🚀 go-pugleaf

In article <2452@iscuva.ISCS.COM> jimc@iscuva.ISCS.COM (Jim Cathey) writes:
}In article <4224@midas.STS.TEK.COM> jeffw@midas.STS.TEK.COM (Jeff Winslow) writes:
}>First of all, if you really mean inrush current as in what you get when you
}>first turn the switch on: Since switching supplies don't try to synchronize
}
}Bad choice of words on my part.  I meant to refer to the peak current through
}the diode during each cycle.  This will be worse for the squarish wave because
}if the capacitors sag during the dead part of the input cycle the voltage
}differential across the diode will be much higher (or at least it'll try to
}be).  This can result in hot spots in the diode, leading to premature failure.
}
}
}>Secondly, rms running current will be lower for a square wave, since the
}>conduction angle is much greater than for a sine wave, even the chopped-off
}>version. The rectifiers will run cooler, last longer. The capacitors will,
}>too - lower rms current -> lower core temperature, and the single most telling
}>variable affecting electrolytic capacitor life (in normal use :-)) is
}>temperature.
}
}I disagree, the conduction angle may well be larger, but the _peak_ current
}will be very high in the early part of the cycle.  Thus higher RMS current,
}and the exact opposite results.
}

I assume this is a full-wave inverter.  If so, the peak voltage is the same as
the averge; the capacitor is quite small; the peak rectifier current is the
same as the average; there is no "sag" in the capacitor voltage.  

If you don't believe this, draw a picture of a full wave rectified square
wave.  
-- 

Norm   (strong@tc.fluke.com)