# What attribute of a capacitor is linked to physical size?

Discussion in 'General Electronics Chat' started by strantor, May 6, 2011.

1. ### strantor Thread Starter AAC Fanatic!

Oct 3, 2010
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Most of the capacitors I have encountered have been in the pico-, nano- or micro- farad range & are small in size. The exception being those used in car audio systems, which are common to be 1F and about the size of 3lbs of ground beef. The other day my friend was telling me about a 3F capacitor 1" long by .5" round. I didn't believe him; I thought it would have to be at least the size of a trash can, he was right. So now I'm puzzled. I though the farad rating determined the size, so what is it actually?

2. ### Veracohr Well-Known Member

Jan 3, 2011
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Probably type is the biggest factor - aluminum electrolytic, mica, ceramic, etc.

There's a 5000F capacitor there that's about 80x150mm and only 8 grams.

3. ### strantor Thread Starter AAC Fanatic!

Oct 3, 2010
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holy balls! that's an "Electrochemical double layer" capacitor; I've never heard of one.

4. ### wmodavis Well-Known Member

Oct 23, 2010
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The main factors that determine capacitance are area of the plates and the dielectric used.
More area means larger capacitance and size.
Higher dielectric constant means more capacitance and smaller size.

Apr 30, 2011
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Physical size increases with capacitance as others have noted but also significantly with voltage. As you can see, those ultracapacitors are rated for less than 3V.

6. ### ErnieM AAC Fanatic!

Apr 24, 2011
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Generally for a given style of cap and a given case size you can make a rough correlation to the product of voltage and capacitance: You can get a lot of capacitance but at a low voltage, or you can get a high voltage rating of a small capacitance.

I've never tried to predict a cap size with this taking it more as a rule of thumb so I know when to stop looking for the cap I want in the case I think it will fit.

strantor likes this.
For a given technology, maximum energy storage capacity may give a better fit with volume. This is given by C*V$^{2}$/2, where C is the capacitance and V is the maximum working voltage.