[sdiy] Xfade *jackpot*

[peter edwards]

Thanks Richie. This is the one.
  1 op amp, 6 resistors and a -3db center gain.
Simple and effective design.
-pete

  On 3/23/2012 7:55 AM, Richie Burnett wrote:
> 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,