Question

A certain super capacitor has a capacitance of 1.0F. The maximum voltage is 2.7 V. It is contained in a 8.00mm diameter can that is 13.00 mm long. It has two leads that are 3.50mm apart. It costs $0.29 when bought in bulk. How can such a large capacitance be made in a small container? The largest dielectric constant for oils is about 90. The largest dielectric constant is about 100,000 in material that costs $100,000/kg. Just describe the type of design (vacuum parallel-plate, dielectric parallel-plate, cylindrical, dielectric cylindrical, rolled parallel-plate, electrolytic, adjustable plate, etc.)

What is the energy density of the capacitor when fully charged?

Answer 1

Super capacitors follow various design methods to achieve such large energy density while keeping the cost low. We should understand that it is not possible to achieve such high capacitance in small sizes using conventional design and material properties. So the two design methods are followed:

1. Double layer Capacitor: This can be estimated as a rolled parallel plate where the distance between the parallel plates are very small (of the order of fraction of nano meters). In such cases the properties of the material is not like carbon elecrodes which ahs a very large surface area provides a large per unit capacitance very low double layer distance.

2. Psudocapacitance: In this case the properties of certain electrolytes are used which displays electrochemical properrty of capacitance by virtue of their movement of electrolyteions to opposite electrodes while presenting a very thin layer of solvent of one molecule thick presenting a very high capacitoance due to extremently small molecular distances.

The energy density of such capactor when fully charged will be given by following equation:

   J/m³

= 5.578 * 10⁶ J/ m³