[sdiy] Xfade

[bbob]

buchla xfade schemo is on Mark Verbos' DIY page:
http://www.simple-answer.com/DIY.html

On Sun, Mar 25, 2012 at 1:21 AM, aankrom <aankrom at bluemarble.net> wrote:
> 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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>
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