Power dissipation in capacitor --||-- - Page 1
Forum section: electronics / projects

author: MoraPol
author_rank: Newbie
timestamp: October 22, 2019, 09:59:09 am
content: Hello,I have problems understanding the maximum ripple current of an electrolytic capacitor. I thought the max ripple current is linked to the thermal destruction of the capacitor, cause the dissipated power is R*I² where R is the ESR of the capacitor.Also I thought the power a capacitor can dissipate should be a fixed number at a given temperature for a certain capacitor.After some calculations I got some really crazy results, my capacitor can dissipate 0.32W at 120Hz and 0.05W at 100KHz.I picked a panasonic FR Type with 220µF and 35V (8x11,5mm)https://industrial.panasonic.com/cdbs/www-data/pdf/RDF0000/ABA0000C1259.pdfMax Dissipation for 120Hz with datasheet values:tan δ= 0.12 (at 120Hz).The max ripple current is 0.95A. (100kHz)Correction factor for Ripple Current at 120Hz = 0.7ESR = tan δ * | Xc|ESR = 0.12 * (1/( 2 * π *120Hz * 220µF) = 0.72 ΩPmax= R*Imax² = 0.72 Ω * (0.7 *0.95A)² =0.32WNow I calculated the max dissipation for 100kHz:Correction factor for Ripple Current at 100Khz= 1.0Total impedance is given |Z| = 0.056 Ω.I now this isn't the ESR, cause in the equivalent circuit we have XCand XLin series with ESR.Xccan be negative and therefore be subtracted of Z, which can lead to an ESR higher than |Z|. However Xcis much smaller than 0.056 Ω and can be neglected.--> ESR ≅ 0.056 ΩP = R*I² = 0.056 Ω * (1.0 * 0.95A)² =0.05WSo what is wrong here? Is there an error in my calculation or an error in my logic? I cant imagine, that the capacitor can dissipate once only 0.05W and on another time 0.32W.Thanks for your help.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 22, 2019, 10:20:34 am
content: Unsurprisingly, the two figures are different measurements.I'm honestly not sure why they wouldn't run it at the same power, give or take where that power is dissipated. But EM forces will tend to make that power dissipate towards the ends of the capacitor, not the center, so it shouldn't be a problem for peak temperature in the middle. And even then, I don't think that would account for an order of magnitude difference.Perhaps this would be a good opportunity to set up a test circuit and see what it really does?Tim

author: sourcecharge
author_rank: Regular Contributor
timestamp: October 22, 2019, 11:23:59 am
content: Ripple current is the current that the cap can safely hold for the rated life expenctancy. The cap can hold more, but temp will increase and the life expectancy decreases. When the life expenctancy decreaces, dissipation factor (DF), or tan delta, increases. The DF is varialbe with frequency, and it is not linear, nor is only increasing or decreasing with freq.If you are using it to filter DC ripple, the lower frequency DF applies.Electrolytic capacitors are usually only good at low frequencies, so I'm guessing that the lower losses are from different components' lower DF or higher Q at higher frequencies.DF is a ratio of the energy lost / the total amount of energy.0.12 tan delta is 0.12 DF or 12% DF, which means, 12 % energy loss right off the bat.Your calculations may be the problem.ESR = DF^2/(1+DF^2) * EPREPR = 1/ (w x C x DF)(EPR is equivilant parallel resistance)If you are interested in decreasing you energy losses, you have to decrease the total DF of the system.It seems as if increasing the frequency, decreases your losses of the power conversion.There are better DF electrolytic capacitors but the best caps are metalized polyester type caps.Metalized Polyester dielectric type caps seems to be the best price to DF ratio.Their DF decreases as frequency increases.Unfortunately, they are quite expensive.

author: MoraPol
author_rank: Newbie
timestamp: October 22, 2019, 02:32:19 pm
content: Thank both of you for your answers.@ T3sl4co1lI also thought that it could be the distribution of the heat, that would be the only plausible reason for me, why in one case the dissipated power is much less. Also, I thought that through the skin effect on high frequencies the ESR might increase, but it’s falling from 120Hz to 100kHz.Yes I think I will test it.This time I thought I calculate something before buying caps, buts it seems to be more difficult than I thought.@ sourcechargeYour hint let me think for a while. I searched some time, but could not find further evidence, that the EPR is responsible for an additional loss and therefore decreased power dissipation capability.In the documents I found the losses are all modelled in one resistor the ESR, so I think Panasonic is doing the same. It should suffice, if I calculate the losses with the ESR?I couldn’t find the origin of your formula, but I calculated the ESR with it: 0.86mΩ, which is much less than my assumption. The losses would be further decreased.So its still unclear for me, where this large discrepancy in the losses comes from. Maybe I should ask Panasonic.Sure I can buy other caps or just some with a higher ripple rating, but from a cost and space perspective I would like to buy one which isn’t much oversized.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 22, 2019, 02:59:20 pm
content: This may be less confusing,https://en.wikipedia.org/wiki/Dissipation_factorSimply put:DF = 1/Q,ESR is Q times less than Xc, orEPR is Q times greater than Xc.Quote from: MoraPol on October 22, 2019, 02:32:19 pmIt will, but first the DF effect needs to drop off -- the result is a bathtub curve for ESR.At low frequencies, the voltage drop across the capacitance is significant; we might model this as a parallel resistance. At higher frequencies, ESR of the electrolyte (at frequencies high enough that ionic motion is negligible) and skin effect of the assembly start to take over, and ESR rises again.So it depends where the ESR is measured, because ESR varies down and up with frequency. It isequivalentafter all! A geometrically distributed resistance at that, i.e., a bunch of R+C's in parallel where the value of each R+C is a constant factor from the last. This gives an impedance that changes more gradually with frequency than a capacitor (Z ~ ω^-1) or inductor (Z ~ ω), but not so gradually as a resistor (Z = R ω^0 = R).Physically, this arises from goopy behaviors like ionic motion in the electrolyte and relaxation of polarization in the dielectric. The motion can be direct diffusion like ionic motion in a viscous liquid, or it can be the collective motion of a lot of very small domains. For example, crystals in the dielectric might change their polarization at a given time constant, but the distribution of those crystals, and their time constants, follows a power law, so their collective average response also follows a power law.QuoteRight, they are equivalent, as the name suggests. Pick whichever is more convenient, just don't use both at once!Tim

author: duak
author_rank: Super Contributor
timestamp: October 23, 2019, 04:31:13 pm
content: Further to what Tim says, I'd expect some of the losses to be localized, ie. hot spots would develop because of differences in electrolyte layer thickness. Even if the average temperature was below the boiling point, some areas could be hot enough to boil and I can't imagine that being good.

author: bson
author_rank: Unknown
timestamp: October 24, 2019, 08:14:44 pm
content: You shouldn't factor reactances into ESR. Reactance is by definition lossless.