ballast Inductor Choke coil (fwd)
From: Mad Coiler [SMTP:tesla_coiler-at-hotmail-dot-com]
Sent: Friday, April 03, 1998 10:13 AM
Subject: Re: 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
>> 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
>> with air as the core.
>NO WAY! The standard equation for calculating the number of turns need
>helical inductor is;
You defenitaly prooved your point here, but I put the figures into a
computer program I have that calculates values for helical coils,
entering 38200uH, 3" radius, and 12" winding length and I got 789.89
turns. Is there something wrong with this program?
> 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
> 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
>the insulation) which would be #33 gauge. Now taking the diameter of
>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
>wire is; 205.7 ohms, multiplying this by 4.634 gives; 953.2 ohms D.C!
>D.C. resistance of this coil is 66 times greater than it's impedance.
>short it is impractical to make a single layer air core with this high
>impedance! It would be neccessary to use a multilayer core, preferable
>an iron core to get the number of turns down far enough so you could us
>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
>Actually you did that in the begining of your calculation! I'll use
>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
>need to put another 14.4 ohms impedance in series with the transformer,
>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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