[sdiy] Xfade

[Richie Burnett]

Hi Pete,

You can get pretty close to a true equal power cross-fade using a passive circuit.

Take a linear potentiometer and ground the wiper.  Feed the two source signals
to opposite ends of the potentiometer track through resistors.  In order to get
-3dB gain at the centre position relative to the gain at the ends you need to
use resistors that are equal to the pot value divided by sqrt(2)  In other words
if you're using a 10k pot, you need to feed the signals to it through 6k8 fixed
resistors.

All that remains to do now is to add together the voltages at the two ends of
the pot track.  This can be done with a simple inverting mixing amp.  Just make
sure you use sufficiently high value resistors that you don't load down the
x-fade pot.  Otherwise the centre point will end up being more than 3dB down
relative to the ends, and you will need to decrease the value of the fixed
resistors to compensate.

Out of interest the attenuation for a signal when the pot is fully at the "loud"
end for that input is 4.6dB.  So you might want to make up some or all of this
signal loss with gain in the mixing amplifier.  (The attentuation for both
signals with the pot at its mid-point is 7.6dB.  3dB down relative to the ends
as expected.)

I hope this helps,

-Richie,