> >> if the load is
> >> inductive (or capacitive) there will be a phase shift between the DC
> >> voltage and DC current. However, the fact remains that a triac will
turn
> >> off when conducting the DC current of a half or full wave rectified
> >signal.
> >
> >It's more than just a phase shift issue.  Series inductance can cause the
> >current to never reach zero (or fall below the triac dropout current).
>
> Well, it obviously *must* pass through zero twice per cycle.

A full wave rectified sine wave with no DC is ABS(SINE()).  This goes to
zero twice per cycle for a short time (instantaneously with perfect
rectifiers).  Now imaging this signal connected to a sufficiently large
inductor and some resistance.  Since this signal has an average DC
component, there will be an average DC current thru the inductor, which will
be AVE(ABS(SINE()))/R.  The current in the inductor will lag the rectified
voltage, but will also "average" it.  With an inductance of 0 the current
will be ABS(SINE())/R.  With an infinite inductor in steady state, the
current will totally flat at AVE(ABS(SINE)))/R.  Real inductances will vary
it between these two extremes, but lowest point in the current waveform will
rise from 0 when the inductance is increased from 0.  For real world parts
some minimum inductance will be required to overcome the finite time the
voltage is 0 due to voltage drops in the rectifiers and to get past the SCR
holding current.


********************************************************************
Olin Lathrop, embedded systems consultant in Littleton Massachusetts
(978) 742-9014, olin@embedinc.com, http://www.embedinc.com

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