Re: Average Current
From: Jack// ani (nospam4u_jack_at_yahoo.com)
Date: 10/20/04
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Date: 20 Oct 2004 05:44:21 -0700
John Popelish <jpopelish@rica.net> wrote in message news:<4175392B.148634D4@rica.net>...
> Jack// ani wrote:
> >
> > John Popelish <jpopelish@rica.net> wrote in message news:<417299D8.2A952AC1@rica.net>...
> >
Thanks a lot.
> > > If the load is primarily capacitive, all the rectifier current is
> > > confined to the brief periods when the transformer voltage is higher
> > > than what is stored in the capacitor, so the RMS current can get quite
> > > a bit higher than the average current. This affects the current
> > > rating required in the transformer.
> >
> > Just two more questions, John
> >
> > 1. Reactance offered by a capacitive load is always higher that a
> > resistive load, then why should it extract more current?
>
> Reactance is a linear concept that applies to sine waves. To get your
> mind around a capacitive input rectifier filter, you have to think in
> the more general differential description of capacitance. I=C*(dv/dt)
> current equals the capacitance times the time rate of change of
> voltage across the capacitor.
>
> As long as the capacitor voltage is equal to or greater than the
> transformer voltage, the rectifier isolates the two. But the moment
> the transformer wave rises above the capacitor voltage, the rectifier
> is essentially a short circuit, and the voltage on the capacitor must
> rise as fast as the transformer wave is rising, regardless of how much
> current that takes. So the current into a capacitive filter is narrow
> sort of half sine wave pulses that occur on the part of the
> transformer voltage wave just before the peak voltage. Since the
> transformer windings are heated by the RMS current, a pulse waveform
> like this has a much higher RMS value than the average of the current
> in those pulses. This is what the 'squared' part of the RMS does. It
> is not unusual to have to double (or more) the transformer RMS current
> rating relative to the DC output average current when using a
> capacitor input rectifier filter to take care of this higher winding
> RMS current. The exact ratio depends on how much leakage reactance
> there is between primary and secondary windings that tends to spread
> out the charging pulses, by sagging the waveform a bit while the cap
> is charging, lowering the slope a bit. If you look at the transformer
> waveform with a scope, you can see the flattened spot on the wave
> where the cap is charging just before peak voltage.
>
> > 2. Why do capacitor charges to peak ac voltage if placed across a
> > bridge rectifier? Without a cap i measured 11v, and it increased to
> > 16v, after hooking a cap?
>
> The diodes act as check valves, pumping the cap up all the way to the
> transformer waveform peak voltage and then turning off, leaving that
> voltage trapped in the capacitor. A resistor load on the rectifier
> keeps it on the whole waveform so the resistor voltage is the same as
> the transformer waveform (except for the inversion of one half). your
> meter reads the average of the rectified transformer waveform instead
> of the peak value.
>
> > Thanks for your help
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