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How to find out the efficiency of an aluminum electrolytic capacitor like this one.

How to find out resistance of such a cap in order to calculate the time it takes for it to charge up or discharge?

The resistance would also allow to account for power loss as stated above. Is there any hint in the datasheet enabling me to calculate the desired values?

@ Christian B: Thank you but the ESR or dissipation factor is given just for 100 Hz. What if I wan to discharge the cap just once a day (-> low frequency)?

$$ESR = tan\delta * \frac{1}{2\pi fC}$$

If I assume f = 1 Hz for example and C = 2 mF, I will get approx. an ESR = 10 Ohm. But assuming f = 1 Hz seams not reasonable to me though.

@ Marcus Müller: The efficiency I am looking for is basically the equivalent ohmic resistance during charging and discharging. Once having that number, together with charge/discharge time and current/voltage, I could calculate the power dissipated in that resistance. Of course the leakage current is important too. At least when the cap stays charged for "longer time".

Or is there a different approach? I simply want to intermediately store energy in that cap.

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  • \$\begingroup\$ take a look at the so called E(quivalent)S(eries)R(esitance). This is not a fixed value but highly dependant on other parameters. \$\endgroup\$ – Christian B. May 13 '19 at 11:58
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    \$\begingroup\$ Efficiency is also not really a feature that you can attach to a capacitor – efficient at what? In an LC low pass filter, the ohmic resistance would be a problem. In a stabilizing role on the output of a power supply, having too little ESR could actually make the thing oscillate, so a "good" resistor might actually be inefficient in this particular role. In your very-slow-discharge example, the efficiency as an energy storage device would be mostly defined by the leakage current. \$\endgroup\$ – Marcus Müller May 13 '19 at 12:05
  • \$\begingroup\$ @stowoda Replying to people in the comments is useful, as people don't get notified of edits to the question, even if you mention them. That "equivalent resistance during charging and discharging" is called ESR, and most capacitors, especially electrolytics, will give a rated ESR. \$\endgroup\$ – Hearth May 13 '19 at 12:32
  • \$\begingroup\$ @Hearth thanks. Was not aware about that.. Reagarding the topic: The ESR I already considered but how to find out the frequency for a one time charge/discharge. Basically I'd like to know how fast I could charge or discharge the cap (I feel like the resistance should be in the mOhms range but I have not read about it in any datasheet). \$\endgroup\$ – stowoda May 13 '19 at 12:39
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    \$\begingroup\$ According to the datasheet I_leakage-after-2-min-at-U_R< 0.01 x C_R x U_R. So with 2.2 mF and 10 V we are talking about I < 0.22 mA (which is actually given in the table as well). So comparing the charge with the discharge current we are talking about seconds to minutes in the worst case scenario. If you want to make it super correct you can use an exp function with a decay constant of 0.01 * 0.0022 if I am not mistaken. \$\endgroup\$ – Christian B. May 13 '19 at 14:23
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You will have two principal sources of energy loss.

1) ESR, Equivalent Series Resistance, which will dissipate power due to \$I^2R\$ losses when you are charging and discharging.

2) Leakage current, which will dissipate power any time the capacitor has a voltage on it.

The data sheet gives the ESR for 100Hz. This is a reasonable frequency to specify, as it's the ripple frequency for a simple full-wave rectifier + capacitor power supply. The ESR at low frequencies will be fairly constant, the figure they give will be a good estimate of the DC and low frequency value. If you want a better figure, you will have to measure it at the frequency of your choice.

Leakage current is a very different beast. It's very dependent on the charge history of the capacitor, the temperature, and the capacitor's age. Once the capacitor has been 'soaked' for a few hours, that is, charged to a higher voltage than you're going to use it, the leakage tends to settle down to a much lower figure than the worst case in the data sheet. You will need to measure it for your capacitors and temperature, by measuring the self-discharge rate. Do not leave a meter connected when you do this, connect it briefly, read, and disconnect. You might want to estimate how long you've connected the meter for, and deduct its effect from the loss of charge you observe.

Dielectric memory can change the voltage orders of magnitude faster than leakage current can, beware its confounding effect. Fun experiment - leave the capacitor charged for 10 minutes, discharge it quickly (a few seconds), connect a DMM and watch its output voltage rise over the next few minutes!

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  • \$\begingroup\$ It's the ripple frequency in most of the world--and close enough to the ripple frequency in 60Hz places that there shouldn't be any real difference, but it's probably worth mentioning that! (there's also that one tiny legacy grid that runs at 16.67Hz but no one cares about that) \$\endgroup\$ – Hearth May 13 '19 at 14:02
  • \$\begingroup\$ @Neil_UK 12 mOhm @ 100 Hz is pretty low. So charging the capacitor will not dissipate much (current should be around 200 mA). \$\endgroup\$ – stowoda May 13 '19 at 19:07
  • \$\begingroup\$ @Christian B Yes, thank you for the hint about the leakage current. It is, compared to the charge/discharge current (approx. 250 mA) pretty low. I wonder though what it will be as long t < 2 min. Maybe Vishay's FAE can help out here. I'll try to contact them. \$\endgroup\$ – stowoda May 13 '19 at 19:10

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