[R-390] Exhumation of dead horse...

[rbethman]

Well, we are after a good start, but no winners yet.

As to:

    > The sinusoidal waveform will be distorted due to the diode forward
    > voltage drop of 0.7 V.

The junction voltage drop is "very" device specific.  Each diode "type" (By
number) has a specific voltage drop for THAT device.  There is NO across the
board "magic" number.  No Kupie Doll Heinz!

As to:

    > Nope, its a bell curve, not sinusoidal
    > Jeff Adams

Another "miss".  The pulse width may or will narrow, but NOT bell shaped.
Drag out your oscilloscopes, a low voltage output transformer, and don't
forget a handful of diodes.  (Lord I hate sand state....)

At least no one is debating the number of pulses per second.  (Smart for a
change...)  Nor the fact that the peak voltage is not "significantly"
changed.  The next concept is current.  It will remain the same for a given
load.

Bob - N0DGN



> rbethman wrote:
> >
> >     If 26 VAC is passed through a diode, then 26 VDC comes out - BUT
only
> > half the time that the 26 VAC did.  The waveform is still sinusoidal in
> > shape, just missing the other half of the complete cycle.  Whatever
voltage
> > drop occurs is caused only by the ohmic losses of the junction of the
diode.
> > That is unless you put in a Zener diode.  Then we have an entirely
different
> > animal.
> >
>
>
> The sinusoidal waveform will be distorted due to the diode forward
> voltage drop of 0.7 V.
>
> The dead horse is stinking worse every day.
>
> 73
> Heinz DH2FA, KM5VT

> >     If 26 VAC is passed through a diode, then 26 VDC comes out - BUT
only
> > half the time that the 26 VAC did.  The waveform is still sinusoidal in
> > shape, just missing the other half of the complete cycle.
>
> Nope, its a bell curve, not sinusoidal
> Jeff Adams