[sdiy] Xfade *jackpot*

[Joshua Lewis]

Could someone please link to a schematic?

Thanks!

On Mar 23, 2012, at 4:00 PM, peter edwards <synth at casperelectronics.com> wrote:

> 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,
>
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