Re: Wiring a Transformer.....Help!!

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Richard W by way of Terry Fritz <twftesla-at-qwest-dot-net>" <potluck-at-xmission-dot-com>

Hi Chris,

VA = (CS / .16)^2

E = (.16 x CS)

where:
VA= Volt Amps
CS = Cross section of the core (center leg only)
E = Volts per turn

5000 = (CS / .16)^2
70.7 = CS / .16
Cross Section = 11.3 square inches

E = (.16 x 11.3)
E = 1.8 volts per turn


PRIMARY

240 / 1.8 = 133 turns
120 / 1.8 = 67 turns

For 240 volts input the wire size needs to be 750 circular mils per amp.
Some recommend 1000 circular mils per amp. I've found that under oil 750 CM
per amp is fine. You'll need this tranny under oil I would think.

5000 / 240 = 20.8 amps

5000 / 120 = 41.7 amps

For 240 volts the wire size needs to be 21 x 750 = 15750 circular mils ( #8
gauge wire)
For 120 volts the wire size needs to be 42 x 750 = 31500 circular mils ( #5
gauge wire)

I've wound parallel primaries to match the CMs necessary. For example:
At 240vac input two primaries of #11 wire at 133 turns in parallel would be
fine. Be very sure each has EXACTLY the same amount of turns.


SECONDARIES

Winding one secondary for that voltage output isn't a good idea. Use two
secondaries, one on each side of the primary.

7500 / 1.8 = 4166 turns

5000 / 15000 = .333 amps (333 MA)

.333 * 750 = 250 circular mils (#26 gauge)



5000va
240vac
15000vac output at 333 ma

1.8 volts per turn
133 turns of #8 (16500 actual CMs for the wire size)
4166 turns of #26 (252.8 actual CMs for the wire size) each secondary


Rick W.




----- Original Message -----
From: Tesla list <tesla-at-pupman-dot-com>
To: <tesla-at-pupman-dot-com>
Sent: Saturday, September 21, 2002 10:30 PM
Subject: Wiring a Transformer.....Help!!


> Original poster: "Chris by way of Terry Fritz <twftesla-at-qwest-dot-net>"
<chrislj-at-earthlink-dot-net>
>
>
> Let s say I have a suitable sized core to wind a transformer on. If I
wanted a
> 5kw transformer
>
> with a 100:1 winding ratio and an output of 15000v. How many primary turns
> would it need to
>
> be able to handle that amount of power? Is there a type of formula for
figuring
> out how much copper needs to go into the primary? How do you know when you
have
> enough turns?
>
>
>
> Chris
>
>
>
>