Saturday, April 9, 2011

Fun With Python and Capacitance Values

In line with my work in modifying my flash unit, I was wondering how much capacitance I could get out of a home made paper foil roll capacitor.

As a goal size, I used a typical soda can:
  • 1.31" (33.274mm) x 4.75"(120.65mm)
For the materials, I'm considering standard kitchen aluminum foil and normal printer paper with the following assumptions about thickness/etc:
  • Aluminum foil assumed to be 0.025 mm in thickness (air/water impermeable)
  • Paper thickness estimated to be 0.10 mm in thickness (rounded up from 0.097)
  • Resulting Layer thickness = 0.25 mm (2 layers of foil + 2 layers of paper)
Running the calculation, assuming concentric layers of diminishing size until the can is filled:
  • ~133 layers
  • ~1691201 mm^2 in surface area
  • ~449 uF in capacitance (estimated with K=0.30)
That's a good chunk of capacitance, especially considering that paper's breakdown voltage is 200V per mil:
  • 1 mils = 0.0254 millimeters
  • 0.10 mm = 3.937 mils
  • ~787V breakdown voltage
So.. a capacitor that can safely handle 700V with a capacitance of 400uF.

If I can reduce the thickness of each layer and of the insulator layer, I can bump the capacitance to 2200uF!! Wild. But it would be rather big and clunky. ^_^;;

In either case, solving little problems like this is also a great way of learning a language like Python.