[sdiy] Xfade

[aankrom]

On Fri, 23 Mar 2012 11:55:23 -0000 (GMT), 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
There is a circuit designed by Buchla ( I think) that is a voltage 
controlled crossfade. It uses Vactrols and the schematic has his 
"syntax."

I'll look for it & email it to you if this sounds interesting.

AA


> 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,
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