# Off Grid Storage Capacitor Calculations

I'm thinking about using a very large home made capacitor as an energy storage device for a light duty off grid application, running some led lights and some 500w loads for a few hours each evening. I've done some calculations and I'm surprised by the results, it's been a while since I've done anything like this and I want to make sure I haven't made a silly mistake. So I'm planning on a flat multi plate capacitor made with tin foil and cling film. I have the permittivity of the dielectric at 3 and it is about 10 microns thick. The plates are 2x.3m and there's 21 of them giving 20 actual capacitors. So the calaculation I made is

e0xe1 x number of plates-1 x plate area / dielectric thickness

8.84x10^-12 x 3 x 20 x 0.6 / 1x10^-11 = 31.824F

Is my math correct?

• There must be a reason why nobody uses capacitors for this. You can try, but I wonder if the calculation is incorrect. What voltage will be stored on the capacitor? My math says that in order to store 1500Wh in a 31.824F capacitor, it needs to be charged up to 582 volts. Can your 10-micron-thick dielectric withstand at least 582 volts? – user253751 Aug 12 '20 at 13:18
• Andy's correction sounds about right. So you'll need about a million of these, at over 500V. You might find a battery is more practical. – user_1818839 Aug 12 '20 at 14:16
• Homemade capacitors are always an order of magnitude or three below the storage density of commercial capacitors, mainly because of the thickness of the conductor. Your alli foil is far thicker than the metalisation manufacturers use on their dielectric. Then they wind it tighter without airgaps. – Neil_UK Aug 12 '20 at 15:00

Plates that have an area of 0.6 sq m and a gap of 10 microns (0.01 mm) produce a capacitance of 1.59 uF when the dielectric constant is 3. With 20 such capacitors in parallel, that's a total capacitance of 31.8 uF. Your result seems to be off by a massive factor.

Is my math correct?

No - this was in error; 1x10^-11 for the gap. It should be 10E-6

running some led lights and some 500w loads for a few hours each evening

Assume a kilowatt and assume the voltage is 240 volts. Assume that the voltage can be discharged to (say) 70% - that's a drop of 72 volts. Say you want the drop to occur over a period of 2 hours (7,200 seconds seems reasonable).

At the half way point (1 hour) the voltage will be about 200 volts and at a kW this is a current of 5 amps. The standard capacitor equation: -

$$I = C\cdot \dfrac{Δv}{Δt}$$

Rearranging for C: -

$$C = I\cdot\dfrac{Δt}{Δv} = 5\cdot\dfrac{\text{7200 seconds}}{\text{72 volts}}$$

$$\boxed{\text{This is a capacitance of 500 farads}}$$

• The OP wrote 10 microns as 10^-11 m (actually 10 picometres) instead of 10^-5 m - a factor of 10^6. This is why they calculated the capacitance as 10^6 times what you calculated. – user253751 Aug 12 '20 at 13:51
• Many thanks for finding the error and the additional info. 31 Farads in a package that size didn't seem right, hence my question. – Bazza14 Aug 12 '20 at 17:10
• @Bazza14 do you know what to do next regarding this site and its conventions? – Andy aka Aug 12 '20 at 17:19
• No I do not. What should I do. – Bazza14 Aug 12 '20 at 18:50
• @Bazza14 - See this – Andy aka Aug 12 '20 at 19:22