[sdiy] RE: cross-fades (was ...)

[David G. Dixon]

> Am 16.01.2009 um 19:45 schrieb David G. Dixon:
> 
> >>> Can't you just feed the two signals into either ends of a pot and
> >>> grab the result from the wiper?
> >
> >> This only works if the two source impedances are zero.  Otherwise
> >> you'll never completely select one or the other source at the ends
> >> of pot travel.
> >
> > ...so this means the signals should be sent through followers
> > first, right?
> > So, effective (manual) cross-fader = 1 TL072 + 1 Pot...?
> >
> 
> The output impedance of this circuit also varies with the wiper
> position. If this matters or not depends on your application.
> 
> Ingo

Yes, if perfectly stiff output were required, then one would have to follow
the output as well.  This would require a TL074, with one opamp to spare.

I've analyzed the single-pot cross-fading circuits suggested by Roy Tellason
and Tom Wiltshire over the last couple of days.  Both have perfectly stiff
outputs, of course.

Roy's method requires one op-amp and five resistors.  However, it is
impossible to get a linear response from this circuit, no matter how the
resistors are selected.  The signals are summed in the middle and/or
attenuated at the ends, and a fair amount of current may flow through the
pot to ground.

Tom's method requires two op-amps and six resistors.  Again, a perfectly
linear response is not possible, although it can be approximated with
careful resistor selection (low value for the pot, high values around the
inverting mixer).  Again, a significant current may flow through the pot to
ground.

It would seem that my circuit is the simplest, with three op-amps and no
resistors, perfectly linear response, perfectly stiff output, and virtually
no current flow.  It would require a 14-pin chip instead of an 8-pin one,
but the lack of resistors would make for a smaller footprint overall.  This
would be a good circuit to design a "pot chiclet" for.