EQ #76

How big would a 10µF capacitor be using the materials in the Answer of EQ #75?

You need to do some basic calculations first. The formula for capacitance is:

where:

εR is the relative permittivity of the dielectric
ε0 is the permittivity of free space
A is the area of one plate
d is the separation between the plates

Let’s say you want to use 1-mil (25.4 µm) waxed paper as a dielectric. Note that this will determine the voltage rating of the capacitor. The dielectric strength of waxed paper is about 35-40 MV/m, so this will give you a capacitor that can theoretically handle almost a kilovolt, but be conservative in how you use it!

The relative permittivity of waxed paper is about 3.7, the permittivity of free space is 8.854e-12 F/m. Solve for the area required:

If you get aluminum foil and waxed paper that’s about 12” (30 cm) wide, you can probably get an overlap of, say, 25 cm, which means that you’ll need a length of about 15.5 m to get the area you need.

If you then roll up your capacitor (using a second layer of waxed paper), the capacitance will be doubled, or about 10 µF. Obviously, this will be physically rather large, more than a foot long and several inches in diameter.

Plastic food wrap has a similar dielectric constant and dielectric strength as waxed paper, but typically comes in a 0.5-mil (12.7 µm) thickness. A capacitor using this would have about half the voltage rating and about half the overall volume.

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EQ #76

by Circuit Cellar Staff time to read: 1 min