Homemade foil capacitors / General Science and Electronics / Forums

Forums

4hv.org :: Forums :: General Science and Electronics

« Previous topic | Next topic »

Homemade foil capacitors

previous 1 2

Move Thread

LAN_403

Ash Small

Fri Sept 14 2012, 11:48PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Yanom wrote ...

What's vacuum potting?

You suck the air out using a vacuum pump. A fridge compressor will do, but make sure it is CFC free before removing the pump.

There are other ways, but they involve rolling the caps under oil, or similar.

Tony Matt

Sat Sept 15 2012, 03:42AM

Registered Member #3700 Joined: Sat Feb 19 2011, 12:59PM
Location:
Posts: 107

C = 8.854 * K * A * (N-1) / D

C = CAPACITANCE IN pF

K = DIELECTRIC CONSTANT

N = NUMBER OF PLATES

A = PLATE AREA [ SQUARE METER ]

D = DIELECTRIC THICHNESS [ METER ]

IF THE TW0 PLATES ARE COILED AS YOU SHOW IN THE BEGIN, THE TOTAL CAPACITANCE SHALL BE TWICED...

Yanom

Sat Sept 15 2012, 04:49PM

Registered Member #4659 Joined: Sun Apr 29 2012, 06:14PM
Location:
Posts: 158

Tony Matt wrote ...

C = 8.854 * K * A * (N-1) / D

C = CAPACITANCE IN pF

K = DIELECTRIC CONSTANT

N = NUMBER OF PLATES

A = PLATE AREA [ SQUARE METER ]

D = DIELECTRIC THICHNESS [METER]

IF THE TW0 PLATES ARE COILED AS YOU SHOW IN THE BEGIN, THE TOTAL CAPACITANCE IS SHALL BE TWICED...

Based on this data, I wrote a Python computer program to calculate the total energy that could be stored in a homemade capacitor based on these values. Is my math right?

#!/usr/bin/python3
K = float(eval(str(input("Dielectric Constant? ") )))
A = float(eval(str(input("Plate area (Square meters)? "))))
D = float(eval(str(input("Dielectric thickness (meters)? "))))
n = 2 #two plates on a regular capacitor

nminusone = n-1
picofarads = (8.854*K*A*nminusone)/D
picofarads = picofarads*2 # rolled capacitor
farads = picofarads*(10**-12)

print(str(picofarads/1000) + " uF capacitor")

S = float(eval(str(input("Dielectric strength (V/m) ? "))))
maxvoltage = S*D

print("Maximum voltage: " + str(maxvoltage) + "V")

J=farads*maxvoltage*maxvoltage

print("Stored Energy: " + str(J) + "J")

I looked up values for wax paper and found Di.Constant=~3, Di.Strength = 3.6E7 V/m. This program tells me I could store upwards of 10KJ in a capacitor made from two rolls of allumninum foil and two rolls of wax paper. Is that right?

EDIT: there was a bug in the code. I fixed it above and it gives sane results now.

Newton Brawn

Mon Sept 17 2012, 05:18AM

Registered Member #3343 Joined: Thu Oct 21 2010, 04:06PM
Location: Toronto
Posts: 311

Yanom:
1 - The info from Tony is the practical and acceptable formula to start a capacitor design.
2- We are skiping the capacitor design procedure and going to abstracts ...
3- There several points that may be not true, or need to be clarified (in the program and in the conclusions).

In order to get a result, lets leave the program at side and try calculate step by step:

info acceptable:
C = 8.854 * K * A * (N-1) / D
C = CAPACITANCE IN pF
K = DIELECTRIC CONSTANT (use 3.00 as you wrote)
N = NUMBER OF PLATES (use two, rolled, ayw )
D = DIELECTRIC THICKNESS [ METER ] ( to be calculated )
A = PLATE AREA [ SQUARE METER ] ( to be calculated )
assume that the waxed paper dielectric is 20kV/mm
assume that a rolled cap has twice the capacitance of flat cap of same area

Problem:
Calculate a rolled capacitor, 10 microfarads, 1000 maximum peak voltage.
a - State the thickness of the available foil. State the thickness of the available paper.
b - first calculate the dielectric thickness.
b - calculate the aluminum foil area and dimensions of each foil.
c - calculate paper area and dimension of each paper sheet.
d - calculate the energy and the energy per sq cm.

I hope you following the above steps it will be easier for us review and providing a oppinion.

Regards

Newton

Newton Brawn

Wed Sept 19 2012, 02:05AM

Registered Member #3343 Joined: Thu Oct 21 2010, 04:06PM
Location: Toronto
Posts: 311

Any comment ?

Tony Matt

Wed Sept 19 2012, 02:57AM

Registered Member #3700 Joined: Sat Feb 19 2011, 12:59PM
Location:
Posts: 107

Hi !

I did some calculations, let me know if Im in the right track

a - Thickness of aluminum foil = 0.018mm (Jerrolds Wrap kitchen foil)... Also assuming that the available waxed paper thickness as 0.1mm.

b- calculating the required paper thickness:

A safety factor on the capacitor voltage will be good practice. Lets use SF as 1.5, So the cap design voltage will be 1500Vpk.

1- assuming that are some small air bubbles 0.1mm thick inside the dielectric, and the bubble air ionizing gradient as 9000/mm. (corona starting voltage)

2- L2=(E/G1-L1)*(K2/K1)

were L2= paper thicness
E= applied pk voltage on the cap = 1500V
G1 = air max gradient starting ionizing
L1= air bubble thickness = 0.1mm
K1= air buble dielectric constant =1
K2 = paper dielectric constant = 3

L2=(1500/9000 - 0.1) * (3/1) = 0.2mm

The minimal paper thickness to be 0.2mm.

Lets use 2 sheet of paper, total dielectric thickness as 0.2mm.

Area of one aluminum electrode (flat electrodes):
C = 8.854 * K * A * (N-1) / D
A = C*D /[ 8.854 *K*(n-1)]
A = 10000000 * 0.0002 / [8.854 *3*(2-1)]
A = 75.3m2
But the cap has coiled electrodes, so each electrode foil to be 37.65 square meter .

Assuming the foil wide as 300mm, and usefull wide as 280mm, each capacitor plate dimension will be 0.28m x 134m.

The waxed paper dielectric shall have similar dimensions.

previous 1 2

Moderator(s): Chris Russell, Noelle, Alex, Tesladownunder, Dave Marshall, Dave Billington, Bjørn, Steve Conner, Wolfram, Kizmo, Mads Barnkob