Calculating how to make a homemade capacitor

[reese houseknecht]

Ok, so I am am trying to make a homemade capacitor that is 4700pf and 15kv. polyester at 125 micrometers thick can withstand 15kv so were good. here is the equation i used
C=ε0 K A / D
Where C = capacitance, ε0 is epsilons constant, K = dielectric constant, A = area of aluminum foil, and D is the space between the 2 conductors (the thickness of my dielectric)
so here is my values
4.7x10-9F = 8.85x10-12 * 4.7 * A / 0.000127M
Farads Epsilon Dielectric Area Meters
i got 0.0142 for A after solving. this is about 3" of area? i don't get it, that is way to small! BTW these capacitors arent being rolled up, they are low inductance so its just flat

 

[sophiecentaur]

Area has units of metres squared.
What is the "M" in that formula? It's not normal to put symbols for the units inside a formula or in the arithmetic. It's assumed (essential) that the formula is dimensionally correct. Also where did the 0.000127 come from?
The answer you get is in m², which comes to something like 0.12m by 0.12m
Unless you can compress the plates together without puncturing the plastic, the average spacing will probably be significantly greater than the 125μm so your C may be less than you planned. No problem as long as you take it into account and find the right size by trial and error. That's one reason why they make many capacitors rolled up.

 

[reese houseknecht]

Area has units of metres squared.
What is the "M" in that formula? It's not normal to put symbols for the units inside a formula or in the arithmetic. It's assumed (essential) that the formula is dimensionally correct. Also where did the 0.000127 come from?
The answer you get is in m², which comes to something like 0.12m by 0.12m
Unless you can compress the plates together without puncturing the plastic, the average spacing will probably be significantly greater than the 125μm so your C may be less than you planned. No problem as long as you take it into account and find the right size by trial and error. That's one reason why they make many capacitors rolled up.

M is meters, my science teacher said it had to be in meters, it is normally 0.005" thick.

 

[sophiecentaur]

M is meters, my science teacher said it had to be in meters, it is normally 0.005" thick.

So just the number is needed. If you want units for everything then you would not be able to read the equation. AND somehow the value changed from one place to another in your post? The symbol for metres is m. "M" usually stands for Mega.
Any comments about the area to side length conversion?

 

[anorlunda]

(15000/0.000127) is more than 118 MV/m voltage gradient.

Is your dialectric breakdown strong enough to withstand that? Some dialectrics can do up to 0.5GV/m, but many can't.

 

[reese houseknecht]

(15000/0.000127) is more than 118 MV/m voltage gradient.

Is your dialectric breakdown strong enough to withstand that? Some dialectrics can do up to 0.5GV/m, but many can't.

yes, my dielectric is able to withstand 15kv

 

[sophiecentaur]

yes, my dielectric is able to withstand 15kv

You have to remember that the local field around an edge or other discontinuity in the plate surfaces can be high than the Infinite Sheet assumption. If you are 'pushing things' a bit then you need to be free of dust and the edges of your plates need to be rounded off (finely sanded). 50% headroom is a good value to work with - two layers of dielectric and twice the area may be worth considering. Be prepared for modifications after your first try.

 

[reese houseknecht]

its going to be very hard to sand down aluminum foil lol. its a piece of polyester with aluminum foil on top, and maybe several layers. all i want to know is why i calculated such a low area, please help me.

 

[anorlunda]

Didn't you get your answer in post #2?

The answer you get is in m2, which comes to something like 0.12m by 0.12m

12 cm x 12 cm . Doesn't that sound right to you?

 

[reese houseknecht]

Didn't you get your answer in post #2?12 cm x 12 cm . Doesn't that sound right to you?

I legit just figured this out and then looked in my email and found this. Lol yes that sounds right thank you!

 

[sophiecentaur]

its a piece of polyester with aluminum foil on top,

You will still need to treat the edges in some way so that they are smooth with no sharp edges to the metal. It would be a good idea to undercut the metal foil so there is a bare border of exposed plastic all the way round. The periphery is where you can expect arcing if you don't do this.
I am pointing these things out because it is better to be aware of possible practical pitfalls before you actually build the capacitor.

 

[reese houseknecht]

This is the capacitor i am basically making but eith my dimensions. I am going to riund the edges of the protruding aluminum but i think i should be fine. So with that equation do i only need aluminum 12 x 12 them dielectric then another piece of aluminum 12x12?

 

[berkeman]

(15000/0.000127) is more than 118 MV/m voltage gradient.

Is your dialectric breakdown strong enough to withstand that? Some dialectrics can do up to 0.5GV/m, but many can't.

yes, my dielectric is able to withstand 15kv

That's not what he asked...

its a piece of polyester with aluminum foil on top

What have you read about the breakdown strength of Mylar (polyester film)? It does not look like it will work for the high voltage you want to use:

http://usa.dupontteijinfilms.com/wp-content/uploads/2017/01/Mylar_Electrical_Properties.pdf

Note that the Wikipedia article on film capacitors has some useful information about construction and breakdown: https://en.wikipedia.org/wiki/Film_capacitor

Finally, what is your source of 15kV? So far I don't have a warm fuzzy feeling for your background and experience with high voltages...