Electronic – How to compute the lifetime of an electrolytic capacitor

WET ALUMINUM ELECTROLYTIC CAPACITOR LIFETIMES

Frameless version of application note here

Save the above document in your master application note collection now ! :-).

That is one marvellous original document that they have built their application note around !!! - with figures photographically added from some long lost source document. A web search failed to turn up a copy of the original.I'll be searching some more in due course as I'd dearly love to know where it originally came from.

As well as the fact that the United Chemi-Con version comes closest to what I have always understood, I'd give it credence because it is accompanied by such a large set of empirically derived data by persons unknown who obviously thought they knew their topic. While "appeal to authority" has its severe risks, this gives every sign of having being compiled with great effort and attention to detail by people who were genuinely well versed in the subject.

My understanding has long been that:

• 1 A wet aluminum electrolytic capacitor should be operated below but near its rated voltage

• 2 Operation well below rated voltage will decrease lifetime at any given temperature.

• 3 Storage at a given temperature without applied voltage will REDUCE lifetime compared to lifetime at the same temperature with rated voltage applied.

2 & 3 may seem counter intuitive. From long (long long) ago recollection they were explained as being due to a combination of accelerated electrolyte dryout without applied voltage and "deforming" of the electrodes with no applied voltage.

I have thought about this on an off over (too many) decades. It is not wholly obvious that heat in the absence of voltage would accelerate leakage across the internal-external seal, but a change in the general chemistry of the "electrolytic cell" is conceivable. Usually the bias voltage serves to build the oxide layer which preserves the aluminum metal from chemical attack. Lack of voltage may well allow a different reaction set, so the "no applied voltage is bad" rule may make sense. The equations below figure 9 in the document, which apply to reverse bias conditions, (arguably) apply equally for zero bias where the combined half cell potentials are no longer offset by the applied voltage.

• 2Al + 3H2O - 6e- -> Al2O3 + 6H+

• 6H+ + 6e- -> 3H2^

I do love their "^" after 2H2 implying, presumably, the Hydrogen floating happily away - the usage may well be well enough known but I've not met it before.

"Forming" is a concept not often met with modern capacitors but well known at the time that the original application note was made. It is just as applicable now as then, but experience in manufacturing processes and designs have led to it largely being able to be carried out during manufacturing and then not needed again for life.

Forming is the process whereby the oxide layer that forms the barrier between the two "plates" of the carrier are formed - one plate being the electrolyte and the other the metallic aluminium with oxide being the separator. The thinness of the oxide layer is what allows such capacitors to attain far higher capacitances per volume than processes which rely on inserted physical materials as separators as capacitance is inversely related to plate separation.

In ye goode olde days large capacitors had to have their plates 'formed" by applying a suitable voltage via a suitable resistor so that the oxide layer could be progressively "formed" by electro-chemical action. Application of rated voltage before forming had occurred could lead to exciting and exiting action. (Exiting as a large capacitor that breaks down under voltage will happily and energetically eject its inner contents - either through the cross scored safety vent in the can top if present, or simply by splattering the can and contents randomly if no preferred venting path is provided. Some older list members, I being no exception, will have memories of being shot in the face with a wad of caustic metal and paper goo on occasion while bending low over a dying electrolytic capacitor.

The fact that modern capacitors form and stay formed across lifetime suggests that the mechanisms that caused low or no voltage dryout or deforming at temperature may now be less significant than when that application note was first penned. However, operation near but below rated voltage still seems most liable to allow maximum lifetime.

Trialing Facebook feed. This address is the same as above and is provided fro Facebook purposes. http://www.tayloredge.com/reference/Electronics/Capacitors/ElectrolyticLife.pdf

You are confusing ESR, that stands for Equivalent Series Resistance, and the leakage. The first is modeled as a series resistor, and take account of leads resistance, leads-internal plates resistance and so on, and is ideally zero. The second is modeled as a resistor in parallel with the capacitor and takes account of small leakage currents in the dielectric, and is ideally infinity.

The formula you use is correct, but the value you come out with is NOT the ESR, is the leakage resistance. Once the capacitor is charged, if you leave it it slowly discharges trough the leakage resistor with a time constant \$R_{leak}\cdot C\$, so \$R_{leak}\$ is what you calculated, approximately \$50M\Omega\$, that is plausible.

To calculate the ESR you need to measure how long does it take the capacitor to discharge through a much smaller resistor, let's call it \$R_{dis}\$. When you discharghe the capacitor through \$R_{dis}\$ the total resistance through which it discharges is actually \$R_{dis}+R_{ESR}\$, so using the very same formula you used for the leakage resistance you can calculate the ESR.

But is it really that easy? Of course not.

The ESR is hopefully quite small, tenths of milliohms if you have a very good capacitor up to a few ohms. Since in the formula you have \$R_{dis}+R_{ESR}\$ you don't want an eccessive \$R_{dis}\$ to mask \$R_{ESR}\$. Ok then! Why don't we choose \$R_{dis}=0\Omega\$? Easy question:

• \$0\Omega\$ resistance does not exist. But i can make it small!
• Time. You need to be capable to measure how long does it take to the capacitor to discharge.

If you charge the capacitor to a certain voltage it will take \$\tau\ln{2}\approx0.7\cdot\tau\$ where \$\tau=RC\$. If \$R=R_{ESR}+R_{dis}=1\Omega+1\Omega=2\Omega\$ and \$C=680\mu F\$ that's less than 1ms. Without proper equipment, that is a properly set oscilloscope, you can't easily measure the ESR.

Last but not least, keep in mind that electrolytic capacitors values have a tolerance of \$\pm10\%\$, that leads to: $$ R_{ESR}=\frac{t_{dis}}{\left( C\pm C/10\right)\ln{2}} - R_{dis} $$ with the above numbers, t=1ms, C=\$680\mu F\$, \$R_{dis}=1\Omega\$, this translates to: $$ R_{ESR}\in\left[0.91,1.33\right]\Omega $$ That's 10% down and over 30% up.