Jack// ani said:
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.