ballast Inductor Choke coil (fwd)
From: Alfred A. Skrocki [SMTP:alfred.skrocki-at-cybernetworking-dot-com]
Sent: Thursday, April 02, 1998 6:39 AM
To: Tesla List
Subject: Re: ballast Inductor Choke coil (fwd)
On Tue, 31 Mar 1998 18:47:49 EST Erik Schulz
> When an inductor is placed in series with an AC source it causes
> resistance governed by this equation. R = 2 * Pi * f * L
> So for a transformer that uses about 1kw you would need 14.4 ohms with a 120
> volt wall socket, so the inductance should be about 0.0382 H,
Up to here everything checks out OK.
> so on a 12 in cylinder with a 6 inch diameter you would need about 780 turns
> with air as the core.
NO WAY! The standard equation for calculating the number of turns need on a
helical inductor is;
N = SQRT (L * ((9 * R) + (10 * H))/R * R)
N = Number of turns needed.
L = inductance in microhenrys.
R = radius (inches).
H = height (inches).
Now in your earlier calculation you determined that you wanted a 0.0382
Henry inductance to convert this into microhenrys you would divide the
value in Henrys by 0.000001 thus giving; 38200 microhenrys! Now pluging
that into the above equation along with your cylinders measurements gives;
N = SQRT (38200 * ((9 * 3) + (10 * 12))/3 * 3)
N = 1475 turns would be required
A far cry greater than 780 turns! Now dividing 12 by 1475 gives 0.008
inches or 8 mils for the maximum possible thickness of the wire (including
the insulation) which would be #33 gauge. Now taking the diameter of that
cylinder; 12 and multiplying this by pi to get the length of one turn
yields; 37.7 inches, Which when multiplied by the 1475 turns yields;
155607.5 inches or apx. 4634 feet! Now the resistance per 1000 fet of #33
wire is; 205.7 ohms, multiplying this by 4.634 gives; 953.2 ohms D.C! The
D.C. resistance of this coil is 66 times greater than it's impedance. In
short it is impractical to make a single layer air core with this high an
impedance! It would be neccessary to use a multilayer core, preferable with
an iron core to get the number of turns down far enough so you could us a
wire such that it's D.C. resistance was significantly smaller than it's
impedance, otherwise it will never pass the needed current!
> I don't know how to figure out the internal resistance of the transformer.
Actually you did that in the begining of your calculation! I'll use your
1 Kilowatt transform. 1000/120 = 8.33 amps drawn by the transformers
primary. Then since Z = E/I we have 120/8.33 = 14.4 ohms impedance. If
you say, wanted to current limit this transformer to 500 watts you would
need to put another 14.4 ohms impedance in series with the transformer, in
other words you would need a 0.0382 Henry choke in series with the
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Alfred A. Skrocki
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