[AMPS] 240VAC Amplifier Wiring

[Bill Fuqua]

Rectifiers followed by a capacitive input filter only draw current during 
only a small portion of each half cycle.
This means that the Peak current is many times the average or RMS .  During 
the charging of the capacitors
the voltage drop is equal to the peak current times the total line 
resistance. This distorts the wave form from
a sine wave to a clipped sine wave. The peak voltage is reduced by the the 
drop due to the recurring charging
current.  Sense the DC output of a capacitive input filter is equal to the 
peak voltage  from the transformer
the distorted sine wave form produces a reduced output.

73
Bill wa4lav



At 03:54 PM 11/15/00 -0500, Paul Christensen wrote:

> > >30A x 100' run on #10AWG CU = 6.9 volts dropped.
> >
> > This is for a resistive load.  C filter DC supplies are different.  The
> > peak to avg current ratio is c. 10 to 1
>
>Rich, wouldn't this only be applicable during the initial inrush of current
>at power-up and that a step-start circuit should limit the inrush of current
>"somewhere close" to that of a resistive load?  Or, is the C-input effect on
>AC current seen on full-power CW mode for instance, where the current demand
>rapidly changes with the activation of the key?
>
>-Paul, W9AC
>
>
>
>--
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William L. Fuqua  III , P.E.  EE
Department of Physics and Astronomy
CP 177 Chemistry Physics Building
University of KY
Lexington, KY 40506-0055
Phone 859 257-4155


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