### It is all about a high dielectric constant and low voltage charging

**Example design parameters:**

The energy stored in the Capacitor: **E=½ CV** where C is the capacitance (F) and V is the charging voltage.

- Energy needed: 85 kWh
- Charging voltage V=600
- Relative permittivity k=16 million (This is the
**highest value**for dielectric constant reported in open literature. In comparison,*EEStor*uses material with a dielectric constant of 18,000 for their solid-state battery development. - Area of a single layer capacitor A=20cm x 20 cm=400 cm
^{2} - Thickness of the dielectric t=20 x10

For a unit of 85 kWh the total capacitance (called geometrical capacitance) is:

- C= 2E/V = 2 x 85 000 x 3600/(600) =1,700 F

**Capacitance ONE layer** (=Ԑ_{0}*kA/t)

0.2832 F/layer

**Energy density:**

5,900 Wh/lit

**Specific Energy Density:**

1,734 Wh/kg

**Gravimetric Power Density | frequency is 1**

7,780 W/kg

**Total weight energy storage device** (only stacked layers | housing not included)

### Discharging depends on load:

There are many advantages of solid state devices over Li-Ion batteries (no environmental pollution, millions of charging and discharging cycles and very fast charging to mention a few) the only disadvantage of super capacitor is the inability to hold the steady voltage during the discharging cycle under load.

The voltage in super capacitor will decrease steadily and the rate of voltage decrease depends on the load.

Higher loads will cause the voltage to drop faster. The **solution** to this problem is to fabricate higher power unit with high energy density.

With such high power density (high kWh/kg) super capacitors, it would be possible to put on board **several units** and discharge one unit after another until the voltage in each unit drops below a certain value.

In this way it will be possible to keep the voltage high during the entire life of the battery set. This can easily be done using today’s modern electronics.