionary (how very dare they!) \$\endgroup\$ – nekomatic Feb 20 at 16:51
6
\$\begingroup\$

An ideal capacitor is lossless, the energy released during discharging is equal to that stored during charging.

However:

  1. Real capacitors are not ideal, the losses in a real capacitor on AC will likely be much higher than on DC, because on DC you only have leakage losses, while on AC you have losses caused by currents flowing in and out of the capacitor. The plates of a capacitor are usually very thin and so can have noticeable resistance.

  2. Even if the capacitor itself was lossless, the current flow caused by the capacitor can change the losses elsewhere in the system. In the simple case consider a capacitor connected to the grid by a long cable, current flow will cause resistive losses in the cable. On the other hand currents from capacitors can cancel with those from inductors, so it's possible for an appropriately sited and sized capacitor to actually reduce losses in the cable feeding an inductive load.

|improve this answer|||||
\$\endgroup\$
4
\$\begingroup\$

No. Pure capacitance does not waste power as heat. Which is why inductive loads such as motors or fluorescent light ballasts are often compensated with capacitors for the loads to look like pure resistance to achieve better power factor. However, non-idealities such as series resistance and dielectric losses do consume energy, so practical capacitors you can buy do waste energy some amount.

|improve this answer|||||
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.