How to estimate electrolytic capacitance from foil area

Question

I opened a failed 1000 μF electrolytic capacitor. It contains 2 foils coiled with paper and some wet dielectric.

With reasonable assumptions (to me), I can not derive marked capitance from this formula:

$$C =\varepsilon _o \varepsilon _d \frac{A} {d}, \mathrm{~where}$$ $$\varepsilon _o = 8.85 \times 10^{-12} \mathrm{~F/m}$$ $$\varepsilon _d = \text{dielectric constant, say 100}$$ $$A = 0.25 \mathrm{~m} \times 0.013 \mathrm{~m} = 0.00325 \mathrm{~m^2}$$ $$ d = 0.00001 \mathrm{~m}, \text{ just a silly guess}$$

But foils are coiled, so do I double area? The foils are etched to increase the effective area, I'll say by a factor of 100.

$$ C = \frac {8.85 \times 10^{-12} \times 100 \times 0.00325 \times 100} {0.00001}$$ $$ C \approx 29 \mathrm{~\mu F}$$

How do they get 1000 μF?

Answer 1

The electrolyte is the cathode, metal is the anode, oxide layer on metal is the insulator.

The d in case of electrolytics has not much to do with distance between the foils, but distance from conductive electrolyte through the oxide layer to aluminum metal. The oxide layer thickness for low voltage capacitors is on the order of 0.1μm..0.01μm.

So, your assumptions were a bit off mark, but quite reasonable given how you thought the capacitor worked. Here are ballpark values:

• Relative permittivity of the oxide layer ε is between 10 and 40.
• Oxide layer thickness d, is obtained during forming - between 1 and 2nm/V of forming voltage.
• The dielectric strength, giving minimum dielectric layer thickness needed, is 700..1000V/μm or 1.3nm/V of rated working voltage.

Source: Wikipedia.

But foils are coiled, so do I double area?

Each side of the aluminum foil contacts the electrolyte retained in the paper(-like) spacer. So indeed, both of the foil surfaces need to be considered.

So, for a 50V rated aluminum electrolytic capacitor with crystalline oxide, relative permittivity 13, having 800V/μm dielectric strength, assuming safety factor 1.5 on the dielectric strength...

... a rough ballpark would be $$C \approx \frac {8.85{\rm\,pF/m} \times 13 \times 2\cdot0.00325{\rm\,m^2} \times 100{\rm\,m^2/m^2}} {50{\rm\,V} \times 1.5 \times 1.3{\rm\,nm/V}} \approx 0.8{\rm\,mF}.$$

Looks better, doesn't it? :)

Your guess of the effective area multiplier was quite close!