On Oct 19, 2005, at 7:54 AM, Jim Brown wrote:
> On Tue, 18 Oct 2005 20:21:39 -0400, Bill Coleman wrote:
>
>> Seems to me that 12.3 A and 13.4 A are both less than 15 A. So a #14
>> circuit seems adequate for your example.
>
> Yes and no. Remember that ALL electronic loads (that is, anything with
> a
> DC power supply) utilize some form of rectifier and capacitor-input
> filter. This includes switching power supplies.
Resonant-choke DC filters are now forbidden?
> Thus current draw will
> be concentrated at the peak of the waveform, and will tend to be quite
> non-linear. So will the IR drop in the wire. And, because the current
> is
> rich in harmonics, IZ drops can enter the picture.
I x Z drops enter the picture at 60Hz.
>
> What does this mean in practical terms? It simply means that the sag
> due to loading will be greater than predicted by simple RMS
> calculations
> assuming sinusoidal 60 Hz current flow. Bigger copper will reduce that
> drop (but not change the fact that the current will be non-sinusoidal).
> The non-sinusoidal current flow causes distortion of the 60 Hz sine
> wave, so bigger copper will reduce that distortion and that drop.
>
> Last year I wrote a "white paper" on power systems for audio
> professionals. It addresses this issue as well as many others.
>
> http://audiosystemsgroup.com/SurgeXPowerGround.pdf
>
> Jim Brown K9YC
>
>
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>
>
Richard L. Measures, AG6K, 805.386.3734. www.somis.org
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